Sequential compactness vs. countable compactness

Asked Jul 28 '12 at 03:19

Active Jul 28 '12 at 14:59

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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:)

edited Apr 13 '17 at 12:19

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asked Jul 28 '12 at 03:19

Paul

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The sequence will have a convergent subnet; but this subnet need not be a subsequence. – David Mitra Jul 28 '12 at 03:24
@David Why the sequence itself is not OK? – Paul Jul 28 '12 at 03:28
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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 $k\ge M$). – David Mitra Jul 28 '12 at 03:34
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:) – Paul Jul 28 '12 at 04:13
See also this question: What's going on with “compact implies sequentially compact”?. – Martin Sleziak Jul 28 '12 at 05:22
t.b., thanks for the editing:) – Paul Jul 29 '12 at 11:24

Sequential compactness vs. countable compactness

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