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This question refers to Rudin's "Principles of Mathematical Analysis", Theorem 3.17, p.56. In particular let $\left\{s_n\right\}$ be a real sequence and let $E$ be the set of all sub-sequential limits with possibly plus and minus infinity included. Denote $s^*=\sup E$. Suppose $x$ is a real number such that $x>s^*$ and that $s_n \ge x$ for infinitely many values of $n$. I want to arrive to a contradiction, but i am having some difficulty doing this very rigorously. More precisely, i can see that we obtain subsequences, which after some index become larger than $x$, but why does that mean that we have a sub-sequential limit greater than $x$ (thus contradicting $s^*$)?

Thanks.

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It relies on the following fact (prove it as an exercise) :

$I \subset \mathbb{N}$ is infinite if and only if there exists an increasing function $\varphi : \mathbb{N} \to I$ which is bijective.

If you put $I$ the set of all integers $n$ such that $s_n \ge x$, if $I$ was infinite, there would exist, by the previous fact, a subsequence $(s_{\varphi(n)})$ whose all terms are $ \ge x$.

Now, extract a converging sub-subsequence from $(s_{\varphi(n)})$. What can be said about its limit ?

(recall that every sequence of real numbers admits a converging subsequence, possibly in the extended real line)

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  • $\begingroup$ Thanks Ahriman, it was the very last statement in parenthesis that i was missing :-) $\endgroup$ – Manos Sep 6 '12 at 13:36
  • $\begingroup$ I think Rudin proved it before in his text, so you can use it. Right ? $\endgroup$ – Ahriman Sep 6 '12 at 13:39
  • $\begingroup$ He proved it in theorem 3.6 in the form "every sequence in a compact metric space contains a convergent subsequence" or "every bounded sequence of $\mathbb{R}$ converges". I can take it from there. Thanks a lot! $\endgroup$ – Manos Sep 6 '12 at 13:42
  • $\begingroup$ Would you know why this sequence was bounded? $\endgroup$ – user135520 May 16 '15 at 17:18

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