Let a, bbesets.Notethat{a, b}isfunction-likeandrelation-like. Let A, Bbesets.Observethat A

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A Result About the Density of Iterated Line Intersections in the Plane

T 2(S) = T (Q2) = Q2 = T (S).
Excluding the sets of ﬁnite order, it follows that iteration of T produces a strictly increasing chain of sets of points in the plane. In light of this observation, a natural question is whether we arrive at a dense set of points by such a procedure. In other words, is i≥0 T i (S) dense in R2? A more diﬃcult but related question is whether we get all of Q2 when S consists of only rational points. We address both of these questions with a complete answer to the ﬁrst in Section 3 and some partial results for the second in Section 4.
arXiv:math/0507472v2 [math.MG] 25 Jul 2005
A RESULT ABOUT THE DENSITY OF ITERATED LINE INTERSECTIONS IN THE PLANE
CHRISTOPHER J. HILLAR AND DARREN L. RHEA
Abstract. Let S be a ﬁnite set of points in the plane and let T (S) be the set of intersection points between pairs of lines passing through any two points in S. We characterize all conﬁgurations of points S such that iteration of the above operation produces a dense set. We also discuss partial results on the characterization of those ﬁnite point-sets with rational coordinates that generate all of Q2 through iteration of T .

Profiling the Brane Drain in a Nonsupersymmetric Orbifold

hep-th/0507190
arXiv:hep-th/0507190v1 20 Jul 2005
Proﬁling the Brane Drain in a Nonsupersymmetric Orbifold
Gregory Moore and Andrei Parnachev Department of Physics, Rutgers University Piscataway, NJ 08854-8019, USA Abstract We study D-branes in a nonsupersymmetric orbifold of type C2 /Γ, perturbed by a tachyon condensate, using a gauged linear sigma model. The RG ﬂow has both higgs and coulomb branches, and each branch supports diﬀerent branes. The coulomb branch branes account for the “brane drain” from the higgs branch, but their precise relation to fractional branes has hitherto been unknown. Building on the results of hep-th/0403016 we construct, in detail, the map between fractional branes and the coulomb/higgs branch branes for two examples in the type 0 theory. This map depends on the phase of the tachyon condensate in a surprising and intricate way. In the mirror Landau-Ginzburg picture the dependence on the tachyon phase is manifested by discontinuous changes in the shape of the D-brane.

A theory of Boolean valued functions and partitions

JOURNAL OF FORMALIZED MATHEMATICSV olume10,Released1998,Published2003Inst.of Computer Science,Univ.of BiałystokA Theory of Boolean Valued Functions and PartitionsShunichi Kobayashi Shinshu UniversityNaganoKui Jia Shinshu UniversityNaganoSummary.In this paper,we deﬁne Boolean valued functions.Some of their algebraic properties are proved.We also introduce and examine the inﬁmum and supremum of Booleanvalued functions and their properties.In the last section,relations between Boolean valuedfunctions and partitions are discussed.MML Identiﬁer:BVFUNC_1.WWW:/JFM/Vol10/bvfunc_1.htmlThe articles[11],[4],[13],[1],[16],[15],[14],[2],[3],[9],[12],[8],[10],[7],[5],and[6]provide the notation and terminology for this paper.1.B OOLEAN O PERATIONSIn this paper Y denotes a set.Let k,l be boolean sets.The functor k⇒l is deﬁned as follows:(Def.1)k⇒l=¬k∨l.The functor k⇔l is deﬁned as follows:(Def.2)k⇔l=¬(k⊕l).Let us note that the functor k⇔l is commutative.Let k,l be boolean sets.Note that k⇒l is boolean and k⇔l is boolean.Let us note that every set which is boolean is also natural.Let k,l be boolean sets.Let us observe that k≤l if and only if:(Def.3)k⇒l=true.We introduce k l as a synonym of k≤l.2.B OOLEAN V ALUED F UNCTIONSLet us consider Y.The functor BVF(Y)is deﬁned as follows:(Def.4)BVF(Y)=Boolean Y.Let Y be a set.Note that BVF(Y)is functional and non empty.Let Y be a set.Note that every element of BVF(Y)is boolean-valued.In the sequel Y is a non empty set.Let a be a boolean-valued function and let x be a set.We introduce Pj(a,x)as a synonym of a(x).1c Association of Mizar UsersLet us consider Y and let a be an element of BVF(Y).Then¬a is an element of BVF(Y).Let b be an element of BVF(Y).Then a∧b is an element of BVF(Y).Let p,q be boolean-valued functions.The functor p∨q yielding a function is deﬁned as follows: (Def.5)dom(p∨q)=dom p∩dom q and for every set x such that x∈dom(p∨q)holds(p∨q)(x)= p(x)∨q(x).Let us note that the functor p∨q is commutative.The functor p⊕q yields a function and is deﬁned by:(Def.6)dom(p⊕q)=dom p∩dom q and for every set x such that x∈dom(p⊕q)holds(p⊕q)(x)= p(x)⊕q(x).Let us note that the functor p⊕q is commutative.Let p,q be boolean-valued functions.One can check that p∨q is boolean-valued and p⊕q is boolean-valued.Let A be a non empty set and let p,q be elements of Boolean A.Then p∨q is an element of Boolean A and it can be characterized by the condition:(Def.7)For every element x of A holds(p∨q)(x)=p(x)∨q(x).Then p⊕q is an element of Boolean A and it can be characterized by the condition:(Def.8)For every element x of A holds(p⊕q)(x)=p(x)⊕q(x).Let us consider Y and let a,b be elements of BVF(Y).Then a∨b is an element of BVF(Y).Then a⊕b is an element of BVF(Y).Let p,q be boolean-valued functions.The functor p⇒q yields a function and is deﬁned as follows:(Def.9)dom(p⇒q)=dom p∩dom q and for every set x such that x∈dom(p⇒q)holds(p⇒q)(x)=p(x)⇒q(x).The functor p⇔q yielding a function is deﬁned as follows:(Def.10)dom(p⇔q)=dom p∩dom q and for every set x such that x∈dom(p⇔q)holds(p⇔q)(x)=p(x)⇔q(x).Let us note that the functor p⇔q is commutative.Let p,q be boolean-valued functions.One can check that p⇒q is boolean-valued and p⇔q is boolean-valued.Let A be a non empty set and let p,q be elements of Boolean A.Then p⇒q is an element of Boolean A and it can be characterized by the condition:(Def.11)For every element x of A holds(p⇒q)(x)=¬Pj(p,x)∨Pj(q,x).Then p⇔q is an element of Boolean A and it can be characterized by the condition:(Def.12)For every element x of A holds(p⇔q)(x)=¬(Pj(p,x)⊕Pj(q,x)).Let us consider Y and let a,b be elements of BVF(Y).Then a⇒b is an element of BVF(Y).Then a⇔b is an element of BVF(Y).Let us consider Y.The functor false(Y)yields an element of Boolean Y and is deﬁned by: (Def.13)For every element x of Y holds Pj(false(Y),x)=false.Let us consider Y.The functor true(Y)yielding an element of Boolean Y is deﬁned as follows: (Def.14)For every element x of Y holds Pj(true(Y),x)=true.One can prove the following propositions:(4)1For every boolean-valued function a holds¬¬a=a.1The propositions(1)–(3)have been removed.(5)For every element a of Boolean Y holds ¬true (Y )=false (Y )and ¬false (Y )=true (Y ).(6)For all elements a ,b of Boolean Y holds a ∧a =a .(7)For all elements a ,b ,c of Boolean Y holds (a ∧b )∧c =a ∧(b ∧c ).(8)For every element a of Boolean Y holds a ∧false (Y )=false (Y ).(9)For every element a of Boolean Y holds a ∧true (Y )=a .(10)For every element a of Boolean Y holds a ∨a =a .(11)For all elements a ,b ,c of Boolean Y holds (a ∨b )∨c =a ∨(b ∨c ).(12)For every element a of Boolean Y holds a ∨false (Y )=a .(13)For every element a of Boolean Y holds a ∨true (Y )=true (Y ).(14)For all elements a ,b ,c of Boolean Y holds a ∧b ∨c =(a ∨c )∧(b ∨c ).(15)For all elements a ,b ,c of Boolean Y holds (a ∨b )∧c =a ∧c ∨b ∧c .(16)For all elements a ,b of Boolean Y holds ¬(a ∨b )=¬a ∧¬b .(17)For all elements a ,b of Boolean Y holds ¬(a ∧b )=¬a ∨¬b .Let us consider Y and let a ,b be elements of Boolean Y .The predicate a b is deﬁned as follows:(Def.15)For every element x of Y such that Pj (a ,x )=true holds Pj (b ,x )=true .Let us note that the predicate a b is reﬂexive.The following propositions are true:(18)For all elements a ,b ,c of Boolean Y holds if a b and b a ,then a =b and if a b andb c ,then a c .(19)For all elements a ,b of Boolean Y holds a ⇒b =true (Y )iff a b .(20)For all elements a ,b of Boolean Y holds a ⇔b =true (Y )iff a =b .(21)For every element a of Boolean Y holds false (Y ) a and a true (Y ).3.I NFIMUM AND S UPREMUMLet us consider Y and let a be an element of Boolean Y .The functor INF a yields an element of Boolean Y and is deﬁned as follows:(Def.16)INF a = true (Y ),if foreveryelement x of Y holds Pj (a ,x )=true ,false (Y ),otherwise.The functor SUP a yields an element of Boolean Y and is deﬁned by:(Def.17)SUP a = false (Y ),if foreveryelement x of Y holds Pj (a ,x )=false ,true (Y ),otherwise.We now state two propositions:(22)For every element a of Boolean Y holds ¬INF a =SUP ¬a and ¬SUP a =INF ¬a .(23)INF false (Y )=false (Y )and INF true (Y )=true (Y )and SUP false (Y )=false (Y )andSUP true (Y )=true (Y ).Let us consider Y.Observe that false(Y)is constant.Let us consider Y.One can check that true(Y)is constant.Let Y be a non empty set.Note that there exists an element of Boolean Y which is constant.We now state several propositions:(24)For every constant element a of Boolean Y holds a=false(Y)or a=true(Y).(25)For every constant element d of Boolean Y holds INF d=d and SUP d=d.(26)For all elements a,b of Boolean Y holds INF(a∧b)=INF a∧INF b and SUP(a∨b)=SUP a∨SUP b.(27)For every element a of Boolean Y and for every constant element d of Boolean Y holdsINF(d⇒a)=d⇒INF a and INF(a⇒d)=SUP a⇒d.(28)For every element a of Boolean Y and for every constant element d of Boolean Y holdsINF(d∨a)=d∨INF a and SUP(d∧a)=d∧SUP a and SUP(a∧d)=SUP a∧d.(29)For every element a of Boolean Y and for every element x of Y holds Pj(INF a,x) Pj(a,x).(30)For every element a of Boolean Y and for every element x of Y holds Pj(a,x) Pj(SUP a,x).4.B OOLEAN V ALUED F UNCTIONS AND P ARTITIONSLet us consider Y,let a be an element of Boolean Y,and let P1be a partition of Y.We say that a is dependent of P1if and only if:(Def.18)For every set F such that F∈P1and for all sets x1,x2such that x1∈F and x2∈F holds a(x1)=a(x2).Next we state two propositions:(31)For every element a of Boolean Y holds a is dependent of I(Y).(32)For every constant element a of Boolean Y holds a is dependent of O(Y).Let us consider Y and let P1be a partition of Y.We see that the element of P1is a subset of Y.Let us consider Y,let x be an element of Y,and let P1be a partition of Y.Then EqClass(x,P1) is an element of P1.We introduce Lift(x,P1)as a synonym of EqClass(x,P1).Let us consider Y,let a be an element of Boolean Y,and let P1be a partition of Y.The functor INF(a,P1)yields an element of Boolean Y and is deﬁned by the condition(Def.19).(Def.19)Let y be an element of Y.Then(i)if for every element x of Y such that x∈EqClass(y,P1)holds Pj(a,x)=true,thenPj(INF(a,P1),y)=true,and(ii)if it is not true that for every element x of Y such that x∈EqClass(y,P1)holds Pj(a,x)= true,then Pj(INF(a,P1),y)=false.Let us consider Y,let a be an element of Boolean Y,and let P1be a partition of Y.The functor SUP(a,P1)yields an element of Boolean Y and is deﬁned by the condition(Def.20).(Def.20)Let y be an element of Y.Then(i)if there exists an element x of Y such that x∈EqClass(y,P1)and Pj(a,x)=true,thenPj(SUP(a,P1),y)=true,and(ii)if it is not true that there exists an element x of Y such that x∈EqClass(y,P1)and Pj(a,x)= true,then Pj(SUP(a,P1),y)=false.Next we state a number of propositions:(33)For every element a of Boolean Y and for every partition P1of Y holds INF(a,P1)is depen-dent of P1.(34)For every element a of Boolean Y and for every partition P1of Y holds SUP(a,P1)is depen-dent of P1.(35)For every element a of Boolean Y and for every partition P1of Y holds INF(a,P1) a.(36)For every element a of Boolean Y and for every partition P1of Y holds a SUP(a,P1).(37)For every element a of Boolean Y and for every partition P1of Y holds¬INF(a,P1)=SUP(¬a,P1).(38)For every element a of Boolean Y holds INF(a,O(Y))=INF a.(39)For every element a of Boolean Y holds SUP(a,O(Y))=SUP a.(40)For every element a of Boolean Y holds INF(a,I(Y))=a.(41)For every element a of Boolean Y holds SUP(a,I(Y))=a.(42)For all elements a,b of Boolean Y and for every partition P1of Y holds INF(a∧b,P1)=INF(a,P1)∧INF(b,P1).(43)For all elements a,b of Boolean Y and for every partition P1of Y holds SUP(a∨b,P1)=SUP(a,P1)∨SUP(b,P1).Let us consider Y and let f be an element of Boolean Y.The functor GPart f yielding a partition of Y is deﬁned by:(Def.21)GPart f={{x;x ranges over elements of Y:f(x)=true},{x ;x ranges over elements of Y: f(x )=false}}\{/0}.The following two propositions are true:(44)For every element a of Boolean Y holds a is dependent of GPart a.(45)For every element a of Boolean Y and for every partition P1of Y such that a is dependent ofP1holds P1isﬁner than GPart a.R EFERENCES[1]Grzegorz Bancerek.Sequences of ordinal numbers.Journal of Formalized Mathematics,1,1989./JFM/Vol1/ordinal2.html.[2]Czesław Byli´n ski.Functions and their basic properties.Journal of Formalized Mathematics,1,1989./JFM/Vol1/funct_1.html.[3]Czesław Byli´n ski.Functions from a set to a set.Journal of Formalized Mathematics,1,1989./JFM/Vol1/funct_2.html.[4]Czesław Byli´n ski.Some basic properties of sets.Journal of Formalized Mathematics,1,1989./JFM/Vol1/zfmisc_1.html.[5]Shunichi Kobayashi and Kui Jia.A theory of partitions.Part I.Journal of Formalized Mathematics,10,1998./JFM/Vol10/partit1.html.[6]Jarosław Kotowicz.Monotone real sequences.Subsequences.Journal of Formalized Mathematics,1,1989./JFM/Vol1/seqm_3.html.[7]Adam Naumowicz and MariuszŁapi´n ski.On T1reﬂex of topological space.Journal of Formalized Mathematics,10,1998.http:///JFM/Vol10/t_1topsp.html.[8]Takaya Nishiyama and Yasuho Mizuhara.Binary arithmetics.Journal of Formalized Mathematics,5,1993./JFM/Vol5/binarith.html.[9]Beata Padlewska.Families of sets.Journal of Formalized Mathematics,1,1989./JFM/Vol1/setfam_1.html.[10]Konrad Raczkowski and PawełSadowski.Equivalence relations and classes of abstraction.Journal of Formalized Mathematics,1,1989./JFM/Vol1/eqrel_1.html.[11]Andrzej Trybulec.Tarski Grothendieck set theory.Journal of Formalized Mathematics,Axiomatics,1989./JFM/Axiomatics/tarski.html.[12]Andrzej Trybulec.Function domains and Frænkel operator.Journal of Formalized Mathematics,2,1990./JFM/Vol2/fraenkel.html.[13]Zinaida Trybulec.Properties of subsets.Journal of Formalized Mathematics,1,1989./JFM/Vol1/subset_1.html.[14]Edmund Woronowicz.Relations and their basic properties.Journal of Formalized Mathematics,1,1989./JFM/Vol1/relat_1.html.[15]Edmund Woronowicz.Interpretation and satisﬁability in theﬁrst order logic.Journal of Formalized Mathematics,2,1990.http:///JFM/Vol2/valuat_1.html.[16]Edmund Woronowicz.Many-argument relations.Journal of Formalized Mathematics,2,1990./JFM/Vol2/margrel1.html.Received October22,1998Published January2,2004。

