I want to prove that the restricted product topology of lcoally compact spaces is locally compact. Let $X_i$ locally compact topological spaces, and $K_i$ corresponding compact open subsets. I consider the restricted product $X$ of the $X_i$ with respect to the $K_i$.
The basis of neighborhood for the restricted product is given by open sets of the form
$$\prod_{i \in S} U_i \times \prod_{i \notin S} K_i$$
where $S$ is a finite set, and $U_i$ is open in $X_i$. We can as usual consider the $S$-restricted product
$$X_S = \prod_{i\in S} X_i \prod_{i\notin S} K_i.$$
It is clear that the $X_S$ cover $X$, and I would like to this to prove the local compactness of $X$. Let $x \in X$, this $x \in X_S$ for a certain $S$. Since $X_S$ is locally compact (essentially a finite product), $x$ has a compact open neighborhood $U$ in $X_S$.
I would like to prove (or disprove?) that $U$ is a compact open in $X$. I can see that an open cover of $U$ in $X$ restricts to an open cover of $U$ in $X_S$, thus there is a finite subcover. But what ensures that this subcover also works in $X$, i.e. that $U$ is covered in $X$ by it?
Best Answer
I'm not sure what you're trying to do here, but the result to go for is this:
The left to right implication seems your concern here. If for some $p=(x_i)_{i \in S}$ has a compact neighbourhood $K$ in $X$ it contains a basic open neighbourhood that only depends on finite many coordinates $F$ and then it follows that, as $X_i = \pi_i[K]$ for $i \notin F$, that all those $X_i$ are compact. The rest is routine.