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?


  • $\begingroup$ @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

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.

  • 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

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