It is known that every compact space is countably compact. These properties are equivalent for metrizable spaces.

So, is it true that every countably compact Hausdorff space is compact?

I think it is true, since every countably compact Hausdorff space is a Polish space and Polish spaces are metrizable. Hence we can deduce that the space is compact.


No, a countably compact space need not be metrizable (and hence not Polish). Consider $\omega_1$ with the order topology (hence a Hausdorff space). Then the cover $$ \omega_1 = \bigcup_{\alpha < \omega_1} [0, \alpha) $$ with open sets does not have a finite (even not a countable) subcover, therefore $\omega_1$ is not compact (even not Lindelöf).

But $\omega_1$ is contably compact, for suppose $A \subseteq \omega_1$ is infinite. Then $A$ contains an increasing sequence $(\alpha_n)_{n < \omega}$, and hence $A$ has the limit point $\beta := \sup_n \alpha_n \in \omega_1$.


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