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I want to show that every bounded sequence in $\mathbb{R}$ has a convergent subsequence. Let $a_n$ be any bounded sequence in $\mathbb{R}$. Since $a_n$ is bounded, $\text{limsup}(a_n)$ exists, call it $k$. By lemma, for any $\epsilon > 0$, the set $\{n \;|\; \text{limsup}(a_n) - \epsilon < a_n < \text{limsup}(a_n) + \epsilon \}$ is infinite. If we let $\epsilon$ become arbitrarily small, and include only the $n$'s from our defined set into our subsequence, we have obtained a subsequence, that, by definition, converges to $k$. Hence, every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

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    $\begingroup$ This can't work because you only use the fact that $a_n$ is bounded above. For example $-n$ has no convergent subsequences but is bounded above $\endgroup$
    – Fishbane
    Sep 23 at 0:08
  • $\begingroup$ Oh well the premise is that $a_{n}$ is bounded above and below. We're working with a sequence that is totally contained within an interval. Would my proof make sense in that context? $\endgroup$ Sep 23 at 0:16
  • $\begingroup$ You never use that assumption though so it can't make the proof valid. And it isn't true that every bounded sequence has a subsequence which converges to its supremum so the result you get is false anyway. $\endgroup$
    – Fishbane
    Sep 23 at 0:19
  • $\begingroup$ Specifically the problem is that you assume that $\{n \;|\; \text{limsup}(a_n) - \epsilon < a_n < \text{limsup}(a_n) + \epsilon \}$ is infinite. In general this is not true. $\endgroup$
    – Fishbane
    Sep 23 at 0:20
  • $\begingroup$ That's a lemma that I was given in class, and is taken to be true for bounded sequences. How is it not true in general? $\endgroup$ Sep 23 at 0:31

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