chpt03

Page139 3Compact and Connected SetsIn this chapter,we study two of the most important and useful kinds ofsets in metric spaces and especially in R n.Intuitively,we want to say thata set in R n is compact when it is closed and is contained in a bounded re-gion,and that a set is connected when it is“in one piece.”Figure3.1givessome examples.As usual,it is necessary to turn these ideas into rigorousdeﬁnitions.In each case the most useful technical deﬁnition appears to bea little removed from our intuition,but in the end we will see that it is ingood accord with it.The fruitfulness of these notions will be revealed inChapter4,where they will be applied to the study of continuous functions.3.1CompactnessIn this section we give the general deﬁnition and properties of compactsets in metric spaces.A criterion for recognizing compact sets,called theHeine-Borel theorem,states that a set in R n is compact iﬀit is closedand bounded.This result,special to the metric space R n,is discussed in§3.2.Recall from our discussion of completeness of R n in Chapter1that everybounded sequence has a convergent subsequence.This can be rephrased:If A⊂R n is a closed and bounded set,then every sequence in A has a subsequence converging to a point of A.Historically,this was recognized tobe an important property of sets,and so was elevated to a deﬁnition.This140Chapter3Compact and Connected Setscompact noncompact noncompactconnectednot connectedpact and connected sets in R2property plays a crucial role in many basic theorems such as the existence of maxima and minima of continuous functions on closed intervals,as we shall see in Chapter4.3.1.1Deﬁnition.Let M be a metric space.A subset A⊂M is called sequentially compact if every sequence in A has a subsequence that con-verges to a point in A.This property is equivalent to another property,called compactness,that we shall now develop.This property is less obvious,and its equivalence to sequential compactness is far from clear,at least atﬁrst.Here is some terminology we need for our formal deﬁnition.Let M be a metric space and A⊂M a subset.A cover of A is a collection{U i}of sets whose union contains A;it is an open cover if each U i is open.A subcover of a given cover is a subcollection of{U i}whose union also contains A or, as we say,covers A;it is aﬁnite subcover if the subcollection contains only aﬁnite number of sets.Open covers are not necessarily countable collections of open sets.For example,the uncountable set of disks{D((x,0),1)|x∈R}in R2covers the real axis,and the subcollection of all disks D((n,0),1)centered at in-teger points on the real line forms a countable subcover.Note that the set of disks D((2n,0),1)centered at even integer points on the real line does not form a subcovering(why?).3.1Compactness141 3.1.2Deﬁnition.A subset A of a metric space M is called compact if every open cover of A has aﬁnite subcover.Here is theﬁrst major result,which links compactness and sequential compactness.3.1.3Theorem(Bolzano-Weierstrass Theorem).A subset of a metric space is compact iﬀit is sequentially compact.Some simple observations will help give a feel for compactness and for this theorem.First,a sequentially compact set must be closed.Indeed,if x n∈A converges to x∈M,then by assumption there is a subsequence converging to a point x0∈A;by uniqueness of limits,x=x0,and so A isclosed.Second,a sequentially compact set A must be bounded,for if not, there is a point x0∈A and a sequence x n∈A with d(x n,x0)≥n.Then x n cannot have any convergent subsequence.To show directly that a compact set is bounded,use the fact that for any x0∈A,the open balls D(x0,n), n=1,2,...,cover A,so there is aﬁnite subcover.Note that in the deﬁnitions,one can take A=M,in which case one just speaks of a compact metric space.We shall develop examples of compact spaces in due course.Another characterization of compactness relates to completeness.It is a useful technical tool used in the proof of the Bolzano-Weierstrass theorem.3.1.4Deﬁnition.A set A⊂M is called totally bounded if for each ε>0there is aﬁnite set{x1,...,x N}in M such that A⊂∪N i=1D(x i,ε).3.1.5Theorem.A metric space is compact if it is complete and totally bounded.Let A⊂M,and assume that M is complete.If we apply this theorem to the metric space A,we conclude that A is compact iﬀit is closed and totally bounded.In Theorem3.1.5,a few things are obvious,others less obvious.First, note that D(x i,ε)⊂D(x1,ε+d(x i,x1)),so that ifR=ε+max{d(x2,x1),...,d(x N,x1)},then A⊂D(x1,R)and so a totally bounded set is bounded.This is consis-tent with our earlier remark that compact sets are bounded.At this stage we do not have eﬀective methods for telling when a given set is compact.We will remedy this in the next section.3.1.6Example.The entire real line R is not compact,for it is un-bounded.Another reason is that{D(n,1)=]n−1,n+1[|n=0,±1,±2,...}is an open cover of R but does not have aﬁnite subcover(why?).142Chapter3Compact and Connected Sets3.1.7Example.Let A=]0,1].Find an open cover with noﬁnite sub-cover.Solution.Consider the open cover{]1/n,2[|n=1,2,3,...}.(Why does the union contain all of A?)It clearly cannot have aﬁnite subcover.This time,compactness fails because A is not closed;the point0is“missing”from A.This collection is not a cover for[0,1];in fact any open cover for [0,1]must have aﬁnite subcover,because,as we prove in the next section, [0,1]is compact.3.1.8Example.Give an example of a bounded and closed set that is not compact.Solution.Let M be any inﬁnite set with the discrete metric:d(x,y)=0if x=y and d(x,y)=1if x=y.Clearly,M⊂D(x0,2)for any x0∈M,and so M is bounded.Since it is already the entire metric space,it is closed. However,it is not compact.Indeed,{D(x,1/2)|x∈M}is an open cover with noﬁnite subcover.3.1.9Example.A collection of closed sets{Kα}in a metric space M is said to have theﬁnite intersection property for A if the intersection of anyﬁnite number of the Kαwith A is nonempty.Show that A⊂M is com-pact iﬀevery collection of closed sets with theﬁnite intersection property for A has nonempty intersection with A.Solution.First,assume A is compact.Let{F i}be a collection of closed sets and let U i=M\F i,so that U i is open.Suppose that A∩(∩∞i=1F i)=∅. Taking complements,this means that the U i cover A.Since the cover-ing is open,there is aﬁnite subcovering,say,A⊂U1∪···∪U N.Then A∩(F1∩···∩F N)=∅,and so{F i}does not have theﬁnite intersection property.Thus,if{F i}is a collection of closed sets with theﬁnite intersec-tion property,then A∩{F i}=∅.Conversely,let{U i}be an open covering of A and let F i=M\U i. Then A∩(∩∞i=1F i)=∅,and so,by assumption,{F i}cannot have theﬁ-nite intersection property for A.Thus,A∩(F1∩···∩F N)=∅for some members F1,...,F N of the collection.Hence,U1,...,U N is the required ﬁnite subcover and thus A is compact.Exercises3.1-1.Show that A⊂M is sequentially compact iﬀevery inﬁnite subset of A has an accumulation point in A.3.2The Heine-Borel Theorem1433.1-2.Prove that{(x,y)∈R2|0≤x<1,0≤y≤1}is not compact.3.1-3.Let M be complete and A⊂M be totally bounded.Show that cl(A)is compact.3.1-4.Let x k→x be a convergent sequence in a metric space and let A={x1,x2,...}∪{x}.(a)Show that A is compact.(b)Verify that every open cover of A has aﬁnite subcover.3.1-5.Let M be a set with the discrete metric.Show that any inﬁnite subset of M is noncompact.Why does this not contradict the statement in Exercise3.1-4?3.2The Heine-Borel TheoremIn Euclidean space we can easily tell if a set is compact from the following theorem:3.2.1Theorem(Heine-Borel Theorem).A set A⊂R n is compact iﬀit is closed and bounded.One half of this was already indicated in§3.1.In fact,a compact set is closed and bounded in any metric space.The converse must be special in view of Example3.1.8.Indeed,it is not even obvious that the closed inter-val[0,1]in R is compact.In fact,[0,1]is compact,and one of the proofs of the Heine-Borel theorem begins by treating this case.3.2.2Example.Determine which of the following are compact:(a){x∈R|x≥0}⊂R(b)[0,1]∪[2,3]⊂R(c){(x,y)∈R2|x2+y2<1}⊂R2Solution.(a)Noncompact,because it is unbounded.(b)Compact,because it is closed and bounded.(c)Noncompact,because it is not closed.3.2.3Example.Let x k be a sequence of points in R n with||x k||≤3for all k.Show that x k has a convergent subsequence.144Chapter3Compact and Connected SetsSolution.The set A={x∈R n|||x||≤3}is closed and bounded,and hence compact.Since x k∈A,we can apply the Bolzano-Weierstrass theo-rem to obtain the conclusion.3.2.4Example.In the deﬁnition of a compact set,can“every”be re-placed by“some”?Solution.No.Let A=R,and let the open cover consist of the single open set R.This has aﬁnite subcover,namely,itself,but being unbounded,R is not compact.3.2.5Example.Let A={0}∪{1,1/2,...,1/n,...}.Show directly thatA satisﬁes the deﬁnition of compactness.Solution.Let{U i}be an arbitrary open cover of A.We must show that there is aﬁnite subcover.The point0lies in one of the open sets—relabeling if needed,we can suppose that0∈U1.Since U1is open and1/n→0,there is an N such that1/N,1/(N+1),...lie in U1.Relabeling again if needed, suppose that1∈U2,...,1/(N−1)∈U N.Then U1,...,U N is aﬁnite sub-cover,since it is aﬁnite subcollection of the{U i}and it includes all of the points of A.Notice that if A were the set{1,1/2,...},then the argument would not work.In fact,this set is not closed,and so it is not compact.Exercises3.2-1.Which of the following sets are compact?(a){x∈R|0≤x≤1and x is irrational}(b){(x,y)∈R2|0≤x≤1}(c){(x,y)∈R2|xy≥1}∩{(x,y)|x2+y2<5}3.2-2.Let r1,r2,r3,...be an enumeration of the rational numbers in[0, 1].Show that there is a convergent subsequence.3.2-3.Let M={(x,y)∈R2|x2+y2≤1}with the standard metric. Show that A⊂M is compact iﬀA is closed.3.2-4.Let A be a bounded set in R n.Prove that cl(A)is compact.3.2-5.Let A be an inﬁnite set in R with a single accumulation point inA.Must A be compact?3.3Nested Set Property1453.3Nested Set PropertyThe next theorem is an important consequence of the Bolzano-Weierstrasstheorem.3.3.1Deﬁnition(Nested Set Property).Let F k be a sequence of com-pact nonempty sets in a metric space M such that F k+1⊂F k for allk=1,2,....Then there is at least one point in∩∞k=1F k.Intuitively,the sets F k are nonempty and decreasing,and so it seemsreasonable that there should be a point in all of them.However,if the F kare not compact,then the intersection can be empty(see Example3.3.4).Thus,the actual proof requires more care.To prove the nested set property using the Bolzano-Weierstrass theorem,pick x k∈F k for each k.The sequence x k has a convergent subsequence,since it lies in the compact set F1.The limit point lies in all of the sets F kbecause they are closed(see Figure3.3.1).An alternative proof is given atthe end of the chapter.some itemsin this nextgraphic aremissing3Figure3.3.1.Nested set propertyOne can rephrase the nested set property in terms of“growing sets”thisway.Let U k=M\F k,so that the U k are open and U k+1⊃U k.Then∪∞k=1U k=M is equivalent to∩∞k=1F k=∅.Thus,if M is a metric spaceand the open sets U k are increasing—i.e.,U k+1⊃U k—and have compact complements,then the union of the U k is not all of M.3.3.2Example.Let M be the unit sphere in R3,M={(x,y,z)|x2+y2+z2=1}with the standard metric.Let U i be the portion ofM strictly below latitude90◦−10/i,i=1,2,3,...,as in Figure3.3.2.The metric space M is compact(why?),and,consistent with the preced-ing remarks,the union of the U i is not all of M,since it excludes the northpole.146Chapter3Compact and Connected Sets° – 10/iMUiFigure3.3.2.An increasing sequence of open sets on the unit sphere3.3.3Example.Verify the nested set property for F k=[0,1/k]⊂R.Solution.Each F k is compact,and F k+1⊂F k.The intersection is{0}, which is nonempty.3.3.4Example.Is the nested set property true if“compact nonempty”is replaced by“open nonempty”or“closed nonempty”?Solution.No.Let F k=]k,∞[or[k,∞[.3.3.5Example.A more exotic family of decreasing compact sets F n, for which∩∞n=1F n is quite complicated,is obtained by removing successive triangles from a given triangle in the plane,as in Figure3.3.3.Exercises3.3-1.Verify the nested set property for F k={x∈R|x≥0,2≤x2≤2+1/k}.3.3-2.Is the nested set property true if“compact nonempty”is replaced by“open bounded nonempty”?3.3-3.Let x k→x be a convergent sequence in a metric space.Verify the validity of the nested set property for F k={x l|l≥k}∪{x}.What happens if F k={x l|l≥k}?3.3-4.Let x k→x be a convergent sequence in a metric space.Let A bea family of closed sets with the property that for each A∈A,there is an N such that k≥N implies x k∈A.Prove that x∈∩A.3.4Path-Connected Sets147Figure3.3.3.Sierpinski’s gasket3.4Path-Connected SetsThe second important topic to be discussed in this chapter is connected-ness.We know intuitively those sets we would like to call“connected.”However,our intuition can fail in judging more complicated sets.For ex-ample,how do we decide whether the set{(x,sin(1/x))|x>0}∪{(0,y)| y∈[−1,1]}⊂R2is connected(see Figure3.4.1)?Therefore,we seek a sound mathematical deﬁnition we can depend on.yFigure3.4.1.Connected?148Chapter3Compact and Connected SetsThere are,in fact,two diﬀerent(but closely related)notions of con-nectedness.The more intuitive and applicable of these is that of path-connectedness,and so we begin with it.Our deﬁnition mustﬁrst deﬁne what is meant by a curve(or path)joining two points.3.4.1Deﬁnition.We call a mapϕ:[a,b]→M of an interval[a,b] into a metric space M continuous if(t k→t)implies(ϕ(t k)→ϕ(t)) for every sequence t k in[a,b]converging to some t∈[a,b].(Students will recall from their calculus courses that,intuitively,a continuous function has no“breaks”or“jumps”in its graph.)A continuous path joining two points x,y in a metric space M is a mappingϕ:[a,b]→M such that ϕ(a)=x,ϕ(b)=y,andϕis continuous.Here x may or may not equal y, and b≥a.A pathϕis said to lie in a set A ifϕ(t)∈A for all t∈[a,b]. See Figure3.4.2.We say that a set is path-connected if every two points in the set can be joined by a continuous path lying in the set.Figure3.4.2.A curve joining x and y in AFor example,it is evident that the region A in Figure3.4.2is path-connected.Another path-connected set is the interval[0,1].To prove this, let x,y∈[0,1]and deﬁneϕ:[0,1]→R byϕ(t)=(y−x)t+x.This is a continuous path connecting x and y,and it lies in[0,1].Using the above deﬁnition of path-connected,a little thought will con-vince the reader that the set in Figure3.4.1is not path-connected,although this fact is not obvious.Most of the time it is easy to determine whether a set is path-connected,by seeing whether any two points can be joined by a continuous curve lying in the set,which is usually clear geometrically.The second notion of connectedness is harder to check directly but will be very useful.It appears in§3.5.3.4Path-Connected Sets 1493.4.2Example.Which of the following sets are path-connected?(a)[0,3](b)[1,2]∪[3,4](c){(x,y )∈R 2|0<x ≤1}(d){(x,y )∈R 2|0<x 2+y 2≤1}Solution.Only (b)is not path-connected,as is clear from a study of Fig-ure 3.4.3.013234Figure 3.4.3.Sets for Example 3.4.23.4.3Example.Must a path-connected set be open or closed?Solution.No;e.g.,[0,1],]0,1[,[0,1[are all path-connected.3.4.4Example.Let ϕ:[0,1]→R 2be a continuous path,and C =ϕ([0,1]).Show that C is path-connected.Solution.This is intuitively clear,for we can use the path ϕitself to join two points in C.Precisely,if x =ϕ(a ),y =ϕ(b ),where 0≤a ≤b ≤1,let c :[a,b ]→R 2be deﬁned by c (t )=ϕ(t ).Then c is a path joining x to y and c lies in C.150Chapter3Compact and Connected SetsExercises3.4-1.Determine which of the following sets are path-connected:(a){x∈[0,1]|x is rational}(b){(x,y)∈R2|xy≥1and x>1}∪{(x,y)∈R2|xy≤1and x≤1}(c){(x,y,z)∈R3|x2+y2≤z}∪{(x,y,z)|x2+y2+z2>3}(d){(x,y)∈R2|0≤x<1}∪{(x,0)|1<x<2}3.4-2.Let A⊂R be path-connected.Give plausible arguments that A must be an interval(closed,open,or half-open).Are things as simple in R2?3.4-3.Letϕ:[a,b]→R3be a continuous path and a<c<d<b.Let C={ϕ(t)|c≤t≤d}.Mustϕ−1(C)be path-connected?3.5Connec t ed Se t sThere is another way of saying that a set has more than one piece that is both more sophisticated and more powerful than using path connectedness.3.5.1Deﬁnition.Let A be a subset of a metric space M.Two open sets U,V are said to separate A if they satisfy these conditions:(a)U∩V∩A=∅.(b)A∩U=∅.(c)A∩V=∅.(d)A⊂U∪V.See Figure3.5.1.We say that A is disconnected if such sets exist,and if such sets do not exist,we say that A is connected.The set in Figure3.4.1can be shown to be connected but not path-connected;thus the two notions are not the same.However,there is a valid relation between the two ideas,which is presented in the next theorem. 3.5.2Theorem.Path-connected sets are connected.Use of this theorem is perhaps the easiest way to identify a connected set.The theorem is reasonable,intuitively.In fact,the(false)converse the-orem is also“reasonable.”Here,then,is an example of two notions that are closely related and that are intuitively almost identical,but whose true relationship requires more care to discern.If a set is not connected(and hence not path-connected),we can divide it up into pieces,or components.More precisely,a component of a set A is a connected subset A0⊂A such that there is no connected set in3.5Connected Sets151Figure3.5.1.A is neither connected nor path-connectedA containing A0other than A0itself.Thus we see that a component is a maximal connected subset.One can deﬁne path component in a similar way,using path-connectedness instead of connectedness.Some properties of components are given in the exercises at the end of the chapter.3.5.3Example.Is Z={...,−2,−1,0,1,2,3,...}⊂R connected? Solution.No,for if U=]1/2,∞[and V=]−∞,1/4[,then Z⊂U∪V, Z∩U={1,2,3,...}=∅,Z∩V={...,−2,−1,0}=∅,and Z∩U∩V=∅. Hence Z is disconnected(i.e.,is not connected).It is also evident that Z is not path-connected,but this fact alone cannot be used to conclude that Z is not connected.3.5.4Example.Is{(x,y)∈R2|0<x2+y2≤1}connected?Solution.As in Example3.4.2(d),we know that this set is path-connected. Hence,by Theorem3.5.2,it is connected.To prove this directly would be more diﬃcult.Exercises3.5-1.Is[0,1]∪]2,3]connected?Prove or disprove.3.5-2.Is{(x,y)∈R2|0≤x≤1}∪{(x,0)|1<x<2}connected? Prove or disprove.3.5-3.Let A⊂R2be path-connected.Regarding A as a subset of the xy-plane in R3,show that A is still path-connected.Can you make a similar argument for A connected?152Chapter3Compact and Connected Sets3.5-4.Discuss the components of(a)[0,1]∪[2,3]⊂R(b)Z={...,−2,−1,0,1,2,...}⊂R(c){x∈[0,1]|x is rational}⊂RTheorem Proofs for Chapter33.1.3Theorem(Bolzano-Weierstrass Theorem).A subset of a metric space is compact iﬀit is sequentially compact.Proof.We begin with two lemmas.Lemma1.A compact set A⊂M is closed.Proof.We will show that M\A is open.Let x∈M\A and consider the following collection of open sets:U n={y|d(y,x)>1/n}.Since every y∈M with y=x has d(y,x)>0,y lies in some U n.Thus,the U n cover A,and so there must be aﬁnite subcover.One of these has a largest index, say,U N.Ifε=1/N,then,by construction,D(x,1/N)⊂M\A,and so M\A is open.Lemma2.If M is a compact metric space and B⊂M is closed,then B is compact.Proof.Let{U i}be an open covering of B and let V=M\B,so that V is open.Thus{U i,V}is an open cover of M.Therefore,M has aﬁnite cover, say,{U1,...,U N,V}.Then{U1,...,U N}is aﬁnite open cover of B.Proof of3.1.3.Let A be compact.Assume there exists a sequence x k∈A that has no convergent subsequences.In particular,this means that x k has inﬁnitely many distinct points,say,y1,y2,....Since there are no con-vergent subsequences,there is some neighborhood U k of y k containing no other y i.This is because if every neighborhood of y k contained another y j, we could,by choosing the neighborhoods D(y k,1/m),m=1,2,...,select a subsequence converging to y k.We claim that the set{y1,y2,...}is closed. Indeed,it has no accumulation points,by the assumption that there are no convergent subsequences.Applying Lemma2to{y1,y2,...}as a subset of A,weﬁnd that{y1,y2,...}is compact.But{U k}is an open cover that has noﬁnite subcover,a contradiction.Thus x k has a convergent subsequence. The limit lies in A,since A is closed,by Lemma1.Theorem Proofs for Chapter 3153Conversely,assume that A is sequentially compact.To prove that A is compact,let {U i }be an open cover of A.We need to prove that this has a ﬁnite subcover.To show this,we proceed in several steps.Lemma 3.There is an r >0such that for each y ∈A,D (y,r )⊂U i for some U i .The number r is called a Lebesgue number for the covering.The inﬁmum of all such r is called the Lebesgue number for the covering.Proof.If not,then for every integer n,there is some y n such that D (y n ,1/n )is not contained in any U i .By hypothesis,y n has a convergent subsequence,say,z n →z ∈A.Since the U i cover A ,z ∈U i 0for some U i 0.Choose ε>0such that D (z,ε)⊂U i 0,which is possible since U i 0is open.Choose N large enough so that d (z N ,z )<ε/2and 1/N <ε/2.Then D (z N ,1/N )⊂U i 0,a contradiction. Lemma 4.A is totally bounded (see Deﬁnition 3.1.4).Proof.If A is not totally bounded,then for some ε>0we cannot cover A with ﬁnitely many disks.Choose y 1∈A and y 2∈A \D (y 1,ε).By as-sumption,we can repeat;choose y n ∈A \[D (y 1,ε)∪···∪D (y n −1,ε)].This is a sequence with d (y n ,y m )≥εfor all n and m,and so y n has no conver-gent subsequence,a contradiction to the assumption that A is sequentially compact.To complete our proof,let r be as in Lemma 3.By Lemma 4we can writeA ⊂D (y 1,r )∪···∪D (y n ,r )for ﬁnitely many y j .By Lemma 3,D (y j ,r )⊂U i j ,j =1,...,n,for some index i j .Then U i 1,...,U i n cover A.3.1.5Theorem.A metric space is compact iﬀit is complete and totallybounded.Proof.First assume that M is compact.By 3.1.3,it is sequentially com-pact.Thus,if x k is a Cauchy sequence,it has a convergent subsequence,and so,as in 1.4.7,the whole sequence converges.Thus M is complete.It is also totally bounded,by Lemma 4.Conversely,assume that M is complete and totally bounded.By 3.1.3,it is enough to show that M is sequentially compact.Let y k be a sequence in M.We can assume that the y k are all distinct,for if y k has inﬁnitely many repetitions,there is a trivially convergent subsequence,and if there are ﬁnite repetitions we may delete them.Given an integer N,cover M with ﬁnitely many balls,D (x L 1,1/N ),...,D (x L N ,1/N ).An inﬁnite number of the y k lie in one of these balls.Start with N =1.Write M =D (x L 1,1)∪···∪D (x L N ,1),and so we can select a subsequence of y k lying entirely in one of these balls.Repeat for N =2,getting a further subsequence lying in a ﬁxed ball of radius 1/2,and so on.Now choose the “diagonal”subsequence,154Chapter3Compact and Connected Setstheﬁrst member from theﬁrst sequence,the second from the second,and so on.This sequence is Cauchy and since M is complete,it converges.3.2.1Theorem.A set A⊂R n is compact iﬀit is closed and bounded.Proof.We have already proved that compact sets are closed and bounded. We must now show that a set A⊂R n is compact if it is closed and bounded. We will give two proofs of this.First Proof.This proof is based on the Bolzano-Weierstrass theorem and the fact that any bounded sequence in R has a convergent subsequence, proved in1.4.3.In fact,we shall prove that a closed and bounded set A is sequentially compact.Let x k=(x1k,x2k,...,x n k)∈R n be a sequence.SinceA is bounded,x1k has a convergent subsequence,say,x1f1(k).Then x2f1(k)has a convergent subsequence,say x2f2(k).Continuing,we get a furthersubsequence x fn(k)=(x1fn(k),...,x n fn(k)),all of whose components con-verge.Thus x fn(k)converges in R n.The limit lies in A since A is closed.Thus A is sequentially compact,and so is compact. Second Proof.This proof uses the deﬁnition of compactness in termsof open covers directly.We begin with a special case:Lemma1.Closed intervals[a,b]in R are compact.Proof.Let U={U i}be an open covering of[a,b].DeﬁneC={x∈[a,b]|the set[a,x]can be covered by aﬁnite collection of the U i}. We want to show that C=[a,b].To this end,let c=sup(C).The sup exists because C=∅(since a∈C)and C is bounded above by b.Since a∈C andb is an upper bound for C,c∈[a,b],by deﬁnition of sup(C).Suppose c∈U i0;such a U iexists,since the U i’s cover[a,b].Since U iis open,there is anε>0such that]c−ε,c+ε[⊂U i0.Since c=sup(C),there exists an x∈Csuch that c−ε<x≤c(see Proposition1.3.2).Because x∈C,[a,x]has a ﬁnite subcover,say,U1,...,U N;then[a,c+ε/2]also has theﬁnite subcoverU1,...,U N,U i0.Thus we conclude that c∈C and moreover that c=b.Indeed,if c<b,we would get a member of C larger than c,since[a,c+ε/2] has aﬁnite subcover.The latter cannot happen,since c=sup(C).=Note.Why does this proof fail for]a,b],[a,b[or[a,∞[?Lemma2.If A⊂R n is compact and x0∈R m,then A×{x0}⊂R n×R m is compact.Theorem Proofs for Chapter 3155Proof.Let U be an open cover of A ×{x 0},andV ={V |V ={y |(y,x 0)∈U },for some U ∈U}.Then V is an open cover of A in R n ,and hence V has a ﬁnite subcover of A,say,V ={V 1,...,V k }.Each V i ∈V corresponds to a U i ∈U ,and U ={U 1,...,U k }is then a ﬁnite subcover in R n ×R m of A ×{x 0}.The next step is an induction argument.Lemma 3.If [−R,R ]n −1⊂R n −1is compact,then[−R,R ]n ⊂R n is compact,where [−R,R ]n =[−R,R ]×···×[−R,R ],n times.Proof.Suppose that [−R,R ]n −1is compact and that U is an open cover of [−R,R ]n .DeﬁneS ={x ∈[−R,R ]|[−R,R ]n −1×[−R,x ]⊂R n has a ﬁnite subcover in U}.Now −R ∈S,since [−R,R ]n −1is compact,by hypothesis,and so,by Lemma 2,[−R,R ]n −1×{−R }has a ﬁnite subcover in U .Since S is bounded above by R,it has a supremum,say,x 0.We will show that x 0=R,which will prove the lemma.Let U ⊂U be a ﬁnite subcover of [−R,R ]n −1×{x 0}.For each point (y,x 0)∈[−R,R ]n −1×{x 0},there exists εy >0such that D ((y,x 0),√2εy )is covered by U .BecauseV y =D (y,εy )×]x 0−εy ,x 0+εy [⊂D ((y,x 0),√y ),it is covered by U .Consider the open cover V ={V y |y ∈[−R,R ]n −1}of [−R,R ]n −1×{x 0}.By Lemma 2,V has a ﬁnite subcover of [−R,R ]n −1×{x 0},say {V y 1,...,V y N }.Let ε=inf {εy 1,...,εy N }.Then[−R,R ]n −1×]x 0−ε,x 0+ε[⊂∞i =1V y i ,and so [−R,R ]n −1×]x 0−ε,x 0+ε[is covered by U .With this ε,there exists x ∈S such that x 0−ε<x ≤x 0.Since x ∈S,there exists a ﬁnite subcover U ⊂U which covers [−R,R ]n −1×[−R,x ],and U ∪U is a ﬁnite cover of [−R,R ]n −1×[−R,x 0+ε[.Thus x 0∈S.Suppose x 0<R ;then choose δsuch that x 0+δ<R and x 0+δ<x 0+ε.Thus [−R,R ]n −1×[−R,x 0+δ]is covered by U ∪U ,and x 0+δ∈S,a contradiction,and therefore x 0=R.To conclude the proof of the Heine-Borel theorem,let A ⊂R n be closedand bounded.Since it is bounded,there is an R >0such that A ⊂。

