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Note. This question concerns steps in a specific proof of the statement from the title, a link to which is (was) provided in the question. That link is now dead, and since the question does not contain all of the information required to accurately answer the question, it will be closed.

I'm reading some notes with a proof that a first countable, countably compact space is sequentially compact.

On page 118 of these notes (third page of the pdf), they are constructing a convergent subsequence. They state that $B_{n(k)+1}\cap T_{n(k)+1}\neq\emptyset$, so you can take an $x_{n(k+1)}\in B_{n(k)+1}$, showing there exists an $x_{n(i)}\in B_{n(i)}$ for all $i\in\mathbb{N}$.

How do they get that? Don't you only have that $x_{n(k+1)}\in B_{n(k)+1}$, not $B_{n(k+1)}$?

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closed as off-topic by arjafi Jul 15 '14 at 9:13

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This question cannot be answered because pertinent details are missing due to a now-dead link. – arjafi Jul 15 '14 at 9:13
up vote 3 down vote accepted

You’re quite right. The proof given in that PDF is defective. Notice that the statement of the induction hypothesis doesn’t even match the first step: he starts with $x_{n(1)}\in B_1$, not $x_{n(1)}\in B_{n(1)}$. Here’s one way to fix it.

Let $n(0)=1$. Since $y\in\operatorname{cl}(T_1)$, $B_{n(0)}\cap T_1 \ne\varnothing$, and there is some $n(1) \ge 1$ such that $x_{n(1)}\in B_{n(0)}$. Suppose that $k \ge 1$ and we have already found $n(0)<n(1)<\dots < n(k)$ such that $x_{n(i)} \in B_{n(i-1)}$ for $1 \le i \le k$. Since $y \in \operatorname{cl}T_{n(k)+1}$, $B_{n(k)} \cap T_{n(k)+1} \ne \varnothing$, and there is an $n(k+1) \ge n(k)+1$ such that $x_{n(k+1)} \in B_{n(k)}$. This completes the induction step of the construction, and we now have a sequence $1 = n(0)<n(1)<n(2)<n(3)<\dots$ such that $x_{n(k)}\in B_{n(k-1)}$ for each $k \ge 1$. Note that by an easy induction $n(k)\ge k$ for each $k$.

Now if $V$ is any nbhd of $y$, there is some positive integer $j$ such that $B_j \subseteq V$, and if $k>j$, then $k-1 \ge j$, and hence $x_{n(k)}\in B_{n(k-1)} \subseteq B_j$.

Added: By the way, this argument isn’t the one that I’d actually use if I were presenting the proof myself; it strikes me as unnecessarily complicated. I’d probably use something similar to the one that LostInMath suggests. But I did want to give you something as close as possible to the faulty proof, since it can be fixed up.

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Thanks for the fix. When you say this argument isn't the one you'd use, are you referring to your answer here, or to the whole proof in the pdf? Is there a nicer proof that first countable countably compact spaces are sequentially compact I'd be gracious for a reference. I just found this googling, and it seems a little messy. – Solange Sep 12 '11 at 0:12
@Solange: Primarily to my answer here. If proving the theorem directly, I’d use the same basic idea, but I’d simply choose $n(k)>n(k-1)$ in such a way that $x_{n(k)}\in B_k$, as LostInMath suggested. If I were teaching the material, I’d probably arrange matters so that this result would be an almost trivial consequence of previous results: (1) Every infinite set in a countably compact space $X$ has a limit point in $X$. (2) If $p\in\operatorname{cl}_X A$, where $X$ is first countable, then there is a sequence in $A$ converging to $p$. ... – Brian M. Scott Sep 12 '11 at 1:21
... But the proof of (2) will be very similar to the argument that we’ve been looking at: Let $\{B_n:n\in\omega\}$ be a nested open base at $p$. For each $n\in\omega$, $B_n\cap A\ne\varnothing$, so we can pick $x_n\in B_n\cap A$, and it should then be clear that $\langle x_n:n\in\omega\rangle$ converges to $p$. – Brian M. Scott Sep 12 '11 at 1:24

Your question is a very good one. Indeed, we only have that $x_{n(k+1)}\in B_{n(k)+1}$ and that is not enough to keep the inductive construction going.

I think that what we want is to construct a strictly increasing sequence $n(k)$ of positive integers such that $x_{n(k)}\in B_k$ for each $k\in\{1,2,\ldots\}$. Then the same argument goes through. Also in the following paragraph, you have to make the correction: "Hence, for $k\in\mathbb{N}$: $k\geq j\Rightarrow x_{n(k)}\in B_k\subset B_j\subset V$."

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I just worked this alternative argument out, and it's much more elegant than that in the pdf. Thanks! – Solange Sep 12 '11 at 1:12

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