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.
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 $A$).