Firstly, I'll give the definitions of sequential compactness and countable compactness.

Sequential compactness: If $X$ is a Hausdorff space and every sequence of points of $X$ has a convergent subsequence.


Countable compactness: if $X$ is a Hausdorff space and every infinite subset of $X$ has a cluster.

My text book gives me some counterexamples which are countably compact, even compact spaces, however they are not sequentially compact; we also can see this link compactness / sequentially compact. They are equal in first countable spaces.

But I think, without the condition of first countability, countable compactness implies sequential compactness. By the definition of countable compactness, every sequence of points of $X$ has a cluster point. Then this sequence has a subsequence (we choose itself) which is convergent, which shows that $X$ is sequentially compact. I don't know where I am wrong. Could anybody point out my mistakes? Thanks ahead:)

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    $\begingroup$ The sequence will have a convergent subnet; but this subnet need not be a subsequence. $\endgroup$ – David Mitra Jul 28 '12 at 3:24
  • $\begingroup$ @David Why the sequence itself is not OK? $\endgroup$ – Paul Jul 28 '12 at 3:28
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    $\begingroup$ $x$ is a cluster point of $(x_n)$ if for every nhood $O$ of $x$ and every $N$, there is an $M\ge N$ with $x_M\in O$. This does not say that the sequence converges to $x$ (there you need $x_k\in O$ for all $k\ge M$). $\endgroup$ – David Mitra Jul 28 '12 at 3:34
  • $\begingroup$ Thanks David for your comments. I just come back after lunch. Maybe I need to review the definitions of "cluster point" and "convergence sequence". Thank you again:) $\endgroup$ – Paul Jul 28 '12 at 4:13
  • $\begingroup$ See also this question: What's going on with “compact implies sequentially compact”?. $\endgroup$ – Martin Sleziak Jul 28 '12 at 5:22

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