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
    $\begingroup$ 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? $\endgroup$
    – t.b.
    Jun 16, 2011 at 2:15
  • $\begingroup$ @Theo Buehler: How would you apply Baire? sorry I meant $X \times Y$ Lindelöf. $\endgroup$
    – user10
    Jun 16, 2011 at 2:17
  • 2
    $\begingroup$ 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. $\endgroup$
    – t.b.
    Jun 16, 2011 at 2:18
  • $\begingroup$ @user10 FYI If $X$ is locally compact Hausdorff, then $X$ is a baire space. $\endgroup$
    – user38268
    Jul 9, 2012 at 13:51
  • $\begingroup$ in last line there is a mistake subsequence converges to z as limit miust be same as that of sequence $\endgroup$
    – user104167
    Oct 29, 2013 at 23:53

1 Answer 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.


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