On the strength of Ramsey’s theorem for pairs

ON THE STRENGTH OF RAMSEY’S THEOREM FORPAIRSPETER A.CHOLAK,CARL G.JOCKUSCH,JR.,ANDTHEODORE A.SLAMANAbstract.We study the proof–theoretic strength and eﬀectivecontent of the inﬁnite form of Ramsey’s theorem for pairs.Let RT n kdenote Ramsey’s theorem for k–colorings of n–element sets,and letRT n<∞denote(∀k)RT n k.Our main result on computability is:Forany n≥2and any computable(recursive)k–coloring of the n–element sets of natural numbers,there is an inﬁnite homogeneousset X with X ≤T0(n).Let IΣn and BΣn denote theΣn inductionand bounding schemes,respectively.Adapting the case n=2ofthe above result(where X is low2)to models of arithmetic enablesus to show that RCA0+IΣ2+RT22is conservative over RCA0+IΣ2forΠ11statements and that RCA0+IΣ3+RT2<∞isΠ11-conservativeover RCA0+IΣ3.It follows that RCA0+RT22does not imply BΣ3.In contrast,we reprove old result of Hirst that RCA0+RT2<∞does imply BΣ3,and so RT2<∞is strictly stronger than RT22overRCA0.1.IntroductionRamsey’s theorem was discovered by Ramsey[1930]and used by him to solve a decision problem in logic.Subsequently it has been an important tool in logic and combinatorics.Deﬁnition1.1.i.[X]n={Y⊆X:|Y|=n}.ii.A k–coloring C of[X]n is a function from[X]n into a set of size k.iii.A set H⊆X is homogeneous for a k–coloring C of[X]n if C is constant on[H]n,i.e.all n–element subsets of H are assigned the same color by C.1991Mathematics Subject Classiﬁcation.Primary03F3503C6203D3003D80.Key words and phrases.Ramsey’s Theorem,conservation,reverse mathematics, recursion theory,computability theory.Research partially supported NSF Grants DMS-96-3465(Cholak),DMS-95-03398and DMS-98-03073(Jockusch),and DMS-97-96121(Slaman).12P.CHOLAK,C.JOCKUSCH AND T.SLAMANRamsey’s Theorem.For all k and n,every k-coloring of[N]n hasan inﬁnite homogeneous set.An extensive treatment of Ramsey’s Theorem,emphasizing itsﬁ-nite version,may be found in Graham,Rothschild and Spencer[1990],where many related results and applications are also discussed.There are(at least)two ways to use the tools of mathematical logicto analyze Ramsey’s theorem.One is via computability theory(orequivalently recursion theory):Study the complexity(in terms of thearithmetical hierarchy or degrees)of inﬁnite homogeneous sets for acoloring C relative to that of C.(For simplicity,we can assume thatC is computable(recursive)and relativize.)The other is via reverse mathematics:Study the proof–theoretic strength of Ramsey’s theorem(and its natural special cases)as a formal statement in second orderarithmetic.There has been much work done along these lines.For example,con-sider the independent work by Jockusch[1972],Seetapun,and Slaman(see Seetapun and Slaman[1995]).Our task in this paper is to reviewbrieﬂy the work that has been done and further this analysis.Before getting into details we mention two themes in this work thatwe would like to make explicit.Theﬁrst is that results in computabilitytheory are sometimes the forerunners of results in reverse mathemat-ics.This is certainly the case for Weak K¨o nig’s Lemma and almostall versions of Ramsey’s Theorem.The second theme is the use ofpaths through trees,more speciﬁcally Weak K¨o nig’s Lemma,the LowBasis Theorem,and Scott sets.Almost all of our results use one ormore of these three items in its statement or proof.Whether this useis necessary is unknown.In Section2,there is a brief summary ofprevious work on the analysis of K¨o nig’s Lemma and the inﬁnite formof Ramsey’s Theorem in terms of computability theory and of reversemathematics.Our starting point is the following result,which refutes an old con-jecture of Jockusch(see Jockusch[1972,Corollary4.7]or the secondparagraph after Theorem2.5).Theorem3.1.For any computable coloring of the unordered pairs ofnatural numbers withﬁnitely many colors,there is an inﬁnite low2homogeneous set X,i.e.,X ≤T0 .The proof is not simply an eﬀectivization of the standard proof ofRamsey’s Theorem.Instead,theﬁrst step is to restrict the given com-putable coloring to a low2r–cohesive set A,which exists by Jockuschand Stephan[1993],Theorem2.5.Since for any a the color of theRAMSEY’S THEOREM3 pair{a,b}is independent of b for suﬃciently large b∈A,the color-ing induces a coloring of[A]1which is∆0,A2.Then the relativizationto A of the following new result easily yields the desired inﬁnite low2 homogeneous set.Theorem3.7.If A1,A2,...,A n are∆02sets and∪ni=1A i=N,thensome A i has an inﬁnite low2subset.We give two proofs of the above result,with the common elements of the two proofs presented in Section3.Theﬁrst proof,which was our original proof,is technically easier since it uses the Low Basis theorem to reduce the problem of control-ling the second jump of the constructed set to the easier problem of controlling theﬁrst jump of the constructed set.(A similar approach was used in Jockusch and Stephan[1993]to construct a low2r–cohesive set.)In Section4we will present thisﬁrst proof and also,for the con-venience of the reader,a construction of a low2r–cohesive set based on control of theﬁrst jump.This“ﬁrst jump”method also yields inter-esting additional information on the jumps of degrees of homogeneous sets(see Section12).Our second proof,which will be presented in Section5,is more direct and also somewhat more complicated.It proceeds by direct control of the second jump of the constructed set.It gives no additional information on degrees of homogeneous sets,and the reader interested only in the computability aspect of this paper could well omit reading it.We also give a construction of a low2r-cohesive set using direct control of the second jump.As above,this is more complicated than the construction used in Jockusch and Stephan[1993].The reason for giving these more involved constructions is that they seem to be more suitable to adapting to models of arithmetic to obtain results in reverse mathematics as described below.In Section6,we quickly introduce the reader to second order arith-metic.(The reader unfamiliar with second order arithmetic may want to start there.)A listing of the needed statements of second order arith-metic and the relationships among them can be found in Section7.We will assume that the reader is somewhat familiar with computability theory;a good introduction is Soare[1987].In Section8,we discuss some results concerning Weak K¨o nig’s Lemma;a reasonable portion of this section was known previously but much of it is new.Sections9–11present our conservation theorems for Ramsey’s The-orem for pairs and related principles.Let X→[X]nk be the statement“every k-coloring of[X]n has an inﬁnite homogeneous set.”Thus,Ram-sey’s Theorem states for all k and n,N→[N]nk .RT nkis the statement4P.CHOLAK,C.JOCKUSCH AND T.SLAMANin the language of second order arithmetic“for all k-colorings of[N]n there is an inﬁnite homogeneous set H.”We adapt the forcing used in the“second jump”constructions to models of arithmetic to produce a notion of forcing for adding inﬁnite homogeneous sets to models of second order arithmetic while preserving the appropriate level of induction.We were led to this notion of forcing by a conjecture of Slaman(see Conjecture2.12or Seetapun and Slaman [1995]).(We do not know how to do this for the“ﬁrst jump”proofs.) Using this notion of forcing we obtain the following result.Theorem10.1.RCA0+IΣ2+RT22+WKL isΠ11-conservative overRCA0+IΣ2.This means that anyΠ11statement provable from RCA0+IΣ2+RT22is provable from just RCA0+IΣ2.The following corollary answers the second part of Seetapun and Slaman[1995,Question4.3].Corollary1.2.RT22does not imply PA over RCA0.This improves Seetapun’s result(see Seetapun and Slaman[1995])that RT22does not imply ACA0over RCA0.In the same paper,Slamanshowed in Theorem 3.6that RCA0+RT22is notΠ04–conservativeover RCA0.We should also mention that Hirst[1987]proved thatRCA0+RT22is notΣ03–conservative over RCA0(see Theorem2.11formore details).It turned out that our proof–theoretic results(but not the corre-sponding results in computability theory)are sensitive to whether our colorings use two colors or an arbitraryﬁnite number of colors.X→[ω]n<∞is the statement that“for all k,for all k-colorings of[X]n there is an inﬁnite homogeneous set.”Ramsey’s Theorem impliesfor all k and n,N→[ω]n<∞.RT n<∞is the statement in the languageof second order arithmetic“for all k,for all k-colorings of[N]n there is an inﬁnite homogeneous set H.”Using a modiﬁcation of the above mentioned notion of forcing(work-ing over a Scott set),we proved the following result.Theorem11.1.RCA0+IΣ3+RT2<∞+WKL isΠ11-conservative overRCA0+IΣ3.Thus anyΠ11statement provable from RCA0+IΣ3+RT2<∞+WKLis provable from just RCA0+IΣ3.So RT2<∞does not imply PA overRCA0.In addition,we reprove the following result of Hirst[1987, Theorem6.11].Our proof is a corollary of a slightly stronger theorem. Corollary11.5(Hirst[1987,Theorem6.11]).RCA0+RT2<∞BΣ3.RAMSEY’S THEOREM5 Since IΣ2is stricter weaker than BΣ3(over RCA0)(see Kaye[1991]or H´a jek and Pudl´a k[1993]),it follows that RT22does not imply RT2<∞over RCA0.Theorem3.1also leads to further results on computability and Ram-sey’s theorem which are covered in Section12.For example,the fol-lowing result is obtained for colorings of n–tuples:Theorem12.1.For each n≥2and each computable2-coloring of [N]n,there is an inﬁnite homogeneous set A with A ≤T0(n).Other results on computability include a characterization of the de-grees d such that every computable2–coloring of[N]2has an inﬁnite homogeneous set with jump of degree d(Corollary12.6)and a result combining cone avoidance with some control of theﬁrst jump of an inﬁnite homogeneous set(Theorem12.2).We do not succeed in obtaining a complete understanding of the proof–theoretic strength of Ramsey’s theorem for pairs or of the degrees of inﬁnite homogeneous sets for computable2–colorings of pairs.A number of open questions are listed in theﬁnal section.One theme of this paper is the close relationship between results in computability theory and results in reverse mathematics.Of course, this relationship has turned up in many other contexts,too.We hope that readers will be interested in both aspects of this paper.However, the reader interested only in the computability aspect need read only Sections1–4and12–13.The reader interested only in the reverse mathematics aspect need read only Sections1–3,5–11,and13.2.HistoryThis paper continues a stream of work on analysis of the eﬀective content of mathematical statements and corresponding work on the strength of these statements within second order arithmetic.Here we give a brief summary of some closely related previous work in this area. For further information,see Simpson[1998].See Section6for a summary of the subsystems of second–order arith-metic we shall consider.More extensive treatments may be found in Friedman[1975]and Simpson[1998].Here we brieﬂy remind the reader that our base theory is RCA0,which is based on algebraic axioms andthe schemes of∆01–comprehension andΣ1induction.Theω–models ofRCA0are those nonempty subsets of P(N)closed under⊕and closed downwards under≤T.The stronger system ACA0includes the arith-metic comprehension scheme ACA.Theω–models of ACA0are theωmodels of RCA0which are closed under the jump operation.6P.CHOLAK,C.JOCKUSCH AND T.SLAMANBefore getting to the analysis of Ramsey’s theorem,we consider K¨o nig’s lemma,which in fact will play an important role in this pa-per.Of course,K¨o nig’s lemma is the assertion that any inﬁnite,ﬁnite branching tree has an inﬁnite path.We shall actually be concerned with the case where there is an eﬀective bound on the branching.Let Weak K¨o nig’s Lemma be the assertion that every inﬁnite tree in2<ωhas an inﬁnite path,and let WKL0be RCA0+Weak K¨o nig’s Lemma. It is easy to construct inﬁnite computable trees in2<ωwith no inﬁnite computable paths,using,for example,the existence of disjoint com-putably enumerable sets which are not separable by any computable set.From this it follows that Weak K¨o nig’s Lemma cannot be proved in RCA0.In the other direction G.Kreisel proved the Kreisel basis theorem: Any inﬁnite computable tree in2<ωhas an inﬁnite path computable from the halting problem0 .The corresponding result in reverse math-ematics,due to Steve Simpson,is that Weak K¨o nig’s Lemma can be proved in the system ACA0.Theorem2.1(Jockusch and Soare[1972]).For any noncomputable sets C0,C1,...and any inﬁnite computable tree T⊆2<ωthere is an inﬁnite path f through T such that(∀i)[C i≤T f].The corresponding result in reverse mathematics is the following. Corollary2.2(Simpson[1998]).Arithmetic Comprehension is not provable in WKL0.The following result,due to Jockusch and Soare,is known as the Low Basis Theorem.Theorem8.1(Jockusch and Soare[1972,Theorem2.1]).Any inﬁ-nite computable tree in2<ωhas an inﬁnite low path f,i.e.,f ≤T0 . The forcing conditions used to prove the above result are trees,and this forcing was adapted by Leo Harrington to obtain the following result.Theorem8.4(Harrington,see Simpson[1998]).AnyΠ11statementprovable from WKL0is provable from just RCA0.SinceΣ2induction(without parameters)is not provable in RCA0(see H´a jek and Pudl´a k[1993]),it follows thatΣ2–induction is not provable from WKL0.We now consider the analysis of Ramsey’s Theorem.Theﬁrst re-sult concerning the eﬀective content of the inﬁnite form of Ramsey’s Theorem was obtained in Specker[1971].RAMSEY’S THEOREM7 Theorem2.3(Specker[1971]).There is a computable2–coloring of [N]2with no inﬁnite computable homogeneous set.Since the family of computable sets is anω–model of RCA0,there is an immediate corollary.Corollary2.4(Specker[1971]).RT22is not provable in RCA0.The next work in the area was due to Jockusch.Theorem2.5(Jockusch[1972]).i.For any n and k,any computable k–coloring of[N]n has an inﬁniteΠ0n homogeneous set.ii.For any n≥2,there is a computable2–coloring of[N]n which hasno inﬁniteΣ0n homogeneous set.iii.For any n and k and any computable k–coloring of[N]n,there is an inﬁnite homogeneous set A with A ≤T0(n).iv.For each n≥2,there is a computable2–coloring of[N]n such that 0(n−2)≤T A for each inﬁnite homogeneous set A.Theﬁrst part was proved by induction on n,using aﬁnite injury priority argument for the case n=2and the Low Basis theorem for the induction step.Note that there is a slight gap between the third and fourth items.Fix a2-coloring of[N]2.The third item tells us that there is an inﬁnite homogeneous set A such that A ≤T0 .Jockusch[1972]con-jectured that this cannot be improved to give the existence of an inﬁnite homogeneous set A such that A ≤T0 .By Theorem3.1,we now know that this conjecture was false.Simpson obtained results in reverse mathematics which are related to Theorem2.5.Corollary2.6(Simpson[1998]).i.For each n≥3and k≥2(both n and kﬁxed),the statementsRT nk and RT n<∞are equivalent to ACA0over RCA0.ii.The statement RT is not provable in ACA0.iii.RT does not prove ATR0.iv.ATR0proves RT.(Actually there are stronger results along this line in Simpson[1998].)Sketch of the proof:i.Fix n≥3.A relativized version of Theorem2.5iv“says”that any model of RCA0+RT n2must be closed under the jumpoperator.Hence any such model must contain all sets arithmetically deﬁnable from the reals in it.A relativized version of Theorem2.5i“says”that every coloring of n-tuples has a homogeneous set which is arithmetic in the coloring.Hence8P.CHOLAK,C.JOCKUSCH AND T.SLAMANin any model of ACA0every coloring of n-tuples has a homogeneous set.ii.A relativized version of Theorem2.5iv“says”that any model ofRT n2is closed under the(n−2)-jump.But one canﬁnd non–standardmodels of ACA0which are not closed under the(n)-jump for any non-standard integer n.iii.The family of all arithmetic sets is anω–model of ACA0+RT. Since this is not a model of ATR0,the claim follows.iv.Any model of ATR0is closed under the(n)-jump,for any n in the model.Theorem2.5ii“says”that every k-coloring of[X]n has a homogeneous set which is Turing reducible to X(n).If X is in M then X(n)is in M and therefore a homogeneous set for the above coloring is in M.This is how things stood for twenty years.During that time,thestrength of RT22remained a mystery.Sometimes this was phrasedas the“3-2”question:is RT22equivalent to RT32(over RCA0)?Inground–breaking work,D.Seetapun answered this question negatively by obtaining the following result.Theorem2.7(Seetapun and Slaman[1995]).For any computable2–coloring C of[N]2and any noncomputable sets C0,C1,...,there is an inﬁnite homogeneous set X such that(∀i)[C i≤T X].This allowed Seetapun to construct anω–model of RCA0+RT22 which was not closed under the jump operator and hence deduce the following corollary.Corollary2.8(Seetapun and Slaman[1995]).In RCA0,RT22doesnot imply ACA0.Hence,over RCA0,RT22is strictly weaker than RT32.In the same paper,Slaman obtained the following result going in the opposite direction.Theorem2.9(Seetapun and Slaman[1995]).RT22is notΠ04-conservativeover RCA0.This is good point to mention some forgotten work of Hirst[1987]. First Hirst focuses on Ramsey’s theory for singletons.He shows that RCA0proves RT1n,for n∈ω.Afterward he shows:Theorem2.10(Hirst[1987,Theorem6.4]).Over RCA0,RT1<∞isequivalent to BΣ2.Proof.Throughout this proof we will use the fact that BΣ2is equiva-lent BΠ1over RCA0(see H´a jek and Pudl´a k[1993]).RAMSEY’S THEOREM 9Let M be a model of RT 1<∞,X be the natural numbers of M and θbe a Σ00formula possibly with set parameters.Fix y and assume that (∀x <y )(∃z )(∀w )θ(x,z,w ).Deﬁne F by F (t )is the least n less than t such that (∀x <y )(∃z <n )(∀w <t )θ(x,z,w ),if such a n exists,otherwise let F (t )=t .Suppose H is an inﬁnite set such that F (H )=l ,for some l .Then (∀x <y )(∃z <l )(∀w )θ(x,z,w )as desired.Suppose,on the other hand that no such homogeneous set exists.By RT 1<∞,the range of F is ing ∆01comprehension,wecan construct a sequence t i i ∈X such that for each i ∈X ,t i <t i +1and F (t i )<F (t i +1).Deﬁne G (i )to be the least x less than y such that (∀z <F (t i )−1)(∃w <t i )¬θ(x,z,w ).Let S be an inﬁnite homogeneous for G .Let x 0=G (S ).Choose z 0.Since S is inﬁnite,there is some i ∈S such that F (t i )−1>z 0.So (∃w <t i )¬θ(x 0,z 0,w ).Hence,(∀z )(∃w )¬θ(x 0,z,w ).This is a contradiction.Now let M be a model of B Π1.Let y ∈X and F be a function from X to y .Suppose that (∀x <y )(∃z )(∀w )(w >z →F (w )=x ).Then by B Π1,(∃t )(∀x <y )(∃z <t )(∀w )(w >z →F (w )=x ).In particular,(∃t )(∀x <y )(∀w )(w >t →F (w )=x ).Let t 0be such a t .Then we have (∀x <y )F (t +1)=x ,contradicting the deﬁnition of F .So it must be the case that (∃x <y )(∀z )(∃w )(w >z ∨F (w )=x ).Let x 0be such an x .Then {t :F (t )=x 0}is the desired inﬁnite homogeneous set.As a corollary,he gets RT 1<∞is independent of WKL 0.He then turns to Ramsey’s theory for pairs.Hirst [1987,Theorem 6.8]shows that RT 22implies RT 1<∞.In Lemma 10.6,we will show a slightly stronger theo-rem:SRT 22implies RT 1<∞.As a result,we get the following theorem.Theorem 2.11(Hirst).i .RT 1<∞is not Σ03-conservative over RCA 0.ii .RT 22proves B Σ2.iii .RT 22is not Σ03-conservative over RCA 0.Proof.(ii)By the above results (Theorems 2.10and 10.6)we know RT 22proves B Σ2.(i)and (iii)However since RCA 0proves the same arithmetic statements as I Σ1,RCA 0does not prove B Σ2[see H´a jek and Pudl´a k,1993,Theorem IV.1.29].Hence there is an instance of B Σ2that is not provable in RCA 0.This instance is an Σ03sentence.As we mentioned earlier Hirst also showed:Corollary 11.5(Hirst [1987,Theorem 6.11]).RCA 0+RT 2<∞ B Σ3.10P.CHOLAK,C.JOCKUSCH AND T.SLAMANWe prove a slightly stronger theorem:Theorem11.4.He also showsthat there is aω-model of WKL0which is not a model of RT22.This is what was known up to the time of our work.Note that the series of results on Ramsey’s theorem for pairs is somewhat paral-lel to the results for Weak K¨o nig’s Lemma.In particular,Seetapun’sTheorem2.7and its corollary that RT22does not imply ACA0(Corol-lary2.8)are analogous,to the Jockusch–Soare cone avoidance theoremforΠ01–classes(Theorem2.1)and its corollary that WKL0is strictlyweaker than ACA0(Corollary2.2),respectively.However,in this his-torical survey there is no analogue for Ramsey’s theorem mentioned forthe Low Basis Theorem and Harrington’sΠ11conservation theorem forWKL0,Theorem8.4.It is the analog between Weak K¨o nig’s Lemma and Ramsey’s theorem,which led Slaman to make the following con-jecture.Conjecture2.12(Seetapun and Slaman[1995]).Any proof that ev-ery computable2-coloring of[N]2has an inﬁnite homogeneous low nset should lead to a proof that RCA0+RT22isΠ11-conservative overRCA0+IΣn.It is the main purpose of this paper to conﬁrm Slaman’s conjecture by supplying the analogues of the Low Basis Theorem and Harrington’sΠ11conservation theorem for WKL0for Ramsey’s Theorem,namelyTheorems3.1,10.2,and11.1.3.Low2Homogeneous SetsThe goal of this section is to outline the structure of the proof of the following theorem.For reasons stated in the introduction,we will ac-tually give two proofs of this result.The two proofs,although diﬀering considerably in their details,will both have the structure outlined in this section.Theorem3.1.For any computable k–coloring of[N]2,there is an in-ﬁnite homogeneous set X which is low2(i.e.,X ≤T0 ).Our proof of this theorem is somewhat indirect.The following deﬁ-nition will play a key role.Deﬁnition3.2.An inﬁnite set X is r-cohesive if for each computable set R,X⊆∗R or X⊆∗R.An inﬁnite set is p-cohesive if the above holds for each primitive recursive set R.Theorem3.3(Jockusch and Stephan[1993]).There exists a low2r-cohesive set.RAMSEY’S THEOREM11 A proof of this result can be found in Jockusch and Stephan[1993], Theorem2.5,although the proof presented there has an error which is corrected in Jockusch and Stephan[1997].We will present a“single jump control”proof of this theorem in Section4and a“double jump control”proof of this theorem in Section5.The reason for considering r–cohesive sets is that if X is r–cohesive and C is a2–coloring of[X]2,then the restriction of C to[X]2is stable in the sense of the following deﬁnition.Deﬁnition3.4.A k–coloring of[X]2is called stable if for each a∈X, the pair{a,b}has aﬁxed color c a for all suﬃciently large b∈X(i.e., there is a d a such that for all b greater than d a with b∈X,the color of{a,b}is c a).Stable colorings were considered in Hummel[1994]and play a crucial role in Hummel and Jockusch[n.d.].Now any computable k–coloring of pairs becomes stable when it is restricted to an r-cohesive set X.(Fix i.The sets R c={j:{i,j} has color c}are computable and partition N−{i}as c ranges over the colors.Since X is r–cohesive,there exists a color c such that X⊆∗R c. Thus the color of{i,j}is independent of j for all suﬃciently large j∈X.)Thus,using Theorem3.3,if we can prove that every stable k-coloring of[N]2has an inﬁnite low2homogeneous set,the result for arbitrary computable k-colorings of[N]2follows by relativization.(Any set which is low2relative to a low2set is low2.)The problem ofﬁnding homogeneous sets for computable stable col-orings of pairs is easily reduced,by the Limit Lemma,to the problem ofﬁnding homogeneous sets for∆02colorings of1-tuples.Lemma3.5.For any computable stable k-coloring C of[N]2,there arek disjoint∆02sets A i such that i<k A i=N and any inﬁnite subset ofany A i computes an inﬁnite homogeneous set for C.Proof.Let A i={a:lim b C({a,b})=i}.Suppose that B is an inﬁnite subset of A i.Deﬁne c k by recursion as the least c∈B such that,for all j<k,c>c j and C({c j,c})=i.Then{c i:i∈N}is the desired inﬁnite set C such that C is homogeneous for C and C≤T B.The following results(relativized to a low2r-cohesive set)will com-plete the proof that each computable k-coloring of pairs has an inﬁnite low2homogeneous set.(Theﬁrst is a special case of the second.)Theorem3.6.For each∆02set A there is an inﬁnite low2set G whichis contained in A or A.12P.CHOLAK,C.JOCKUSCH AND T.SLAMANTheorem 3.7.Let {A i }i<k be k disjoint ∆02sets such thati<k A i =N .Then for some k ,there is an inﬁnite low 2set G which is contained in A k .These results will be proved by “single jump control”in Section 4and by “double jump control”in Section 5.Before we proceed,we should note that Theorem 3.7follows by in-duction from Theorem 3.6.(Let {A i }i<k +1be k +1disjoint ∆02sets such that i<k +1A i =N .Let A =A k .Apply Theorem 3.6.If there isa low 2subset of A =A k ,we are done.Otherwise apply the relativized (to the set G )version of the induction hypothesis (i.e.,Theorem 3.7)to {A i ∩G }i<k .)But as we will later see (Theorem 11.4)this does not hold for models of arithmetic;the statement of Theorem 3.6,D 22,in second order arithmetic does not imply the statement of Theorem 3.7,D 2<∞.For this reason we will show,in Section 5.3,how to alter the forcing proof of Theorem 3.6to get a proof of Theorem 3.7.We now complete the proof of Theorem 3.1,assuming Theorems 3.3and 3.7.The idea is that the existence of a low 2r–cohesive set allows us to restrict attention to computable stable partitions of pairs,which are basically the same as ∆02partitions of 1–tuples,and these have inﬁnite low 2homogeneous sets by Theorem 3.7.In more detail,let a computable k –coloring C of [N ]2be given.Let X be a low 2r–cohesive set,and let f be the unique increasing function with range X .Deﬁne an X –computable coloring C 1of [N ]2by C 1({a,b }=C ({f (a ),f (b )}).Then,since the restriction of C to [X ]2is stable,as remarked above,the coloring C 1is a stable k –coloring of [N ]2.By Theorem 3.5relativized to X there are sets A 0,...A k −1with ∪i<k A i =N such that each A i is ∆0,X 2and for any inﬁnite set B contained in any A i ,there is an inﬁnite homogeneous set H for C 1such that H ≤T X ⊕B .By Theorem 3.7relativized to X ,there exists i <k such that A i has an inﬁnite subset B with (X ⊕B ) ≤T X .Let H be a homogeneous set for C 1with H ≤T X ⊕B ,and let H ∗=f (H ).Then H ∗is inﬁnite and homogeneous for C ,and (H ∗) ≤T (X ⊕B ) ≤T X ≤T 0 ,so H ∗is the desired inﬁnite low 2homogeneous set for C .4.Constructing low 2sets by first jump controlIn this section,we prove Theorems 3.3and 3.6by constructing sets A with A of degree at most d where d is an appropriately chosen degree satisfying d ≤0 .Here “appropriately chosen”means that d >>0 ,where the relation >>is deﬁned as follows.RAMSEY’S THEOREM13 Deﬁnition4.1.Let a and b be degrees.Then a>>b means that every b–computable{0,1}–valued partial function has a total a–computable extension.The notation>>was deﬁned and studied in Simpson[1977,pp.652–653].Actually,Simpson deﬁned a>>b to mean that each inﬁnite b–computable tree in2<ωhas an inﬁnite a–computable path.We will see in Section8that this is equivalent to the above deﬁnition.We immediately have the following implications:a≥b ⇒a>>b⇒a>bAlso,for each degree b there is a degree a>>b such that a =b .To prove this,consider the case where b=0and then relativize the result to b.Let P be the class of all{0,1}–valued(total)functions f such that f( e,i )=ϕe(i)wheneverϕe(i)↓≤1.Then P is a nonemptyΠ01subset of2ω,so by the Low Basis Theorem there is a low degree bwhich contains a function f∈P.Clearly b>>0.Of course,it is possible to decide the truth of a givenΠ02sentence inthe integers using a0 –oracle.The following lemma shows that a d–oracle has a somewhat weaker property,which will however be suﬃcient for our construction.It is related to the concept of semirecursiveness studied in Jockusch[1968].Lemma4.2.Suppose that d>>0 and that(γe,0,γe,1)e∈ωis an ef-fective enumeration of all ordered pairs ofΠ02sentences ofﬁrst–orderarithmetic.Then there is a d–computable{0,1}–valued(total)function f such thatγe,f(e)is true wheneverγe,0orγe,1is true.Proof.Let R(e,i,s)be a0 -computable predicate such that,for all e∈ωand i≤1,γe.i is true iﬀ(∀s)R(e,i,s)holds.Letδ(e)be the least s such that either R(e,0,s)or R(e,1,s)is false,if such an s exists,and otherwiseδ(e)is undeﬁned.Letθ(e)=1−i,where i is minimal such that R(e,i,δ(e))is false,providedδ(e)is deﬁned,andθ(e)is undeﬁned otherwise.Thenθis a0 –computable{0,1}–valued partial function and so has a b–computable total extension f.This f satisﬁes the conclusion of the lemma.4.1.Constructing a low2r-cohesive set usingﬁrst jump con-trol.The following theorem easily implies Theorem3.3(see Corol-lary4.5).Theorem4.3(Jockusch and Stephan[1993]).Suppose that the sets R0,R1,...are uniformly computable,and suppose that d>>0 .Then there is an inﬁnite set G such that G has degree at most d,and for all e,either G⊆∗R e or G⊆∗R e.。

