# Why are closed balls in the $p$-adic topology compact?

I was skimming through some of this paper Measurable Dynamics and Simple $p$-adic Polynomials out of curiosity.

A few pages in, the author claims that closed balls are both open and compact sets in the $p$-adic topology on $\mathbb{Q}_p$. I have not been able to verify this, and would like to understand it before proceeding further.

For clarity, let a closed ball $B(x,r)=\{y\in\mathbb{Q}_p:|x-y|_p\leq p^{-r}\}$. Then why is $B(x,r)$ both open and compact in the $p$-adic topology?

I have been able to show that since $\mathbb{Q}_p$ has a non-Archimedian absolute value, then any point inside the ball can be taken to be the center, and from that, that any two closed balls $B(x,r)$ and $B(y,s)$ are either disjoint or one is contained in the other.

Thanks!

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Note that $B(x,r)$ is the same as the open ball arouns $x$ of radius $p^{-(r-0.5)}$, so $B(x,r)$ is open. – Srivatsan Nov 3 '11 at 3:04
Thanks for that, @Srivatsan. – Buble Nov 3 '11 at 5:51
Since it is a closed subset of a complete metric space, it suffices that you show that a closed ball is totally bounded. Show that your closed ball can be covered by finitely many closed balls of identical, arbitrarily small radius (the number of balls needed only depends on the radius of the large ball and the radius you required of the small ones). I think you can achieve this by taking some ball with small enough radius and taking translations of it. – Mark Nov 3 '11 at 11:49

It’s sufficient to look at balls centred at $0$. It’s easiest to see if you represent the $p$-adic rationals in the form $$x = \sum_{k\ge k_0}\frac{x_k}{p^k},$$ where the $x_k$ are the $p$-adic digits of $x$, so that $$B(0,r) = \left\{\sum_{k\ge r} \frac{x_k}{p^k}:\forall k\bigg(x_k\in\{0,1,\dots,p-1\}\bigg)\right\}.$$

Now let $D = \{0,1,\dots,p-1\}$, and give $D$ the discrete topology; I claim that $B(0,r)$, as a subspace of $\mathbb{Q}_p$, is homeomorphic to the product space $D^\omega$, which is of course compact. (In fact it’s a Cantor set.) The homeomorphism is the obvious one: $$h:D^\omega\to B(0,r):\langle x_k:k\in\omega\rangle\mapsto \sum_{k\ge r}\frac{x_{k-r}}{p^k}.$$

Clearly $h$ is a bijection: $$h^{-1}:B(0,r)\to D^\omega:x=\sum_{k\ge r}\frac{x_k}{p^k}\mapsto \langle x_{k+r}:k\in\omega\rangle\;.$$

$B(0,r)$ has a base of clopen sets of the form $B(x,s)$, where $x\in B(0,r)$ and $s\ge r$. Fix such a $B(x,s)$, with $x=\sum_{k\ge r}\frac{x_k}{p^k}$. If $y=\sum_{k\ge r}\frac{y_k}{p^k}\in B(0,r)$, then $|x-y|_p = p^{-m}$, where $m=\min\{k\ge r:x_k\ne y_k\}$, so $y \in B(x,s)$ iff $m\ge s$. In other words, $$B(x,s) = \bigg\{y\in B(0,r):\min\{k\ge r:y_k\ne x_k\}\ge s\bigg\},$$ and therefore $$h^{-1}[B(x,s)] = \bigg\{\langle y_{k+r}:k\in\omega\rangle:(\forall k<s-r) \big[y_{k+r}=x_{k+r}\big]\bigg\},$$ which is a basic open set in the product $D^\omega$. Thus, $h$ is a continuous bijection. The sets of the form $h^{-1}[B(x,s)]$ are a base for $D^\omega$, so the same calculation shows that $h^{-1}$ is continuous and hence that $h$ is a homeomorphism.

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Thanks Brian, this is really clever. Is there a particular reason why it's fine to only look at balls centered at $0$? – Buble Nov 3 '11 at 23:44
@Buble: The balls of a given radius $r$ are all just translates of $B(0,r)$: $B(x,r)=x+B(0,r)=\{x+y:y\in B(0,r)\}$. – Brian M. Scott Nov 3 '11 at 23:49
Ah, I think this was what Mark Schwarzmann was suggesting. Again, thanks! – Buble Nov 3 '11 at 23:50
I don't think you want reciprocals of prime powers if these are to be expansions of $p$-adic numbers. (Even though the valuations will be reciprocals.) – anon Oct 5 '12 at 14:58

It's probably easiest to show sequential compactness... that every sequence of points ion the ball has a convergent subsequence (This is equivalent to compactness for metric spaces).

This can be done very much as you would show using decimal expansions that every sequence of real numbers between 0 and 1 has a convergent subsequence, using Cantor diagonalization for example.

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Thanks for the idea, Zarrax. – Buble Nov 3 '11 at 23:50