Question on countably compact, sequentially compact and compact

Is there a countably compact space which is neither sequentially compact nor compact?

If not, does that mean countably compact implies sequentially compact or compact?

I know there are some weakly countably compact spaces that are neither sequentially compact nor compact. But I can't find a countably compact space.

Many thanks!

• Welcome to Math.SE! Please expand your question by adding context: what do you think is the correct answer, why are you interested in this question etc. – Hrodelbert Dec 10 '15 at 14:10

Let $D=\{0,1\}$ be the discrete two-point space, and let $Y=D^{[0,1]}$, the product of uncountably many copies of $D$; $Y$ is compact but not sequentially compact. (There is a proof here.) You could instead use $\beta\omega=\beta\Bbb N$, the Čech-Stone compactification of the natural numbers, for $Y$: it is also compact but not sequentially compact.
The ordinal space $\omega_1$ with the order topology is sequentially compact (and hence countably compact) but not compact.
Now let $X=\omega_1\times Y$. $X$ has a closed subspace homeomorphic to $\omega_1$, so $X$ is not compact, and $X$ has a closed subspace homeomorphic to $Y$, so $X$ is not sequentially compact. However, $X$ is the product of a countably compact space and a compact space, so $X$ is countably compact.