# How can I prove that a sequence such that every converging subsequence coverges to the same limit, converges?

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

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

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.)

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