Closed Compact Subset of Product Space Must Have Empty Interior Suppose that $\{X_{\alpha}\}_{\alpha\in A}$ is a nonempty family of topological spaces. Suppose also that there is an infinite index set $B\subseteq A$ such that $X_{\alpha}$ is not compact for any $\alpha\in B$. Furthermore, assume that


*

*$K\subseteq\prod_{\alpha\in A}X_{\alpha}$;

*$K$ is compact (in the product topology);

*$K$ is closed (in the product topology).


$\mathbf{Claim:}\quad$ $K$ has no interior.
How to show this? Should I assume that $K$ has nonempty interior and try to find an open cover of it that has no finite subcover (using the fact that infinitely many of the $\{X_{\alpha}\}_{\alpha\in A}$ are not compact) to derive a contradiction? Or a net in $K$ that does not have a cluster point? Or use the finite intersection property? I don't need a complete proof, just a hint to know where it is right place to begin. Thank you in advance for your help.
 A: I think I figured it out, but I would appreciate any feedback confirming it is correct.
Suppose that $K$ has a nonempty interior. Then, there exists an open set $U\subseteq\prod_{\alpha\in A} X_{\alpha}$ in the product topology such that


*

*$U$ is not empty;

*$U=\prod_{\alpha\in A}U_{\alpha}$, where $U_{\alpha}$ is open in $X_{\alpha}$ for all $\alpha\in A$, and $U_{\alpha}=X_{\alpha}$ for all but finitely many $\alpha\in A$;

*$U\subseteq K$.


Then, since $K$ is closed, $\overline U\subseteq K$ (the bar denoting closure in the product topology). Since $\overline U$ is closed and contained in a compact set, it must be compact itself. Moreover, since the canonical projection map $\pi_{\alpha}: X\to X_{\alpha}$ is continuous for all $\alpha\in A$ by the construction of the product topology and the continuous image of any compact set is compact, $\pi_{\alpha}(\overline U)$ is compact for all $\alpha\in A$. But, for all but finitely many $\alpha\in A$, we have (given also that $U\neq\varnothing$; otherwise $\pi_{\alpha}(U)=\varnothing$ vacuously for all $\alpha\in A$):
\begin{align*}
X_{\alpha}=\pi_{\alpha}(U)\subseteq \pi_{\alpha}(\overline U)\subseteq X_{\alpha}.
\end{align*}
Hence, for all but finitely many $\alpha\in A$, $\pi_{\alpha}(\overline U)=X_{\alpha}$, which is compact. This contradicts there being infinitely many $\alpha\in A$ for which $X_{\alpha}$ is not compact.
A: Hint: Suppose $\mathbf{x} = \langle x_\alpha \rangle_{\alpha \in A} \in \mathrm{Int} ( K )$ and take a basic open neighbourhood $U = \prod_{\alpha \in A} U_\alpha$ of $\mathbf{x}$ such that $U \subseteq K$.  Then $U_\alpha = X_\alpha$ for some $\alpha$ where $X_\alpha$ is not compact.  From this construct an open cover of $K$ with no finite subcover.
