I know this question has been asked to death, but I wish to prove that $\mathbb{Q}$ is not locally compact with a use of a lemma I have not seen used in the other proofs.


Given $(X, \mathfrak{T})$ Hausdorff, then it is locally compact iff every point is contained in an open set with compact closure

My attempt:

Since $\mathbb{Q}$ is a subspace of $\mathbb{R}$, therefore $\mathbb{Q}$ is Hausdorff given that Hausdorffness is arbitrarily hereditary.

To show that $\mathbb{Q}$ is not locally compact, we wish to produce a point $x \in \mathbb{Q}$ that is contained in an open set without a compact closure.

Let $U = (a,b) \subset \mathbb{R}$. Then $U \cap \mathbb{Q}$ is an open set in the subspace topology. Pick $x \in U \cap \mathbb{Q}$, then $x \in U \cap \mathbb{Q} \subseteq \overline {U \cap \mathbb{Q}}$. (Alarm bells: $\overline {A \cap B}$ might not be equal to $\overline A \cap \overline B$)

Now we need to show that $ \overline {U \cap \mathbb{Q}}$ is not compact...however intuitively this set feels like $[a,b] \subset \mathbb{R}$...


Does anyone see how to resolve this? Thanks so much

Also there is another answer using a much simpler characterization Why Q is not locally compact, connected, or path connected? Can we see that the lemma I am using and the one given https://math.stackexchange.com/a/650270/174904 is equivalent?

  • $\begingroup$ All compact sets in $\mathbb Q$ have empty interior. $\endgroup$ – user42761 Aug 9 '16 at 9:37

You need to find a point so that none of its open neighborhoods has compact closure.

Let's consider $0$. Suppose $U$ is an open neighborhood of $0$ (in $\mathbb{Q}$) that has compact closure. Then $U\supseteq(-1/n,1/n)\cap\mathbb{Q}$, for some integer $n>0$. Therefore the closure of $(-1/n,1/n)\cap\mathbb{Q}$ (in $\mathbb{Q}$) is compact as well.

Let $C$ be the closure (in $\mathbb{Q}$) of $(-1/n,1/n)\cap\mathbb{Q}$. Take your favorite irrational number $r\in(0,1/n)$ and an increasing sequence $(r_n)$ in $(0,1/n)\cap\mathbb{Q}$ that converges to $r$. Then $(r_n)$ is a sequence in $C$, but no subsequence is convergent (in $\mathbb{Q}$), a contradiction.


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