QuestionDetails: 13) Prove that a is an accumulation point of S if and only if a isthe limit of a
sequence of other points in S.
QuestionDetails:
QuestionDetails: 13) Prove that a is an accumulation point of S if and only if a isthe limit of a
sequence of other points in S.
Solution
a is an accumulation point of S.
where S is a sequence.
then by definition,. to each > 0 , thereexists a member of S , let it be x1 such that |x1 - a| < .
so, a is the limit point of the sequence..
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QuestionDetailsCompactness in the metric spaceTheorem 1) X .pdf
1. QuestionDetails:
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 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.
QuestionDetails:
Compactness in the metric space:
2. 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 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
3. 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
given X is a compact metric space.
i.e ., every open cover in X has a finite subcover.
suppose { Sn} is a sequence in X.
then {Sn} is in an open cover which has a finitesubcover.
i.e. { Sn} has a convergent subsequence included in the finitesubcover.
let the finite subcover be K.
so, a is in K such that | xn - a| < where{xn} is the subsequence of {Sn}.
since K is the finite subcover , ( a- , a+) - { a}is non empty in K.
so, every limit point of K is a point of K.
so, K is a closed bounded subset of X.
