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Show that a monotone sequence is bounded if it has a bounded subsequence.


Proof:

Let $\{a_n\}$ be monotone sequence and $\{a_{n_i}\}$ is the subsequence. Since $\{a_{n_i}\}$ is bounded, then there exists an $M\in\mathbb{R}$ such that $|a_{n_i}|\leq M$.

Suppose that for any $k\in n$ such that $k\leq n_i$ for all $i$, then we can get $|a_k|\leq M$ since $|a_{n_i}|\leq M$; thus $\{a_n\}$ is bounded.


I am not sure my second paragraph is right or not, because I just say we can pick a random $k$ which is less than $n_i$ since $a_{n_i}$ is a subsequence. Can anyone check my solution? Thanks in advanced.

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    $\begingroup$ Consider the two cases separately: $(a_n)$ monotonically increasing or $(a_n)$ monotonically decreasing. $\endgroup$ – Ujan Gangopadhyay Oct 16 '15 at 4:35
  • $\begingroup$ @UjanGangopadhyay Why we $|\{a_{n_i}\}\leq M$ isn't enough to show the sequence is bounded? $\endgroup$ – Simple Oct 16 '15 at 4:41
  • $\begingroup$ Note that $-100 <1$ does not imply $|-100| <1$. $\endgroup$ – user99914 Oct 16 '15 at 4:45
  • $\begingroup$ @JohnMa I see, thanks $\endgroup$ – Simple Oct 16 '15 at 4:46
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You are on the right track. We would like to use the bound on the subsequence as a bound on the entire sequence. However, we may have to slightly adjust the bound a bit.

Suppose that $\{a_n\}$ is monotonically increasing. The decreasing case is similar. Note that because this sequence is increasing, it is bounded below by $a_1$. Let $\{a_{n_i}\}$ be a bounded subsequence so that $|a_{n_i} | \leq M$ for all $i$. Let $k \in \mathbb{N}$. Then for some $i$, $k \leq n_{i}$, so $a_1 \leq a_k \leq a_{n_i} \leq M.$ Therefore $ |a_k| \leq \max\{ M,\left|a_1\right| \}$, so the sequence is bounded.

So in adjusting our bound, we really only need to also account for how small $a_1$ can be, since our bound on $\{a_{n_i}\}$ does not account for this.

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