The Necessity of the Tube Lemma

Question

Munkres proves the following fact:

The product of fintely many compact spaces is compact.

He does this via the tube lemma. When I thought about this before reading his proof I argued in the following way:

The components are compact, so there are finte open covers. So for instance $X_1 \times ... \times X_N$ will for each coordinate have open covers $\mathcal{A}_{X_1},...,\mathcal{A}_{X_1}$. A member of the product cover will then be: $U_{X_1} \times U_{X_2} \times .. U_{X_N}$. There will be finitely many such open sets because a finite product of finite numbers is finite. Thus the product space is compact.

EDIT: As I'm typing, Math.SE is linking a bunch of relevant posts: Product of two compact spaces is compact

Can someone elaborate on the answer in this post? Am I producing something that isn't a member of any open cover of the product?

Answer 1

Every space has a finite open cover; that's not the issue.

Compactness means we can reduce any open cover to a finite one. So you have to start with any old cover and reduce that and that's where the tube lemma could come in handy.