# Rationals are not locally compact and compactness

I was wondering if someone can please help me with the following problems:

1. Show that $\mathbb{Q}$ is not locally compact.
2. Prove that if $X$ is Lindelöf and $Y$ is compact then $X \times Y$ is Lindelöf.

I think I got 1, here's my work:

Suppose $\mathbb{Q}$ is locally compact and let $x \in \mathbb{Q}$. Then by definition of local compactness (for Hausdorff spaces) we can find an open set $U \subset \mathbb{Q}$ such that $\overline{U}$ is compact. Since the open intervals form a basis for the usual topology then we can find an open interval $(a,b)$ such that $x \in (a,b) \cap \mathbb{Q} \subset U$. Then observe $[a,b] \cap \mathbb{Q} \subseteq \overline{U}$. But $[a,b] \cap \mathbb{Q}$ is closed so we have a closed subset of the compact set $\overline{U}$ , thus $[a,b] \cap \mathbb{Q}$ is compact. Hence it suffices to show this set is not compact. Now pick an irrational number $z \in (a,b)$ then we can find a sequence $\{x_{n}\}$ of consisting of rational numbers in $(a,b)$ such that $x_{n} \rightarrow z$. But $[a,b] \cap \mathbb{Q}$ is compact so sequentially compact. Therefore the sequence $\{x_{n}\}$ must have a subsequence converging to a point in $(a,b) \cap \mathbb{Q}$, which is impossible because every subsequence converges to $p$ and $p$ is irrational. Is this OK?

2) Stuck in this one for a while. How to prove this?

• 1) I'd apply Baire but your argument seems ok. 2) This seems wrong. Is $\mathbb{N} \times [0,1]$ compact? Or do you want Lindelöf $\times$ compact is Lindelöf? – t.b. Jun 16 '11 at 2:15
• @Theo Buehler: How would you apply Baire? sorry I meant $X \times Y$ Lindelöf. – user10 Jun 16 '11 at 2:17
• A point in $\mathbb{Q}$ is nowhere dense hence $\mathbb{Q}$ is a countable union of nowhere dense sets, thus it can't be locally compact. – t.b. Jun 16 '11 at 2:18
• @user10 FYI If $X$ is locally compact Hausdorff, then $X$ is a baire space. – user38268 Jul 9 '12 at 13:51
• in last line there is a mistake subsequence converges to z as limit miust be same as that of sequence – user104167 Oct 29 '13 at 23:53

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.