# If the sequence is monotone increasing and unbounded, then is any subsequence of that sequence also unbounded?

If the sequence is monotone increasing and unbounded, then is any subsequence of that sequence also unbounded? I haven't yet seen such statement in my textbook but I am curious if this is true. I know that there is unbounded sequence that has bounded subsequence, but I think this is only possible when the sequence is not monotone increasing nor decreasing. For example, I can have a sequence that behaves like this $$s = 1,2,1,3,1,4,1,5,1,6, \dots$$ Then, the sequence is clearly unbounded but we have a subsequence $s_{2k-1}$ for $k\in\mathbb{N}$ that is bounded aboved by 1. But if the sequence is monotone increasing, then wouldn't all of its subsequences also be unbounded?

• Yes, it's true, and not that hard to prove.
– bof
Feb 19, 2017 at 2:20

Yes, because if a subsequence $(a_{n_k})_{k=1}^{\infty}$ would be bounded (say, by $M$) then we would also have
$$a_n \leq a_{n_n} \leq M$$
for any $n \in \mathbb{N}$ (because $n_n \geq n$) so $(a_n)_{n=1}^{\infty}$ would also be bounded by $M$, a contradiction.
• How to show $a_{n}$ is bounded below by $-M$ as well? Mar 20, 2017 at 14:33
• @LittleRookie: If $a_n$ is monotone increasing then it is bounded by $a_0$ from below. Mar 20, 2017 at 16:14