$X_\alpha$ locally compact and all but a finite number of factors are compact $ \implies\prod_\alpha X_\alpha$ is locally compact

331 Views Asked by At

How can I prove that if $X_\alpha$ are locally compact and all but a finite number of factors are compact then $\prod_\alpha X_\alpha$ is locally compact?

What needs to be proved is

the existence of $K\subset \prod X_\alpha$ where $K$ is compact and $x\in int(K),x\in\prod_\alpha X_\alpha$.

As each $X_\alpha$ is locally compact, then we have that $\exists K_\alpha\subset X_\alpha$ where $K_\alpha$ are compact sets and $y_\alpha\in int(K_\alpha)$, for all $y_\alpha \in X_\alpha$ .

We also have that a finite number of $X_\alpha$ are compact.

What can I do now?

How to construct the compact set $K$?

Any hints about how I should proceed would be greatly appreciated.

1

There are 1 best solutions below

16
On BEST ANSWER

Let $I=\{\alpha: X_{\alpha}~\text{is compact}\}$, given $x\in\displaystyle\prod_{\alpha}X_{\alpha}$, then for each $\alpha\in I^{c}$, look up a compact $K_{\alpha}$ such that $x_{\alpha}\in\text{int}(K_{\alpha})$, then $\displaystyle\prod_{\alpha\in I^{c}}{K_{\alpha}}\times\prod_{\alpha\in I}X_{\alpha}$ is a compact neighbourhood of $x$.