Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove that $\mathbb{Q} \subset \mathbb{R}$ is not locally compact directly? That is, how to construct a cover of an arbitrary neighborhood (e.g. $[0, 1] \cap \mathbb{Q}$) that does not have a finite subcover?

share|cite|improve this question
possible duplicate of Rationals are not locally compact and compactness – t.b. Jul 6 '11 at 6:46
up vote 9 down vote accepted

Let $X:= \mathbb Q \cap [0,1]$. I'll show that if $U \subset X$ then $U$ doesn't have compact closure. Since $X$ is hausdorff this shows that it's not locally compact. It suffices to show that for any epsilon an open ball in X doesn't have compact closure. Given $x \in X$ and some $\epsilon>0$ pick an irrational $y \in B(x,\epsilon)$. Set $a=x-\epsilon$ and $b=x+\epsilon$, then the following is an open cover of $B(x,\epsilon)$ with no finite subcover:

$$\mathcal U := \{ ( (a,y-1/n)\cup(y+1/n,b) ) \cap \mathbb Q | n \in \mathbb N\}.$$

To see this note that $\mathcal U$ is a nested covering and that for any $n$ it doesn't cover $B(x,\epsilon)$.

share|cite|improve this answer
That's an incredible trick! 8) – Alexei Averchenko Jul 6 '11 at 5:03

Choose irrational point $i$ inside $[0,1]$. Take the class $O = \{[0, q) \cup (r, 1] \quad: \quad(q,r) \in \mathbb Q^2 \cap(0,1)^2,\quad q < i < r\}$ as your covering. Now we have a cover for the rationals in $[0,1]$ without a finite subcover, showing that the space is not compact.

For the sake of rigor:

$O$ is an open cover in the subspace topology relative to the compact set $[0,1]$.

$\cup O = [0,1]- \{i\}$, showing that we do, in fact, cover the rationals in $[0,1]$.

Consider the union of any finite subcover: $\cup_{k=1}^N[0,q_k)\cup(r_k,1]$ where $(q_k,r_k) \in \mathbb Q^2 \cap(0,1)^2,\quad\mathrm{and}\quad q < i < r$. This set is equivalent to $[0,\mathrm{max}_k\{q_k\})\cup(\mathrm{min}_k\{r_k\},1]$, where $k$ is understood to range from $1$ to $N$.

We conclude that any finite subcover misses the points in the interval $[\mathrm{max}_k\{q_k\},\mathrm{min}_k\{r_k\}]$

In general, choose any neighborhood in $\mathbb R$. We must be able to find a segment $(a,b)$ within the neighborhood. Repeat the argument using segment $(a,b)$ instead of $[0,1]$.

We conclude that any neighborhood of $\mathbb R$ has an open cover for which no subcover can cover the rationals in that neighborhood.

share|cite|improve this answer
For the sake of the readers, you should probably state that you are taking an enumeration $(q_n,r_n)$ of $(\mathbb{Q}\cap [0,i])\times (\mathbb{Q}\cap [i,1])$ and considering the family $U_n = [0,q_n)\cup (r_n,1]$. This family gives a countable open cover with no finite subcover. I suspect the down-votes have more to do with other users not understanding your argument (which is identical to the one Jacob Schlather gave in the accepted answer).... – Willie Wong Jul 6 '11 at 10:22
There is no need for an enumeration. That is, we don't need to know the cover is countable. Even if the cover is not countable, still it is an open cover that has no finite subcover. – GEdgar Jul 6 '11 at 14:34
I feel that Jacob's construction is a little more intricate and involves an increasing sequence of sets. As GEdgar said, we have no need to create a sequence and we consider the family of open sets [0,q)U(r,1] indexed by the pair of rationals (q,r). – Mark Jul 7 '11 at 3:29

If $\mathbb Q$ were locally compact, then for every rational q we would be able to find a 'hood (neighborhood) $U_q$ of q in which every sequence has a convergent subsequence. But if we take an irrational x in $U_q$ (which always exists, since the irrationals are dense in $\mathbb R $), then a sequence ${x_n}$ approximating $x$ will not have a convergent subsequence. (e.g., if the interval contains $\sqrt 2$ , there cannot be a sequence of rationals converging to $\sqrt 2$), because the rationals are not order-complete.

share|cite|improve this answer
@Theo: Sorry I could not rewrite my entry on A,B disjoint subsets of metric space, etc., since I could not access my old account. – gary Jul 7 '11 at 4:24
@Theo: I just set up a formal account; hopefully won't have more troubles. – gary Jul 7 '11 at 4:32
I've merged gary's accounts (thanks for the flag, Theo). gary, registering your account should prevent this issue from happening further, so I recommend you do so. – Zev Chonoles Jul 7 '11 at 4:50
@All: seriously sorry for my thickness; I think this time I did register; I received a confirmation e-mail and everything. – gary Jul 7 '11 at 5:09
Yes, this looks good now :) I removed my earlier comments in order to reduce meta-noise. – t.b. Jul 7 '11 at 5:17

Let's use that if a space (here $[0,1]\cap\mathbb Q$) has an open infinite partition, then it is not compact since the partition is an open covering that has no subcovering at all .

Choose an increasing sequence of irrational numbers $0\lt a_1\lt \ldots \lt a_n\lt\ldots\lt \sqrt 2/2$ converging to $\sqrt 2/2$ and a decreasing sequence of irrational numbers $1\gt b_1\gt \ldots \gt b_n\gt\ldots\gt \sqrt 2/2$ also converging to $\sqrt 2/2$. Then contemplate the open partition (where $[[x,y]]$ means $[x,y]\cap\mathbb Q$)

$$[[0,1 ]]= [[0,a_1 ]]\sqcup [[a_1,a_2 ]] \sqcup \ldots \bigsqcup \; \;\; [[b_1,1 ]]\sqcup [[b_2,b_1 ]] \ldots $$

share|cite|improve this answer
+1 but i guess you mean $[0,a_1)$ and not $[0,a_1]$ and similarly for others... – Praphulla Koushik Feb 18 '15 at 17:49
Dear @Praphulla: please read my definition of $[[x,y]]$ carefully and notice that $[0,a]\cap \mathbb Q=[0,a)\cap \mathbb Q$ if $a\notin \mathbb Q$. – Georges Elencwajg Feb 18 '15 at 19:05
Oh yes yes.... I could not realize it at that time.... :) – Praphulla Koushik Feb 18 '15 at 20:39

Q is any set from negative infinite up to positive infinity,thus it is not bounded from both side and also it is not closed Q is not compact because it fail heines therom, Heines therom stated as any set S is said to be compact if and only if the set is both bounded and closed.

share|cite|improve this answer
You misread the question. The question is not whether or not $\mathbb Q$ is compact but rather locally compact. – Asaf Karagila Dec 6 '12 at 12:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.