The title is the question : Is there a non-metrizable topological space for which any countably compact subset is compact ?
EDIT : non-metrizable and Hausdorff
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$\begingroup$ To the OP: if you register your original account you will be able to make edits without waiting for approval. If you no longer have that cookie set, you can make a new account and ask for it to be merged with the original one. $\endgroup$– Nate EldredgeOct 3, 2016 at 15:58
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3 Answers
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Any Lindelöf non-metrizable Hausdorff space will do (EDIT: you need that the space is C-closed as well, see below), but more generally, a space is called isocompact iff every closed countably compact subset of X is compact, cl-isocompact iff the closure of a countably compact subset is compact, and C-closed iff any countably compact subset is closed. This paper by Cho and Park contains various results on (cl)-isocompactess, as well a other references:
http://www.mathnet.or.kr/mathnet/kms_tex/978637.pdf
And this one by Ismail and Nyikos has good informations on C-closed spaces:
http://www.sciencedirect.com/science/article/pii/0166864180900279
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In addition to the above examples, it might be of interest that functional analysis is awash with such spaces. For example, the space of distributions on a compact interval or that of functions analytic on a closed domain in the complex plane, both with their natural topologies. More generally, any Silva space has this property. A further ubiquitous source of examples is the dual of a separable Banach space with the bounded weak $\ast$ topology. A suitable reference would be Köthe's classic on topological linear spaces.
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$\begingroup$ And $\ell_1$ (or even its unit ball) with the weak topology. $\endgroup$ Oct 3, 2016 at 23:27
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$\begingroup$ The space $\mathscr D'(\Omega)$ of distributions on an open set $\Omega \subseteq \mathbb R^n$ also has the property because it is a strict (locally convex) inductive limit of Frechet spaces. $\endgroup$ Oct 4, 2016 at 6:30
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$\begingroup$ Warning: both of these comments contain false statements. $\endgroup$– vltavaOct 4, 2016 at 11:11
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$\begingroup$ The unit ball of $\ell^1$ is metrisable and so not a counterexample, the space of distributions is not a strict $LF$-space. $\endgroup$– vltavaOct 4, 2016 at 13:11
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I think any compact, first countable, non-metrizable Hausdorff space suffices, since then any countably compact set is closed and hence compact. In Steen and Seebach's Counterexamples in Topology, #95, the weak parallel line topology, has these properties.
answered Oct 3, 2016 at 16:43