# Does my alternative proof of Bolzano-Weierstraß make sense, or am I missing something?

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.

• 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 Sep 23 at 0:08
• 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? Sep 23 at 0:16
• 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. Sep 23 at 0:19
• 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. Sep 23 at 0:20
• 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? Sep 23 at 0:31