Compactness in the metric spaceTheorem 1) X ix compact .pdf

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Compactness in the metric space: Theorem: 1) X ix compact 2) Every sequence has a convergent subsequence converging to apoint in K. 3)Then K is closed and bounded. I just want you to prove the 2) => 3) in metricspace. For example I am showing that 1) =>2) By contradiction there exist sequence subset K with noconvergent subsequence. For all xn in the sequence there existB(xn , rn) so that B(xn ,rn) contains only finitely many xj\'s. By hypothesis, 2 finitely many such balls cover K.. Only finitely many xj\'s are covered. There are infinitely many xj\'s. This is a contradiction. For example 2) => 1) Let { U } be an open cover of k. Claim:- There exist >0 for all x belongs tok B(xn, 1/n) subset to U for some. proof of claim:- Suppose not. Let (xn) be a sequence such that B(xn,1/n) not subset to U for any . by 2) there exist U such that B(a,r) subset toU. since xn--->aeventually (xn, a) < r/2 futhermore eventually 1/n < r/2 Let both these happen and m be big enough. ie,(xm, a) < r/2 and 1/m < r/2. So let Z belongs to B(xm, 1/m) (Z,a) (Z,xm)+(xm, a) < 1/m+r/2 < r/2+r/2 =r So B(xm, 1/m) subset to B (a,r) subset toU. Choose a sequence a points yn such that the ballsB(yn, /2) are disjoint. Note: This cannot be an infinite because noticeB(xn,xm) /2. So { yn } does not converge, nor do anysubsequence. By hypothesis all sequence have a convergentsubsequence. Compactness in the metric space: Theorem: 1) X ix compact 2) Every sequence has a convergent subsequence converging to apoint in K. 3)Then K is closed and bounded. I just want you to prove the 2) => 3) in metricspace. For example I am showing that 1) =>2) By contradiction there exist sequence subset K with noconvergent subsequence. For all xn in the sequence there existB(xn , rn) so that B(xn ,rn) contains only finitely many xj\'s. By hypothesis, 2 finitely many such balls cover K.. Only finitely many xj\'s are covered. There are infinitely many xj\'s. This is a contradiction. For example 2) => 1) Let { U } be an open cover of k. Claim:- There exist >0 for all x belongs tok B(xn, 1/n) subset to U for some. proof of claim:- Suppose not. Let (xn) be a sequence such that B(xn,1/n) not subset to U for any . by 2) there exist U such that B(a,r) subset toU. since xn--->aeventually (xn, a) < r/2 futhermore eventually 1/n < r/2 Let both these happen and m be big enough. ie,(xm, a) < r/2 and 1/m < r/2. So let Z belongs to B(xm, 1/m) (Z,a) (Z,xm)+(xm, a) < 1/m+r/2 < r/2+r/2 =r So B(xm, 1/m) subset to B (a,r) subset toU. Choose a sequence a points yn such that the ballsB(yn, /2) are disjoint. Note: This cannot be an infinite because noticeB(xn,xm) /2. So { yn } does not converge, nor do anysubsequence. By hypothesis all sequence have a convergentsubsequence. Let { U } be an open cover of k. Claim:- There exist >0 for all x belongs tok B(xn, 1/n) subset to U for some. proof of claim:- Suppose not. Let (xn) be a sequence such that B(xn,1/n) not subset to U for any . by 2) there exist U such that B(a,r) subset to.

Education

3)Then K is closed and bounded.
I just want you to prove the 2) => 3) in metricspace.
For example I am showing that 1) =>2)
By contradiction there exist sequence subset K with noconvergent subsequence.
For all xn in the sequence there existB(xn , rn) so that B(xn ,rn) contains only finitely many xj's.
By hypothesis, 2 finitely many such balls cover K..
Only finitely many xj's are covered.
There are infinitely many xj's.
This is a contradiction.
For example 2) => 1)
Let { U } be an open cover of k.
Claim:- There exist >0 for all x belongs tok B(xn, 1/n) subset to U for some.
proof of claim:- Suppose not.
Let (xn) be a sequence such that B(xn,1/n) not subset to U for any .
by 2) there exist U such that B(a,r) subset toU.
since xn--->aeventually (xn, a) < r/2
futhermore eventually 1/n < r/2
Let both these happen and m be big enough. ie,(xm, a) < r/2 and 1/m < r/2.
So let Z belongs to B(xm, 1/m)
(Z,a) (Z,xm)+(xm, a)
< 1/m+r/2
< r/2+r/2
=r
So B(xm, 1/m) subset to B (a,r) subset toU.
Choose a sequence a points yn such that the ballsB(yn, /2) are disjoint.
Note: This cannot be an infinite because noticeB(xn,xm) /2.
So { yn } does not converge, nor do anysubsequence.
By hypothesis all sequence have a convergentsubsequence.
Let { U } be an open cover of k.
Claim:- There exist >0 for all x belongs tok B(xn, 1/n) subset to U for some.
proof of claim:- Suppose not.
Let (xn) be a sequence such that B(xn,1/n) not subset to U for any .
by 2) there exist U such that B(a,r) subset toU.
since xn--->aeventually (xn, a) < r/2
futhermore eventually 1/n < r/2
Let both these happen and m be big enough. ie,(xm, a) < r/2 and 1/m < r/2.

So let Z belongs to B(xm, 1/m)
(Z,a) (Z,xm)+(xm, a)
< 1/m+r/2
< r/2+r/2
=r
So B(xm, 1/m) subset to B (a,r) subset toU.
Choose a sequence a points yn such that the ballsB(yn, /2) are disjoint.
Note: This cannot be an infinite because noticeB(xn,xm) /2.
So { yn } does not converge, nor do anysubsequence.
By hypothesis all sequence have a convergentsubsequence.
Solution
2) Every sequence has a convergent subsequence converging to apoint in K.
we know that the point of convergence is the limit point.
every limit point of K is a point of K.
K is closed.
further, every convergent sequence is bounded.
every sequence in K is bounded ==> K is bounded.
K is closed bounded.

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