Sets with more sums than differences

= {0, 2} ∪ {3, 7, 11} ∪ (14 − {0, 2}) ∪ {4}. Hegarty and Roesler (personal communication) observed that this set belongs to the following inﬁnite sequence of MSTD sets: For k ≥ 3, let A∗ = {0, 2} ∪ {3, 7, 11, . . . , 4k − 1} ∪ {4k, 4k + 2} A = A∗ ∪ {4}. and To see that the set A is an MSTD set of integers, we note thatA∗ is symmetric and so |A∗ + A∗ | = |A∗ − A∗ |. Since 8 ∈ (A + A) \ (A∗ + A∗ ), it suﬃces to prove that |A − A| = |A∗ − A∗ |. For this, we only need to show that A∗ − {4} ⊆ A∗ − A∗ , and this is true because 1 = 3 − 2 ∈ A∗ − A∗ 4k − 4 = (4k − 1) − 3 ∈ A∗ − A∗ 4k − 2 = (4k ) − 2 ∈ A∗ − A∗ . We can generalize this construction as follows. Theorem 1. Let m, d, and k be integers such that m ≥ 4, 1 ≤ d ≤ m − 1, d = m/2, and k ≥ 3 if d < m/2 and k ≥ 4 if d > m/2. Let B = [0, m − 1] \ {d} L = {m − d, 2m − d, . . . , km − d} 4 = 7 − 3 ∈ A∗ − A∗

Algebras and modules in monoidal model categories

ALGEBRAS AND MODULES IN MONOIDALMODEL CATEGORIESSTEFAN SCHWEDE and BROOKE E.SHIPLEY[Received4September1998;revised1December1998]1.IntroductionIn recent years the theory of structured ring spectra(formerly known as A1-and E1-ring spectra)has been signi®cantly simpli®ed by the discovery of categories of spectra with strictly associative and commutative smash products.Now a ring spectrum can simply be de®ned as a monoid with respect to the smash product in one of these new categories of spectra.In order to make use of all of the standard tools from homotopy theory,it is important to have a Quillen model category structure[20]available here.In this paper we provide a general method for lifting model structures to categories of rings,algebras,and modules.This includes,but is not limited to,each of the new theories of ring spectra.One model for structured ring spectra is given by the S-algebras of[11].This example has the special feature that every object is®brant,which makes it easier to form model structures of modules and algebras.There are other new theories such as`symmetric ring spectra'[13],`functors with smash product'[2,3,16]or `diagram ring spectra'[19]which do not have this special property.This paper provides the necessary input for obtaining model categories of associative structured ring spectra in these contexts.Categories of commutative ring spectra appear to be intrinsically more complicated,and they are not treated system-atically here.Our general construction of model structures for associative monoids also gives a uni®ed treatment of previously known cases(simplicial sets, simplicial abelian groups,chain complexes,S-modules)and applies to other new examples(G-spaces and modules over group algebras).We discuss these examples in more detail in§5.Technically,what we mean by an`algebra'is a monoid in a symmetric monoidal category,for example,a ring in the category of abelian groups under tensor product.To work with this symmetric monoidal product it must be compatible with the model category structure,which leads to the de®nition of a monoidal model category;see De®nition3.1.To obtain a model category structure of algebras we have to introduce one further axiom,the monoid axiom(De®nition 3.3).A®ltration on certain pushouts of monoids(see Lemma6.2)is then used to reduce the problem to standard model category arguments based on Quillen's `small object argument'.The case of modules also uses the monoid axiom,but the argument here is straightforward.Our main result is stated in Theorem4.1.The research of the second author was partially supported by an NSF Postdoctoral Fellowship.1991Mathematics Subject Classi®cation:primary55U35,secondary18D10.Proc.London Math.Soc.(3)80(2000)491±511.q London Mathematical Society2000.492stefan schwede and brooke e.shipley Organization.We assume that the reader is familiar with the language of homotopical algebra(cf.[20,10,12])and with the basic ideas concerning monoidal and symmetric monoidal categories(cf.[17,VII;4,Chapter6])and triples(also called monads,cf.[17,VI.1;4,Chapter4]).In§2we consider the general question of lifting model categories to categories of algebras over a triple. This forms a basis for the following study of the more speci®c examples of algebras and modules in a monoidal category.In§3we discuss the compatibility that is necessary between the monoidal and model category structures.In§4we state our main results which construct model categories for modules and algebras and compare the homotopy categories of modules or algebras over weakly equivalent monoids.In§5we list examples to which our theorems apply.Then ®nally,in§6we prove the main theorem,Theorem4.1.Acknowledgements.We would®rst like to thank Charles Rezk for conversa-tions which led us to the®ltration that appears in Lemma6.2.We also bene®ted from several conversations about this project with Bill Dwyer,Mark Hovey and Manos Lydakis.We would also like to thank Bill Dwyer,Phil Hirschhorn,and Dan Kan for sharing the draft of[9]with us.In§2we recall the notion of a co®brantly generated model category from their book.2.Co®brantly generated model categoriesIn this section we review a general method for creating model category structures;we will later apply this material to the special cases of module and algebra categories.We need to transfer model category structures to categories of algebras over triples.In[20,II,p.3.4],Quillen formulates his small object argument,which is now the standard device for such purposes.In our context we will need a trans®nite version of the small object argument,so we work with the `co®brantly generated model category'of[9],which we now recall.This material also appears in more detail in[12,2.1].If a model category is co®brantly generated,its model category structure is completely determined by a set of co®brations and a set of acyclic co®brations. The trans®nite version of Quillen's small object argument allows functorial factorization of maps as co®brations followed by acyclic®brations and as acyclic co®brations followed by®brations.Most of the model categories in the literature are co®brantly generated,for example,topological spaces and simplicial sets,as are all the examples that appear in this paper.The only complicated part of the de®nition of a co®brantly generated model category is formulating the de®nition of relative smallness.For this we need to consider the following set-theoretic concepts.The reader might keep in mind the example of a compact topological space with isÀ0-small relative to closed inclusions.Ordinals and cardinals.An ordinal g is an ordered isomorphism class of well ordered sets;it can be identi®ed with the well ordered set of all preceding ordinals.For an ordinal g,the same symbol will denote the associated poset category.The latter has an initial object0=,the empty ordinal.An ordinal k is a cardinal if its cardinality is larger than that of any preceding ordinal.A cardinal k is called regular if for every set of sets f X j g j P J indexed by a set J of cardinalityless than k such that the cardinality of each X j is less than that of k ,then the cardinality of the union J X j is also less than that of k .The successor cardinal (the smallest cardinal of larger cardinality)of every cardinal is regular.Trans®nite composition .Let C be a cocomplete category and g a well ordered set which we identify with its poset category.A functor V :g À3C is called a g -sequence if for every limit ordinal b <g the natural map colim V j b À3V b is an isomorphism.The map V 0= À3colim g V is called the trans®nite composition of the maps of V .A subcategory C 1ÌC is said to be closed under trans®nite composition if for every ordinal g and every g -sequence V :g À3C with map V a À3V a 1 in C 1for every ordinal a <g ,the induced map V 0= À3colim g V is also in C 1.Examples of such subcategories are the co®brations or the acyclic co®brations in a closed model category.Relatively small objects .Consider a cocomplete category C and a subcategory C 1ÌC closed under trans®nite composition.If k is a regular cardinal,an object C P C is called k -small relative to C 1if for every regular cardinal l >k and every functor V :l À3C 1which is a l -sequence in C ,the mapcolim l Hom C C ;V À3Hom C C ;colim l V is an isomorphism.An object C P C is called small relative to C 1is there exists a regular cardinal k such that C is k -small relative to C 1.I-injectives,I-co®brations and regular I-co®brations .Given a cocomplete category C and a class I of maps,we use the following notation.By I -inj we denote the class of maps which have the right lifting property with respect to the maps in I .Maps in I -inj are referred to as I -injectives .By I -cof we denote the class of maps which have the left lifting property with respect to the maps in I -inj.Maps in I -cof are referred to as I -co®brations.By I -cof reg ÌI -cof we denote the class of the (possibly trans®nite)compositions of pushouts (cobase changes)of maps in I .Maps in I -cof reg are referred to as regular I-co®brations .Quillen's small object argument [20,p.II 3.4]has the following trans®nite analogue.Note that here I has to be a set ,not just a class of maps.The obvious analogue of Quillen's small object argument would seem to require that coproducts are included in the regular I -co®brations.In fact,any coproduct of regular I -co®brations is already a regular I -co®bration;see [12,2.1.6].L emma 2.1[9;12,2.1.14].Let C be a cocomplete category and I a set of maps in C whose domains are small relative to I -cof reg .Then(i)there is a functorial factorization of any map f in C as f qi with q P I -inj and i P I -cof reg ,and thus(ii)every I-co®bration is a retract of a regular I-co®bration.D e®nition 2.2[9].A model category C is called co®brantly generated if it is complete and cocomplete and there exist a set of co®brations I and a set of acyclic co®brations J such that(i)the ®brations are precisely the J -injectives;493algebras and modules in monoidal model categories494stefan schwede and brooke e.shipley(ii)the acyclic®brations are precisely the I-injectives;(iii)the domain of each map in I or J is small relative to I-cof reg or J-cof reg, respectively.Moreover,here the co®brations are the I-co®brations,and the acyclic co®brations are the J-co®brations.For a speci®c choice of I and J as in the de®nition of a co®brantly generated model category,the maps in I will be referred to as generating co®brations,and those in J as generating acyclic co®brations.In co®brantly generated model categories,a map may be functorially factored as an acyclic co®bration followed by a®bration and as a co®bration followed by an acyclic®bration.Let C be a co®brantly generated model category and T a triple on C.We want to form a model category on the category of algebras over the triple T,denoted T-alg.De®ne a map of T-algebras to be a weak equivalence or a®bration if the underlying map in C is a weak equivalence or a®bration,respectively.De®ne a map of T-algebras to be a co®bration if it has the left lifting property with respect to all acyclic®brations.The forgetful functor T-algÀ3C has a left adjoint`free' functor.The following lemma gives two different situations in which one can lift a model category on C to one on T-alg.We make no great claim to originality for this lemma.Other lifting theorems for model category structures can be found in [1,Theorem4.14;6,Theorem2.5;8,Theorem3.3;11,VII,Theorems4.7,4.9; 21,Lemma B.2;9].Let X be a T-algebra.We de®ne a path object for X to be a T-algebra X I together with T-algebra mapsXÀ3,X IÀ3X´Xfactoring the diagonal map,such that the®rst map is a weak equivalence and the second map is a®bration in the underlying category C.L emma2.3.Assume that the underlying functor of T commutes with®ltered direct limits.Let I be a set of generating co®brations and J be a set of generating acyclic co®brations for the co®brantly generated model category C.Let I T and J T be the images of these sets under the free T-algebra functor.Assume that the domains ofI T and J T are small relative to I T-cof reg and J T-cof reg respectively.Suppose that(1)every regular J T-co®bration is a weak equivalence,or(2)every object of C is®brant and every T-algebra has a path object. Then the category of T-algebras is a co®brantly generated model category with I T a generating set of co®brations and J T a generating set of acyclic co®brations.Proof.We refer the reader to[10,3.3]for the numbering of the model category axioms.All those kinds of limits that exist in C also exist in T-alg, and limits are created in the underlying category C[4,Proposition 4.3.1]. Colimits are more subtle,but since the underlying functor of T commutes with ®ltered colimits,they exist by[4,Proposition4.3.6].Model category axioms MC2(saturation)and MC3(closure properties under retracts)are clear.One half of MC4(lifting properties)holds by the de®nition of co®brations of T-algebras.The proof of the remaining axioms uses the trans®nite small object argument (Lemma 2.1),which applies because of the hypothesis about the smallness of the domains.We begin with the factorization axiom,MC5.Every map in I T and J T is a co®bration of T -algebras by adjointness.Hence any I T -co®bration or J T -co®bration is a co®bration of T -algebras.By adjointness and the fact that I is a generating set of co®brations for C ,a map is I T -injective precisely when the map is an acyclic ®bration of underlying objects,that is,an acyclic ®bration of T -algebras.Hence the small object argument applied to the set I T gives a (functorial)factorization of any map in T -alg as a co®bration followed by an acyclic ®bration.The other half of the factorization axiom,MC5,needs hypothesis (1)or (2).Applying the small object argument to the set of maps J T gives a functorial factorization of a map in T -alg as a regular J T -co®bration followed by a J T -injective.Since J is a generating set for the acyclic co®brations in C ,the J T -injectives are precisely the ®brations among the T -algebra maps,once more by adjointness.In case (1)we assume that every regular J T -co®bration is a weak equivalence on underlying objects in C .We noted above that every J T -co®bration is a co®bration in T -alg.So we see that the factorization above is an acyclic co®bration followed by a ®bration.In case (2)we can adapt the argument of [20,II,p.4.9]as follows.Let i :X À3Y be any J T -co®bration.We claim that it is a weak equivalence in the underlying category.Since X is ®brant and ®brations are J T -injectives,we obtain a retraction r to i by lifting in the squareX ÀÀÀ3id Xi ÀÀÀÀ3r ÀÀÀÀ3Y ÀÀÀ3ÃHere Y possesses a path object and i has the left lifting property with respect to ®brations.So a lifting exists in the squareX ÀÀÀ3i Y ÀÀÀ3Y IÀÀÀ3ÀÀÀ3Y ÀÀÀÀÀÀÀÀÀ3 id ;i ±r Y ´Y This shows that in the homotopy category of C ,i ±r is equal to the identity map of Y .Since maps in C are weak equivalences if and only if they become isomorphisms in the homotopy category of C ,this proves that i is a weak equivalence,and it completes the proof of model category axiom MC5under hypothesis (2).It remains to prove the other half of MC4,that is,that any acyclic co®bration A À3,B has the left lifting property with respect to ®brations.In other words,we need to show that the acyclic co®brations are contained in the J T -co®brations.The small object argument provides a factorizationA À3,W À3B with A À3W a J T -co®bration and W À3B a ®bration.In addition,W À3B is a 495algebras and modules in monoidal model categories ´ÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁ´´ÁÁÁÁÁÁÁÁÁÁÁÁÁÁweak equivalence since A À3B is.Since A À3B is a co®bration,a lifting inA ÀÀÀ3WÀÀÀ3ÀÀÀ3,B ÀÀÀ3id B exists.Thus A À3B is a retract of a J T -co®bration;hence it is a J T -co®bration.R emark 2.4.To simplify the exposition,we will assume that every object ofC is small relative to the whole category C when we apply Lemma 2.3in the rest of this paper.This holds for G -spaces and symmetric spectra based on simplicial sets.These two categories are in fact examples of the very general notion of a `locally presentable category'[4,5.2].Category theory takes care of the smallness conditions here since every object of a locally presentable category is small [4,Proposition 5.2.10].As a rule of thumb,diagram categories involving sets or simplicial sets are locally presentable,but categories involving actual topological spaces are not.If the underlying functor of the triple T on C commutes with ®ltered direct limits,then so does the forgetful functor from T -algebras to C .Hence by adjointness,if every object of C is small relative to C ,then every free T -algebra is small relative to the whole category of T -algebras,so the smallness conditions of Lemma 2.3hold.Of course,if one is interested in a category where not all objects are small with respect to all of C one can verify those smallness conditions directly.So by adding hypotheses about smallness of the domains of the new generators to each of the statements in the rest of the paper,we could remove the condition that all objects are small.3.Monoidal model categoriesA monoidal model category is essentially a model category with a compatible closed symmetric monoidal product.The compatibility is expressed by the pushout product axiom below.In this paper we always require a closed symmetric monoidal product,although for expository ease we refer to these categories as just `monoidal'model categories.One could also consider model categories enriched over a monoidal model category with certain compatibility requirements analogous to the pushout product axiom or the simplicial axiom of [20,II.2].For example,closed simplicial model categories [20,II.2]are such compatibly enriched categories over the monoidal model category of simplicial sets.See [12,Chapter 4]for an exposition on this material.We also introduce the monoid axiom which is the crucial ingredient for lifting the model category structure to monoids and modules.Examples of monoidal model categories satisfying the monoid axiom are given in §5.D e®nition 3.1.A model category C is a monoidal model category if it is endowed with a closed symmetric monoidal structure and satis®es the following pushout product axiom.We will denote the symmetric monoidal product by ^,the unit by I and the internal Hom object by [±,±]Pushout product axiom .Let A À3B and K À3L be co®brations in C .Then 496stefan schwede and brooke e.shipley´ÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁthe mapA ^L ÈA ^KB ^K À3B ^L is also a co®bration.If in addition one of the former maps is a weak equivalence,so is the latter map.R emark 3.2.Mark Hovey has pointed out that an extra condition is needed to ensure that the monoidal structure on the model category induces a monoidal structure on the homotopy category;see [12,4.3.2].The pushout product axiom guarantees that for co®brant objects the smash product is an invariant of the weak equivalence type,so it passes to a product on the homotopy category.However,if the unit of the smash product is not co®brant,then it need not represent a unit on the homotopy category level.The following additional requirement ®xes this problem:let c I À3I be a co®brant replacement of the unit.Then for any co®brant X the map c I ^X À3I ^X >X should be a weak equivalence (or equivalently:for any ®brant Y the map Y > I ;Y À3 c I ;Y should be a weak equivalence).This extra property holds in all of our examples;for G -spaces,symmetric spectra and simplicial functors the unit is co®brant,and for S -modules this condition is in [11,III,3.8].However this extra condition is irrelevant for the purpose of the present paper since we always work on the model category level.If C is a category with a monoidal product ^and I is a class of maps in C ,we denote by I ^C the class of maps of the formA ^Z À3B ^Zfor A À3B a map in I and Z an object of C .Recall that I -cof reg denotes the class of maps obtained from the maps of I by cobase change and composition (possibly trans®nite;see §2).D e®nition 3.3.A monoidal model category C satis®es the monoid axiom if every map inf acyclic cofibrationsg ^C -cof regis a weak equivalence.R emark 3.4.Note that if C has the special property that every object is co®brant,then the monoid axiom is a consequence of the pushout product axiom.To see this,®rst note that the initial object acts like a zero for the smash product since ^preserves colimits in each of its variables.So the pushout product axiom says that for an acyclic co®bration A À3B and for co®brant (that is,for all)Z ,the map A ^Z À3B ^Z is again an acyclic co®bration.Since the acyclic co®brations are also closed under cobase change and trans®nite composition,every map in the class (f acyclic co®brations g ^C )-cof reg is an acyclic co®bration.In co®brantly generated model categories ®brations can be detected by checking the right lifting property against a set of maps,the generating acyclic co®brations,and similarly for acyclic ®brations.This is in contrast to general model categories where the lifting property has to be checked against the whole class of acyclic co®brations.Similarly,in co®brantly generated model categories,the pushout 497algebras and modules in monoidal model categories498stefan schwede and brooke e.shipleyproduct axiom and the monoid axiom only have to be checked for a set of generating(acyclic)co®brations.L emma3.5.Let C be a co®brantly generated model category endowed with a closed symmetric monoidal structure.(1)If the pushout product axiom holds for a set of generating co®brations anda set of generating acyclic co®brations,then it holds in general.(2)Let J be a set of generating acyclic co®brations.If every map in J^C -cof reg is a weak equivalence,then the monoid axiom holds.Proof.For the®rst statement consider a map i:AÀ3B in C.Denote by G i the class of maps j:KÀ3L such that the pushout productA^LÈA^K B^KÀ3B^Lis a co®bration.This pushout product has the left lifting property with respect to a map f:XÀ3Y if and only if j has the left lifting property with respect to the mapp: B;X À3 B;Y ´ A;Y A;X :Hence,a map is in G i if and only if it has the left lifting property with respect to the map p for all f:XÀ3Y which are acyclic®brations in C.Thus G i is closed under cobase change,trans®nite composition and retracts.If i:AÀ3B is a generating co®bration,G i contains all generating co®brations by assumption;because of the closure properties it thus contains all co®brations;see Lemma2.1.Reversing the roles of i and an arbitrary co®bration j:KÀ3L,we thus know that G j contains all generating co®brations.Again by the closure properties,G j contains all co®brations,which proves the pushout product axiom for two co®brations.The proof of the fact that the pushout product is an acyclic co®bration when one of the constituents is,follows in the same manner.For the second statement note that by the small object argument,Lemma2.1, every acyclic co®bration is a retract of a trans®nite composition of cobase changes along the generating acyclic co®brations.Since trans®nite compositions of trans®nite compositions are trans®nite compositions,every map in(f acyclic co®brations g^C -cof reg is thus a retract of a map in J^C -cof reg.4.Model categories of algebras and modulesIn this section we state the main theorem,Theorem4.1,which constructs model categories for algebras and modules.The proof of this theorem is delayed to§6. Examples of model categories for which this theorem applies are given in§5.We end this section with two results which compare the homotopy categories of modules or algebras over weakly equivalent monoids.We consider a symmetric monoidal category with product^and unit I.A monoid is an object R together with a`multiplication'map R^RÀ3R and a `unit'IÀ3R which satisfy certain associativity and unit conditions(see[17, VII.3]).Note that R is a commutative monoid if the multiplication map is unchanged when composed with the twist,or the symmetry isomorphism,of R^R.If R is a monoid,a left R-module(`object with left R-action'in[17, VII.4])is an object N together with an action map R^NÀ3N satisfyingassociativity and unit conditions (see again [17,VII.4]).Right R -modules are de®ned similarly.Assume that C has coequalizers.Then there is a smash product over R ,denoted M ^R N ,of a right R -module M and a left R -module N .It is de®ned as the coequalizer,in C ,of the two maps M ^R ^N À3À3M ^N induced by the actions of R on M and N respectively.If R is a commutative monoid,then the category of left R -modules is isomorphic to the category of right R -modules,and we simply speak of R -modules.In this case,the smash product of two R -modules is another R -module and smashing over R makes R -mod into a symmetric monoidal category with unit R .If C has equalizers,there is also an internal Hom object of R -modules, M ;N R .It is the equalizer of two maps M ;N À3À3 R ^M ;N .The ®rst map is induced by the action of R on M ,the second map is the composition ofR ^À: M ;N À3 R ^M ;R ^Nfollowed by the map induced by the action of R on N .For a commutative monoid R ,an R-algebra is de®ned to be a monoid in the category of R -modules.It is a formal property of symmetric monoidal categories (cf.[11,VII,1.3])that specifying an R -algebra structure on an object A is the same as giving A a monoid structure together with a monoid map f :R À3A which is central in the sense that the following diagram commutes:R ^A ÀÀÀÀÀ3switch A ^R ÀÀÀÀÀ3id ^f A ^Af ^id ÀÀÀ3ÀÀÀ3mult :A ^A ÀÀÀÀÀÀÀÀÀÀÀÀÀÀÀÀÀÀÀ3mult :A Now we can state our main theorem.It essentially says that monoids,modules and algebras in a co®brantly generated,monoidal model category C again form a model category if the monoid axiom holds.To simplify the exposition,we assume that all objects in C are small relative to the whole category;see §2.This last assumption can be weakened as indicated in Remark 2.4.The proofs will be delayed until the last section.In the categories of monoids,left R -modules (when R is a ®xed monoid),and R -algebras (when R is a ®xed commutative monoid),a morphism is de®ned to be a ®bration or weak equivalence if it is a ®bration or weak equivalence in the underlying category C .A morphism is a co®bration if it has the left lifting property with respect to all acyclic ®brations.In part (3)of the following theorem we can take R to be the unit of the smash product,in which case we see that the category of monoids in C forms a model category.Note that this theorem does not treat the case of commutative R -algebras.See Remark 4.5for examples of categories C satisfying the hypotheses but where the category of commutative monoids in fact does not have a model category structure with ®brations and weak equivalences de®ned in the underlying category.T heorem 4.1.Let C be a co®brantly generated,monoidal model category.Assume further that every object in C is small relative to the whole category and that C satis®es the monoid axiom.(1)Let R be a monoid in C .Then the category of left R-modules is a co®brantly generated model category.499algebras and modules in monoidal model categories。

