Need help understanding two steps in the proof that limit point compactness implies sequential compactness

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 $$\forall k$$?

• What is $x$ in $B_{1/k}(x)$? Jun 2 '20 at 23:00
• 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
– user789450
Jun 2 '20 at 23:03
• Yes, but you haven't defined what $x$ is. Jun 2 '20 at 23:04
• Ah yes, my bad, I missed that line. I will edit the post, but x is the limit point of A
– user789450
Jun 2 '20 at 23:06

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.]
• 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 $A$. Jun 3 '20 at 6:39
• @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). Jun 3 '20 at 6:45