If $X=\prod_{\alpha \in I}X_{\alpha}$ is a $k$-space then each $X_{\alpha}$ is a $k$-space? My definition
Let $X$ be a topological space and $A\subset X$. We say that $A$ is a $\textbf{$k$-closed}$ in $X$ if for all compact $K\subset X$ it happens that $A\cap K$ is closed at $K$. So we have that $X$ is a $\textbf{$k$-space}$ if all $k$-closed from $X$ is closed at $X$.
My question is
If $X=\prod_{\alpha \in I}X_{\alpha}$ is a $k$-space then each $X_{\alpha}$ is a $k$-space? If this is not true, could you mention an example or a bibliographic reference where you can consult that example?
 A: There are trivial counterexamples: if a single $X_\alpha$ is empty then $X$ will be empty regardless of what the other $X_\alpha$'s are.  But these are the only counterexamples: if every $X_\alpha$ is nonempty and $X$ is a $k$-space then so is each $X_\alpha$.  To prove it, suppose $A\subseteq X_{\alpha_0}$ is $k$-closed.  Write $p_\alpha:X\to X_\alpha$ for the projection maps.  To show $A$ is open in $X_{\alpha_0}$, it suffices to show that $B=A\times\prod_{\alpha\neq\alpha_0}X_\alpha$ is closed in $X$, since $p_{\alpha_0}$ is an open map and $p_{\alpha_0}(X\setminus B)=X_{\alpha_0}\setminus A$ (this last equality is where we are using the assumption that each $X_\alpha$ is nonempty).
So, suppose $K\subseteq X$ is compact.  Then $K_\alpha=p_\alpha(K)$ is compact for each $\alpha$, and $K\subseteq \prod_\alpha K_\alpha$.  So, it suffices to show that $B\cap \prod_\alpha K_\alpha$ is closed in $\prod_\alpha K_\alpha$.  But $B\cap \prod_\alpha K_\alpha=(A\cap K_{\alpha_0})\times\prod_{\alpha\neq\alpha_0}K_\alpha$, which is closed in $\prod_\alpha K_\alpha$ since $A$ is $k$-closed so $A\cap K_{\alpha_0}$ is closed in $K_{\alpha_0}$.
