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On page 19 of Counterexamples in Topology, the authors say "sequential compactness clearly implies countable compactness" without explaining.

I feel dumb, why is this so obvious? Does anyone have nice explanation or reference for this fact?

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Earlier on that page, the authors give a few equivalent conditions for a space to be countably compact. One of these is "every sequence has an accumulation point". Does that clear things up? – Dylan Moreland Sep 11 '11 at 23:51
I think it has to see with the fact that any 'hood (neighborhood) containing the limit point will "swallow" most of the elements, i.e., you are given a countable cover {$S_n$}, then take an element $a_n$ for each $S_n$; this gives you an infinite set, which has a limit point $a$; the 'hood containing a will contain all-but-finitely-many elts. of the sequence. – gary Sep 11 '11 at 23:57
@Dylan, mind if I ask some additional questions about the prior explanation? (i) If a sequence has a convergent subsequence, why is that limit point an accumulation point of the original sequence? (ii) I get why $CC_2$ and $CC_3$ are equivalent. But in showing that $CC_2\implies CC_1$, the book states every one of the $O_i$ intersects only finitely many points of the set. Why only finitely many? If you post it as an answer, I'd be happy to accept. – Gotye Sep 11 '11 at 23:59
(i) Because every nbhd of the point contains infinitely many terms of the sequence. Some of those terms may actually be the same point of $X$, but that’s all right: an accum. pt. of a sequence $\langle x_n:n\in\omega\rangle$ is not necessarily an accum. pt. of $\{x_n:n\in\omega\}$. E.g., $0$ is an accum. pt. of the sequence $\langle 0,1,0,2,0,3,0,4,\dots\rangle$ but not of the set $\{0,1,2,3,4,\dots\}$. – Brian M. Scott Sep 12 '11 at 0:04
(ii) I don’t have the book, but I can make a good guess that the set in question is something like $\{x_i:i\in\omega\}$, where $x_i$ is chosen so that $x_i \notin \bigcup_{j\le i}O_j$. If so, you have $x_i \notin O_j$ whenever $i\ge j$, and therefore any given $O_i$ can contain at most $x_0,x_1,\dots$, and $x_{i-1}$. – Brian M. Scott Sep 12 '11 at 0:10
up vote 6 down vote accepted

Suppose $X$ is sequentially compact but $\{U_n: n \in {\mathbb N}\}$ is a countable open cover with no finite subcover. For every $N$, since $\{U_n: n \le N\}$ is not a subcover, there must be $x_N \notin \bigcup_{n \le N} U_n$. By sequential compactness, some subsequence $x_{N_j}$ converges to some point $x \in X$. But $x \in U_m$ for some $m$, and since $x_{N_j} \notin U_m$ when $N_j > m$ we get a contradiction.

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This is much quicker than trying to chase all those equivalent statements back. Thanks Prof. Israel. – Gotye Sep 12 '11 at 0:05
This is probably a dumb question, but what if $N_j\leq m$ for all $N_j$ if all convergent subsequences of $\{x_N\}$ happen to be finite? – Gotye Sep 12 '11 at 1:38
@Gotye: A subsequence is by definition an infinite sequence. – Robert Israel Sep 12 '11 at 2:40

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