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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.

And

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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    The sequence will have a convergent [*subnet*](http://en.wikipedia.org/wiki/Subnet_%28mathematics%29); but this subnet need not be a subsequence.2012-07-28
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    @David Why the sequence itself is not OK?2012-07-28
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    $x$ 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 $k\ge M$).2012-07-28
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    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:)2012-07-28
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    See also this question: [What's going on with “compact implies sequentially compact”?](http://math.stackexchange.com/questions/44907/whats-going-on-with-compact-implies-sequentially-compact).2012-07-28
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    t.b., thanks for the editing:)2012-07-29

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