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
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.)