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One way to define compactness in metric spaces is to note that in compact metric space each sequence has a convergent subsequence.

Understanding compactness is difficult for me from this perspective. Let's take the set of real numbers as an example. Say my sequence is something like $(1,2,3,...)$ that isn't convergent. But why can't I pick, say the subsequence $(1,1,1,...)$, i.e. a sequence that is constant, and so convergent? And similarly for every possible sequence in $\mathbb{R}$?

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$(1,1,1,\dots)$ is not a subsequence of $(1,2,3,\dots)$.

If you have a sequence $a_1,a_2,\dots$, then a subsequence of that is any sequence $$a_{i_1},a_{i_2},a_{i_3},\dots$$ where $i_1,i_2,i_3,\dots$ are some natural numbers such that $$i_1<i_2<i_3<\dots.$$

This means that for the sequence $(1,2,3,\dots)$, you can take $i_1=1$ so the first element of the subsequence will be $1$, but you cannot take $i_2=1$ because $i_2$ must be larger than $i_1$.


The subsequence must be a "proper" subsequence, meaning you can only take each element of the original sequence one (or zero) times.

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  • $\begingroup$ This is it, thank you! I had forgotten there was a strict inequality between the indices. So the subsequences are always "evolving to the same direction" with the original sequence for certain. Thanks a lot! $\endgroup$ – souf Apr 29 '14 at 7:11
  • $\begingroup$ You are welcome. If this answer is indeed all you need, consider upvoting and accepting it so the question will not be listed as unanswered... $\endgroup$ – 5xum Apr 29 '14 at 7:14
  • $\begingroup$ Yes, I'll do that. I think there is a timelimit in how fast can one accept an answer :) $\endgroup$ – souf Apr 29 '14 at 7:16

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