Is the Product of a Compact Space and a Countably Compact Space Countably Compact?

Question

I know the statement does not hold for the product of two countably compact spaces, but I was wondering if it holds if one of the spaces is compact.

My idea was to use a cover consisting of basic open sets and show that there is a finite subcover for each space via projections on the basic open sets, and then take the corresponding products of these sets to yield a finite subcover for the product.

However, if this is true, then the same technique would show the result for products of countably compact spaces, which I know isn't true. Can anyone shed some light on this?

Answer 1

Your proposed argument is wrong because if you have subsets of $X\times Y$ whose projections cover $X$ and $Y$ that does not mean the sets cover $X\times Y$. Moreover, to test countable compactness you cannot restrict to basic open sets, because given a countable open cover, it may not have a refinement by basic open sets that is still countable.

It is true that a product of a compact space $X$ and a countably compact space $Y$ is countably compact. To prove it, suppose $(U_n)_{n\in\mathbb{N}}$ is a countable open cover of $X\times Y$. For each finite subset $F\subset\mathbb{N}$, let $$V_F=\left\{y\in Y:X\times \{y\}\subseteq \bigcup_{n\in F}U_n\right\}.$$ Note that $V_F$ is open in $Y$: if $y\in V_F$, then $X\times\{y\}$ is covered by finitely many open rectangles contained in the $U_n$ for $n\in F$ (by compactness of $X$), and thus the intersection of the second factors of those rectangles is an open neighborhood of $y$ contained in $V_F$. Moreover, every $y\in Y$ is in $V_F$ for some finite $F\subset\mathbb{N}$, since $X\times\{y\}$ is compact and thus covered by finitely many of the $U_n$. Thus $(V_F)$ is a countable open cover of $Y$, so by countable compactness $Y$ is covered by finitely many of them $V_{F_1},\dots,V_{F_m}$. But then $X\times Y$ is covered by the sets $U_n$ for $n\in\bigcup_{i=1}^m F_i$, so our original open cover has a finite subcover.

Answer 2

Here's another proof using projections. If $X$ is compact and $Y$ countably compact, then the projection $\pi:X\times Y\to Y$ is closed. Note that countable compactness is equivalent to any countable family of closed sets with finite intersection property having non-empty intersection. If $F_n\subseteq X\times Y$ are such sets, we can assume without loss of generality that $F_n$ are decreasing. Then $\pi[F_n]$ are decreasing, non-empty closed sets in $Y$, so their intersection is non-empty. Thus $F_n\cap X\times \{y\}\neq \emptyset$ for some $y\in Y$. Using that $X\cong X\times\{y\}$ is compact, we have $\bigcap_{n=1}^\infty F_n \cap X\times\{y\}\neq \emptyset$ so that $\bigcap_{n=1}^\infty F_n\neq \emptyset$ as had to be shown.