2
$\begingroup$

I want to claim that if $(x_n)_{n\in N}$ is a sequence, and there is $a$ such that if $(x_{n_k})$ converges, so $\lim x_{n_k} = a$ (it means that all converging subsequences have the same limit), then $(x_n)$ converges. (I don't really mind sequence of what.. could be numbers, could be a sequence in any Hilbert space).

Is my proposition even right?

Assume that a converging subsequence exists, if it helps. I think it should.

My intuition is YES, using some how that $\liminf =\limsup$ (Why is that right exactly? as explicitly as you could).

Does it also hold for weak convergence?

Thanks!

Added: assume it's bounded. I understood it is false if not bounded

$\endgroup$
  • 1
    $\begingroup$ Looks incorrect: how about $a_n = (-1)^n$ $\endgroup$ – Simon S Nov 20 '14 at 20:41
  • 2
    $\begingroup$ Not if the sequence is unbounded. $\endgroup$ – David Mitra Nov 20 '14 at 20:41
  • $\begingroup$ @SimonS: it is not a good counterexample because i asked that all converging subsequences converges to the same limit. $\endgroup$ – user188400 Nov 20 '14 at 20:42
  • $\begingroup$ @DavidMitra: could you explain how the boundness is used/required? $\endgroup$ – user188400 Nov 20 '14 at 20:43
  • 2
    $\begingroup$ What about a sequence with no convergent subsequence. Then vacuously, all convergent subsequences converge to the same limit, but the sequence does not converge. $\endgroup$ – Nishant Nov 20 '14 at 20:44
3
$\begingroup$

Here is a counterexample: $$ a_n = \begin{cases} 0, & n \mbox{ even }\\ n, & n \mbox{ odd } \end{cases} $$

However, as David Mitra points out, if you require boundedness, then the result should hold.


Let $a_n$ be a bounded sequence of real numbers such that every convergent subsequence converges to the same number $a$. Suppose toward contradiction it does not converge to $a$. Then there is an $\epsilon>0$ such that $|a_n - a|>\epsilon$ for infinitely many $n$.

Index these as $b_k$. This is a bounded sequence of real numbers, hence has a convergent subsequence. By construction, this convergent subsequence cannot converge to $a$. However, it is a subsequence of $a_n$. Contradiction.

(The proof is the same for such sequences in any compact metric space.)

$\endgroup$
  • $\begingroup$ Thanks. could you prove it while using boundeness? $\endgroup$ – user188400 Nov 20 '14 at 20:46
  • $\begingroup$ @Functional_Analysis Yep! $\endgroup$ – Neal Nov 20 '14 at 20:49

Your Answer

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