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I'm reading Elementary Analysis by Ross and working through the proof of Theorem 11.7 (see below). I don't understand the claim in the proof that "... ($s_n$) is bounded above, so that lim sup $s_n$ is finite." For example, consider the sequence {$-n^2$} for all $n \in \mathbb{N}$. This is bounded above by $0$, but isn't lim sup $s_n$ = $-\infty$ in this case, and therefore not finite? What am I missing?

See various theorem references and proof below from the text:

Theorem 11.7:

Let ($s_n$) be any sequence. There exists a monotonic subsequence whose limit is $\limsup s_n$ and there exists a monotonic subsequence whose limit is $\lim inf s_n$.

Proof:

If ($s_n$) is not bounded above, then by Theorem 11.2(ii), a monotonic subsequence of ($s_n$) has limit $+\infty = \limsup s_n$. Similarly, if ($s_n$) is not bounded below, a monotonic subsequence has limit $-\infty = \liminf s_n$.

The remaining cases are that ($s_n$) is bounded above or is bounded below. These cases are similar, so we only consider the case that ($s_n$) is bounded above, so that lim sup $s_n$ is finite. Let $t = \limsup s_n$, and consider $\epsilon > 0$. There exists $N_0$ so that

$\sup\{s_n : n > N\} < t + \epsilon$ for $ N \geq N_0$.

In particular, $s_n$ $ < t + \epsilon $ for all $n > N_0$. We now claim

$\{n \in \mathbb{N} : t - \epsilon $ < $s_n$ < $t + \epsilon\}$ is infinite.

Otherwise, there exists $N_1 > N_0$ so that $s_n$ $\leq t - \epsilon$ for $n > N_1$. Then $\sup \{s_n : n > N\} \leq t - \epsilon$ for $N \geq N_1$, so that $\limsup s_n < t$, a contradiction. Since (1) holds for each $\epsilon > 0$, Theorem 11.2(i) shows that a monotonic subsequence of $(s_n)$ converges to $t = \limsup s_n$.


Theorem 11.2(i):

If $t$ is in $\mathbb{R}$, then there is a subsequence of $(s_n)$ converging to $t$ if and only if the set $\{n \in \mathbb{N} : |s_n - t| < \epsilon\}$ is infinite for all $\epsilon > 0$.


Theorem 11.2 (ii):

If the sequence $(s_n)$ is unbounded above, it has a subsequence with limit $+\infty$.

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  • $\begingroup$ Alternatively, and this might be obvious, but I thought it was worth mentioning, if $(s_n)_{n\in\mathbb{N}}$ is a sequence, then $\limsup(s_n)$ and $\liminf(s_n)$ are finite if and only if $(s_n)_{n\in\mathbb{N}}$ is bounded above and below. $\endgroup$
    – C Squared
    Jun 4 at 9:36
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You are correct that there is a slight error here. The proof should probably say "so that $\limsup s_n<\infty$; if $\limsup s_n=-\infty$ then also $\liminf s_n=-\infty$ and we already have a suitable sequence, so we may assume $\limsup s_n$ is finite."

An alternative would be to start with the observation that if $\limsup s_n=\liminf s_n$ then any monotonic subsequence works for both, so we may assume they are different.

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Let me first state a lemma:

Lemma: If $\limsup (x_n)$ is finite then $(x_n)$ is bounded above.

Proof: Let $s=\limsup (x_n)$ and so there exists $N\in \mathbb N$ such that we have $x_n\lt s+1$ for all $n\gt N$. It follows that $x_n\lt\max\{x_1,x_2,\cdots,x_N,s+1\}$. This proves our lemma.

Now, coming back to your question. You are right about observing that $\limsup (-n^2)=-\infty$ and this shows that converse to the lemma stated above is not true, which you probably thought should be true.

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  • $\begingroup$ Could you please explain why $\limsup(-n^2) = -\infty$? $\endgroup$
    – fwd
    Jun 4 at 11:25
  • $\begingroup$ @fwd: Use definition of limsup. Every real no. bounds the sequence $(-n^2)$ from above when n is large enough so what's the lowest upper bound? $\endgroup$
    – Koro
    Jun 4 at 12:09
  • $\begingroup$ You're right, thanks. For some reason I was stupid and kept assuming only non-negative reals bound the sequence above. $\endgroup$
    – fwd
    Jun 4 at 12:15
  • $\begingroup$ @fwd: It's absolutely no problem. It's part of learning. Glad that you understand it now :) $\endgroup$
    – Koro
    Jun 4 at 12:17

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