# Prob. 1, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: The subspace $\mathbb{Q}$ of $\mathbb{R}$ is not locally compact

Here is Prob. 1, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:

Show that the rationals $$\mathbb{Q}$$ are not locally compact.

My Attempt:

Here the topology on the set $$\mathbb{Q}$$ of rational numbers is the same as the subspace topology that $$\mathbb{Q}$$ inherits from the standard topology on the set $$\mathbb{R}$$ of real numbers.

As the standard topology on $$\mathbb{R}$$ has as a basis the collection of all the open intervals of the form $$(a, b)$$, where $$a, b \in \mathbb{R}$$ and $$a < b$$, so the subspace topology on $$\mathbb{Q}$$ has as a basis the collection of all the intersections $$(a, b) \cap \mathbb{Q}$$, where $$a, b \in \mathbb{R}$$ and $$a < b$$.

Let $$q$$ be any point of $$\mathbb{Q}$$. Let us suppose that $$\mathbb{Q}$$ is locally compact at $$q$$. Then there exists a neighborhood $$U$$ of $$q$$ in $$\mathbb{Q}$$ and a compact subspace $$C$$ of $$\mathbb{Q}$$ such that $$U \subset C. \tag{0}$$

As $$U$$ is open in $$\mathbb{Q}$$, so $$U = V \cap \mathbb{Q} \tag{1}$$ for some open set $$V$$ in $$\mathbb{R}$$.

Now as $$q \in V$$ and as $$V$$ is open in $$\mathbb{R}$$, so there exists an open interval $$(a, b)$$ on the real line such that $$q \in (a, b) \subset V. \tag{2}$$ Thus there exist real numbers $$a$$ and $$b$$ such that $$a < q < b.$$ Let us choose some irrational numbers $$c$$ and $$d$$ such that $$a < c < q < d < b. \tag{3}$$ Then by (2) above we have $$q \in (c, d) \subset (a, b) \subset V,$$ and then by (1) above we also have $$q \in (c, d) \cap \mathbb{Q} \subset (a, b) \cap \mathbb{Q} \subset V \cap \mathbb{Q} = U,$$ that is $$q \in (c, d) \cap \mathbb{Q} \subset U, \tag{4}$$ and also $$(c, d) \cap \mathbb{Q} = [c, d] \cap \mathbb{Q}, \tag{5}$$ because the endpoints $$c$$ and $$d$$ are not in either of the two intervals involved in the preceding identity.

Thus from (0), (4), and (5) above we also have $$q \in [c, d] \cap \mathbb{Q} \subset C. \tag{6}$$

Now $$C$$ is a compact subspace of $$\mathbb{Q}$$; moreover $$\mathbb{Q}$$, being a metric space, is also a Hausdorff space. So $$C$$ is also closed in $$\mathbb{Q}$$. Therefore using (6) above we can also conclude that $$[c, d] \cap \mathbb{Q} = \big( [c, d] \cap \mathbb{Q} \big) \cap C$$ is also closed in $$C$$.

Thus we have seen that $$[c, d] \cap \mathbb{Q}$$ is a closed subset of the compact space $$C$$. So $$[c, d] \cap \mathbb{Q}$$ is also compact.

But let us consider the collection $$\left\{ \ \big( c + \frac{d-c}{n+2}, d - \frac{d-c}{n+2} \big) \cap \mathbb{Q} \ \colon \ n \in \mathbb{N} \ \right\}.$$ This collection is an open covering of $$[c, d]\cap \mathbb{Q}$$ such that no finite subcollection of this collection can cover $$[c, d] \cap \mathbb{Q}$$, thus showing that $$[c, d] \cap \mathbb{Q}$$ or $$(c, d) \cap \mathbb{Q}$$ [Refer to (5) above.] is not compact.

Thus we have reached a contradiction. Therefore our supposition that $$\mathbb{Q}$$ is locally compact at $$q$$ is wrong. Hence $$\mathbb{Q}$$ is not locally compact at any point $$q \in \mathbb{Q}$$.

Is my proof correct in each and every detail? If so, then is my presentation clear enough too? If not, then where are the issues?

• Or use that $\mathbb{Q}$ is not a Baire space. Locally compact Hausdorff spaces are Baire. And $\mathbb{Q}$ is clearly Hausdorff. – Henno Brandsma Apr 12 '19 at 21:25
• @HennoBrandsma thank you for your comment, but I wasn't aware of the concept of a Baire space. – Saaqib Mahmood Apr 13 '19 at 6:33
• p 295 of Munkres defines them. He notes in example 2 that $\Bbb Q$ is not a Baire space and exercise 3 on p 299 asks to show a locally compact Hausdorff space is Baire. – Henno Brandsma Apr 13 '19 at 6:39

Your proof is both correct and clear.

I think that a simpler approach would consist in proving that there are sequences of elements of $$C$$ without convergent subsequences. That is easy, of course: you just take a sequence of elements of $$C$$ which converges (in $$\mathbb R$$) to an irrational number.

A briefer method: A nbhd of $$x$$ is a set $$N$$ such that there exists an open set $$U$$ with $$x\in U\subset N$$. Let $$x\in N\subset \Bbb Q$$ where $$N$$ is a nbhd of $$x$$ in the space $$\Bbb Q.$$ Then $$N$$ is not compact.

Proof: There exists $$r\in \Bbb R^+$$ such that $$\Bbb Q\cap (-r+x,r+x)\subset N.$$ There exists $$y\in (\Bbb R \setminus \Bbb Q)\cap (-r+x,r+x).$$

Let $$C_0=\{\Bbb Q\cap (-\infty,z): y>z\in \Bbb R\}.$$ Let $$C_1=C_0 \cup \{\Bbb Q\cap (y,\infty)\}.$$

Then $$\cup C_1=\Bbb Q$$ (because if $$q\in \Bbb Q \cap (-\infty,y)$$ then $$q\in \Bbb Q \cap (-\infty,(y+q)/2 )\in C_0\subset C_1 ,$$ so $$q\in \cup C_1)$$.

So $$C_1$$ is an open cover of $$N$$ in the space $$\Bbb Q.$$

Let $$F$$ be a finite subset of $$C_1.$$ Then:

(i). If $$F \cap C_0=\emptyset$$ then $$N\cap (-r+x,y)=\Bbb Q\cap (-r+x,y)$$ is a non-empty subset of $$N$$ that is not covered by $$F.$$

(ii). If $$F\cap C_0\ne \emptyset$$ then (as $$F$$ is finite) there exists $$z_0=\max \{z

Let $$z_1=\max (z_0,-r+x).$$ Then $$-r+x\le z_1 so $$N\cap (z_1,y)=\Bbb Q\cap (z_1,y)$$ is a non-empty subset of $$N$$ not covered by $$F.$$