Let $x_n$ be a bounded sequence of real numbers. Show that, if all the convergent subsequence converges to the same limit $x^⋆$ , then $x_n$ is convergent and converges to $x^⋆$ .

EDIT: Would this be sufficient?

Suppose $x_n$ does not converge, then given $\epsilon \gt 0$, $|x_{n} - x^*| \ge \epsilon$ for infinitely many $n$ therefore there exists a subsequence $x_{n_k}$ with $|x_{n_k} - x^*| \ge \epsilon$ for all $k \in \Bbb N$. This is a condtradiction, becaue $(x_{n_k})\subset (x_n)$.

If not, what would be sufficient?


Suppose there were some $\epsilon$ for which infinitely many entries in the sequence are at least distance $\epsilon$ away from $x^*$. Then you could build another convergent subsequence from those "far away" points which converges to something else.

  • $\begingroup$ I edited my post with what I came up with. Is that what you meant? $\endgroup$ – Meecolm Feb 23 '15 at 16:16
  • $\begingroup$ Close, but there's one more step. The subsequence of far away points, $x_{n_k}$, does not necessairily converge. But since it is bounded, a subsequence of it ("sub-sub-sequence") does converge. Furthermore the sub-sub-sequence must converge to something other than $x^*$ since all it's elements are greater than distance $\epsilon$ from $x^*$. $\endgroup$ – Nick Alger Feb 23 '15 at 17:49

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