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?

  • $\begingroup$ Yes, it's true, and not that hard to prove. $\endgroup$ – bof Feb 19 '17 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.

  • $\begingroup$ How to show $a_{n}$ is bounded below by $-M$ as well? $\endgroup$ – Little Rookie Mar 20 '17 at 14:33
  • $\begingroup$ @LittleRookie: If $a_n$ is monotone increasing then it is bounded by $a_0$ from below. $\endgroup$ – levap Mar 20 '17 at 16:14
  • $\begingroup$ oh yes, silly me $\endgroup$ – Little Rookie Mar 20 '17 at 17:14

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