1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\beta := \sup_n \alpha_n \in \omega_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.