Convergence of sequence where every subsequence of specific type converges Let $(x_n)$ be a sequence in $\mathbb{R}$. Suppose that every infinite subsequence of $(x_n)$ which omits infinitely many terms of $(x_n)$ converges.
Does this imply that $(x_n)$ converges?
I have failed so far to come up with a counterexample. But I can't see why this is true. It seems that there could be such a sequence where two of the subsequences converge to different points.
Any hints are welcomed. 
 A: The odd and even subsequences $E = x_{2n}$ and $O = x_{2n+1}$ converge, by assumption. It suffices to show that they converge to the same limit.
Now consider the subsequence of the form $x_1, x_2, x_5, x_6, x_9, x_{10}, \ldots$ which alternately includes and omits adjacent pairs. This converges by assumption to some limit $L$, hence so do its subsequences  $E' = x_2, x_6, x_{10}, \ldots$ and $O' = x_1, x_5, x_9, \ldots$. But $E'$ is a subsequence of $E$ and therefore must converge to the same limit as $E$, hence $E$ converges to $L$. By the same reasoning, $O$ also converges to $L$.
A: Suppose the sequence $x_n$ has this property, but does not converge.
The subsequence $x_{2n}$ (omitting all the odd-numbered terms) must converge, let's say to $L$.  But since $x_n$ does not converge to $L$, there is some $\epsilon > 0$
such that infinitely many $x_n$ have $|x_n - L| > \epsilon$.  Those $x_n$ form a subsequence, say $x_{m_j}$, which by assumption must converge, say to $M$, where $M \ne L$.  Now take a subsequence which alternates between members of the first subsequence and the second, but leaves out infinitely many terms (e.g. the sequence is $x_{r_j}$ where $r_1 = 2$ and for even $j$, $r_j$ is the first $m_k$ greater than $r_{j-1}$, while for odd $j$, $r_j$ is the second even number greater than $r_{j-1}$).  Again this subsequence leaves out infinitely many terms, but it's clear that it can't converge.
