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Convergence of subsequence in metric space

by kingwinner
Tags: convergence, metric, space, subsequence
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kingwinner
#1
Feb9-10, 12:30 AM
P: 1,270
1. The problem statement, all variables and given/known data


2. Relevant equations
N/A

3. The attempt at a solution
I'm really not having much progress on this question. My thoughts are as shown above.
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boboYO
#2
Feb9-10, 02:18 AM
P: 104
1) yes you are right

2) you construct it. ('let the first term be ___, let the next term be ____').

once you have constructed one, call it w, then there is a contradicion because the hypotheses is that EVERY subsequence has a subsequence that converges to a. clearly w is a subsequence that does not have a subsequence that converges to a since it stays away from a, hence contradiction
kingwinner
#3
Feb9-10, 05:02 PM
P: 1,270
2) Any hints about how to construct the subsequence? I'm always struggling with this type of question about constructing subsequences becuase I just don't know how to begin...

3) "clearly w does not have a subsequence that converges to a since it stays away from a"
I understand that by construction w stays away from a, but why does it imply that ALL subsequences stay away from a? How can we prove this? Or is there a theorem that guarantees this?

Thank you!!

boboYO
#4
Feb9-10, 05:48 PM
P: 104
Convergence of subsequence in metric space

2) since xn does not converge to a, then for some epsilon, say e, there are infinitely many points in the sequence xn such that |x-a|>e. So, make the first element of our subsequence the first such point, the 2nd element the 2nd such point, and so on. We won't run out of points because there are infinitely many of them.




3) just making sure i'm clear here, i mean that all subsequences of w can not converge to a. think about it, draw a diagram if you have to: if you have a sequence that is always at least a certain distance away from a then obviously no subsequence of it can converge to a. I edited my original post, hopefully a bit clearer now.
kingwinner
#5
Feb9-10, 11:59 PM
P: 1,270
Quote Quote by boboYO View Post
2) since xn does not converge to a, then for some epsilon, say e, there are infinitely many points in the sequence xn such that |x-a|>e. So, make the first element of our subsequence the first such point, the 2nd element the 2nd such point, and so on. We won't run out of points because there are infinitely many of them.




3) just making sure i'm clear here, i mean that all subsequences of w can not converge to a. think about it, draw a diagram if you have to: if you have a sequence that is always at least a certain distance away from a then obviously no subsequence of it can converge to a. I edited my original post, hopefully a bit clearer now.
2) I'm trying to figure out why when xn does not converge to a, then there are infinitely many points in the sequence xn such that |xn-a|>e
To negate the definition of convergence,
xn does NOT converge to a iff
there exists e>0 s.t. for all N, there exists n s.t. n>N, but |xn -a|>=e.

Does this imply that for some e, there are infinitely many points in the sequence xn such that |xn-a|>=e? Why?

And if there are infinitely many points, then we can always choose the subsequence s.t. the indices are always strictly increasing (as required in the definition of subsequence), right?


3) OK, now I can see intuitively that no subsequence of w can converge to a. But how can we prove it formally? Would it be a "proof by contradiction" kind of thing?

Thank you!!
boboYO
#6
Feb10-10, 01:12 AM
P: 104
both 2) and 3) follow almost immediately from the definition of convergence :)

no need to use contradiction for 3) either, using your

xn does NOT converge to a iff
there exists e>0 s.t. for all N, there exists n s.t. n>N, but |xn -a|>=e.
it's quite straightforward. just find a suitable e.


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