Noetherian lattices

JOURNAL OF FORMALIZED MATHEMATICSVolume11,Released1999,Published2003Inst.of Computer Science,Univ.of BiałystokNoetherian LatticesChristoph SchwarzwellerUniversity of TuebingenSummary.In this article we deﬁne noetherian and co-noetherian lattices and show how some properties concerning upper and lower neighbours,irreducibility and density canbe improved when restricted to these kinds of lattices.In addition we deﬁne atomic lattices.MML Identiﬁer:LATTICE6.WWW:/JFM/Vol11/lattice6.htmlThe articles[8],[4],[12],[13],[3],[14],[1],[5],[6],[11],[10],[2],[7],and[9]provide the notation and terminology for this paper.Let us observe that there exists a lattice which isﬁnite.Let us observe that every lattice which isﬁnite is also complete.Let L be a lattice and let D be a subset of L.The functor D yields a subset of Poset(L)and is deﬁned as follows:(Def.1)D ={d ;d ranges over elements of L:d∈D}.Let L be a lattice and let D be a subset of Poset(L).The functor D yields a subset of L and is deﬁned as follows:(Def.2) D={ d;d ranges over elements of Poset(L):d∈D}.Let L be aﬁnite lattice.Note that Poset(L)is well founded.Let L be a lattice.We say that L is noetherian if and only if:(Def.3)Poset(L)is well founded.We say that L is co-noetherian if and only if:(Def.4)Poset(L) is well founded.One can check that there exists a lattice which is noetherian,upper-bounded,lower-bounded, and complete.One can verify that there exists a lattice which is co-noetherian,upper-bounded,lower-bounded, and complete.One can prove the following proposition(1)For every lattice L holds L is noetherian iff L◦is co-noetherian.Let us mention that every lattice which isﬁnite is also noetherian and every lattice which isﬁnite is also co-noetherian.Let L be a lattice and let a,b be elements of L.We say that a is upper neighbour of b if and only if:(Def.5)a=b and b⊑a and for every element c of L such that b⊑c and c⊑a holds c=a or c=b.1c Association of Mizar UsersWe introduce b is lower neighbour of a as a synonym of a is upper neighbour of b.One can prove the following propositions:(2)Let L be a lattice,a be an element of L,and b,c be elements of L such that b=c.Then(i)if b is upper neighbour of a and c is upper neighbour of a,then a=c⊓b,and(ii)if b is lower neighbour of a and c is lower neighbour of a,then a=c⊔b.(3)Let L be a noetherian lattice,a be an element of L,and d be an element of L.Suppose a⊑dand a=d.Then there exists an element c of L such that c⊑d and c is upper neighbour of a.(4)Let L be a co-noetherian lattice,a be an element of L,and d be an element of L.Supposed⊑a and a=d.Then there exists an element c of L such that d⊑c and c is lower neighbour of a.(5)For every upper-bounded lattice L it is not true that there exists an element b of L such thatb is upper neighbour of⊤L.(6)Let L be a noetherian upper-bounded lattice and a be an element of L.Then a=⊤L if andonly if it is not true that there exists an element b of L such that b is upper neighbour of a.(7)For every lower-bounded lattice L it is not true that there exists an element b of L such thatb is lower neighbour of⊥L.(8)Let L be a co-noetherian lower-bounded lattice and a be an element of L.Then a=⊥L ifand only if it is not true that there exists an element b of L such that b is lower neighbour of a.Let L be a complete lattice and let a be an element of L.The functora=⌈−⌉L{d;d ranges over elements of L:a⊑d∧d=a}.The functor∗a yielding an element of L is deﬁned by:(Def.7)∗a=L{d;d ranges over elements of L:d⊑a∧d=a}.Let L be a complete lattice and let a be an element of L.We say that a is completely-meet-irreducible if and only if:(Def.8)a and∗a⊑a.(10)For every complete lattice L holdsa is upper neighbour of a and for every element c of L such that c is upperneighbour of a holds c=(15)Let L be a co-noetherian complete lattice and a be an element of L .Then a is completely-join-irreducible if and only if there exists an element b of L such that b is lower neighbour of a and for every element c of L such that c is lower neighbour of a holds c =b .(16)Let L be a complete lattice and a be an element of L .If a is completely-meet-irreducible,then a is meet-irreducible.(17)Let L be a complete noetherian lattice and a be an element of L .Suppose a =⊤L .Then ais completely-meet-irreducible if and only if a is meet-irreducible.(18)Let L be a complete lattice and a be an element of L .If a is completely-join-irreducible,then a is join-irreducible.(19)Let L be a complete co-noetherian lattice and a be an element of L .Suppose a =⊥L .Thena is completely-join-irreducible if and only if a is join-irreducible.(20)Let L be a ﬁnite lattice and a be an element of L such that a =⊥L and a =⊤L .Then(i)a is completely-meet-irreducible iff a is meet-irreducible,and (ii)a is completely-join-irreducible iff a is join-irreducible.Let L be a lattice and let a be an element of L .We say that a is atomic if and only if:(Def.10)a is upper neighbour of ⊥L .We say that a is co-atomic if and only if:(Def.11)a is lower neighbour of ⊤L .One can prove the following two propositions:(21)Let L be a complete lattice and a be an element of L .If a is atomic,then a is completely-join-irreducible.(22)Let L be a complete lattice and a be an element of L .If a is co-atomic,then a is completely-meet-irreducible.Let L be a lattice.We say that L is atomic if and only if the condition (Def.12)is satisﬁed.(Def.12)Let a be an element of L .Then there exists a subset X of L such that for every element x of L such that x ∈X holds x is atomic and a = L X .One can verify that there exists a lattice which is strict,non empty,and trivial.Let us observe that there exists a lattice which is atomic and complete.Let L be a complete lattice and let D be a subset of L .We say that D is supremum-dense if and only if:(Def.13)For every element a of L there exists a subset D ′of D such that a =L D ′.We say that D is inﬁmum-dense if and only if:(Def.14)For every element a of L there exists a subset D ′of D such that a =⌈−⌉L D ′.Next we state three propositions:(23)Let L be a complete lattice and D be a subset of L .Then D is supremum-dense if and only if for every element a of L holds a = L {d ;d ranges over elements of L :d ∈D ∧d ⊑a }.(24)Let L be a complete lattice and D be a subset of L .Then D is inﬁmum-dense if and only if for every element a of L holds a =⌈−⌉L {d ;d ranges over elements of L :d ∈D ∧a ⊑d }.(25)Let L be a complete lattice and D be a subset of L .Then D is inﬁmum-dense if and only ifD is order-generating.Let L be a complete lattice.The functor MIRRS L yields a subset of L and is deﬁned by: (Def.15)MIRRS L={a;a ranges over elements of L:a is completely-meet-irreducible}.The functor JIRRS L yields a subset of L and is deﬁned by:(Def.16)JIRRS L={a;a ranges over elements of L:a is completely-join-irreducible}.We now state two propositions:(26)For every complete lattice L and for every subset D of L such that D is supremum-denseholds JIRRS L⊆D.(27)For every complete lattice L and for every subset D of L such that D is inﬁmum-dense holdsMIRRS L⊆D.Let L be a co-noetherian complete lattice.Observe that MIRRS L is inﬁmum-dense.Let L be a noetherian complete lattice.Observe that JIRRS L is supremum-dense.R EFERENCES[1]Grzegorz plete lattices.Journal of Formalized Mathematics,4,1992./JFM/Vol4/lattice3.html.[2]Grzegorz Bancerek and Krzysztof Hryniewiecki.Segments of natural numbers andﬁnite sequences.Journal of Formalized Mathematics,1,1989./JFM/Vol1/finseq_1.html.[3]Czesław Byli´n ski.Functions and their basic properties.Journal of Formalized Mathematics,1,1989./JFM/Vol1/funct_1.html.[4]Czesław Byli´n ski.Some basic properties of sets.Journal of Formalized Mathematics,1,1989./JFM/Vol1/zfmisc_1.html.[5]Agata Darmochwał.Finite sets.Journal of Formalized Mathematics,1,1989./JFM/Vol1/finset_1.html.[6]Beata Madras.Irreducible and prime elements.Journal of Formalized Mathematics,8,1996./JFM/Vol8/waybel_6.html.[7]Piotr Rudnicki and Andrzej Trybulec.On same equivalents of well-foundedness.Journal of Formalized Mathematics,9,1997.http:///JFM/Vol9/wellfnd1.html.[8]Andrzej Trybulec.Tarski Grothendieck set theory.Journal of Formalized Mathematics,Axiomatics,1989./JFM/Axiomatics/tarski.html.[9]Andrzej Trybulec.Finite join andﬁnite meet,and dual lattices.Journal of Formalized Mathematics,2,1990./JFM/Vol2/lattice2.html.[10]Wojciech A.Trybulec.Partially ordered sets.Journal of Formalized Mathematics,1,1989./JFM/Vol1/orders_1.html.[11]Wojciech A.Trybulec.Groups.Journal of Formalized Mathematics,2,1990./JFM/Vol2/group_1.html.[12]Zinaida Trybulec.Properties of subsets.Journal of Formalized Mathematics,1,1989./JFM/Vol1/subset_1.html.[13]Edmund Woronowicz.Relations and their basic properties.Journal of Formalized Mathematics,1,1989./JFM/Vol1/relat_1.html.[14]Stanisław˙Zukowski.Introduction to lattice theory.Journal of Formalized Mathematics,1,1989./JFM/Vol1/lattices.html.Received June9,1999Published January2,2004。

Exact Penalty Principle

One of the purposes of this paper is to extend Clarke’s exact penalty principle to the case where f is vector-valued. Our result in the special case where the objective function f is scalar-valued proved the following improved Clarke Exact Penalty Principle which is a corollary of Theorem 3.1. Theorem 1.2 (Improved Clarke’s exact penalty principle) Let X be a normed space, C ⊂ S ⊂ X and f : X → R be Lipschitz of rank Lf on S . Then for L > Lf , f attains a minimum over C at x if and only if the function g (y ) = f (y ) + LdC (y ) attains a minimum over S at x. Unfortunately when local optimal solutions are considered, the reverse statement of Clarke’s exact penalty principle does not hold without additional conditions. In [17], Scholtes and St¨ ohr gave some conditions which ensure the reverse statements hold for the distance function and the error bound function. In this paper we extend these results to the vector optimization case. Under the assumption that S is compact and the local (global) optimal solutions of the problem (P) lies in the interior of the set S , Di Pillo and Grippo [4, 5] showed that the extended Mangasarian Fromovitz constraint qualiﬁcation (EMFCQ) can be used to insure that the local (global) minimizers of the penalized problem with penalty function ψ (x) := ρ( h(x)

Let Dbeanonemptysubsetof R.Onecancheckthateveryelementof D

FORMALIZED MATHEMATICSVolume9,Number3,2001University of BiałystokThe Urysohn LemmaJózef Białas ŁódźUniversity Yatsuka Nakamura Shinshu UniversityNaganoSummary.This article is the third part of a paper proving the funda-mental Urysohn Theorem concerning the existence of a real valued continuousfunction on a normal topological space.The paper is divided into two parts.Intheﬁrst part,we describe the construction of the function solving thesis of theUrysohn Lemma.The second part contains the proof of the Urysohn Lemma innormal space and the proof of the same theorem for compact space.MML Identiﬁer:URYSOHN3.The notation and terminology used here have been introduced in the following papers:[15],[10],[7],[8],[4],[1],[9],[6],[12],[16],[17],[13],[14],[2],[3],[11], and[5].Let D be a non empty subset of R.One can check that every element of D is real.One can prove the following proposition(1)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Let n be a natural number.Then there exists a function G fromdyadic(n)into2the carrier of T such that for all elements r1,r2of dyadic(n)if r1<r2,then G(r1)is open and G(r2)is open and G(r1)⊆G(r2)andA⊆G(0)and B=ΩT\G(1).Let T be a non empty topological space,let A,B be subsets of T,and let n bea natural number.Let us assume that T is a T4space and A=∅and A is closedand B is closed and A∩B=∅.A function from dyadic(n)into2the carrier of T is said to be a drizzle of A,B,n if it satisﬁes the condition(Def.1).(Def.1)Let r1,r2be elements of dyadic(n).Suppose r1<r2.Then it(r1)is open and it(r2)is open and it(r1)⊆it(r2)and A⊆it(0)and B=ΩT\it(1).631c 2001University of BiałystokISSN1426–2630632józef białas and yatsuka nakamuraOne can prove the following propositions:(2)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Let n be a natural number and D be a drizzle of A,B,n.Then A⊆D(0)and B=ΩT\D(1).(3)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Let n be a natural number and G be a drizzle of A,B,n.Then there exists a drizzle F of A,B,n+1such that for every element rof dyadic(n+1)if r∈dyadic(n),then F(r)=G(r).Let A,B be non empty sets,let F be a function from N into A˙→B,and let n be a natural number.Then F(n)is a partial function from A to B.Next we state the proposition(4)Let T be a non empty topological space,A,B be subsets of T,andn be a natural number.Then every drizzle of A,B,n is an element ofDYADIC˙→2the carrier of T.Let A,B be non empty sets,let F be a function from N into A˙→B,and let n be a natural number.Then F(n)is an element of A˙→B.One can prove the following proposition(5)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closedand A∩B=∅.Then there exists a sequence F of partial functions fromDYADIC into2the carrier of T such that for every natural number n holdsF(n)is a drizzle of A,B,n and for every element r of dom F(n)holdsF(n)(r)=F(n+1)(r).Let T be a non empty topological space and let A,B be subsets of T.Let us assume that T is a T4space and A=∅and A is closed and B is closed and A∩B=∅.A sequence of partial functions from DYADIC into2the carrier of T is said to be a rain of A,B if it satisﬁes the condition(Def.2).(Def.2)Let n be a natural number.Then it(n)is a drizzle of A,B,n and for every element r of dom it(n)holds it(n)(r)=it(n+1)(r).Let x be a real number.Let us assume that x∈DYADIC.The functor InfDyadic x yields a natural number and is deﬁned by:(Def.3)x∈dyadic(0)iﬀInfDyadic x=0and for every natural number n such that x∈dyadic(n+1)and x/∈dyadic(n)holds InfDyadic x=n+1.The following propositions are true:(6)For every real number x such that x∈DYADIC holds x∈dyadic(InfDyadic x).(7)For every real number x such that x∈DYADIC and for every naturalnumber n such that InfDyadic x n holds x∈dyadic(n).the urysohn lemma633(8)For every real number x such that x∈DYADIC and for every naturalnumber n such that x∈dyadic(n)holds InfDyadic x n.(9)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closedand A∩B=∅.Let G be a rain of A,B and x be a real number.Ifx∈DYADIC,then for every natural number n holds G(InfDyadic x)(x)=G(InfDyadic x+n)(x).(10)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Let G be a rain of A,B and x be a real number.Supposex∈DYADIC.Then there exists an element y of2the carrier of T such thatfor every natural number n if x∈dyadic(n),then y=G(n)(x).(11)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Let G be a rain of A,B.Then there exists a function F fromDOM into2the carrier of T such that for every real number x holds(i)if x∈R<0,then F(x)=∅,(ii)if x∈R>1,then F(x)=the carrier of T,and(iii)if x∈DYADIC,then for every natural number n such that x∈dyadic(n)holds F(x)=G(n)(x).Let T be a non empty topological space and let A,B be subsets of T.Let us assume that T is a T4space and A=∅and A is closed and B is closed and A∩B=∅.Let R be a rain of A,B.The functor Tempest R yielding a function from DOM into2the carrier of T is deﬁned by the condition(Def.4).(Def.4)Let x be a real number such that x∈DOM.Then(i)if x∈R<0,then(Tempest R)(x)=∅,(ii)if x∈R>1,then(Tempest R)(x)=the carrier of T,and(iii)if x∈DYADIC,then for every natural number n such that x∈dyadic(n)holds(Tempest R)(x)=R(n)(x).Let X be a non empty set,let T be a topological space,let F be a function from X into2the carrier of T,and let x be an element of X.Then F(x)is a subset of T.One can prove the following three propositions:(12)Let T be a non empty topological space and A,B be subsets of T.Suppose T is a T4space and A=∅and A is closed and B is closed andA∩B=∅.Let G be a rain of A,B and r be a real number.If r∈DOM,then for every subset C of T such that C=(Tempest G)(r)holds C isopen.(13)Let T be a non empty topological space and A,B be subsets of T.Suppose T is a T4space and A=∅and A is closed and B is closed andA∩B=∅.Let G be a rain of A,B and r1,r2be real numbers.Suppose634józef białas and yatsuka nakamurar1∈DOM and r2∈DOM and r1<r2.Let C be a subset of T.IfC=(Tempest G)(r1),then C⊆(Tempest G)(r2).(14)Let T be a non empty topological space,A,B be subsets of T,G be arain of A,B,and p be a point of T.Then there exists a subset R of R suchthat for every set x holds x∈R if and only if the following conditions aresatisﬁed:(i)x∈DYADIC,and(ii)for every real number s such that s=x holds p/∈(Tempest G)(s).Let T be a non empty topological space,let A,B be subsets of T,let R bea rain of A,B,and let p be a point of T.The functor Rainbow(p,R)yielding asubset of R is deﬁned by:(Def.5)For every set x holds x∈Rainbow(p,R)iﬀx∈DYADIC and for every real number s such that s=x holds p/∈(Tempest R)(s).Let T,S be non empty topological spaces,let F be a function from the carrier of T into the carrier of S,and let p be a point of T.Then F(p)is a point of S.One can prove the following propositions:(15)Let T be a non empty topological space,A,B be subsets of T,G be arain of A,B,and p be a point of T.Then Rainbow(p,G)⊆DYADIC.(16)Let T be a non empty topological space,A,B be subsets of T,and Rbe a rain of A,B.Then there exists a map F from T into R1such thatfor every point p of T holdsif Rainbow(p,R)=∅,then F(p)=0and for every non empty subset S ofR such that S=Rainbow(p,R)holds F(p)=sup S.Let T be a non empty topological space,let A,B be subsets of T,and let R be a rain of A,B.The functor Thunder R yielding a map from T into R1is deﬁned by the condition(Def.6).(Def.6)Let p be a point of T.Then if Rainbow(p,R)=∅,then(Thunder R)(p)= 0and for every non empty subset S of R such that S=Rainbow(p,R)holds(Thunder R)(p)=sup S.Let T be a non empty topological space,let F be a map from T into R1, and let p be a point of T.Then F(p)is a real number.One can prove the following propositions:(17)Let T be a non empty topological space,A,B be subsets of T,G be arain of A,B,p be a point of T,and S be a non empty subset of R.SupposeS=Rainbow(p,G).Letℓ1be an extended real number.Ifℓ1=1,then sup S and sup S ℓ1.R(18)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Let G be a rain of A,B,r be an element of DOM,and p be athe urysohn lemma635point of T.If(Thunder G)(p)<r,then p∈(Tempest G)(r).(19)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closedand A∩B=∅.Let G be a rain of A,B and r be a real number.Supposer∈DYADIC∪R>1and0<r.Let p be a point of T.If p∈(Tempest G)(r),then(Thunder G)(p) r.(20)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closedand A∩B=∅.Let G be a rain of A,B,n be a natural number,and r1be an element of DOM.If0<r1,then for every point p of T such thatr1<(Thunder G)(p)holds p/∈(Tempest G)(r1).(21)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Let G be a rain of A,B.Then(i)Thunder G is continuous,and(ii)for every point x of T holds0 (Thunder G)(x)and(Thunder G)(x) 1and if x∈A,then(Thunder G)(x)=0and if x∈B,then(Thunder G)(x)=1.(22)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A=∅and A is closed and B is closed andA∩B=∅.Then there exists a map F from T into R1such that(i)F is continuous,and(ii)for every point x of T holds0 F(x)and F(x) 1and if x∈A,then F(x)=0and if x∈B,then F(x)=1.(23)Let T be a non empty topological space.Suppose T is a T4space.LetA,B be subsets of T.Suppose A is closed and B is closed and A∩B=∅.Then there exists a map F from T into R1such that(i)F is continuous,and(ii)for every point x of T holds0 F(x)and F(x) 1and if x∈A,then F(x)=0and if x∈B,then F(x)=1.(24)Let T be a non empty topological space.Suppose T is a T2space andcompact.Let A,B be subsets of T.Suppose A is closed and B is closedand A∩B=∅.Then there exists a map F from T into R1such that(i)F is continuous,and(ii)for every point x of T holds0 F(x)and F(x) 1and if x∈A,then F(x)=0and if x∈B,then F(x)=1.References[1]Grzegorz Bancerek.The fundamental properties of natural numbers.Formalized Mathe-matics,1(1):41–46,1990.636józef białas and yatsuka nakamura[2]Józef Białas.Inﬁmum and supremum of the set of real numbers.Measure theory.For-malized Mathematics,2(1):163–171,1991.[3]Józef Białas.Series of positive real numbers.Measure theory.Formalized Mathematics,2(1):173–183,1991.[4]Józef Białas.Properties of the intervals of real numbers.Formalized Mathematics,3(2):263–269,1992.[5]Józef Białas and Yatsuka Nakamura.Dyadic numbers and T4topological spaces.Forma-lized Mathematics,5(3):361–366,1996.[6]Leszek Borys.Paracompact and metrizable spaces.Formalized Mathematics,2(4):481–485,1991.[7]Czesław Byliński.Functions and their basic properties.Formalized Mathematics,1(1):55–65,1990.[8]Czesław Byliński.Functions from a set to a set.Formalized Mathematics,1(1):153–164,1990.[9]Czesław Byliński.Partial functions.Formalized Mathematics,1(2):357–367,1990.[10]Agata Darmochwał.Compact spaces.Formalized Mathematics,1(2):383–386,1990.[11]Agata Darmochwałand Yatsuka Nakamura.Metric spaces as topological spaces-funda-mental concepts.Formalized Mathematics,2(4):605–608,1991.[12]Beata Padlewska and Agata Darmochwał.Topological spaces and continuous functions.Formalized Mathematics,1(1):223–230,1990.[13]Beata Perkowska.Functional sequence from a domain to a domain.Formalized Mathe-matics,3(1):17–21,1992.[14]Zinaida Trybulec.Properties of subsets.Formalized Mathematics,1(1):67–71,1990.[15]Zinaida Trybulec and HalinaŚwięczkowska.Boolean properties of sets.Formalized Ma-thematics,1(1):17–23,1990.[16]Edmund Woronowicz.Relations and their basic properties.Formalized Mathematics,1(1):73–83,1990.[17]Edmund Woronowicz.Relations deﬁned on sets.Formalized Mathematics,1(1):181–186,1990.Received February16,2001。

Abstract

Because the functions N (m, t), Nd (m, t) and Nreg (r, m, t) are monotone nondecreasing for ﬁxed r and t, it is useful to deﬁne m(N, t) = max{m : N (m, t) ≤ N }; md (N, t) and mreg (r, N, t) are then deﬁned in the obvious way. Many constructions of splitting systems are conveniently described using the incidence matrix representation of a splitting system, which we deﬁne now. Deﬁnition 1.8 Let (X, B ) be an (N ; m, t)-SS, where X = {xj : 1 ≤ j ≤ m} and B = {Bi : 1 ≤ 1 ≤ N }. The incidence matrix of (X, B ) is the N × m matrix A = (ai,j ) where ai,j = The following result is obvious. Lemma 1.9 Suppose A = (ai,j ) is an N × m matrix having entries in the set {0, 1}. Then A is the incidence matrix of an (N ; m, 3)-SS if and only if, for all choices of three columns c1 , c2 , c3 of A, the following property is satisﬁed: There is a row r such that (ar,c1 , ar,c2 , ar,c3 ) ∈ {(0, 0, 1), (0, 1, 0), (1, 0, 0)}. Furthermore, we have the following: 1. A is the incidence matrix of an (N ; m, 3)-disjunct SS if and only if, for all choices of two columns c1 , c2 of A, there is a row r such that ar,c1 = 1 and ar,c2 = 0. 2. A is the incidence matrix of an (N ; m, 3)-uniform SS if and only if every row of A has hamming weight equal to m 2 , and 3. A is the incidence matrix of an (N ; m, t)-r-regular SS if and only if every column of A has hamming weight equal to r. We now give a few small examples of splitting systems. Example 1.1 The blocks {1, 5}, {2, 5} and {3, 5} form a (3; 5, 3)-uniform SS on the set {1, 2, 3, 4, 5}. The incidence matrix of this splitting system is 1 0 0 0 1 0 1 0 0 1 . 0 0 1 0 1 (1) 1 0 if xj ∈ Bi otherwise.

