集合论悖论的解决V7.5

集合论悖论的解决V7.5
2010.12.25 QQ:165442523
摘要：实数集R的所有幂集：P(R),P(P(R)),P(P(P(R))),...,Pn(R),...因为所有Pn(R)都是不包含自身的集合，罗素悖论中“所有不包含自身的集合”必包含所有Pn(R),也就是包含广义连续统假设中的全部基数{X0,X1,...Xn...}，从而无意义。

简而言之，集合可以包含自身，但集合不可以包含自身的幂集，这就是我与公理集合论最大不同点。

虽然我知道公理集合论是为了解决罗素悖论而产生的,但我认为公理集合论是在走弯路,甚至是误入岐路了.如果不包含下列的理论,我认为<<集合论>>是不完整的.
广义连续统假设:无限集合的基数必是X0,X1,...Xn...之一.
其中的基数X就是阿列夫,因为我找不到这个字符,所以用英文字母X表示了. 无意义公理:一个无限集的基数是极限limXn(n→∞),则这个集合是没有什么意义的.
这个公理是我引入的，我还没在别处见到过。

这个公理是易理解的，它就相当于公理集合论中的真类的概念，但公理集合论引入这个类的概念后就误入岐路了，至少作者是这样认为的。

李均宇第一定理：如果一个集合包含广义连续统假设中全部的基数，也就是集合{X0,X1,...Xn...},则这个集合的基数是limXn(n→∞)
这个定理是显而易见的，用反证法不难证明的。

李均宇第二定理:如果一个无限集合又包含自身的幂集,也就是集合
A={......,P(A)),则这个集合A的基数是limXn(n→∞)
证明:设无限集合A的基数是Xn,n是固定不变的.因为无限集合A又包含自身的所有子集或幂集,而幂集的基数是 X(n+1)=2^Xn,所以无限集合A的势变成
X(n+1),这与原先假设无限集合A的基数是Xn,n是固定不变的相矛盾,所以无限集合A的基数是limXn(n→∞).
李均宇第三定理: 如果一个集合包含一个无穷集的所有幂集，也就是集合
B={P(A),P(P(A)),P(P(P(A))),...,Pn(A),...},则这个集合B的基数是
limXn(n→∞),尤其是当A为实数集R时，集合
B={P(R),P1(R),P2(R),...,Pn(R),...},则这个集合B的基数是limXn(n→∞)
所有幂集,假设无穷集A，则其幂集P(A),幂集的幂集P(P(A)),幂集的幂集的幂集P(P(P(A))),...Pn(A).....称为其所有幂集。

因为一个无穷集的所有幂集的基数就是广义连续统假设中全部的基数，所以由李均宇第一定理知此定理成立。

李均宇第四定理: 假设集合P'n(A)与幂集Pn(A)等势，也就是基数一样，则P'n(A)也相当于幂集Pn(A)一样适用于李均宇第二和第三定理中。

一。

基数悖论
定理1:所有集合的集合的基数是limXn(n→∞).
这个显而易见，这在<<集合论>>中早已有之，这里重述而已。

因为所有集合的集合包含自身幂集，由李均宇第二定理知其基数是limXn(n→∞).所以这种集合在公理集合论中称为真类。

二。

罗素悖论
李均宇第五定理:假设A是不包含自身的集合(不是真类)，则A的幂集P(A)也是不包含自身的集合.
证明:用反证法.假设A是不包含自身的集合(不是真类)，B=P(A),如果B包含自身，则B的元素包含自身B，B的元素的元素也包含自身B，而B的元素是A的子集，B的元素的元素是A的子集的元素，A的子集的元素与A的元素是一样的，也就是说，A包含B，根据李均宇第二定律知A是真类，矛盾。

这点不难理解的，例如集合{1,2,3}不包含自身，则其所有元素和子集和幂集也是不包含自身的，这很易理解的，只是推广到无限集合中去而已。

再如实数集R 不包含自身，则R的任一子集和幂集也是不包含自身的。

定理2:所有不包含自身的集合的基数也是limXn(n→∞).
证明:因为实数集R是不包含自身的集合，由李均宇第五定理所以R的所有幂集也是不包含自身的，也就是R的幂集R1，R的幂集的幂集R2，R的幂集的幂集的幂集R3。

