TITLE: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.
QUESTION [73 upvotes]: Is the following true?
Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a
further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.
I guess that it is true but I am not sure how to prove this.

REPLY [7 votes]: This is much easier to prove than the other answers here might suggest, if you remember the handy little fact (given, for instance, in Baby Rudin but which aught to be more well-known) that a sequence $x_n$ does not converge to $x$ if, and only if, there exists $\epsilon>0$ such that $|{x_n-x}| \geq \epsilon$ for infinitely many $n$.
On the one hand, if $x_n \to x$ then this fact gives that any subsequence $(x_{n_i}) \subset (x_i)$ can also have finitely many terms a distance $\epsilon$ from $x$, and therefore also converges to $x$. In particular, if $x_n$ converges to $x$ then every subsequence has a subsequence converging to $x$, namely itself
On the other hand, if $x_n \not\to x$ then there exist infinitely many $n$ such that $|x_n-x| \geq \epsilon$ and the subsequence of these terms has no subsequence converging to $x$ as any subsequence has all terms a distance of at least $\epsilon$ away from $x$
Note, however, that it is necessary that the sub-subsequnces converge to the same limit: for instance, any subsequence of the alternating sequence $0,1,0,1,0,1,\ldots$ has a convergent subsequence by the pigeonhole principle, but such sub-subsequences can converge to either 0 or 1