Again, the goal is to prove that prove that a monotone sequence is bounded IF AND ONLY IF it has a bounded subsequence.

The similar question: Show that a monotone sequence is bounded if it has a bounded subsequence has been asked Show that a monotone sequence is bounded if it has a bounded subsequence., but, I'm not sure about the if and only if part of this variation of the question.

My idea is follows, based on the linked proof,


For the forward direction, I'll quickly quote the top answer from the similar post: "Suppose that {an} is monotonically increasing. The decreasing case is similar. Note that because this sequence is increasing, it is bounded below by a1. Let {ani} be a bounded subsequence so that |ani|≤M for all i. Let k∈ℕ. Then for some i, k≤ni, so $a_1≤a_k≤a _n {_i}≤M$. Therefore |ak|≤max{M,|a1|}, so the sequence is bounded." As is said, clearly, the decreasing case is very similar. But this part isn't a problem.

My question is how to prove the "reverse direction." I really have been blanking on how to prove that. If I understand what I must do, I must show that a bounded subsequence implies a monotone sequence. My thought for the reverse direction, is that I have to show that if $ {a _n {_k}} $ is bounded, then $ {a _n} $ is also bounded.

This is what I'm struggling with. It seems to me that there is no way to show that a sequence is bounded just by knowing that its subsequence is bounded.


Let $(a_{n})$ be monotone increasing, so that in particular $a_n \geqslant a_1$ for all $n\in\mathbb N$. Let $(a_{n_k})$ be a bounded subsequence, say $a_{n_k}\leqslant M$ for all $k\in\mathbb N$. If $(a_{n})$ were not bounded, then we would have that for some $m\in\mathbb N$, $a_m\geqslant M+1$.

But since $a_m$ is increasing, $a_{n}\geqslant M+1$ for all $n\geqslant m$. In particular, for $k$ large enough $a_{n_k}\geqslant M+1$, a contradiction.


In fact, the forward direction is trivial. You don't need to do anything, just choose that subsequence to be the original sequence.

The key of the reverse direction is the monotonicity. WLOG, it is increasing. So if there is a term $a_n$ in the middle of the original sequence, there is always some terms behind it which belongs to the subsequence, say $a_{n_k}$. But since the sequence is increasing, $a_n\leq a_{n_k} \leq M$, where $M$ is a upper bound of the subsequence.

Moreover, since the sequence is increasing, a lower bound is the first term of the sequence.

Remark: There is a common skill in studying monotone sequence. You don't need to consider both cases of increasing and decreasing. If the sequence is decreasing, then you can simply consider negative of such sequence to get an increasing sequence.

  • $\begingroup$ If there are always some terms behind it which belong to the subsequence, then since $a_n$ is increasing, shouldn't it be $a_{n_k}\leq a_n$? $\endgroup$ – Slavik Egorov Jul 2 '18 at 13:47
  • $\begingroup$ $(a_n)$ being increasing means $a_m\geq a_n$ whenever $m \geq n$. Now, $a_{n_k}$ is behind $a_n$ means $n_k \geq n$, so $a_{n_k} \geq a_n$. $\endgroup$ – Jerry Jul 2 '18 at 17:42

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