I have to prove that these two statements are equivalent:
(a) Any bounded sequence must have a converging subsequence,
(b) Any bounded monotone sequence must converge.

Now clearly (b) implies (a), since any bounded monotone sequence is a subsequence of itself, which means that there is a converging subsequence.

Now for (a) implies (b) I've tried drawing out a bounded, increasing (monotone) sequence {$a_n$} with converging subsequence {$b_n$}, which converges to $B$. Now how does this convergence of {$b_n$} to $B$ imply that {$a_n$} converges?

(I know that a bounded monotone sequence converges, but that's not what the question is about.)

The fact that this bounded monotone sequence has a converging subsequence should imply that {$a_n$} itself converges but I don't see how it does. My drawing is below. enter image description here

  • 1
    $\begingroup$ Your subsequence $b_n$ can't converge to $B$. What happens after $b_7$? Keep in mind that subsequences are infinitely long, by definition. They don't terminate. $\endgroup$ – Bungo Nov 22 '16 at 19:23

When proving $a) \Rightarrow b) $ you have a really neat trick at your disposal.

First, understand that you are assuming that any bounded sequence has a converging subsequence (by assuming the truthfullness of $a) $). Say that $\{b_n\} $ is a monotone, bounded sequence. Now note that any subsequence of it is also monotone and is bounded. So you have a monotone converging subsequence of $\{b_n\} $, which is monotone itself...

I will sketch the proof now, and will leave the details with you as an exercise.

Remember that a sequence is said to converge to $L $ if, for any epsilon greater than 0, there is an order $N $ for which all elements after $a_N $ are within a distance epsilon of $L $.

Take your subsequence. It is monotone, so for every element in it, there is an order in the main sequence for which they are all bigger (or smaller, depending on if you have a monotone increasing/decreasing sequence). Let us assume it is increasing.

Say the subsequence converges to $L $ and take some epsilon greater than 0. On one hand, there is a point starting from which every element of the subsequence is within that epsilon distance of $L $. On the other hand, the main sequence lies within the elements of the subsequence thus they must have a distance to $L $ smaller than the elements of the subsequence that come before. This means the distance from the sequence and $L $ is getting smaller and smaller.

  • $\begingroup$ Well I have a feeling that both the subsequence and the sequence {$b_n$} should have the same limit. But it seems in my picture that they don't. So I don't see what the subsequence tells us about the complete sequence. $\endgroup$ – QuestionMaker Nov 22 '16 at 19:22
  • $\begingroup$ @QuestionMaker your feeling is right. Your picture is not a counter example because that is not a subsequence! A subsequence must be infinite. You just picked 7 elements from the sequence ;) $\endgroup$ – RGS Nov 22 '16 at 19:23
  • $\begingroup$ But a sequence can be finite right? I mean $a_1$ $a_2$ $a_3$ is a sequence? $\endgroup$ – QuestionMaker Nov 22 '16 at 19:25
  • $\begingroup$ @QuestionMaker it is a finite sequence. It is what a layman would call a sequence. Mathematicians mean infinite sequences when they just say sequences. Otherwise your counter-example would be right. $\endgroup$ – RGS Nov 22 '16 at 19:26
  • $\begingroup$ @QuestionMaker No, a sequence is always infinite (in real analysis, at least). Formally, it is a function from $\mathbb N$ to $\mathbb R$ (assuming it's real-valued). Subsequences are sequences, so they must also be infinite. $\endgroup$ – Bungo Nov 22 '16 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.