全不包含自身，则所有不包含自身的集合必含R的所有幂集，由李均宇第三定理所以其基数也就是limXn(n→∞).
所以罗素悖论中的“所有不包含自身的集合”，这个集合的基数就是
limXn(n→∞)，也就是公理集合论中的真类。

三。

序数悖论
定理3:任何序数的非空集合都有最小数,从而任何序数的集合在小于等于关系
下都是良序集.
定理3是<<集合论>>已有的定理,所以这里无须证明.
李均宇第六定理:任何序数的集合的幂集也是序数.
证明:因为任何序数的集合的子集也是序数的集合,所以由定理3知其子集也是良序数,所以子集也是一个序数,则所有子集组成的幂集也就是序数的集合,由定理3知此幂集也是良序集,所以此幂集也是一个序数.
定理4:所有序数的集合的基数也是limXn(n→∞).
证明:设所有序数的集合为集合A,由李均宇第六定理知此集合A的幂集也是序数,所以也应包含在集合A中,则集合A包含自身的幂集,由李均宇第二定理知此集合的基数是limXn(n→∞).
基数悖论的问题在于"所有集合的集合",序数悖论的问题在于"所有序数的集合",罗素悖论的问题在于"所有不包含自身的集合组成的集合".因为根据上面证明,这三个集合的基数都是limXn(n→∞).则这三个集合是没有什么意义的,所以集合论悖论没有动摇现有科学的基础.
作者认为公理集合论引进了类的概念是正确的，但随后是把简单的问题复杂化,作者把集合论悖论的解决用最简单的语言讲明白出来,抛弃了公理集合论这个科学上的怪胎,意义是十分重大的。

四。

下面深入讨论下一些集合的性质
命题一：所有不包含自身幂集的集合是真类吗？是的。

因为实数集R的所有幂集都是不包含自身幂集的集合。

所以所有不包含自身幂集的集合必包含实数集R的所有幂集，由李均宇第三定理知其为真类。

为什么实数集R的所有幂集都是不包含自身幂集的集合呢，因为假设其任一幂集Rn包含自身幂集，则由李均宇第二定理知其为真类，这与Rn有固定Xn矛盾的。

命题二：所有不包含1的集合是真类吗？是的。

因为不包含1的集合的幂集也是不包含1的，这用反证法不难证明，因为它根本没有元素1了，所以其幂集也不可能包含有元素1.则其所有幂集也不包含元素1，假设实数集R去掉1后为数集r,则r的所有幂集r1,r2,...rn,...也不包含元素1，由李均宇第三定理知其为真类。

命题三：所有包含1的集合是真类吗？是的。

因为实数集R的幂集必包含元素{1},将括号去掉后就是元素1，去掉括号后的幂集与原幂集一一对应，仅仅{1}变成1，所以去掉括号后的幂集与原幂集等势，
也就是相同基数，同理，实数集R的所有幂集都有等势幂集包含元素1，由李均宇第四定理和第三定理知其为真类。

