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

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.

• Consider the two cases separately: $(a_n)$ monotonically increasing or $(a_n)$ monotonically decreasing. – Ujan Gangopadhyay Oct 16 '15 at 4:35
• @UjanGangopadhyay Why we $|\{a_{n_i}\}\leq M$ isn't enough to show the sequence is bounded? – Simple Oct 16 '15 at 4:41
• Note that $-100 <1$ does not imply $|-100| <1$. – user99914 Oct 16 '15 at 4:45
• @JohnMa I see, thanks – Simple Oct 16 '15 at 4:46

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.