Rationals are not locally compact and compactness

Solution 1:

The statement of the second theorem should remind you of the fact that the product of two compact spaces is compact. In fact, one can adopt the proof of this theorem given in chapter 3 of Munkres to prove your claim. Here's how it's done.

Let $\mathcal{U}$ be an open cover of $X\times Y.$ Then for each $x$ in $X$ the collection $\mathcal{U}$ is an open cover of $\{x\}\times Y.$ Hence, as $\{x\}\times Y$ is compact, there exists for each $x\in X$ a finite subcollection $\mathcal{U}_x$ of $\mathcal{U}$ which covers $\{x\}\times Y.$ Choose such a collection and let $U_x$ be the set obtained by unioning the elements of $\mathcal{U}_x.$ The set $U_x$ is open in $X\times Y$ and contains $\{x\}\times Y.$ Appealing once more to the compactness of $Y,$ it follows by the tube lemma, that for each $x\in X$ there exists an open neighborhood $N_x$ of $x$ such that $N_x \times Y \subset U_x.$ Consider the collection $\{N_x: x\in X\}.$ As $X$ is Lindelof, there exists a countable subset $I\subset X$ such that $\{N_x: x\in I\}$ covers $X.$ It follows that the set $\bigcup_{x\in I} \mathcal{U}_x$ is a countable subcollection of $\mathcal{U}$ which covers $X\times Y.$ We conclude $X\times Y$ is Lindelof.