Locally Compact Question This question is from Munkres' text Topology:

Let $\{X_\alpha\}$ be an indexed family of nonempty spaces. First, show that if $\prod X_\alpha$ is locally compact, then each $X_\alpha$ is locally compact and $X_\alpha$ is compact for all but finitely many values of $\alpha$. Second, prove the converse assuming the Tychonoff theorem.

For the first one, I know that a continuous, open map of $X$ onto $Y$ where $X$ is locally compact implies $Y$ is locally compact, so I figured projections here would be helpful, but I wasn't sure how to prove that only finitely many $X_\alpha$ are locally compact while the others are compact. How can I conclude this?
 A: Pick a point $x\in\prod X_\alpha$. Because of local compactness we can find a compact neighbourhood $U$ of $x$. The images of $U$ with the projections will be compact but, since $U$ contains a basic open set, only finitely many of these images will be strictly smaller than the corresponding $X_\alpha$.
A: HINT: Let $A$ be the index set, and let $X=\prod_{\alpha\in A}X_\alpha$. Let $$A_0=\{\alpha\in A:X_\alpha\text{ is not compact}\}\;;$$ your problem is to show that $A_0$ is finite.
Let $x=\langle x_\alpha:\alpha\in A\rangle\in X$. If $X$ is locally compact, there is a compact $K\subseteq X$ such that $x\in\operatorname{int}K$. The projection maps $\pi_\alpha:X\to X_\alpha$ are continuous, so $K_\alpha=\pi_\alpha[K]$ is compact in $X_\alpha$ for each $\alpha\in A$. In particular, $K_\alpha\ne X_\alpha$ for each $\alpha\in A_0$.
Recall that a basic open nbhd of $x$ in $X$ is a set of the form $\prod_{\alpha\in A}U_\alpha$ such that 


*

*$U_\alpha$ is open in $X_\alpha$ for each $\alpha\in A$;  

*$x_\alpha\in U_\alpha$ for each $\alpha\in A$;  

*$\{\alpha\in A:U_\alpha\ne X_\alpha\}$ is finite.


If $x\in\operatorname{int}K$, there is a basic open set $B=\prod_{\alpha\in A}U_\alpha$ such that $x\in B\subseteq K$. Can you see how to get a contradiction from the assumption that $A_0$ is infinite?