2.3 Functions

Definition 6:
A function f whose domain and codomain are subsets of the set of real numbers is called strictly increasing if f(x)<f(y) whenever x<y and x and y are in the domain of f. Similarly, f is called strictly decreasing if f(x)>f(y) whenever x<y and x and y are in the domain of f.
a ns
Example 9:
Determine whether the function f(x)=x2 from the set of integers is one-toone.
Example 10:
Determine whether the function f(x)=x+1 is one-to-one.
Definition 5
A function f is said to be one-to-one, or injective(单射的), if and only if f(a)=f(b) implies that a=b for all a and b in the domain of f. A function is said to be an injection if it is one-to-one. A function f is one-to-one if and only if f(a)≠f(a) whenever a≠b.
Discrete Mathematics

On the finite generation of a family of Ext modules

a rXiv:089.268v1[mat h.A C]11Se p28ON THE FINITE GENERATION OF A F AMILY OF EXT MODULES TONY J.PUTHENPURAKAL Dedicated to Prof.L.L.Avramov on the occasion of his sixtieth birthday Abstract.Let Q be a Noetherian ring with ﬁnite Krull dimension and let f =f 1,...f c be a regular sequence in Q .Set A =Q/(f ).Let I be an ideal in A ,and let M be a ﬁnitely generated A -module with projdim Q M ﬁnite.Set R (I )=L n ≥0I n ,the Rees-Algebra of I .Let N =L j ≥0N j be a ﬁnitely generated graded R (I )-module.We show that M j ≥0M i ≥0Ext i A (M,N j )is a ﬁnitely generated bi-graded module over S =R (I )[t 1,...,t c ].We give two applications of this result to local complete intersection rings.1.introduction Let A be a Noetherian ring.Let I be an ideal in A and let M be a ﬁnitely gener-ated A -module.M.Brodmann [4]proved that the set Ass A M/I n M is independent of n for all large n .This result is usually deduced by proving that Ass A I n M/I n +1M is independent of n for all large n .Some Generalizations of Brodmann’s result Fix i ≥0.The following sets are independent of n for all large n .1(L.Melkerson and P.Schenzel)[10,Theorem 1](a)Ass A Tor A i (M,I n /I n +1).(b)Ass A Tor A i (M,A/I n ).2(same argument as in 1(a)).Ass A Ext i A (M,I n /I n +1).3(D.Katz &E.West;[8,3.5])Ass A Ext i A (M,A/I n A ).An example of A.Singh [12]shows thatAss A lim →Ext i A (A/I n,M )need not be ﬁnite.So in this examplen ≥1Ass A Ext i A (A/I n ,M )is not even ﬁnite .2TONY J.PUTHENPURAKALWe state some questions in this area which motivated me.(1)(W.Vasconcelos:[13,3.5])Is the seti≥0Ass A Ext i A(M,A)ﬁnite?(2)(L.Melkerson and P.Schenzel:[10,page936])Is the seti≥0 n≥0Ass A Tor A i(M,A/I n)ﬁnite?The motivation for the main result of this paper came from a Vasconcelos’s ques-tion.The author now does not believe that Vasconcelos’s question has a positive answer in this generality.However he is unable to give a counter-example.Note that if A is a Gorenstein local ring then Vasconcelos’s question has,trivially,a positive answer.If we change the question a little then we may ask:if M,D are twoﬁnitely generated A-modules then is the seti≥0Ass A Ext i A(M,D)ﬁnite?This is not known for Gorenstein rings in ing Melkerson and Schenzel’s question as a guidepost the questions I was interested to solve was Let(A,m)be a local complete intersection of codimension c.Are the sets(a) i≥0 j≥0Ass A Ext i A(M,D/I j D)ﬁnite?(b) i≥0 j≥0Ass A Ext i A(M,I j D)ﬁnite?In Theorem5.9we prove that(b)holds.I have been unable to verify whether(a) holds.The main result in this paper is the following regardingﬁnite generation of a family of Ext modules.Let R(I)= n≥0I n t n be the Rees algebra of I. Theorem1.Let Q be a Noetherian ring withﬁnite Krull dimension and let f= f1,...f c be a regular sequence in Q.Set A=Q/(f).Let M be aﬁnitely generated A-module with projdim Q Mﬁnite.Let I an ideal in A and let N= n≥0N n be a ﬁnitely generated R(I)-module.ThenE(N)= i≥0 n≥0Ext i A(M,N n)is aﬁnitely generated bi-graded S=R(I)[t1,...,t c]-module.An easy consequence of this result is that(b)holds(by taking N= n≥0I n D); see Theorem5.1.A complete local complete intersection ring is a quotient of a regular local ring mod a regular sequence.So in this case(b)holds from Theorem 5.1.The proof of(b)for local complete intersections in general is a little technical; see Theorem5.9We next discuss a surprising consequence of Theorem1.Let(A,m)be a local complete intersection of codimension c.Let M,N be twoﬁnitely generated A-modules.Deﬁnecx A(M,N)=inf b∈N|n b−1<∞ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES3 In this section6we prove,see Theorem6.1,that(†)cx A(M,I j N)is constant for all j≫0.We now describe in brief the contents of this paper.In section one we give a module structure to E(N)over S(as in Theorem1).We also discuss a few preliminaries.The local case of Theorem1is proved in section2while the global case is proved in section3.In section4we prove our results on asymptotic primes. In section5we prove(†).Acknowledgements:The author thanks Prof.L.L.Avramov and Prof.J.Herzog for many discussions regarding this paper2.module structureLet Q be a Noetherian ring and let f=f1,...f c be a regular sequence in Q. Set A=Q/(f).Let M be aﬁnitely generated A-module with projdim Q Mﬁnite. We will not change M throughout our discussion.Let I an ideal in A and let N= n≥0N n be aﬁnitely generated R(I)= n≥0I n t n-module.SetE(N)= i≥0 n≥0Ext i A(M,N n).In this section we show E(N)is a bi-graded S=R(I)[t1,...,t c]-module.We also discuss two preliminary results that we will need later in this paper.2.1.Let F:···F n→···F1→F0→0be a free resolution of M as a A-module.Let t1,...t c:F(+2)→F be the Eisenbud-operators[6,section1.]Then(1)t i are uniquely determined up to homotopy.(2)t i,t j commute up to homotopy.2.2.Set T=A[t1,...,t c]with deg t i=2.In[7]Gulliksen shows that if projdim Q M isﬁnite then i≥0Ext i A(M,L)is a ﬁnitely generated T-module.2.3.Let N= n≥0N n be a f.g module over R(I).Let u=xt s.The mapN n u−→N n+1yieldsHom(F,N n)uHom(F,N n)(+2)uHom(F,N n+s)(+2)Taking homology gives thatE(N)= i≥0 n≥0Ext i A(M,N n)is a bi-graded S=R(I)[t1,...,t c]-module. Remark2.4.1.For each i,we have n≥0Ext i A(M,N n)is aﬁnitely generated R(I)-module.2.For each n,we have i≥0Ext i A(M,N n)is aﬁnitely generated A[t1,...,t c]-module.We state two Lemma’s which will help us in proving Theorem1.4TONY J.PUTHENPURAKAL2.5.Notation(1)Let N= n≥0N n be a graded R(I)-module.Fix j≥0.SetN≥j= n≥j N n.E(N≥j)is naturally isomorphic to the submoduleE(N)≥j= i≥0 n≥j E(N)ijof E(N).(2)If A→A′is a ring extension and if D is an A-module then set D′=D⊗A A′. Notice that if D isﬁnitely generated A-module then D′is aﬁnitely generated A′-module.(3)Set S′=S⊗A A′.Notice S′is aﬁnitely generated bi-graded A′-algebra.Let U= i≥0 n≥0U i,n be a graded S-module.ThenU′=U⊗A A′= i≥0 n≥0U′i,nis a graded S′-module.Lemma2.6.If E(N≥j)is aﬁnitely generated S-module then E(N)is aﬁnitely generated S-module.Proof.Set D=E(N)/E(N≥j)We have the following exact sequence of S-modules0−→E(N≥j)−→E(N)−→D−→0.Using Gulliksen’s result it follows that D is aﬁnitely generated T=A[t1,...,t c]-module.Since T is a subring of S,we get that D is aﬁnitely generated S-module. Thus if E(N≥j)is aﬁnitely generated S-module then E(N)is aﬁnitely generated S-module Lemma2.7.[with notation as in2.5(3)]Let A→A′be a faithfullyﬂat extension of rings and let U= i≥0 n≥0U i,n be a graded S-module such that U i,n is aﬁnitely generated A-module for each i,n≥0.If U′is aﬁnitely generated S′-module.then U is aﬁnitely generated S-module.Proof.Choose aﬁnite generating set L of U′.Suppose deg i u≤a and deg n u≤b for each u∈L.For0≤i≤a and0≤n≤b choose aﬁnite generating set C i,n of U i,n.SetC=ai=0b n=0C i,n.Set C′={u⊗1|u∈C}.Let V be the submodule of U generated by C.By construction C′generates U′.So U′=V′.Thus(U/V)⊗A A′=0.Since A′is a faithfullyﬂat A-algebra we get U=V.So U is aﬁnitely generated S-module.3.The local caseIn this section we prove Theorem1when(Q,n)is local.Let m be the maximal ideal of A.Set k=A/m.Let I be an ideal in A.Set F(I)=R(I)⊗A k= n≥0I n/m I n theﬁber cone of I.ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES5 3.1.Assume N= n≥0N n is aﬁnitely generated R(I)-module.NoticeF(N)=N⊗A k= n≥0N n/m N nis aﬁnitely generated F(I)-module.Setspread(N):=dim F(I)N/m N the analytic spread of N.Proof of Theorem1in the local case.Case1.The residueﬁeld k=A/m is inﬁnite.We induct on spread(N).First assume spread(N)=0.This implies that N n/m N n=0for all n≫0. By Nakayama Lemma,N n=0for all n≫0;say N n=0for all n≥j.Then E(N≥j)=0and it is obviously aﬁnitely generated S-module.By2.6we get that E(N)is aﬁnitely generated S-module.When spread(N)>0then there exists u=xt∈R(I)1which is N⊕F(N)-ﬁlter regular,i.e.,there exists j such that(0:N u)n=0and(0:F(N)u)=0for all n≥j.Set N≥j= n≥j N n and U=N≥j/uN≥j.Notice we have an exact sequence of R(I)-modules0−→N≥j(−1)u−→N≥j−→U−→0.For each n≥j the functor Hom A(M,−)induces the following long exact sequence of A-modules0−→Hom A(M,N n)u−→Hom A(M,N n+1)−→Hom A(M,U n+1)−→Ext1A(M,N n)u−→Ext1A(M,N n+1)−→Ext1A(M,U n+1)···························−→Ext i A(M,N n)u−→Ext i A(M,N n+1)−→Ext i A(M,U n+1)···························Using the naturality of Eisenbud operators we have the following exact sequence of S-modules−−−→E(N≥j)−→E(U)E(N≥j)(−1,0)(u,0)By constructionspread(U)=spread(N≥j)−1=spread(N)−1.By induction hypothesis E(U)is aﬁnitely generated S-module.Therefore by Lemma3.2we get E(N≥j)is aﬁnitely generated ing2.6we get that E(N)isﬁnitely generated S-module.Case2.The residueﬁeld k isﬁnite.In this case we do the standard trick.Let Q′=Q[X]n Q[X].Set A′=A⊗Q Q′. Notice A′=Q[X]m Q[X]is aﬂat A-algebra with residueﬁeld k(X)which is inﬁnite. Set I′=IA′and M′=M⊗Q Q′=M⊗A A′.Notice projdim Q′M′isﬁnite.Set R(I)′=R(I′)the Rees algebra of I′.Notice that N′=N⊗A A′is aﬁnitely generated R(I)′-module.Also note that E(N′)=E(N)⊗A A′.By Case1we have that E(N′)is aﬁnitely generated S′-module.So by Lemma 2.7we get that E(N)is aﬁnitely generated S-module.The next Lemma is a bi-graded version of Lemma2.8(1)from[11].6TONY J.PUTHENPURAKALLemma 3.2.Let R be a Noetherian ring(not necessarily local)and let B= i,j≥0B i,j be aﬁnitely generated bi-graded R-algebra with B0,0=R.SetB x= i≥0B(i,0)and B y= j≥0B(0,j).Let V= i,j≥0V i,j be a bi-graded B-module such that(1)V i,j is aﬁnitely generated R-module for each i,j≥0.(2)For each i≥0,V i= j≥0V i,j isﬁnitely generated as a B y-module(3)For each j≥0,V j= i≥0V i,j isﬁnitely generated as a B x-module(4)There exists z∈B(r,0)(with r≥1)such that we have the following exactsequence of B-modulesV(−r,0)z−→Vψ−→Dwhere D is aﬁnitely generated bi-graded B-module.Then V is aﬁnitely generated B-moduleProof.Step1.We begin by reducing to the case whenψis surjective.Notice D′=imageψis aﬁnitely generated bi-graded B-module.Ifψ′:V→D′is the map induced byψthen we have an exact sequenceV(−r,0)z−→Vψ′−→D′−→0.Thus we may assumeψis surjective.Step2.Choosing generators:2.1:Choose aﬁnite set W in V of homogeneous elements such thatψ(W)={ψ(w)|w∈W}is a generating set for D.2.2:Assume all the elements in W have x-co-ordinate≤c.2.3:For each i≥0,by hypothesis V i is aﬁnitely generated B y-module.So we may choose aﬁnite set P i of homogeneous elements in V i which generates V i as aB y-module.2.4:SetG=W c i=0P i .Clearly G is aﬁnite set.Claim:G is a generating set for V.Let U be the B-submodule of V generated by G.It suﬃces to prove that U i,j=V i,j for all i,j≥0.By construction we have that for0≤i≤c(*)U i,j=V i,j for each j≥0Let be the lex-order on X=Z≥0×Z≥0.It is well-known that is a total order on X.So we can prove our result by induction on X with respect to the total order .The base case is(0,0).In this case U0,0=V0,0by(*).Let(i,j)∈X\{(0,0)}and assume that for all(r,s)≺(i,j);we have U i,j=V i,j.Subcase1.i≤c.By(*)we have U i,j=V i,j.ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES7 Subcase2.i>c.Let p∈V i,j.By construction,note that there exists w1,...,w m∈W⊆C suchthatψ(p)=ml=0h lψ(w l)where h i∈B.We may assume that deg h l w l=(i,j)for each l Set p′= m i=0h i w i∈V i,j.Notice1.p′∈U i,j.2.p−p′∈kerψ.Sop−p′=z•q where q∈V(i−r,j).If q=0then p=p′∈U i,j.Otherwise note that(i−r,j)≺(i,j).So by induction hypothesis q∈U(i−r,j).It follows that p∈U i,j.Thus V i,j⊆U i,j.Since U i,j⊆V i,j by construction it follows that U i,j=V i,j. The result follows by induction on X.4.The global case:We need quite a few preliminaries to prove the global case.See4.2for the diﬃculty in going from local to the global case.Note that in the local case we proved the result by inducting on spread(N).This is unavailable to us in the global situation as there are usually inﬁnitely many maximal ideals in a global ring.Most of this section we will discuss two invariants of a graded R(I)-module N= n≥0N n.We will use these invariants to prove Theorem1by induction.4.1.Notation and Conventions:We take dimension of the zero module to be−1. We deﬁne the zero-polynomial to have degree−1.Let P∈Spec Q.If P⊇f then set p=P/f.If P f then any A-module localized at P is zero.So assume P⊇f.Notice(1)R(I)p∼=R(IA p)and S p∼=R(I)p[t1,...,t c](2)M p=M P hasﬁnite projective dimension as a Q P-module.(3)E(N)p∼=E(N p).4.2.The diﬃculty in going from local to global:For each p∈Spec A it follows from4.1that E(N p)is aﬁnitely generated A p-module. Usually Supp A E(N)will be an inﬁnite set.So we cannot apply the local case and conclude.The situation when Supp A E(N)is aﬁnite set will help in the base step of our induction argument to prove Theorem1.So we show it separately.Lemma4.3.If Supp A E(N)is aﬁnite set then E(N)is aﬁnitely generated S-module.Proof.We may choose aﬁnite subset C of E(N)such that its image in E(N)p generates E(N)p for each p∈Supp A E(N).Set U to be theﬁnitely generated submodule of E(N)generated by C.Set D=E(N)/U.Notice that D p=0for each p∈Spec A.So D=0.Therefore E(N)=U is aﬁnitely generated A-module.8TONY J.PUTHENPURAKAL4.4.First Inductive device:Since N is aﬁnitely generated R(I)-module we have ann A N i⊆ann A N i+1for all i≫0.Since A is Noetherian it follows that ann A N n is constant for all n≫0. Call this stable value L N.This enables us to deﬁne Limit dimension of N.limDim N=limn→∞dim A N n=dim A/L N.Since A hasﬁnite Krull-dimension we get that limDim N isﬁnite.4.5.Let P be a prime in ideal in A.If D is aﬁnitely generated A-module thenann AP D P=(ann A D)P=(ann A D)A P.Therefore(L N)P=L NP.4.6.Note that if limDim(N)=−1then N j=0,say for all j≥j0.So E(N≥j)=0. Using2.6it follows that E(N)is aﬁnitely generated S-module.Theﬁrst non-trivial case is the followingProposition4.7.If limDim(N)=0then E(N)is aﬁnitely generated A-module. Proof.This implies that A/L N is Artinian.Say dim N n=0for n≥r.Clearly Supp A E(N≥r)⊆Supp A A/L N aﬁnite set of maximal ideals in A.It follows from4.3that E(N≥r)is aﬁnitely generated ing2.6we get that E(N)is aﬁnitely generated S-module.4.8.higher degreeﬁlter-regular elementWe do not haveﬁlter regular elements of degree1in the global situation.However we can do the following:Set E=N/H0R+(N).Assume E=0.As H0R+(E)=0there exists homogeneousu∈R+such that u is E-regular,[5,1.5.11].Say deg u=s.Since E n=N n for all n≫0it follows that the map N i→N i+s induced by multiplication by u is injective for all i≫0.We will say that u is a Nﬁlter-regular element of degree s.4.9.The second inductive device:We now discuss a global invariant of N which patches well with local ones.4.10.The local invariantLet(A,m)be local and let W= n≥0W n be aﬁnitely generated R(I)-module. For convenience we assume that L W=ann A W n for all n≥0.Let a⊆L W be an ideal.Fix j≥0.Setd a(W,j)= 0,if dim W j<dim A/a,e(m,W j);otherwise.Note that W j is an A/a-module and that d a(W,j)is the modiﬁed multiplicity function on the A/a-module W j.Remark4.11.Notice if dim W j=dim A/a thend a(W,j)=d LW(W,j).Letµ(D)denote the minimal number of generators of an A-module D. Lemma4.12.The function d a(W,−)is polynomial of degree≤µ(I)−1.ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES 9Proof.We may assume that the residue ﬁeld of A is inﬁnite.Set T =R (I )/a R (I )= n ≥0T n .Notice T 0=A/a .Let x =x 1,...,x r be a minimal reduction of m (A/a ).So e (m ,−)=e (x ,−),cf.[5,4.6.5].Then by a result due to Serre,cf.,[5,4.7.6],we get that e (x ,W j )=r i =0(−1)i ℓ(H i (x ,W j ))Notice H i (x ,W )= j ≥0H i (x ,W j )is a ﬁnitely generated T/x T -module.Notice (T/x T )0=A/(a +x )is Artinian.Furthermore (T/x T )1is a quotient of R (I )1and so can be generated by µ(I )elements.Therefore the function j →ℓ(H i (x ,W j ))is polynomial of degree ≤µ(I )−1.The result follows. Deﬁnition 4.13.θ(a ,W )=degree of the polynomial function d a (W,−).Clearly θ(a ,W )is non-negative if and only if limDim W =dim R/a and is −1otherwise.4.14.The global invariantLet A be a Noetherian ring with ﬁnite Krull dimension.Let I =(x 1,...,x s )be an ideal in A .Let W = n ≥0W n be a ﬁnitely generated R (I )-module.For convenience we assume that L W =ann A W n for all n ≥0.Let a ⊆L W be an ideal.SetC (a )={m |m ∈m-Spec(A ),m ⊇a &dim(A/a )m =dim A/a }.Let I =(x 1,...,x s ).If m ∈C (a )then note that (a)W m = n ≥0(W n )m .(b)L W m =(L W )m .So a m ⊆L W m (c)θ(a m ,W m )≤s −1.Deﬁneθ(a ,W )=max {θ(a m ,W m )|m ∈C (a )}.By (c)above we get that θ(a ,W )is deﬁned and is ≤s −1.4.15.Properties of θ(a ,W ).We describe some properties of θ(a ,W )we need for the proof of global case of Theorem 1.Let I =(x 1,...,x s ).(i)θ(a ,W )≤s −1.This is clear.(ii)If L W =A then θ(L W ,W )≥0.It suﬃces to consider the local case.Note that then d L W (W,j )>0for allj ≥0.It follows that θ(L W ,W )≥0.(iii)θ(a ,W )=−1if and only if limDim W <dim A/a .This follows from the following four facts:(a)By deﬁnition of C (a )we have thatdim A/a =dim(A/a )m for each m ∈C (a ).(b)If θ(a ,W )=−1then θ(a m ,W m )=−1for all m ∈C (a ).This is equivalentto saying that limDim W m <dim(A/a )m for all m ∈C (a )(c)Note that since a ⊆L W we get thatlimDim W =max {limDim W m |m ∈C (a )}10TONY J.PUTHENPURAKAL(iv)Ifθ(a,W)≥0thenθ(L W,W)≤θ(a,W).By previous item we get that limDim W=dim A/a.By hypothesis we also have a⊆L W.Since dim A/a=dim A/L W it follows that C(L W)⊆C(a).Using4.11it follows thatθ(L W,W)≤θ(a,W).(v)u∈R(I)+be homogeneous of degree b.Assume u is W-ﬁlter regular.Set E=W/uW.Thenθ(L W,E)≤θ(L W,W)−1.Supposeθ(L W,E)=θ((L W)p,E p)for some p∈C(a).Since u is Wﬁlter regular;multiplication by u induces the following exact sequence0→W j−b→W j→E j→0for all j≫0.Localization at p yields an exact sequence0→(W j−b)p→(W j)p→(E j)p→0for all j≫0.(−,−)is an additive functor on(A/L W)p)-modules we get that Since d LW pθ((L W)p,E p)=θ((L W)p,W p)−1.The result follows sinceθ((L W)p,E p)=θ(L W,E)andθ((L W)p,W p)≤θ(L W,W).We now give a proof of our main resultProof of Theorem1.We induct on limDim N.If limDim N=−1,0then the result follows from4.7.Assume limDim N≥1and assume the result holds for all R(I)-modules E with limDim E≤limDim N−1.Let x∈R(I)+be homogeneous and a N-ﬁlter regular element.Set D=N/xD.By Lemma2.6it suﬃces to assume the case when x is N-regular.We now induct onθ(L N,N)Ifθ(L N,N)=0thenθ(L N,D)≤−1,by4.15(v). Using4.15(iii)we get thatlimDim D<dim A/L N=limDim N.By induction hypothesis(on limDim)the module E(D)isﬁnitely generated S-module.The short exact sequence of R(I)-modules0→N(−r)x−→N→D→0induces an exact sequence of S-modulesE(N)(−r)x−→E(N)→E(D).By Lemma3.2we get that E(N)is aﬁnitely generated S-module.We assume the result ifθ(L N,N)≤i and prove whenθ(L N,N)=i+1.Let D be as above.Soθ(L N,D)≤i,by4.15(v).Ifθ(L N,D)=−1then the argument as above yields E(N)to be aﬁnitely generated S-module.Ifθ(L N,D)≥0then by4.15(iv)we get thatθ(L D,D)≤θ(L N,D)≤i.So by induction hypothesis onθ(−,−)we get that E(D)is aﬁnitely generated S-module. By an argument similar to the above we get that E(N)is aﬁnitely generated S-module.ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES115.Application IAsymptotic Associated primesIn this section we give a proof of our main motivating question Theorem5.9. We also give two proofs of Theorem5.1.Theorem5.1.Let Q be a Noetherian ring withﬁnite Krull dimension and let f= f1,...f c be a regular sequence in Q.Set A=Q/(f).Let M be aﬁnitely generated A-module with projdim Q Mﬁnite.Let I an ideal in A and let N= n≥0N n be a ﬁnitely generated R(I)-module.Thenn≥0 i≥0Ass Ext i A(M,N n)is aﬁnite set.Furthermore there exists i0,n0such that for all i≥i0and n≥n0we haveAss Ext2i A(M,N n)=Ass Ext2i0A (M,N n)Ass Ext2i+1A (M,N n)=Ass Ext2i0+1A(M,N n)The following example shows that two set of stable values of associate primes can occurExample 5.2.Let Q=k[[u,x]],A=Q/(ux).Let M=Q/(u),I=A and N=M[t](so N n=M for all n).For i≥1one hasExt2i−1A(M,M)=0Ext2i A(M,M)=k.5.3.We now state special case of a result due to E.West[14,3.2;5.1].Let R=A[x1,...,x r;y1,...y s]be a bi-graded A-algebra with deg x i=(2,0)and deg y j=(0,1).Let M= i,n≥0M(i,n)be aﬁnitely generated R-module.Then(1) i≥0 n≥0Ass A M(i,n)is aﬁnite set.(2)∃i0,n0such that for all i≥i0and n≥n0we haveAss A M(2i,n)=Ass A M(2i,n0)Ass A M(2i+1,n)=Ass A M(2i+1,n0)First proof of Theorem5.1.The result follows from our main Theorem1and5.3.For the convenience of the readers we give a self-contained second proof of Theo-rem5.1.We need the following exercise problem from Matsumara’s text(6.7,page 42)[9].Fact5.4.Let f:A→B be a ring homomorphism of Noetherian rings.Let U be aﬁnitely generated B-module.ThenAss A U={P∩A|P∈Ass B U}In particular Ass A U is aﬁnite set.We will also need the followingLemma5.5.Let R=A[x1,...,x r;y1,...y s]be a bi-graded A-algebra with deg x i= (1,0)and deg y j=(0,2).Let D= i,j≥0D(i,j)be aﬁnitely generated R-module. Set l=r•s.Then there exists i0,j0such that there exists inclusionsD(i,j)֒→D l(i+1,j+2)for all i≥i0&j≥j012TONY J.PUTHENPURAKAL Proof.Let R++= i≥1,j≥1R(i,j)be the irrelevant ideal of R.Set(D).E=D/H0R++It is easy to check that D(i,j)=E(i,j)for all i,j≫0.So if E=0then we have nothing to prove.We consider the case when E=0.Notice that(E)=0.So grade(R++,E)>0.H0R++Now R++is generated by the l elements x i y j;where1≤i≤r and1≤j≤s. Notice also that deg x i y j=(1,2)for all i,j.Consider the mapφ:E→(E(1,2))ld→x i y j d.Since grade(R++,E)>0we get thatφis injective.Since E(i,j)=D(i,j)for all i,j≫0;the result follows. Second proof of Theorem5.1.Let S=R(I)[t1,...,t c].Then D=E(N)is a ﬁnitely generated S-module.1.By5.4,it follows that Ass A E(N)is aﬁnite set.Notice thatAss A E(N)= n≥0 i≥0Ass Ext i A(M,N n).2.By5.5it follows that we have inclusions(M,N n+1)for all i,n≫0.Ext i A(M,N n)֒→Ext i+2AIt follows that(M,N n+1)for all i,n≫0.Ass A Ext i A(M,N n)⊆Ass A Ext i+2AUsing1.we get the result.To prove an analog of Theorem5.1for a local complete intersection we need the following result.Lemma5.6.Let(A,m)be a Noetherian local ring.Let A be the completion of A with respect to m.Let B be aﬁnitely generated A-algebra containing A.Let E be an A-module such that E⊗A A is aﬁnitely generated B-module.Let D be any ﬁnitely generated A-module.Then(a)Ass bE⊗A A is aﬁnite set.A(b)Ass A E⊗A A is aﬁnite set.(c)Ass A E=Ass A E⊗A A.In particular Ass A E is aﬁnite set.(d)Ass A D=Ass A D⊗A A.To prove this result we need Theorem23.3from[9].Unfortunately there is a typographical error in the statement of Theorem23.3in[9].So we state it here. Theorem5.7.Letϕ:A→B be a homomorphism of Noetherian rings,and let E be an A-module and G a B-module.Suppose that G isﬂat over A;then we have the following:(i)if p∈Spec A and G/p G=0thenaϕ(Ass B(G/p G))=Ass A(G/p G)={p}.ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES13 (ii)Ass B(E⊗G)= p∈Ass A(E)Ass B(G/p G).Remark5.8.In[9]Ass A(E⊗G)is typed instead of Ass B(E⊗G).Proof of Theorem5.6.We consider the natural ring homomorphismsα:A֒→ Aβ: A֒→Bγ:A֒→BClearlyγ=β◦α.(a)We use the mapβand fact5.4to get our result.(b)We use the mapγand fact5.4to get our result.(c)On the spectrum of the rings we have:aγ=aα◦aβ.Therefore using fact5.4we getAss A E⊗ A=aα(Ass b A E⊗ A)E⊗ A}.={P∩A|P∈Ass bAWe consider theﬂat extensionα:A→ A.By Theorem5.7.(ii)(with G= A) we have(*)Ass bA E⊗A A = P∈Ass A E Ass b A A/P A.Now A is faithfullyﬂat.So if P∈Spec A then A/P A=0.By Theorem5.7.(i)we getaα Ass b A A14TONY J.PUTHENPURAKALﬁnitely generated B=R(I A)[t1,...t c]-algebra.By Lemma5.6we get that Ass A E is aﬁnite set.NoticeAss A E(N)= n≥0 i≥0Ass A Ext i A(M,N n)2.Set E=E(N).By5.5we have injective maps(*)E(i,n)⊗ A֒→ E(i+1,n+2)⊗ A l for all i,n≫0. Using Lemma5.6with B= A and E=E(i,n)we get that for each i≥0and n≥0Ass A E(i,n)=Ass A E(i,n)⊗A AThe result follows from1.and(*).6.Application IISupport VarietiesLet(A,m)be a local complete intersection of codimension c.Let M,N be two ﬁnitely generated A-modules.Deﬁnecx A(M,N)=inf b∈N|n b−1<∞In this section we prove the following theoremTheorem6.1.Let(A,m)be a local complete intersection,M,N twoﬁnitely gen-erated A-modules and let I be a proper ideal in A.Thencx A(M,I j N)is constant for all j≫0.6.2.Reduction to the case when A is complete and residueﬁeld of A is algebraically closed6.3.Suppose A′is aﬂat local extension of A such that m′=m A′is the maximal ideal of A′.If E is an A-module then set E′=E⊗A A′.Notice I′∼=IA′and we consider it as an ideal in A′.By[1,7.4.3]A′is also a complete intersection.It can be easily checked thatcx A′(M′,(I′)j N′)=cx A(M,I j N)for all n≥0.We now do our reduction in two stepsBy[3,App.Th´e or´e me1,Corollaire],there exists aﬂat local extension A⊆ A such that m=m A is the maximal ideal of A and the residueﬁeld k of A is an algebraically closed extension of k.By6.3it follows that we may assume k to be algebraically closed.We now complete A.Note that A is aﬂat extension of A which satisﬁes6.3.Thus we may assume that our local complete intersection A(1)is complete.So A=Q/(f1,...,f c)where(Q,n)is regular local and f1,...,f c∈n2is a regular sequence.(2)The residueﬁeld of k is algebraically closed.Of course there exists many Q and f1,...,f c of the type as indicated above.We simplyﬁx one such representation of AON THE FINITE GENERATION OF A FAMILY OF EXT MODULES15 6.4.Let U,V be twoﬁnitely generated A-modules.Let E(U,V)= n≥0Ext n A(U,V)be the total ext module of U and V.We con-sider it as a(ﬁnitely generated)module over the ring of cohomological operators A[t1,...,t c].Since projdim Q U isﬁnite E∗(U,V)is aﬁnitely generated A[t1,...,t c]-module.6.5.Let C(U,V)=E(U,V)⊗A k.Clearly C(U,V)is aﬁnitely generated T= k[t1,...,t c]-module.(Here degree of t i is2for each i=1,...,c).Seta(U,V)=ann T C(U,V).Notice that a(U,V)is a homogeneous ideal.6.6.We now forget the grading of T and consider the aﬃne space A c(k).LetV(U,V)=V(a(U,V))⊆A c(k).Since a(U,V)is graded ideal we get that V(U,V)is a cone.6.7.By a result due to Avramov and Buchweitz[2,2.4]we get thatdim V(U,V)=cx A(M,N)Lemma6.8.If I is an ideal in A then there exists j0≥0such thatV(U,I j V)=V(U,I j0V)for all j≥j0.We give a proof of Theorem6.1assuming Lemma6.8.Proof of Theorem6.1.By6.3we may assume that A is also complete and has an algebraically closed residueﬁeld.The result now follows from6.7and6.8.6.9.Let N= j≥0I j V.Set M=U.By hypothesis A=Q/(f1,...,f c)where (Q,n)is a regular local ring and f1,...,f c⊆n2is a regular sequence.6.10.Set C(N)=E(N)⊗A k.By Theorem1E(N)is aﬁnitely generated S= R(I)[t1,...,t c]-module.It follows that C(N)is aﬁnitely generated,bi-graded, G=F(I)[t1,...,t c]-module.Recall that F(I),theﬁber-cone of I,is aﬁnitely generated k-algebra.So we may as well consider C(N)as a bi-graded R=k[X1,...,X m,t1,...t c]-module(of course here X1,...,X m are variables). Furthermore deg X l=(1,0)for l=1,...m and deg t s=(0,2)for s=1,...,c.Set T=k[t1,...,t c].6.11.Advantages of coarsening the grading on C(N)By forgetting the degree on t’s we may consider R=T[X1,...,X m].Notice that the correspondingly we obtainC(N)= n≥0C(U,I j V)We now give aProof of Lemma6.8.We make the constructions as in6.11.So C(N)is aﬁnitely generated R=T[X1,...,X m]-module.Notice that R is N-standard graded.So there exists j0such thatann T C(N)j=ann T C(N)jfor all j≥0.The results follows.。

