2
$\begingroup$

I have been thinking about this for a while now.

Clearly if a sequence converges then also it will also have a convergent subsequence (take for example the whole sequence). However, I have been told the the opposite it not true. Could someone give an explicit example on a sequence which have a convergent subsequence but which do not converge?

Also we know that on a compact set, any bounded sequence has a convergent subsequence. Is there any criteria so that bounded sequences converges on compact set? Or more generally, is there a criteria so that if $x_n$ has a convergent subsequence then it will also converge?

Thanks.

$\endgroup$
  • $\begingroup$ @Henrick : what you do mean by "bounded"? What kind of space does the sequence live in? Is it a metric space? $\endgroup$ – Stefan Smith Apr 30 '13 at 22:23
  • $\begingroup$ When I mean bounded sequence I mean that the set of all points in the sequence is bounded. So if (X,d) is a metric space and $x_n$ is a sequence in X, then I would say that the sequence is bounded if for any elements in the sequnece $x_i, x_j$ we have $d(x_i, x_j) < \infty$. $\endgroup$ – Henrik Finsberg May 1 '13 at 15:26
  • $\begingroup$ @Henrick : you didn't specify whether you were using a metric space. I'm not sure if "bounded" has a meaning in an arbitrary topological space. And your notion of "bounded" in your comment is faulty: any sequence is bounded: e.g. take $(1,2,3,\ldots) \subset \mathbb{R}$. The distance between any two elements of the sequence is finite. $\endgroup$ – Stefan Smith May 1 '13 at 21:58
  • $\begingroup$ Yes, that true. That was a bad definition. Thanks for pointing that out. Usually I would be interested in normed vector spaces, and I would say that the sequence is bounded if there exist an $K \in \mathbb{R}$ such that $\| x_n \| \leq K$ for all elements in the sequence. For metric spaces I'm not quite sure how to define a bounded sequence. $\endgroup$ – Henrik Finsberg May 2 '13 at 14:03
  • $\begingroup$ for a metric space I'm quite sure the definition is $\sup \{d(x_i,x_j) \mid i, j \geq 1\} < \infty$, which I think is what you were trying to say. $\endgroup$ – Stefan Smith May 3 '13 at 12:47
1
$\begingroup$

You could take the sequence $a_n = (-1)^n$. What are the convergent subsequences?

It is necessary that a convergent sequence be a Cauchy sequence. Conversely, a Cauchy sequence with a converging subsequence converges.

$\endgroup$
  • 1
    $\begingroup$ Yes, that was nice! Thank you. The convergent subsequence of $a_n$ is the constant sequence of 1^s and (-1)'s. ;) $\endgroup$ – Henrik Finsberg Apr 30 '13 at 9:54
  • $\begingroup$ Indeed. Rather, those that eventually only consist of $1$s or only of $-1$s. E.g. $-1,1,1,1\ldots$ converges too :) $\endgroup$ – Lord_Farin Apr 30 '13 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.