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 $+\infty$.
Best Answer
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.