A Note On The Kadison-Singer Problem

A Note On The Kadison-Singer Problem Charles A. Akemann, Betül Tanbay and Ali Ülger Abstract. Let H be a separable Hilbert space with a fixed orthonormal basis e n n≥1 and BH be the full von Neumann algebra of the bounded linear operators T : H → H. Identifying ℓ CN with the diagonal operators, we consider CN as a subalgebra of BH. For each t ∈ N, let t be the set of the states of BH that extend the Dirac measure t . Our main result shows that, for each t in N, the set t either lies in a finite dimensional subspace of BH ∗ or else it must contain a homeomorphic copy of N. Introduction. Let H be a separable infinite dimensional Hilbert space with a fixed orthonormal basis e n n≥1 . Let BH be the full von Neumann algebra of the bounded linear operators T : H → H. Identifying each bounded sequence with a diagonal operator, we can consider ℓ as a C ∗ −subalgebra of BH. The famous Kadison-Singer problem asks the following: Problem (KS): Does every pure state of ℓ extend in a unique way to a pure state of BH? This 50 year old problem has turned out to be a basic problem related to a dozen other important problems [Cz-Tr]. An apparently more general problem, which is extensively studied in the papers [Ar1], [Ar2], [Ar3], [Ch-Ku] and [Bu-Ch], is the following one: Problem (A). Let A ⊆ B be two C ∗ -algebras, A being a C ∗ -subalgebra of B. When does every pure state of A extend in a unique way to a pure state of B? Problem (KS), and also partially problem (A), may be generalized as follows. Problem (B). Let K be a compact Hausdorff space and Y be a Banach space such that CK ⊆ Y (i.e. Y contains CK as a closed subspace). When does every Dirac measure t (t ∈ K) extend in a unique way to a functional on Y ? Hoping that this general problem may shed some light on the Kadison-Singer problem, we want to study Problem (B) under various hypotheses. Suppose for a moment that, for each t ∈ K, the Dirac measure t has a unique Hahn-Banach extension to an element t of Y ∗ . Then || t || 1 and, as one can easily see (see Lemma 1 in [Ch-Ku]), the mapping t t is a weak ∗ continuous function from K into Y ∗ . Whether t extends to Y uniquely or not, the set t of all the Hahn-Banach ———————————

Growth and generation in SL_2(ZpZ)

let alone for general A, uniformly on A or not. 1.2. Results. We prove the conjecture for G = SL2 (Z/pZ). Main Theorem. Let p be a prime. Let A be a set of generators of G = SL2 (Z/pZ). Then the Cayley graph Γ(G, A) has diameter O((log p)c ), where c and the implied constant are absolute. The theorem is a direct consequence of the following statement. Key Proposition. Let p be a prime. Let A be a subset of SL2 (Z/pZ) not contained in any proper subgroup. (a) Assume that |A| < p3−δ for some ﬁxed δ > 0. Then where c > 0 and ǫ > 0 depend only on δ. (b) Assume that |A| > pδ for some ﬁxed δ > 0. Then there is an integer k > 0, depending only on δ, such that every element of SL2 (Z/pZ) can be expressed as a product of at most k elements of A ∪ A−1 . The crucial fact here is that the constants c, ǫ and k do not depend on p or on A. It follows immediately from the main theorem (via [DSC], §2, Lem. 2, §3, Cor. 3.1, and §3, Cor. 3.2) that the mixing time of Γ(SL2 (Z/pZ), A) is O(|A|(log p)2c+1 ), where c and the implied constant are absolute, and c is as in the main theorem. (The mixing time is the least t for which a lazy random walk of length t starting at the origin of the Cayley graph has a distribution of destinations close to the uniform distribution in the ℓ1 norm; vd. §6) If A equals the projection of a ﬁxed set of generators of a free group in SL2 (Z) (take, e.g., A as in (1.2) or (1.3)) it follows by a simple argument that A must grow rapidly at ﬁrst when multiplied by itself. In such a situation, we obtain a bound of where the implied constant depends on the elements of SL2 (Z) of which A is a projection. For (1.3) and most other examples, this bound is new; for A as in (1.2), it is, of course, known, and the novelty lies in the proof2.

