Lee's proof that limit point compactness implies sequential compactness in 1:st countable Hausdorff spaces

I'm working through Lee's Introduction to Topological Manifolds and am finding the proof of the lemma that limit point compactness implies sequential compactness on page 99 somewhat confusing. Where do we use that $$p_n \neq p$$ (bolded part)?

Lemma 4.43. For first countable Hausdorff spaces, limit point compactness implies sequential compactness.

Proof. Suppose $$X$$ is first countable, Hausdorff, and limit point compact, and let $$(p_n)_{n \in N}$$ be any sequence of points in $$X$$. If the sequence takes on only finitely many values, then it has a constant subsequence, which is certainly convergent. So we may suppose it takes on infinitely many values.

By hypothesis the set of values $$\{ p_n \}$$ has a limit point $$p \in X$$. If $$p$$ is actually equal to $$p_n$$ for infinitely many values of $$n$$, again there is a constant subsequence and we are done; so by discarding finitely many terms at the beginning of the sequence if necessary we may assume $$p_n \neq p$$ for all $$n$$. Because $$X$$ is first countable, Lemma 2.47 shows that there is a nested neighborhood basis at $$p$$, say $$(B_n)_{n \in N}$$. For such a neighborhood basis, it is easy to see that any subsequence $$(p_{n_i})$$ such that $$p_{n_i} \in B_i$$ converges to $$p$$.

Since $$p$$ is a limit point, we can choose $$n_1$$ such that $$p_{n_1} \in B_1$$. Suppose by induction that we have chosen $$n_1 < n_2 < \cdots < n_k$$ with $$p_{n_i} \in B_i$$. By Proposition 2.39, the sequence takes on infinitely many values in $$B_{k+1}$$, so we can choose some $$n_{k+1} > n_k$$ such that $$p_{n_{k+1}} \in B_{k+1}$$. This completes the induction, and proves that there is a subsequence $$p_{n_i}$$ converging to $$p$$.

• It isn’t actually necessary in order to prove the result, but it may be necessary in order for the auther to use Proposition 2.39; exactly how is that proposition stated? Apr 16 '20 at 0:38
• Proposition 2.39: Suppose $X$ is a Hausdorff space and $A \subseteq X$. If $p \in X$ is a limit point of $A$, then every neighborhood of $p$ contains infinitely many points of $A$. I don't think you need $p_n \neq p$ to use it. Apr 16 '20 at 1:28

The bolded assumption is indeed unnecessary at that point. The crucial point is that $$p$$ is a limit point of the set $$S=\{p_n:n\in\Bbb Z^+\}$$, which is guaranteed by the fact at this point in the argument we’re dealing with the case in which $$S$$ is infinite. The recursive construction of the subsequence in the last paragraph could just as well be carried out at this point, though the resulting subsequence might turn out to be constant (or eventually constant) at $$p$$: that could happen, since $$\{n\in\Bbb Z^+:p_n=p\}$$ could be infinite even when $$\{n\in\Bbb Z^+:p_n\ne p\}$$ is infinite. For some reason Lee has chosen to distinguish the case in which $$\{n\in\Bbb Z^+:p_n=p\}$$ is infinite from the case in which it’s finite (and can then be assumed to be empty).
I don’t know why he did this, unless he wanted to make the point that if the original sequence does not have a subsequence constant at $$p$$, then it has a subsequence in $$X\setminus\{p\}$$ that converges to $$p$$. But in that case I’d expect him to point that out explicitly, either in the proof or immediately after it. Possibly he had that idea in the back of his mind when writing the proof and simply didn’t notice that he’d made the argument a little more complicated than necessary.
• Thanks Brian. But even if $p \in A$, wouldn't Proposition 2.39 still apply? Where would the argument fail in that case? Apr 16 '20 at 2:19