# Visualizing Balls in Ultrametric Spaces

I've been reading about ultrametric spaces, and I find that a lot of the results about balls in ultrametric spaces are very counter-intuitive. For example: if two balls share a common point, then one of the balls is contained in the other.

The reason I find these results so counter-intuitive is that I can easily picture "counter-examples," the problem being that these "counter-examples" are balls in Euclidean space.

My issue is not that I cannot prove these results. My issue is that I don't know how to think about/picture balls in ultrametric spaces, which makes it more difficult for me to actually come up with the proofs. Hence, does anyone have any hints as to how to think about/picture 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.

Think of the Cantor set and its basic closed-and-open intervals, pictured below. Note that for any two such intervals, they either do not intersect, or one is contained in the other.

Some of the other answers here are technically good, but (for me at least) not very intuitive.

So here's what I suggest. Think about ultrametric balls as being infinitely large. Or somewhat more precisely, as containing everything that is finitely close.

As you point out, two ultrametric balls that intersect must be nested. As a special case of this, you can also show that two ultrametric balls with the same radius that intersect must be identical. Pretty weird, right? But here's a simple example. Let $$d_D(x, y) = \left\{ \begin{array}{ccc} 1 & & x \neq y \\ \\ 0 & & x = y \end{array} \right.$$

be the discrete metric on $\mathbb{R}$. This is an ultrametric, but don't take my word for it -- make sure you can prove it!

Now let's see what balls in this space look like. If you play with it, you'll see that they're either singletons, or else they're all of $\mathbb{R}$. For example, for any $x$,

• $B\left(x, \tfrac{1}{2}\right) = \{ x \}$
• $B\left(x, \tfrac{3}{2} \right) = \mathbb{R}$

This is a simple example of ultrametric balls containing everything that's "finitely close". But why would this be true?

Let's say we wanted an ultrametric, but we started off with the usual metric on $\mathbb{R}$, given by $d(x, y) = \vert x - y \vert$. If we wanted to mess around with this until we had an ultrametric, we could start by thinking about the balls in the metric space $(\mathbb{R}, d)$, and how to modify them so that they'll satisfy the "ultrametric ball" property you mentioned in your post.

For any $x \in \mathbb{R}$, the balls $B(x, 2)$ and $B(x+1, 2)$ intersect but are not identical. But as we mentioned before, in an ultrametric, intersecting balls with the same radius must be the same. So this means that we need to make sure our ultrametric balls $\widetilde{B}$ satisfy $$\widetilde{B}(x, 2) = \widetilde{B}(x+1, 2)$$

for all $x$. So the ball centered at $x$ is the same as the ball centered at $x+1$, which (by the same logic) has to be the same as the ball centered at $x+2$, and all the way on up (and also on down). The only possible choice for our modified ball is $$\widetilde{B}(x, 2) = \mathbb{R}$$

This is why ultrametric balls have to be "infinite". Say you start with a regular metric and want to turn it into an ultrametric. Then imagine drawing any collection of circles (balls under the regular metric), all of the same radius, and where every circle you draw has to overlap (at least a little) with one of the other circles you've drawn previously. Whatever your ultrametric ball ends up being, it has to include any of these "circle sequences" that you could possibly draw -- because that's what the ultrametric ball property says.

To finish tying this up, let's go back to the case of our regular metric $d$ on $\mathbb{R}$, and the fact that any ultrametric ball has to contain all "circle sequences" of that radius drawn under the original metric. That means that, in $\mathbb{R}$, there are only two types of ultrametric ball you can end up with.

• Singleton sets, where $\widetilde{B}(x, r) = \{x\}$. If your original ball $B(x, r)$ was also just a single point, the "circle sequences" you draw are just the same point over and over.
• The entire space, where $\widetilde{B}(x, r) = \mathbb{R}$. If the original ball $B(x, r)$ was larger than a single point, "circle sequences" can cover all of the real line.

This means if you're going to create an ultrametric that gives you the modified $\widetilde{B}$ balls, it can only take on two different values. Since the distance of $0$ is reserved for when the point are the same, there's only one value left for every other combination of points, which people usually choose to be 1. This gives us the ultrametric $d_D$ that we started with.

Finally, this perspective of "ultrametric balls are infinite" might have given you the idea that ultrametric balls always have to be the whole space. Of course this isn't true, or else people wouldn't study them! This is why a better term than "infinite" (although less catchy) is "everything finitely close", or most accurately, "everything you can get from circle sequences".

I'll leave you with an example to play with. Let's consider the set $$X = \left\{ (x, y) \in \mathbb{R}^2 \; : \; y \in \mathbb{Z} \right\}$$

of all horizontal lines in the plane with integer $y$-values. But don't think about this as having the same metric that we usually have on $\mathbb{R}^2$. Here's our new metric.

$$\rho\Big( (x_1, y_1), (x_2, y_2) \Big) = \left\{ \begin{array}{ccc} 2^{\max\{y_1, y_2\}} & & (x_1, y_1) \neq (x_2, y_2) \\ \\ 0 & & \textrm{else} \end{array} \right.$$

Play around with this, and convince yourself that it's a metric, and in fact an ultrametric. (Can you prove it?) What do balls under this ultrametric look like? (Hint: they're "infinite", but they aren't the whole space!)