# Example of a subspace of the p-adic numbers which is not locally compact

Let $$p$$ be a prime number, and let $$\mathbf{Q}_p$$ be the field of $$p$$-adic numbers together with its topology derived from the $$p$$-adic valuation. Can someone give me an example (together with an argument) of a subspace of $$\mathbf{Q}_p$$ which is not locally compact? Thank you in advance.

P.S. My motivation arises from the fact that given the field of real numbers $$\mathbb{R}$$ together with its euclidean topology, the subspace of rational numbers $$\mathbb{Q}$$ is not locally compact. Maybe the same happens if we complete $$\mathbb{Q}$$ with respect to the metric arising from the $$p-$$adic valuation?

• Any open or closed subset is going to inherit $\mathbf Q_p$'s local compactness
– D_S
Commented Nov 10, 2020 at 17:02

A subspace $$A\subseteq X$$ of a locally compact Hausdorff space $$X$$ is locally compact if and only if it is locally closed. That is if and only if $$A$$ is open in its closure $$\overline A$$.

Since $$\mathbb{Q}_p$$ is Hausdorff and locally compact the statement applies. The subspace $$\mathbb{Q}$$ is dense in $$\mathbb{Q}_p$$, but not open. Hence it cannot be locally compact in the subspace topology.

Note that the same arguement shows that $$\mathbb{Q}$$ is not locally compact in the subspace topology inherited from $$\mathbb{R}$$.

The metric space $$\Bbb{Z}$$ (with the $$p$$-adic metric) is covered by the collection of open sets $$U_n = n+p^{|n|+3} \Bbb{Z},n\in\Bbb{Z}$$ but no finite subcollection $$\bigcup_{n=-N}^N U_n$$ covers it, because $$U_n$$ covers $$p^{N-|n|}$$ residue classes modulo $$p^{N+3}$$, so $$\bigcup_{n=-N}^N U_n$$ covers at most $$\sum_{n=-N}^N p^{N-|n|}$$ residue classes modulo $$p^{N+3}$$, ie. not them all.

• Can you please explain the line $\text{$U_n$covers$p^{N-|n|}$residue classes modulo$p^{N+3}$}$ ?
– MAS
Commented Nov 10, 2020 at 18:12
• $a\in U_n$ iff $a\equiv n+p^{|n|+3} b\bmod p^{N+3}$ for some $b$. There are $p^{N-|n|}$ possible values for $b$. Commented Nov 10, 2020 at 18:29
• Is $|n|$ a absolute norm or $p$-adic norm ? I think both will work here
– MAS
Commented Nov 11, 2020 at 13:44
• Euclidean norm, I'm just saying that for $s \ge r$, $a+p^r \Bbb{Z}=\bigcup_{b=0}^{p^{s-r}-1} (a+p^r b+p^s\Bbb{Z})$ Commented Nov 11, 2020 at 13:49
• ok, thank you very much
– MAS
Commented Nov 11, 2020 at 14:26