compact subset of product topology is nowhere dense

Question

Let $\{X_i\}$ be a family of non-empty, Hausdorff spaces. Prove that if infinitely many of them are non-compact then every compact subset of the topological product $\prod X_i$ is nowhere dense.

In order to show that some compact $K\subset \prod_i X_i$ is nowhere dense, we need to show first that $\text{int}(\overline{K})=\emptyset$. Now since $\prod X_i$ is Hausdorff, we can say that $\overline{K}=K$. So the problem reduces to show that $\text{int}(K)=\emptyset$. Now one information that we still didn't use is $\color{red}{\text{infinitely many of them (i.e.$ \ X_i$ for infinitely many $i$) are non-compact}}$. But how to conclude the result from there is not clear to me. Any help is appreciated.

Answer 1

If $K \subseteq \prod_i X_i$ is compact, it's already closed (as all $X_i$, and thus the product, is Hausdorff) so we only need to check it has empty interior.

To this end it suffices that any non-empty basic open subset $\prod_i O_i$ of the product (so all $\emptyset \neq O_i$ are open in $X_i$ and almost all $O_i$ are the whole space, or $F:=\{i \in I: X_i \neq O_i \}$ is finite; this is the standard product base) cannot be a subset of $K$.

So suppose (for a contradiction) that $\prod_i O_i \subseteq K$. By assumption $J:=\{ i \in I: X_i \text{ not compact }\}$ is infinite. So we have some $i_0 \in J\setminus F$ ($F$ defined above for this basic open set). So we have from the inclusion that $$X_{i_0} = O_{i_0} = \pi_{i_0}[\prod_i O_i] \subseteq \pi_{i_0}[K] \subseteq X_{i_0}$$

That $X_{i_0}=O_{i_0}$ follows from $i_0 \notin F$, the rest is obvious. But then $X_{i_0} = \pi_{i_0}[K]$ which is compact as a continuous image of a compact set, contradicting $i_0 \in J$.

This contradiction shows that $\operatorname{int}(K)=\emptyset$ for any compact $K \subseteq \prod_i X_i$ and we're done.