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Why is it that:

If $A$ is a compact set and $\left ( a_{n} \right )$ a sequence in $A$, then there is a subsequence $\{a_{n_k}\}$ such that $\lim_{k\to\infty} a_{n_k}=a$ with $a\in A$.

I get that the sequence is bounded and thus we can extract a convergent subsequence, but it looks to me to be stating that the limit can be whatever you would like it to be! It's a pre-fixed sequence, so I can extract all sorts of subsequences. I'm confused.

Reading this question brought up this confusion: If $A$ is compact and $B$ is closed, show $d(A,B)$ is achieved

Can anyone help explain this. Thanks in advance.

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John11
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  • If $a_n \to a$ and $b_n$ is any subsequence of $a_n$, $b_n \to a$ also. – Matias Heikkilä May 01 '16 at 20:01
  • If the original sequence does not converge, you may have different limits. E.g. with $a_n = (-1)^n$. – Matias Heikkilä May 01 '16 at 20:02
  • Isn't that the definition of a compact set? – ahorn May 01 '16 at 20:05
  • @MatiasHeikkilä the original sequence is just bounded. With your example, I can extract sequences with limits $-1$ or $1$. The statement sounds to me like I could say there is a subsequence of $a_n = (-1)^n$ with limit $1/2$ for example – John11 May 01 '16 at 20:05
  • I was trying to explain that you can't have an arbitrary limit with these sequences. – Matias Heikkilä May 01 '16 at 20:07
  • The statement of the theorem is just that there is a subsequence with a limit. Not the other way around. You can't first fix the limit and then necessarily find a subsequence that converges to it. – Matias Heikkilä May 01 '16 at 20:08
  • @MatiasHeikkilä yes I realise that, that's precisely the source of my confusion with the statement in my question. Why is it legitimate to say there is a subsequence with $a$ as a limit? – John11 May 01 '16 at 20:09
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    Ah, it's just some $a \in A$. You don't know which. The point is that it is in $A$. – Matias Heikkilä May 01 '16 at 20:11
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    @John11 I think $a$ doesn't have any particular significance. The point is that you can find a convergent subsequence with a limit in $A$. – ahorn May 01 '16 at 20:12
  • @MatiasHeikkilä ohhhh ok I thought it was THE $a$ closest to the other set. – John11 May 01 '16 at 20:13
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    @MatiasHeikkilä Thank you!! – John11 May 01 '16 at 20:13

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The definition of a (sequentially) compact set $A \subseteq\mathbb{R}^n$ is that, for every sequence in $A$, you can find a convergent subsequence with the limit in $A$. By writing $\lim_{k\to\infty} a_{n_k}=a$, the author is saying that the limit exists, but more information is needed to understand what is special about this limit ($a$ could be anything if we don't say something about it). $a\in A$ gives the necessary information, and because this is written straight afterwards, the two pieces of information go hand-in-hand.

Indeed, there can be many subsequences, but we are only interested in those that converge (and those that converge will have their unique limit in $A$).

ahorn
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