数学分析英文课件13-1

Then for sufficiently large n ,
a n bn ,
cn d n .
This shows that Fn is covered by . This is the desired contradiction.
Homework Page 142: 2; 4; 5; 7
Chapter 13
Lecture 1
Title: §1 Point sets in the plane R 2 (I) Date: Teaching type: Theory Aim: Notions of point sets in R 2 and their basic properties Emphasis: Four basic properties Difficulty: The proofs of the four basic properties Procedure: 1 1.1 Neighbourhood Let M 0 ( x0 , y 0 ) be a fixed point and 0 , and let O( M 0 , ) M ( x, y ) : r ( M 0 , M ) , where r ( M 0 , M ) ( x x0 ) 2 ( y y 0 ) 2 . Then O( M 0 , ) is called an -neigh bourhood of M 0 . 1.2 Limit of a point sequence Let M n ( x n , y n ) be a sequence of points in R 2 and M 0 ( x0 , y 0 ) be a fixed point. If for any 0 , there is some N 0 such that for all n N , M n O( M 0 , ) , i.e., r (M 0 , M n ) , then we call M n is convergent and denote it by Neighbourhood and limit of point sequences
is a subsequence {x nk } of {x n } which is convergent. Since { y nk } is also bounded, we see that there is a subsequence { y nr } which is convergent. The sequence {M nr ( x nr , y nr )} is desired.
If the sequence {M n ( x n , y n )} is bounded, then there exists a
subsequence {M nk } of {M n } which is convergent.
Proof The hypothesis implies that both {x n } and { y n } are bounded. Hence there
Proposition 2.5.1 If M 3 is a cluster point of E , then there exists a sequence
{M n } of E such that lim M n M 0 .
n
2.6 Closed sets
If each point of E is a cluster point, then E is closed.
n n
Then there is a unique M 0 ( x0 , y 0 ) such that for each n , M 0 ( x0 , y 0 ) E n . The proof is obvious.
3.2 Weierstrass’s theorem Theorem 3.2.1
(2) bn a n
Theorem 3.1.1 implies that there exists some M 0 ( x0 , y 0 ) such that for each n , a n x0 bn , cn y0 d n .
Since M 0 D , there is at least one E such that M 0 . Suppose that {( x, y ) : x , y } .
F {( x, y ) : a x b, c y d }
such that D F . By using the two lines x
ab cd and y , we know that 2 2
F is equally divided into four rectangles F1 , F2 , F3 , F4 . It is obvious that there is at
lim M n M 0 .
n
Proposition 1.2.1
( x n , y n ) ( x0 , y 0 ) if and only if x n x0 and y n y 0 .
Proof Necessity It follows from ( x n , y n ) ( x0 , y 0 ) that for any 0 , there is some N 0 such that for all n N ,
For M 2 R 2 , if for any 0 ,
O( M 2 , ) E and O( M 2 , ) ( R 2 E ) ,
then M 2 is called a boundary point.
2.4 Open sets
If each point of E is interior, then E is open.
there is some N 0 such that for all n N ,
xn x0
whence

2
and
yn y0

2
,
xn x0 yn y0
2
2
.
This shows that lim M n M 0 .
n
Proposition 1.2.2 If lim M n exists, then it is unique.
n
The proof easily follows from Proposition 1.2.1.
2 Open sets, closed sets and domains
Suppose E is a point set in R 2 .
2.1 Interior points
For M 0 E , if there is some 0 such that O( M 0 , ) E , then M 0 is called an interior point of E .
least one rectangle, say for example F1 , such that F1 D can not be covered by finitely many elements in E . By repeating this procedure, we will obtain a sequence of rectangles {Fn } such that (1) Fn {( x, y ) : a n x bn , c n y d n } ; ba d c , d n cn n . n 2 2
3.3 Finite covering theorem Theorem 3.3.1 Suppose E {} , where each is an open rectangle and D is a
bounded and closed domain in R 2 .
If D ,
then there are finitely many
s : 1 , , k such that
D i .
i 1
k
Proof
We prove this result by contradiction.
Suppose not. Since D is a bounded and closed domain in R 2 , we see that there is a rectangle
2.2 Exterior points
For M 1 R 2 , if there is some 0 such that O( M 1 , ) E , then M 1 is called an exterior point of E .
2.3 Boundary points
2.5 Cluster (or accumulation) points Definition 2.5.1 For M 3 R 2 , if for any 0 the following
O( M 3 , ) ( E {M 3 })
holds, then M 3 is called a cluster (or accumulation) point of E .
2.7 Domains
If E is connected and open, then E is called a domain. We always call E E the closure of E which is a closed domain, where E denotes the boundary of E .
3 Four basic properties (I)