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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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    Thanks for answering my questions, Professor Scott.2011-09-12

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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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    @Gotye: A subsequence is by definition an infinite sequence.2011-09-12