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 by virtue of Lemma 16.1 in Munkres 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 we have $$ a < q < b. $$
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 $d$ either interval are not in either of the two sets involved.
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 metrizable space, is also a Hausdorff space, by the discussion in the third paragraph of Sec. 21 in Munkres. So by Theorem 26.3 in Munkres $C$, being a compact subspace of the Hausdorff space $\mathbb{Q}$, 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 by Theorem 26.2 in Munkres the subspace $[c, d] \cap \mathbb{Q}$ is also compact (as a subspace of $C$); but since $C$ is a subspace of $\mathbb{Q}$, therefore $[c, d] \cap \mathbb{Q} = (c, d) \cap \mathbb{Q}$ is also a compact subspace of $\mathbb{Q}$.
Therefore every open covering of $[c, d] \cap \mathbb{Q}$ has a finite sub-collection that also covers $[c, d] \cap \mathbb{Q}$.
However, as $c$ and $d$ are irrational numbers [Please refer to (3) above.], so we now show that what has stated in the preceding paragraph is not true.
In what follows, the set $\mathbb{N}$ denotes the set of all the positive integers, namely, $1, 2, 3, \ldots$.
Let us consider the collection
$$ \left\{ \ \left( c + \frac{d-c}{n+2},\, d – \frac{d-c}{n+2} \right) \cap \mathbb{Q} \ \colon \ n \in \mathbb{N} \ \right\}. \tag{A} $$
This collection is an open covering of $[c, d]\cap \mathbb{Q}$ such that no finite subcollection of this collection can cover $[c, d] \cap \mathbb{Q}$; for if $$ \left\{ \ \left( c + \frac{d-c}{n_1+2}, \, d – \frac{d-c}{n_1+2} \right) \cap \mathbb{Q},\ \ldots, \ \left( c + \frac{d-c}{ n_r + 2 }, \, d – \frac{d-c}{n_r +2} \right) \cap \mathbb{Q} \ \right\} \tag{B} $$
is any finite sub-collection of the collection in (A) above, where $n_1, \ldots, n_r \in \mathbb{N}$ such that $n_1 < \cdots < n_r$, then
\begin{align}
& \ \ \ \bigcup_{j=1}^r \left[ \left( c + \frac{d-c}{n_j + 2}, \, d – \frac{d-c}{n_j +2} \right) \ \cap \ \mathbb{Q} \right] \\
&= \left[ \bigcup_{j=1}^r \left( c + \frac{d-c}{n_j + 2}, d – \frac{d-c}{n_j +2} \right) \right] \ \cap \ \mathbb{Q} \\
&= \left( c + \frac{d-c}{n_r + 2}, \, d – \frac{d-c}{n_r +2} \right) \ \cap \ \mathbb{Q} \\
&\subsetneqq [c, d] \cap \mathbb{Q}.
\end{align}
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?
Best Answer
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.