Showing no point in the rationals (as a subspace of $\mathbb{R}$) has a precompact neighborhood

Question

I'm studying for a Topology exam and I'm reviewing my professor's notes. In the notes, he writes that no point $p \in \mathbb{Q}$ has a precompact neighborhood $U$ (where we define $A \subset\mathbb{R}$ to be precompact if $cl(A)$ is compact). This does not seem right to me. Let $U=(-1,1) \cap \mathbb{Q}$. This is a neighborhood of $0 \in \mathbb{Q}$. And $cl(U) = [-1,1]$, which is compact. Am I missing something?

Answer 1

You have to take the closure in $\Bbb Q$, not in $\Bbb R$. And in the rationals the closure is $[-1,1] \cap \Bbb Q$ which is not compact, as there is a rational sequence in it (converging in $\Bbb R$ to $\frac{1}{\sqrt{2}}$, say) that has no convergence subsequence (in $\Bbb Q$ nor in $[-1,1] \cap \Bbb Q$).