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?

  • $\begingroup$ Or use that $\mathbb{Q}$ is not a Baire space. Locally compact Hausdorff spaces are Baire. And $\mathbb{Q}$ is clearly Hausdorff. $\endgroup$ – Henno Brandsma Apr 12 '19 at 21:25
  • $\begingroup$ @HennoBrandsma thank you for your comment, but I wasn't aware of the concept of a Baire space. $\endgroup$ – Saaqib Mahmood Apr 13 '19 at 6:33
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $z_0=\max \{z<y:\Bbb Q \cap (-\infty,z)\in F\}.$

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


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