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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.

Saaqib Mahmood
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Manos
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2 Answers2

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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)

Ahriman
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  • Thanks Ahriman, it was the very last statement in parenthesis that i was missing :-) – Manos Sep 06 '12 at 13:36
  • I think Rudin proved it before in his text, so you can use it. Right ? – Ahriman Sep 06 '12 at 13:39
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    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! – Manos Sep 06 '12 at 13:42
  • Would you know why this sequence was bounded? – user135520 May 16 '15 at 17:18
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How about this.

Suppose there are infinite $s_{n}$'s in the sequence $\{s_{n}\}$ such that $s_{n} \geq x$.

Let $\{s_{n_{k}}\}$ be the subsequence of $\{s_{n}\}$ consisting of the $s_{n}$'s that are greater or equal to $x$. Then there are two possibilities:

(i). If $\{s_{n_{k}}\}$ is bounded above, then there exists a real number $y$ such that $y \in E$ and $y \geq x$.(By theorem 3.6-[b] and the fact that we cannot have a subsequence of $\{s_{n_{k}}\}$ that converges to a number less than $x$.) Notice that we have $x > s^{*}$, which implies $y \geq x > s^{*}$, then $s^{*}$ is no longer an upper bound.(A contradiction).

(ii). If $\{s_{n_{k}}\}$ is not bounded above, then we must have $+\infty \in E$.(See the lemma below). Since $+\infty \geq x$ by definition, we get $+\infty \geq x > s^{*}$, which implies, again, that $s^{*}$ is not an upper bound. A contradiction.

Therefore, there are only finite $s_{n}$'s in the sequence $\{s_{n}\}$ such that $s_{n} \geq x$. That is, there exists an integer $N$ such that $n \geq N$ implies $s_{n} < x$.

You may find this result useful:

Lemma: If a sequence $\{s_{n}\}$ is not bounded above, then there is a subsequence of $\{s_{n}\}$ that converges to $+\infty$.

We can construct a subsequence $\{s_{n_{k}}\}$ that converges to $+\infty$ like this: Let $s_{n_{k}} > k$ for all $k \in \mathbb{N}$ with $n_{1} < n_{2}<\dots$. (Can we do this? If there was no $n_{1}$ such that $s_{n_{1}} > 1$, then $s_{n} \leq 1$ for all $n$, a contradition. After we have chosen $s_{n_{1}}, s_{n_{2}}, \dots$, choose $s_{n_{k}}$ such that $n_{k} > n_{k-1}> \dots > n_{1}$ and $s_{n_{k}}>k$. If there was no such $s_{n_{k}}$ available, then $s_{n_{k}} \leq k$ for all $n > n_{k-1}$. Therefore, $\{s_{n}\}$ is bounded by $\max \{s_{1}, \dots, s_{n_{k-1}}, k\}$, a contradiction. Therefore, we can indeed construct such an sequence $\{s_{n_{k}}\}$ as desired.) Now, notice that for all $M \in \mathbb{R}$, there exists an integer $N$ such that $n_{k} \geq N$ implies $s_{n_{k}} \geq M$. Hence $s_{n_{k}} \to +\infty$, as desired.

jlchen
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