那么所有不包含1的集合就真的无意义了吗？不是的。

这就是全集的问题。

如果全集是某个有固定基数Xn的集合，在这个全集内的所有子集中再讨论所有不包含1的集合，这就有意义了，不是真类了。

如果全集是真类所有集合的集合，基数是limXn(n→∞)，那么才会可能是真类的。

也就是说，任何将“所有集合的
集合”划分为有限个子集的集合，都必定有一个子集是真类。

再论罗素悖论中的“所有不包含自身的集合”，也是因为它的全集是所有集合的集合，才会无意义的，如果是某个集合内的“所有不包含自身的集合”，则有意义矣。

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Solve the paradox of set theory V7.5
by LiJunYu 2010.12.25 email: myvbvc@ or 165442523@
Brief:All power sets of real number set
R:P(R),P(P(R)),P(P(P(R))),...,Pn(R),...Because all Pn(R) does not contain its own,in Russell's paradox,"all sets which does not contain its own" must contain all Pn(R),that is to say,it contains all of the cardinality of generalized continuum hypothesis {X0,X1,...Xn...},so became meaningless.
Although I know that axiomatic set theory is to solve the paradox arising from, but I think the axiom of set theory in the detours, and even strayed into the manifold road. If the theory does not contain the following, I think <<set theory>> is incomplete.
Generalized continuum hypothesis: the cardinality of an infinite set must be one of X0, X1, ... Xn ....
Where X is the Greece character aleph, because I can not find the character, so the letter X was expressed in English.
Meaningless axioms: If the cardinality of an infinite set is limit limXn (n-->infinite), then this set is meaningless.
I can not find the sign expressed infinite in english computer,so I use the character "infinite".
This axiom is my introduction, I have not seen elsewhere.
This axiom is easy to understand, it is equivalent to axiomatic set theory in the concept of the true class, but after the introduction of the concept of this class of axiomatic set ,the theory straying into the manifold road,
at least I think so.
LiJunYu first theorem: If a set contains all of the cardinality of generalized continuum hypothesis , that is the set {X0,X1,...Xn...},the cardinality of this set is limXn (n-->infinite)
This theorem is obvious, by reduction to absurdity is not difficult to prove.
LiJunYu second theorem: If an infinite set also contains its own power set, that is the set A={......,P(A)),then the cardinality of this set A is limXn (n-->infinite)
Proof: Let the cardinality of infinite sets A is Xn, n be fixed. Because they contain an infinite set A subset or all of its power set, while the power set of the cardinality is X (n +1) = 2 ^ Xn, Therefore, the cardinality becomes X (n +1), which assumes an infinite set A, the original cardinality is Xn, n is a constant contradiction,antinomy, so the cardinality of infinite sets A is limXn (n-->infinite).
LiJunYu third theorem: If a set contains all power set of an infinite set , that is the set B={P(A),P(P(A)),P(P(P(A))),...,Pn(A),...} ,then the cardinality of this set B is limXn (n-->infinite),for example,the real number set R,B={P(R),P1(R),P2(R),...,Pn(R),...},then the cardinality of this set B is limXn (n-->infinite).
All the power set, assuming infinite set A, then the power set P (A), power set of the power set P (P (A)), the power set of the power set of the power set P (P (P (A))), ... Pn (A )..... as all of its power set.
Because all power set of infinite set is the cardinality of the generalized continuum hypothesis in all of the cardinality, so by the LiJunYu first theorem know this theorem.
LiJunYu fourth Theorem: If the set P'n(A) have the same cardinality with the power set Pn(A), that is, then the set P'n(A) equivalent to the power set Pn(A) for LiJunYu second and LiJunYu third theorem.
****I. cardinality Paradox
Theorem 1: The set of all sets of the cardinality is limXn (n-->infinite). The obvious, that in the "<set theory>> has long been, here repeat it. Because the set of all sets contains its own power set,by the LiJunYu second theorem the cardinality is limXn (n-->infinite). So this set is referred to as the true class of axiomatic set theory.
****II. Russell's paradox
LiJunYu Fifth Theorem: If the set A does not contain its own,then the power set of A is P(A),it is also not contain its own.
Proof: by contradiction. To assume that any one does not contain its own set of is A, assume that power sets of set A is B=P(A),if B is the set that contains itself, then there is a element B in the set B which contain its own, and also there is a element B of a element B in the set B which contain its own, the element of B is subset of A,the element of subset of A is the same element to the element of A,so the element of A is B,
by the LiJunYu second theorem the cardinality is limXn (n-->infinite).
A is a proper class.That is not true.
This is understandable, for example, the set {1,2,3} does not contain itself, then its elements and its power set and all subsets, also does not contain itself, it is very easy to understand, but extended to an infinite set to it. Then real number set R does not contain its own , then any child set and power set of R does not contain itself.
Theorem 2: If the set B is all sets which does not contain its own,the cardinality of B is limXn (n-->infinite).