2012Exact Combinatorial Branch-and-Bound for Graph Bisection

Exact Combinatorial Branch-and-Bound for Graph Bisection Daniel Delling∗Andrew V.Goldberg†Ilya Razenshteyn‡Renato F.Werneck§AbstractWe present a novel exact algorithm for the minimum graph bisection problem,whose goal is to partition a graph into two equally-sized cells while minimizing the number of edges between them.Our algorithm is based on the branch-and-bound framework and,un-like most previous approaches,it is fully combinatorial. We present stronger lower bounds,improved branching rules,and a new decomposition technique that contracts entire regions of the graph without losing optimality guarantees.In practice,our algorithm works particu-larly well on instances with relatively small minimum bisections,solving large real-world graphs(with tens of thousands to millions of vertices)to optimality.1IntroductionWe consider the minimum graph bisection problem.Its input is an undirected,unweighted graph G=(V,E), and its goal is to partition V into two sets A and B such that|A|,|B|≤ |V|/2 and the number of edges between A and B(the cut size)is minimized.This fundamental combinatorial optimization problem is a special case of graph partitioning,which asks for arbitrarily many cells. It has numerous applications,including image process-ing[45,51],computer vision[33],divide-and-conquer algorithms[35],VLSI circuit layout[6],distributed com-puting[36],and route planning[14].Unfortunately,the bisection problem is NP-hard[21]for general graphs, with a best known approximation ratio of O(log n)[40]. Only some restricted graph classes,such as grids with-out holes[18]and graphs with bounded treewidth[28], have known polynomial-time solutions.In practice,there are numerous general-purpose heuristics for graph partitioning,such as METIS[31], SCOTCH[12,38],Jostle[50],and KaFFPaE[42]. Successful heuristics tailored to particular graph classes, such as DibaP[37](for meshes)and PUNCH[15](for road networks),are also available.These algorithms ∗Microsoft Research Silicon Valley.dadellin@ †Microsoft Research Silicon Valley.goldberg@ ‡Lomonosov Moscow State University.ilyaraz@. This work was done while the author was at Microsoft Research Silicon Valley.§Microsoft Research Silicon Valley.renatow@ are quite fast(often running in near-linear time)and can handle very large graphs,with tens of millions of vertices.They cannot,however,prove optimality or provide approximation guarantees.Moreover,most of these algorithms only perform well if a certain degree of imbalance is allowed.There is also a vast literature on practical exact al-gorithms for graph bisection(and partitioning),mostly using the branch-and-bound framework[34].Most of these algorithms use sophisticated machinery to obtain lower bounds,such as multicommodityﬂows[43,44],or linear[3,7,20],semideﬁnite[1,3,30],and quadratic programming[25].Computing such bounds is quite ex-pensive,however,in terms of time and space.As a result,even though the branch-and-bound trees can be quite small for some graph classes,published algorithms can only solve instances of moderate size(with hundreds or a few thousand vertices)to optimality,even after a few hours of processing.(See Armbruster[1]for a sur-vey.)Combinatorial algorithms[19]can oﬀer a diﬀerent tradeoﬀ:they provide weaker lower bounds,but com-pute them much faster(often in sublinear time).This works well for random graphs with up to100vertices, but does not scale to larger instances.This paper introduces a new exact algorithm for graph bisection.We use novel combinatorial lower bounds that can be computed in near-linear timein Figure1:Example of a minimum bisection.practice.Even so,these bounds are quite strong,and can be used to ﬁnd optimum solutions to real-world graphs with remarkably many vertices (more than a million for road networks,or tens of thousands for VLSI and mesh instances).See Figure 1for an example.To the best of our knowledge,our method is the ﬁrst to ﬁnd exact solutions for instances of such scale.In fact,it turns out that the running time of our algorithm depends more on the size of the bisection than on the size of the graph.Our paper has four main contributions.First,we introduce (in Section 3)new and improved combinato-rial lower bounds that signiﬁcantly strengthen previous bounds.Second,we propose (in Section 4)elaborate branching rules that help to exploit the full potential of our bound.Third,Section 5introduces a new decompo-sition technique that boosts performance substantially:it ﬁnds the optimum solution to the input by solving a small number of (usually much easier)subproblems in-dependently.Finally,we present (in Section 6)a careful experimental analysis of our techniques.2PreliminariesWe use G =(V,E )to denote the input graph,with n =|V |vertices and m =|E |edges.Each vertex v ∈V may have an associated integral weight ,denoted by w (v ),and each edge e ∈E has an associated integral cost c (e ).Let W = v ∈V w (v ).A partition of G is a partition of V ,i.e.,a set of subsets of V which are disjoint and whose union is V .We say that each such subset is a cell ,whose weight is deﬁned as the sum of the weights of its vertices.The cost of a partition is the sum of the costs of all edges whose endpoints belong to diﬀerent cells.A bisection is a partition into two cells.A bisection is -balanced if each cell has weight at most (1+ ) W/2 .If =0,we say the partition is perfectly balanced (or just balanced ).The minimum graph bisection problem is that of ﬁnding the minimum-cost balanced bisection.To simplify exposition,unless otherwise noted we consider the unweighted,balanced version of the prob-lem,where w (v )=1for all v ∈V ,c (e )=1for all e ∈E ,and =0.We must therefore partition G into two cells,each with weight at most n/2 ,while minimizing the number of edges between cells.(Section 3.4shows how we can handle less restrictive settings.)A standard technique for ﬁnding exact solutions to NP-hard problems is branch-and-bound [22,34].It performs an implicit enumeration by dividing the original problem into two or more slightly simpler subproblems,solving them recursively,and picking the best solution found.Each node of the branch-and-bound tree corresponds to a distinct subproblem.Ina minimization context,the algorithm keeps a global upper bound U on the solution of the original problem,which can be updated as the algorithm ﬁnds improved solutions.To process a node in the tree,we ﬁrst compute a lower bound L on any solution to the corresponding subproblem.If L ≥U ,we prune the node:it cannot lead to a better solution.Otherwise,we branch ,creating two or more simpler subproblems.In the concrete case of graph bisection,each node of the branch-and-bound tree corresponds to a partial assignment (A,B ),where A,B ⊆V and A ∩B =∅.We say the vertices in A or B are assigned ,and all others are free (or unassigned ).This node implicitly represents all valid bisections (A +,B +)that are extensions of (A,B ),i.e.,such that A ⊆A +and B ⊆B +.In particular,the root node,which represents all valid bisections,has the form (A,B )=({v },∅).(Note that the root can ﬁx an arbitrary node v to one cell to break symmetry.)To process an arbitrary node (A,B ),we must com-pute a lower bound L (A,B )on the value of any ex-tension (A +,B +)of (A,B ).The fastest exact algo-rithms [1,3,7,20,25,30]usually apply mathematical programming techniques to ﬁnd lower bounds.In this paper,we use only combinatorial bounds.In particu-lar,our basic algorithm uses the well-known [9,15]ﬂow bound :the minimum s –t cut between A and B .It is a valid lower bound because any extension (A +,B +)must separate A from B .If the minimum cut happens to be balanced,we can prune (and update U ,if possible).Otherwise,we choose a free vertex v and branch on it,generating subproblems (A ∪{v },B )and (A,B ∪{v }).Note that the ﬂow lower bound can only work well when A and B have similar sizes;even in this case,the corresponding minimum cuts are often far from balanced,with one side containing many more vertices than the other.This makes the ﬂow bound rather weak by itself.To overcome these issues,we introduce a new packing lower bound .3The Packing Lower BoundLet (A,B )be a partial assignment.To make it a balanced bisection,at least n/2 −|A |free vertices must be assigned to A ,obtaining an extended set A +.(A similar argument can be made for B .)Suppose that,for each possible extension A +of A ,we could compute the maximum ﬂow f (A +)between B and A +.Let f ∗be the minimum such ﬂow value (over all possible A +);f ∗is clearly a lower bound on the value of any bisection consistent with (A,B ).Finding f ∗exactly seems expensive;instead,we propose a fast algorithm to compute a lower bound for f ∗.It works as follows (see Figure 2).Let G =G \(A ∪B )be the subgraph of G induced by the vertices that areFigure2:The packing bound.Filled circles are vertices in B;their free neighbors(ﬁlled squares)form a set R.Left:Weﬁrst partition the free vertices into connected cells,each with at most one element of R.Middle: Given any extension A+(hollow circles),the number of nontrivial cells it touches(eight)is a lower bound on the minimum(B,A+)cut.Right:The extension A+that hits the fewest cells(three)is a lower bound on any valid extension.currently unassigned,and let R be the set of vertices of G with at least one neighbor in B(in G).We partition the vertices in G into connected cells,each containing at most one element of R.(Any such partition is valid; as we shall see,we get better lower bounds if the cells containing elements in R are as large as possible and have roughly the same size.)We say that a cell C is nontrivial if it contains exactly one element from R;we call this element the root of the cell and denote it by r(C).Cells with no element from R are trivial. Lemma3.1.Let A+be a valid extension of A,and let c(A+)be the number of nontrivial cells hit by A+.Then c(A+)is a lower bound on the maximumﬂow f(B,A+) from B to A+.Proof.We claim we canﬁnd c(A+)disjoint paths be-tween A+and B,each in a diﬀerent nontrivial cell.Take a nontrivial cell C containing an element v from A+.Be-cause the cell is connected,there is a path P within C between v and its root r(C).Because r(C)belongs to R,there is an edge e(in the original graph G)between r(C)and a vertex w in B.The concatenation of P and e is a path from A+to B.Since any valid extension must contain at least one edge from each of the c(A+) disjoint paths,the lemma follows.Recall that we need a lower bound on any possible extension A+of A.We get one byﬁnding the extension for which Lemma3.1gives the lowest possible bound (for aﬁxed partition into connected cells).We can build this extension with a greedy packing algorithm. First,pick all vertices in trivial cells;because we cannot associate these cells with paths,they do not increase the lower bound.From this point on,we must pick vertices from nontrivial cells.Since the lower bound increases by one regardless of the number of vertices picked in a cell,it makes sense to pick entire cells at once(after one vertex is picked,others in the cell are free—they do not increase the bound).The optimal strategy is to pick cells in decreasing order of size,stopping when the sum of the sizes of all picked cells(trivial and nontrivial)is at least n/2 −|A|.We have thus shown the following: Theorem3.1.The greedy packing algorithmﬁnds alower bound on the value of any bisection consistent with (A,B).3.1Computing Packing Lower Bounds.Note that the packing lower bound is valid for any partition, but its quality depends strongly on which one we pick. We should choose the partition that forces the worst-case extension A+to hit as many nontrivial cells as possible.This means minimizing the total size of the trivial cells,and ensuring all nontrivial cells have the same number of vertices.This problem is hard[13,11], but we propose two heuristics that work well in practice.Theﬁrst heuristic is a constructive algorithm that builds a reasonable initial partition from scratch.Start-ing from|R|unit cells(each with one element of R),in each step it adds a vertex to a cell whose current size is minimum.This algorithm can be implemented in linear time by keeping with each cell C a list E+(C)of poten-tial expansion edges,i.e.,edges(v,w)such that v∈C and w∈C.Vertices that are not reachable from R are assigned to trivial cells.As the algorithm progresses, some cells will run out of expansion edges,as all neigh-boring vertices will already be taken.This may lead to very unbalanced solutions.To improve the partition,we use our second heuris-tic:a local search routine that makes neighboring cells more balanced by moving vertices between them.To do so eﬃciently,it maintains a spanning tree for each nontrivial cell C,rooted at r(C).Initially,this is the tree built by the constructive algorithm.⇓Figure3:Packing local search.Left:The boundaryedge(v,w)determines a path between cells C1and C2(triangles are subtrees).Right:Splitting the path at adiﬀerent point leads to a more balanced partition.The local search works by moving entire subtreesbetween neighboring cells.It processes one boundaryedge at a time.Consider one such edge(v,w),with v∈C1and w∈C2.Without loss of generality,assumecell C1has more vertices than C2.To improve thesolution,we attempt to move an entire subtree from C1to C2.Weﬁnd the best subtree to switch by traversingthe path(in the spanning tree of C1,the largest cell)from v to r(C1).(See Figure3.)Each vertex u on thepath is associated with a possible move:removing thesubtree rooted at u from C1and inserting it into C2.Among these,let u∗be the vertex corresponding to the most balancedﬁnal state(in which the sizes of C1andC2would be closest).If this is more balanced than thecurrent state,we switch.The local search runs until a local optimum,whenno improving switch exists.To implement it eﬃciently,we keep track of boundary edges and subtree sizes ex-plicitly.This ensures the algorithm runs in polynomial(but superlinear)worst-case time.In practice,however,we reach a local optimum after very few moves,and thelocal search is about as fast as the constructive algo-rithm.(Note that theoretical improvements would bepossible using a dynamic-tree data structures[46],butin practice they are quite costly for graphs of moderatediameter[49],as in our case.)3.2Combining Packing and Flows.Although we have two lower bounds,based on packing andﬂows,we cannot simply add them to obtain a uniﬁed lower bound, since they may interfere with one another.It is easy to see why:each methodﬁnds(implicitly)a set of edge-disjoint paths such that at least one edge from each such path must be in the solution.(For theﬂow bound,these are the paths in theﬂow decomposition.)Simply adding the bounds would require the sets of paths found by each algorithm to be mutually disjoint,which is usually not the case.To combine the bounds properly,we must ﬁrst compute theﬂow bound,then a packing bound that takes theﬂow into account.More precisely,weﬁrst compute aﬂow bound f as usual.We then remove allﬂow edges from G,obtaining a new graph G f.Finally,we compute the packing lower bound p on G f.Now f+p is a valid lower bound on the cost of the best bisection extending the current assignment(A,B),since there is no overlap between the paths considered by each method(ﬂow and packing).This algorithm provides valid lower bounds regard-less of theﬂow decomposition used,but its packing por-tion is better if it has more edges to work with.We therefore favorﬂow decompositions with as few edges as possible:instead of using the standard push-relabel approach[24],we prefer an augmenting-path algorithm that greedily sendsﬂows along shortest paths.Our im-plementation uses a simpliﬁed version of the IBFS(in-cremental breadthﬁrst search)algorithm[23],which is about as fast as push-relabel on our test instances.3.3Forced Assignments.Assume we have already computed theﬂow bound f followed by an additional packing lower bound p(using the neighbors of B as roots).For a free vertex v,let N(v)be its set of neighbors in G f(the graph withoutﬂow edges),let deg Gf(v)=|N(v)|,and let C be the cell(in the packing partition)containing v.We can often use logical implications to assign v to one of the sides(A or B)without actually branching on it.The idea is simple:if we can show that assigning v to one side would increase the lower bound to at least match the upper bound,we can safely assign v to the other side.First,consider what would happen if v were added to A.Let x(v),the expansion of v,be the number of nontrivial cells(from the packing bound)that contain vertices from N(v).Note that0≤x(v)≤deg Gf(v). Assigning v to A would create x(v)disjoint paths from A to B,eﬀectively increasing theﬂow bound to f = f+x(v).(See Figure4.)Note,however,that f +p may not be a valid lower bound,since the new paths may interfere with the“pure”packing bound.Instead,we compute a restricted packing lower bound p ,taking as trivial the cells that intersect N(v)(we just assume they belong to A+).If f +p is at least as high as the current upper bound,we have proven that v must be assigned to B.In general,this test will succeed only when the cells are unevenly balanced(otherwise the increase in ﬂow is oﬀset by a decrease in the packing bound).Conversely,consider what would happen if v were added to B:we could split C into deg Gf(v)cells,oneFigure4:Illustration of forced assignments.Left:State after initial bounds have been computed(withﬂow edges already removed).Filled circles belong to B,and v is free.Middle:Assigning v to A would increase the ﬂow bound.Right:Assigning v to B can increase the packing bound by splitting a cell into several ones. rooted at each neighbor of v.The size of each newcell can be computed in constant time,since we knowthe subtree sizes within the original spanning tree ofC.1We then recompute the packing lower bound(usingthe original cells,with C replaced by the newly-createdsubcells)and add it to the originalﬂow bound f.Ifthis at least matches the current upper bound,thenwe have proven that v must actually be assigned to A.This works particularly well for trivial cells(the packingbound is unlikely to change for nontrivial ones).Note that these forced assignments only work whenlower and upper bounds are very close.Their main ben-eﬁt is to eliminate vertices that are not good candidatesfor branching.Since the tests are very fast,they arestill worth running.3.4Extensions.We can easily generalize the pack-ing bound to handle -balanced partitions.In this case,cells must have size at most M+= (1+ ) n/2 and atleast M−=n−M+;the packing bound must distributeM−vertices instead of n/2 .Dealing with weightedvertices is also quite simple.The packing bound is theminimum number of cells containing at least half of thetotal weight.When creating the packing partition,weshould therefore strive to make cells balanced by weightinstead of number of vertices;this can easily be incor-porated into the local search.To handle small integraledge weights,we can simply use parallel edges.Addi-tional extensions(such as arbitrary edge weights or par-titions into more than two cells)are possible,but morecomplicated.4BranchingIf the lower bound for a given subproblem(A,B)isnot high enough to prune it,we must branch on an1Note that,if the cell containing v is nontrivial,we couldsplit it into deg Gf (v)+1cells by keeping the original root.Forsimplicity and performance,our implementation does not do this.unassigned vertex v,creating subproblems(A∪{v},B) and(A,B∪{v}).Our experiments show that the choice of branching vertices has a signiﬁcant impact on the size of the branch-and-bound tree(and the total running time).Intuitively,we should branch on vertices that lead to higher lower bounds on the child subproblems. Given our lower-bounding algorithms,we can infer some properties the branching vertex v should have.First,theﬂow and packing bounds would both ben-eﬁt from having the assigned vertices evenly distributed (on both sides of the optimum bisection).Since we do not know what the bisection is,a reasonable strategy is to spread vertices over the graph by branching on vertices that are far from both A and B.(Note that a single BFS canﬁnd the distances from A∪B to all vertices.)We call this the distance criterion.Second,we prefer to branch on vertices that appear in large cells(from the packing bound).By branching on a large cell,we allow it to be split,thus improving the packing bound.Finally,to help ourﬂow bound,we would like to send a large amount ofﬂow from a branching vertex v to A or B.This suggests branching on vertices that are well-connected to the rest of the graph.A proxy for connectivity is the degree of v,a trivial upper bound on anyﬂow out of v.In practice,connectivity tends to be more impor-tant than the other criteria,so we branch on the ver-tex v that maximizes q(v)=dist(v)·csize(v)·conn(v)2, where dist(v)indicates the distance from v to the closest assigned vertex,csize(v)is the size of the cell containing v,and conn(v)is the connectivity(degree)of v.4.1Filtering.For some graph classes(notably road networks),high-degree vertices are often separated by a small cut from the rest of the graph[15].This makes degrees poor proxies for connectivity.We could obtain a more robust measure of connectivity by reusing the packing algorithm described in Section3.For eachvertex v,we can run the algorithm with A=∅and B={v}toﬁnd a partition of V\{v}into deg(v) cells.If v is well-connected,all cells should have roughly the same size;if not,some cells will be much smaller than others.Unfortunately,computing this bound for every vertex in the graph would be quite expensive, particularly for large road networks.Instead of explicitly computing the packing bound for every vertex in the graph,we propose aﬁltering routine.Its goal is more modest:determine if some of the most promising vertices(those with the highest degrees)are actually well-connected to the rest of the graph.This is done in two stages.First,we determine whether each vertex is sepa-rated by a cut with exactly one or two edges from the re-mainder of the graph.We use the algorithm of Pritchard and Thurimella[39]toﬁnd all1-cuts and2-cuts in the graph in linear time.(These cuts are quite numerous in road networks[15].)For a vertex v,deﬁne cut(v)=1 if it is inside a1-cut of size at most n/10.Otherwise, if v is contained in a2-cut of size at most n/10,let cut(v)=2.For all other vertices v,let cut(v)=deg(v).The second stage ofﬁltering computes the packing bound for the set S containing the2U vertices v with the highest cut(v)values(recall that U is the best known upper bound),with ties broken at random.Let pack(v) be the corresponding values.Letδbe theﬂoor of the average value of pack(v)over all vertices v∈S.For all vertices w∈S,we set pack(w)=δ.The branch-and-bound algorithm then uses the standard criterion(dist(v)·csize(v)·conn(v)2)to choose the next vertex to branch on,but using a modiﬁed deﬁnition of connectivity:conn(v)=cut(v)·pack(v). 5ContractionBoth lower bounds we consider depend crucially on the degrees of the vertices already assigned.More precisely,let D A and D B be the sum of the degrees of all vertices already assigned to A and B,respectively, with D A≤D B(without loss of generality).It is easy to see that theﬂow bound cannot be larger than D A, and that the packing bound is at most D B/2(when the regions are perfectly balanced).If the maximum degree in the graph is a small constant(which is often the case on meshes,VLSI instances,and road networks, for example),our branch-and-bound algorithm cannot prune anything until deep in the tree.Arguably,the dependency on degrees should not be so strong.The fact that increasing the degrees of only a few vertices could make a large instance substantially easier to solve is counter-intuitive.A natural approach to deal with this is branching on entire regions(connected subgraphs)at once.We would like to pick a region and add all of its vertices to A in one branch,and all toB in the other.Since the“degree”of the region(i.e.,the number of neighbors outside the region)is substantially higher,lower bounds should increase much faster as we traverse the branch-and-bound tree.The obvious problem with this approach is that the optimal bisection may actually split the region in two.Assigning the entire region to A or to B does not exhaust all possibilities.One way to overcome this is to make the algorithm probabilistic.Intuitively,if we contract a small number of random edges,with reasonable probability none of them will actually be cut in the minimum bisection.If this is the case,the optimum solution to the contracted problem is also the optimum solution to the original graph.We can boost the probability of success by repeating this entire procedure multiple times(with multiple randomly selected contracted sets)and picking the best result found.With high probability,it will be the optimum.Probabilistic contractions are a natural approach for cut problems,and indeed known.For example,they feature prominently in Karger and Stein’s randomized global minimum-cut algorithm[29],which uses the fact that contracting a random edge is unlikely to aﬀect the solution.This idea has been used for the minimum bisection problem as well.Bui et al.[9]use contraction within a polynomial-time method which,for any input graph,either outputs the minimum bisection or halts without output.They show the algorithm has good average performance on the class of d-regular graphs with small enough bisections.Since our goal is toﬁnd provably optimum bi-sections,probabilistic solutions are inadequate.In-stead,we propose a contraction-based decomposition al-gorithm,which is guaranteed to output the optimum so-lution for any input.It is(of course)still exponential, but for many inputs it has much better performance than our standard branch-and-bound algorithm.The algorithm is as follows.Let U be an upper bound on the optimum bisection.First,partition E into U+1disjoint sets(E0,E1,...,E U).For each subset E i, create a corresponding(weighted)graph G i by taking the input graph G and contracting all the edges in E i. Then,use our standard algorithm toﬁnd the optimum bisection U i of each graph G i independently,and pick the best.Theorem5.1.The decomposition algorithmﬁnds the minimum bisection of G.Proof.Let U∗≤U be the minimum bisection cost. We must prove that min(U i)=U∗.First,note that U i≥U∗for every i,since any bisection of G i can betrivially converted into a valid bisection of G.Moreover,we argue that the solution of at least one G i willcorrespond to the optimum solution of G itself.LetE∗be the set of cut edges in an optimum bisection of G.(If there is more than one optimum bisection,pickone arbitrarily.)Because|E∗|=U∗and the E i sets are disjoint,E∗∩E i can only be nonempty for at most U∗sets E i.Therefore,there is at least one j such that E∗∩E j=∅.Contracting the edges in E j does not change the optimum bisection,proving our claim.The decomposition algorithm solves U+1inde-pendent subproblems,but the high-degree vertices in-troduced by contraction should make each subproblemmuch easier for our branch-and-bound routine.Besides,the subproblems are not completely independent:theycan all share the same best upper bound.In fact,wecan think of the algorithm as a single branch-and-boundtree with a special root node that has U+1children,each responsible for a distinct contraction pattern.Thesubproblems are not necessarily disjoint(the same par-tial assignment may be reached in diﬀerent branches),but this does not aﬀect correctness.5.1Finding a Decomposition.The decomposition algorithm is correct regardless of how edges are parti-tioned among subproblems,but its performance may vary signiﬁcantly.To make all subproblems have com-parable degree of diﬃculty,our edge partitioning algo-rithm should allocate roughly the same number of edges to each subproblem.Moreover,the choice of which edges to allocate to each subproblem G i also matters. The eﬀect on the branch-and-bound algorithm is more pronounced if we can create vertices with much higher degree.We can achieve this by making sure the edges assigned to E i induce relatively large connected com-ponents(or clumps)in G.(In contrast,if all edges in E i are disjoint,the degrees of the contracted vertices in G i will not be much higher that those of the remaining vertices.)Finally,the shape of each clump matters:all else being equal,we would like its expansion(number of neighbors outside the clump)to be as large as possible.To achieve these goals,we perform the decomposi-tion in two stages:the clump generation partitions allthe edges in the graph into clumps,while the allocationstage ensures that each subproblem is assigned a well-spread subset of the clumps of comparable total size.The goal of the generation routine is to build a set Fof clumps(initially empty)that partition all the edges inthe graph.It does so by maintaining a set C of candidateclumps,which are not necessarily disjoint and may notinclude all edges in the graph.The clumps in C are high-expansion subpaths extracted from BFS trees grownfrom random vertices.(Because they minimize the number of internal edges,such paths tend to have high expansion.)Once there are enough candidates in C,the algorithm transfers a few clumps from C to theﬁnal set F.The clumps are picked from C greedily,according to their expansion,and observing the constraint that clumps in F must be edge-disjoint.Once C no longer has suitable clumps with high enough expansion(higher than a certain thresholdτ),a new iteration of the algorithm starts:it repopulates C by growing new BFS trees,then transfers some of the resulting candidate clumps to F.This algorithm stops when F is complete, i.e.,when every edge in the graph belongs to a clump in F.To ensure convergence,the algorithm gradually decreases the thresholdτbetween iterations:initially only clumps with very high expansion are added to F,but eventually even single-edge clumps are allowed. (The interested reader willﬁnd additional details in Section A,in the appendix.)The allocation phase distributes the clumps to the U+1subproblems(E0,E1,...,E U),which are initially empty.It allocates clumps one at a time,in decreasing order of expansion(high-expansion clumps are allocated ﬁrst).In each step,a clump c is assigned to the set E i whose distance to c is maximum,with ties broken arbitrarily.The distance from E i to c is deﬁned as the distance between their vertex sets,or inﬁnity if E i is empty.(For eﬃciency,we keep the Voronoi diagram of each E i explicitly,updating it whenever a new clump is added.)This ensures clumps are well spread in each subproblem.6ExperimentsWe implemented our algorithms in C++using Visual Studio2010.We ran most experiments on one core of an Intel Core2Duo E8500running Windows7 Enterprise at3.16GHz with4GB of RAM.For a couple of particularly hard instances(clearly marked),we ran a distributed version of the code using the DryadOpt framework[8].DraydOpt is written in C#,and calls our native C++code to solve individual nodes of the branch-and-bound tree.The distributed version was run on a cluster where each machine has two2.6GHz dual-core AMD Opteron processors,16GB of RAM, and runs Windows Server2003.We used100machines only.We always report the total CPU time,the sum of the times spent by our C++code on all402cores (400on the cluster plus2on our standard machine). Note that this excludes the communication overhead, which is negligible.Unless otherwise mentioned,weﬁnd perfectly balanced partitions( =0).6.1Parameter Evaluation.We start by consider-ing the eﬀects of each improvement we propose on per-。

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FORMALIZED MATHEMATICSVolume7,Number2,1998University of BiałystokLattice of Substitutions is a HeytingAlgebraAdam GrabowskiUniversity of BiałystokMML Identiﬁer:HEYTING2.The terminology and notation used in this paper have been introduced in the following articles:[2],[15],[1],[7],[13],[9],[3],[4],[10],[18],[5],[16],[17],[11], [14],[8],[12],and[6].1.PreliminariesWe adopt the following convention:V,C,x are sets and A,B are elements of SubstitutionSet(V,C).Let a,b be sets.Note that{ a,b }is function-like and relation-like.Let A,B be sets.Observe that A˙→B is functional.Next we state several propositions:(1)For all non empty sets V,C there exists an element f of V˙→C such thatf=∅.(2)For all sets a,b such that b∈SubstitutionSet(V,C)and a∈b holds a isaﬁnite function.(3)For every element f of V˙→C and for every set g such that g⊆f holdsg∈V˙→C.(4)V˙→C⊆2[:V,C:].(5)If V isﬁnite and C isﬁnite,then V˙→C isﬁnite.One can check that there exists a set which is functional,ﬁnite,and non empty.323c 1998University of BiałystokISSN1426–2630324adam grabowski2.Some Properties of Sets of SubstitutionsOne can prove the following four propositions:(6)For everyﬁnite element a of V˙→C holds{a}∈SubstitutionSet(V,C).(7)If A B=A,then for every set a such that a∈A there exists a set bsuch that b∈B and b⊆a.(8)Ifµ(A B)=A,then for every set a such that a∈A there exists a setb such that b∈B and b⊆a.(9)If for every set a such that a∈A there exists a set b such that b∈Band b⊆a,thenµ(A B)=A.Let V be a set,let C be aﬁnite set,and let A be an element of Fin(V˙→C).The functor Involved A is deﬁned by:(Def.1)x∈Involved A iﬀthere exists aﬁnite function f such that f∈A and x∈dom f.In the sequel C denotes aﬁnite set.The following propositions are true:(10)For every set V and for everyﬁnite set C and for every element A ofFin(V˙→C)holds Involved A⊆V.(11)For every set V and for everyﬁnite set C and for every element A ofFin(V˙→C)such that A=∅holds Involved A=∅.(12)For every set V and for everyﬁnite set C and for every element A ofFin(V˙→C)holds Involved A isﬁnite.(13)For everyﬁnite set C and for every element A of Fin(∅˙→C)holdsInvolved A=∅.Let V be a set,let C be aﬁnite set,and let A be an element of Fin(V˙→C).The functor−A yielding an element of Fin(V˙→C)is deﬁned as follows: (Def.2)−A={f;f ranges over elements of Involved A˙→C: g:element of V˙→C(g∈A⇒f≈g)}.One can prove the following propositions:(14)A −A=∅.(15)If A=∅,then−A={∅}.(16)If A={∅},then−A=∅.(17)For every set V and for everyﬁnite set C and for every element A ofSubstitutionSet(V,C)holdsµ(A −A)=⊥SubstLatt(V,C).(18)For every non empty set V and for everyﬁnite non empty set C and forevery element A of SubstitutionSet(V,C)such that A=∅holdsµ(−A)=⊤SubstLatt(V,C).lattice of substitutions is a heyting (325)(19)Let V be a set,C be aﬁnite set,A be an element ofSubstitutionSet(V,C),a be an element of V˙→C,and B be an elementof SubstitutionSet(V,C).Suppose B={a}.If A B=∅,then thereexists aﬁnite set b such that b∈−A and b⊆a.Let V be a set,let C be aﬁnite set,and let A,B be elements of Fin(V˙→C).The functor A֌B yielding an element of Fin(V˙→C)is deﬁned as follows: (Def.3)A֌B=(V˙→C)∩{ {f(i)\i;i ranges over elements of V˙→C: i∈A};f ranges over elements of A˙→B:dom f=A}.Next we state two propositions:(20)Let A,B be elements of Fin(V˙→C)and s be a set.Suppose s∈A֌B.Then there exists a partial function f from A to B such that s= {f(i)\i;i ranges over elements of V˙→C:i∈A}and dom f=A.(21)For every set V and for everyﬁnite set C and for every element A ofFin(V˙→C)such that A=∅holds A֌A={∅}.We adopt the following convention:u,v are elements of the carrier of SubstLatt(V,C),a is an element of V˙→C,and K,L are elements of SubstitutionSet(V,C).The following proposition is true(22)For every set X such that X⊆u holds X is an element of the carrier ofSubstLatt(V,C).ttice of Substitutions is ImplicativeLet us consider V,C.The functor pseudo compl(V,C)yielding a unary operation on the carrier of SubstLatt(V,C)is deﬁned as follows:(Def.4)For every element u′of SubstitutionSet(V,C)such that u′=u holds (pseudo compl(V,C))(u)=µ(−u′).The functor StrongImpl(V,C)yielding a binary operation on the carrier of SubstLatt(V,C)is deﬁned by:(Def.5)For all elements u′,v′of SubstitutionSet(V,C)such that u′=u and v′=v holds(StrongImpl(V,C))(u,v)=µ(u′֌v′).Let us consider u.The functor2u yielding an element of Fin(the carrier of SubstLatt(V,C))is deﬁned by:(Def.6)2u=2u.The functor \u yielding a unary operation on the carrier of SubstLatt(V,C) is deﬁned by:(Def.7)( \u )(v)=u\v.326adam grabowskiLet us consider V,C.The functor Atom(V,C)yielding a function from V˙→C into the carrier of SubstLatt(V,C)is deﬁned as follows:(Def.8)For every element a of V˙→C holds(Atom(V,C))(a)=µ{a}.Next we state a number of propositions:(23) f K Atom(V,C)=FinUnion(K,singleton V˙→C).(24)For every element u of SubstitutionSet(V,C)holds u= f u Atom(V,C).(25)( \u )(v)⊑u.(26)For every element a of V˙→C such that a isﬁnite and for every set csuch that c∈(Atom(V,C))(a)holds c=a.(27)For every element a of V˙→C such that K={a}and L=u and L K=∅holds(Atom(V,C))(a)⊑(pseudo compl(V,C))(u).(28)For everyﬁnite element a of V˙→C holds a∈(Atom(V,C))(a).(29)Let u,v be elements of SubstitutionSet(V,C).Suppose that for every setc such that c∈u there exists a set b such that b∈v and b⊆c∪a.Thenthere exists a set b such that b∈u֌v and b⊆a.(30)Let a be aﬁnite element of V˙→C.Suppose for every element b ofV˙→C such that b∈u holds b≈a and u⊓(Atom(V,C))(a)⊑v.Then(Atom(V,C))(a)⊑(StrongImpl(V,C))(u,v).(31)u⊓(pseudo compl(V,C))(u)=⊥SubstLatt(V,C).(32)u⊓(StrongImpl(V,C))(u,v)⊑v.Let us consider V,C.Observe that SubstLatt(V,C)is implicative.One can prove the following proposition(33)u⇒v= f2u((the meet operation of SubstLatt(V,C))◦(pseudo compl(V,C),(StrongImpl(V,C))◦( \u ,v))).References[1]Grzegorz Bancerek.Filters-part I.Formalized Mathematics,1(5):813–819,1990.[2]Czesław Byliński.Binary operations.Formalized Mathematics,1(1):175–180,1990.[3]Czesław 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