Visualizing Balls in Ultrametric Spaces

Let’s keep things simple by considering only the balls $B(x,2^{-n})$ for $x\in X$ and $n\in\Bbb N$: after all, these are enough to give us a base for the topology. Let

$$\mathscr{B}=\{B(x,2^{-n}):x\in X\text{ and }n\in\Bbb N\}\;,$$

and for each $n\in\Bbb N$ let

$$\mathscr{B}_n=\{B(x,2^{-n}):x\in X\}\;,$$

so that $\mathscr{B}=\bigcup_n\mathscr{B}_n$.

The first thing to notice is that each $\mathscr{B}_n$ is a partition of $X$ into clopen sets. Moreover, if $m\ge n$, the partition $\mathscr{B}_m$ is a refinement of the partition $\mathscr{B}_n$: for each $B\in\mathscr{B}_m$ there is a unique $B\,'\in\mathscr{B}_n$ such that $B\subseteq B\,'$. In fact, if $B=B(x,2^{-m})$, then $B\,'=B(x,2^{-n})$. Conversely, if $B\in\mathscr{B}_n$ and $m\ge n$, $\{B\,'\in\mathscr{B}_m:B\,'\cap B\ne\varnothing\}$ is a clopen partition of $B$. In fact, if $B=B(x,2^{-n})$, then $\{B\,'\in\mathscr{B}_m:B\,'\cap B\ne\varnothing\}=\{B(y,2^{-m}):y\in B(x,2^{-n})\}$.

What all of this says is that $\langle\{X\}\cup\mathscr{B},\supseteq\rangle$ is a tree of height $\omega$, with root $X$ and levels $\{X\}$, $\mathscr{B}_0$, $\mathscr{B}_1$, and so on. Each point of $X$ corresponds to a branch through the tree, though if the space is not complete, there will also be branches that don’t correspond to any point of the space. For a relatively familiar example, let $X=\{0,1\}^{\Bbb N}$, where $\{0,1\}$ has the discrete topology. Points of $X$ are infinite sequences of zeros and ones. If $x=\langle x_n:n\in\Bbb N\rangle$ and $y=\langle y_n:n\in\Bbb N\rangle$ are distinct points of $X$, let $m(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}$. We can now define a metric $d$ on $X$ by

$$d(x,y)=\begin{cases} 0,&\text{if }x=y\\ 2^{-m(x,y)},&\text{otherwise}\;. \end{cases}$$

This is an ultrametric, and the space is homeomorphic to the familiar middle-thirds Cantor set. If $p_0$ and $p_1$ are the sequence of all zeros and the sequence of all ones, respectively, then $\mathscr{B}_0=\{B(p_0,1),B(p_1,1)\}$; $B(p_0,1)$ contains every sequence whose first term is $0$, and $B(p_1,1)$ contains every sequence whose first term is $1$. For simplicity let me call these two sets simply $B(0)$ and $B(1)$, respectively. Now let $B(00)$ be the set of all sequences that begin with $\langle 0,0\rangle$, $B(01)$ the set of all sequences that begin with $\langle 0,1\rangle$, and so on. Then it’s not hard to check that $\mathscr{B}_1=\{B(00),B(01),B(10),B(11)\}$, where $B(00)$ and $B(01)$ partition $B(0)$, and $B(10)$ and $B(11)$ partition $B(1)$. So far, then, we have the following tree of sets:

This particular tree base happens to be binary: each set splits in two at the next level. (And if this reminds you of the usual construction of the Cantor set, that’s good: it should.)

Suppose that we kept only the sequences that are eventually zero; call this subset $D$. $D$ is dense in $X$, so every member of $\mathscr{B}$ contains a point of $D$. Thus, if $\mathscr{B}_D=\{B\cap D:B\in\mathscr{B}\}$, then $\mathscr{B}_D$ is a base for the subspace topology on $D$ that also forms the complete binary tree of height $\omega$. The difference is that $D$ is countable, so most of the branches through the tree no longer correspond to points of $D$.

For another example, the space of irrational numbers is well-known to be homeomorphic to the product space $X=\Bbb N^{\Bbb N}$, where $\Bbb N$ has the discrete topology. Points of $X$ are infinite sequences of non-negative integers, and a metric $d$ can be defined on $X$ exactly as I defined $d$ above on $\{0,1\}^{\Bbb N}$. This is again an ultrametric; it’s equivalent to the usual metric but obviously not the same. We can look at the same base of sets of the form $B(x,2^{-n})$, and again we get a tree in which each level is a clopen partition of the space, but this tree splits countably infinitely many ways at each node instead of just two ways.

We can do this for any ultrametric space, and I find it extremely helpful to picture the base in terms of the resulting tree and the associated finer and finer partitions of the space.

This answer to an earlier question may also be useful.

Don’t forget that a ball is a coset of a subgroup of the additive group of your field. And that if $B_1, B_2$ are any two balls centered at the origin, one of them contains the other. When you keep these facts in mind, your “counterintuitive” properties of balls become more than reasonable, they become necessary.