Proof: because the real numbers set R is not contain its own ,by LiJunYu Fifth theorems ,so all power set of R is not contain its own, that is, the power set of R is R1, the power set of the power set of R is R2, the power set of the power set of the power set of R is R3. . . . ALL the power set Rn does not contain itself, then All the set does not contain its own ,that is the set B,containing all the power set of R, by the theorem of LiJunYu third,the cardinality so is limXn (n-->infinite).That is set B contain {P(R),P1(R),P2(R),...,Pn(R),...}.
So Russell's paradox in "All sets which do not contain its own set ",the cardinality of this set is limXn (n-->infinite), in axiomatic set theory call as the true class.
****III. Ordinal number paradox
Theorem 3: Any ordinal number set has a minimum order, so any ordinal number set on less than or equal relations are well-ordered set. Theorem 3 is the <<Set Theory>> there's theorem, so there need not be proved.
LiJunYu sixth Theorem: Any set of ordinals ,its power set is also ordinal number .
Proof: for any ordinal number of set subset is ordinal set, so by Theorem 3 knowing subset is also a well-ordering set ,so the subset is an ordinal number, then all subsets of the power set is ordinal number of the set, by Theorem 3 know that this power set is well-ordered set, so this power set is a ordinal number.
Theorem 4: The cardinality of the set of all ordinals is limXn
(n-->infinite).
Proof: Let all order number of the set named A, by the LiJunYu sixth theorem, this set A power set is ordinal, it should also be included in the set A, then the set A contains its own power set ,by LiJunYu second theorem ,The cardinality of this set is limXn (n-->infinite).
The problem of the cardinality paradox is "a set of all sets", the problem of Ordinal number paradox is "the set of all ordinals," the problem of Russell's paradox is "All the set does not contain its own." Because according to the above shows that this the cardinality of the three sets
are limXn (n-->infinite). then this is meaningless three sets, so the paradoxes of set theory did not shake the existing science cardinality. Axiomatic set theory that the introduction of the concept of class is correct, but then the issue is to complicate the simple, I solve the paradoxes of set theory with the most simple language to understand them, abandoned the scientific axiom of set theory , meaning is very important. ****IV. The following in-depth discussion of the nature of some of the set
Proposition I: All the set which does not contain its own power set is proper class? Yes.
Because all power set of the real numbers R does not contain its own power set . So all the set which does not contain its own power set must contain all power set of real numbers R , by the third theorem of LiJunYu knowing it is proper class. Why do all power set of the real numbers R does not contain its own power set, because assumption any power set Rn contains its own power set , by knowing LiJunYu second theorem it is proper class, which have a fixed Xn Rn, contradictory.
Proposition II: all the set which do not contain number 1 is proper class? Yes.
Because the power set which do not contain number 1 is also a set which do not contain number 1, it is not difficult to prove by contradiction, because it does not element 1, so its power set and can not contain element 1. So all of its power set does not contain elements 1. Assuming the real number set R after removing a number of set is named r, then the power set of all r is r1, r2, ... rn, ... all does not contain element 1, the third by the theorem of LiJunYu knowing it is proper class. Proposition III: all the set which do contain number 1 is proper class? Yes.
Because the power set of real number set R must contain elements of {1}, after removing the brackets is the element 1, the power set which remove parentheses is one by one corresponding the power set of the original, only {1} into 1, so the brackets removed power set is same cardinality with the original set , which is the same cardinality, the same token, all the power set of real numbers R, exists corresponding same cardinality power set,which contains element 1, by the theorem of LiJunYu fourth and third theorems know it is proper class .
Then all the set which do not contain number 1 really meaningless it? Not. This is the problem of the complete works . If the set is a set of a fixed cardinality Xn, all within this subset of complete works and then discuss all the set does not contain 1, which makes sense, is not proper class. If the set is a set of proper class of all sets, the cardinality is limXn (n-->infinite), then the will be proper class . That is, any of "the set of all sets" made into a limited number count of subset , there must be one subset l is the proper class. On Russell's Paradox, "a set of all do
not contain themselves", also because it's complete works is a set of all sets, will be meaningless, if it is within a set which has fixed cardinality Xn, then meaningful carry on.。