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.
Best Answer
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 $L$.