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So I want to learn the proof that a compact metric space $(X,d)$ is also sequentially compact. The proof goes as follows:

(X,d) is compact, so it is also limit point compact. Let $\{x_k\}_{k=1}^\infty$ be a sequence in X and put $A = \{x_k : k\in \mathbf{N}\}$. Then A is either finite or infinite. If A is finite, we are done, so assume A is infinite. Then A has a limit point x. Since we are in a metric space, $B_{1/k}(x) \cap A$ is infinite. Choose $n_{k} > n_{k-1}$ s.t $x \in B_{1/k}(x) \cap A$. Then $\{x_{n_{k}}\}_{k=1}^\infty$ is a subsequence with $x_{n_{k}} \rightarrow x$, $k \rightarrow \infty$

I'm confused about the step where we intersect the ball of radius $1/k$ with $A$. Why is the intersection infinite? And why do we choose $n_k > n_{k-1}$? Is it just to say that the sequence is increasing $\forall k$?

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  • $\begingroup$ What is $x$ in $B_{1/k}(x)$? $\endgroup$ Jun 2 '20 at 23:00
  • $\begingroup$ They never specify what it is, which contributes to my confusion. But I believe that it is an open ball of radius 1/k centered around x $\endgroup$
    – user789450
    Jun 2 '20 at 23:03
  • $\begingroup$ Yes, but you haven't defined what $x$ is. $\endgroup$ Jun 2 '20 at 23:04
  • $\begingroup$ Ah yes, my bad, I missed that line. I will edit the post, but x is the limit point of A $\endgroup$
    – user789450
    Jun 2 '20 at 23:06
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We know $x$ is a limit point of $A$. So by definition, every open neighborhood of $x$ must intersect $A$ at a point other than $x$. If there were only finitely many points in $B_\epsilon(x) \cap A$ for any $\epsilon>0$, where $B_\epsilon(x)$ is the open ball of radius $\epsilon$ centered at $x$, then take a point distinct from $x$ closest to $x$ and call this $P$. Then define the distance from $x$ to $P$ using your metric $d$ as $D= d(x,P)$. But then $B_{D/2}(x) \cap A$ is either empty or simply $x$ itself. But then $x$ is not a limit point of $A$, contradiction.

As for the next question, the choice of $n_k$ are merely so that you choose a point in $B_{1/k}(x) \cap A$ 'further down' in the sequence $\{x_n\}$ so that the point you choose will be even closer to $x$ (because the distance is at most $1/k$ for that choice of $k$). Then you are choosing points further and further down your sequence, i.e. forming a subsequence from your given sequence, where the points are getting closer and closer to $x$. Then because you can choose them arbitrarily close, you know that $x_{n_k} \to x$. Then you have found a convergent subsequence. [Not necessarily the only one, but one at least.]

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  • $\begingroup$ The first paragraph seems to be generally true in Hausdorff spaces: $x$ is a limit point of infinite $A \subseteq X$ iff each open set containing $x$ contains infinitely many elements of $A$. $\endgroup$ Jun 3 '20 at 6:39
  • $\begingroup$ @JordanMitchellBarrett You don't even need Hausdorff but rather something slightly weaker. If $X$ is a space satisfying the $T_1$ axiom (slightly weaker than Hausdorff), and $A \subseteq X$, then $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$. You do need metric space for the notion of 'closeness' here. Hausdorff also gets you the sequence of points converging to at most one point (here only one). $\endgroup$ Jun 3 '20 at 6:45

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