# Ultrametric ball

I am trying to prove that any ball in ultrametric space is the union of some finite number of smaller balls. I can't find proof and don't unerstand how to do it. I thought to bring the explicit form of splitting into 2 balls, but I did not succeed. I understand that ultrametric balls don't intersect. We can introduce an equivalence relation and our ball will be a union of disjoint balls. $$\mathrm{B} \left(a, \, R \right) = \bigcup_{x \, \in \, \mathrm{B} \, \left(a, \, R \right)} \mathrm{B} \left(x, \, \varepsilon \right)$$ But how to prove that their number is finite? Thank you!

Added. (for ultrametric spaces, in which all closed ball is compact):

We know that all balls in the ultrametric space are clopen. Then we can introduce a binary relation so that $x \sim_{r} y$ if $y \in \mathrm{B} \left(x, \, r \right)$, where $\mathrm{B} \left(a, \, r \right) = \left\lbrace x \in \mathrm{X}| \, \mathrm{d}(x, a) \leq r \right\rbrace$. This is a binary relation, because the properties of reflexivity, symmetry, and transitivity are satisfied. Since every equivalence relation generates a partition of the space into disjoint equivalence classes, then $$\mathrm{X} = \bigcup_{x \, \in \, \mathrm{X}} \mathrm{B} \left(x, \, r \right)$$ for any $r$. So any ball will be a union of disjoint balls, i.e $$\mathrm{B} \left(a, \, R \right) = \bigcup_{x \, \in \, \mathrm{B} \, \left(a, \, R \right)} \mathrm{B} \left(x, \, r \right).$$ If any ball is compact, then in each of its open covers has a finite subcover. In this space balls are clopen, then in our cover of balls has finite subcover. Thus any ball is the union of a finite number of smaller balls. Is it correctly?

• So you are not really talking about a property of ultrametric spaces. you are talking about a property of compact sets, ultrametric or not. – GEdgar Apr 7 '17 at 22:34
• In ultrametric spaces we can introduce this relation. In other space this does not work – Ann Apr 7 '17 at 22:40
• You do not really need that equivalence relation. All you need is a metric space in which all bounded sets are totally bounded. Without that property, having an ultrametric does not help. WIth that property, you do not need an ultrametric. For example, any metric subspace of a Euclidean space will do. – Niels J. Diepeveen Apr 7 '17 at 23:05
• That is, we can prove that any balls consist of smaller ones in any metric space, where all bounded sets are totally bounded? – Ann Apr 7 '17 at 23:11
• Sorry, I had overlooked that you wanted an open ball to be equal to the union of finitely smaller balls, rather than just contained in the union. To get equality, I think you do need an ultrametric. I will try to add a proof to my answer. – Niels J. Diepeveen Apr 8 '17 at 2:00

It is not true. To see this, consider the metric on the positive integers where $d(m, n) = d(n, m) = 1 - 2^{-m}$ when $m > n$. This is an ultrametric because we have $d(m, n) = d(m, p) > d(n, p)$ whenever $m > n > p$.

The open unit ball about any point is the whole space, but for any radius $0 < r < 1$ we have $B(n, r) = \{n\}$ for all $n > -\log_2 (1-r)$. It follows that every ball smaller than a unit ball is a finite set, so no finite number of them can cover a unit ball.

Proof: For a given open ball $B_0 = B(x, r)$ let $C = \{B(y, r/2) \mid x \in B_0\}$. By the above property, $C$ is a partition of $B_0$. On one hand, $C$ is a maximal cover of $B_0$ by balls of radius $r/2$, since it contains such a ball about every point of $B_0$. On the other hand, $C$ is a minimal cover because it is a partition. Hence $C$ is the only cover of $B_0$ by balls of radius $r/2$ and since total boundedness of $B_0$ implies that there is such a cover that is finite, $C$ must be finite.
Note: If "smaller ball" is interpreted in the stricter sense of "ball that is a proper subset", then we need the extra assumption that the space has no isolated points and $C$ can be defined as $\{B(y, \delta(B_0)/2 \mid x \in B_0 \}$.