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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?

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

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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 $\mathbb{R}$.

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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.

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

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