Suppose $X_{\alpha}$ is nonempty for each $\alpha\in A$. Then $\prod X_{\alpha}$ is locally compact iff
each $X_{\alpha}$ is locally compact and all but finitely many $X_{\alpha}$ are compact.
I understand the proof except when showing that all but finitely many $X_{\alpha}$ are compact. Here is where I'm at:
Let $x\in\prod X_{\alpha}$ and $U\subseteq\prod X_{\alpha}$ open for which $x\in U$. As $\prod X_{\alpha}$ is locally compact, there exists an open set $K$, with $\overline{K}$ compact, so that
$$x\in K\subseteq\overline{K}\subseteq U.$$
As $K$ is open it contains a basic open set of the form
$$\pi^{-1}_{\alpha_1}\left(U_{\alpha_1}\right)\cap\cdots\cap
\pi^{-1}_{\alpha_n}\left(U_{\alpha_n}\right).$$
The author (Willard) then goes on to say; "it follows that if $\alpha\neq\alpha_1,\dots,\alpha_n$, then $\pi_{\alpha}\left(K\right)=X_{\alpha}$."
I'm having trouble understanding this. I know that if $\alpha\neq\alpha_1,\dots,\alpha_n$, then $U_{\alpha}=X_{\alpha}$ so that
$$\pi_{\alpha}^{-1}(U_{\alpha}) = \pi_{\alpha}^{-1}(X_{\alpha}) = \prod X_{\alpha}.$$
Though, I feel this is irrelevant.
Best Answer
So we know that $$x \in \pi_{\alpha_1}^{-1}[U_{\alpha_1}] \cap \pi_{\alpha_2}^{-1}[U_{\alpha_2}] \cap \pi_{\alpha_n}^{-1}[U_{\alpha_n}]\subseteq K$$
Now, if $\alpha \notin \{\alpha_1, \ldots, \alpha_n\}$, we know that $X_\alpha \subseteq \pi_\alpha[K]$: suppose that $p \in X_\alpha$ then I define the point $x'$ of $\prod_{\alpha} X_\alpha$ that equals $x$ on all coordinates, except on coordinate $\alpha$ where $x'_\alpha= p$ and I note that $x' \in \pi_{\alpha_1}^{-1}[U_{\alpha_1}] \cap \pi_{\alpha_2}^{-1}[U_{\alpha_2}] \cap \pi_{\alpha_n}^{-1}[U_{\alpha_n}]$, because it obeys the finitely many conditions on the $\alpha_i$ coordinates and so $x' \in K$ by the above inclusion, and in particular $p = \pi_\alpha(x') \in \pi_\alpha[K]$.
So $$X_\alpha \subseteq \pi_\alpha[K] \subseteq \pi_\alpha[\overline{K}] \subseteq X_\alpha$$ for all such coordinates $\alpha$. In particular we have that $X_\alpha$ must be compact as the continuous image of $\overline{K}$, which is compact.
So indeed all but finitely many $X_\alpha$ must be compact.