3

A metric space $(X,d)$ where $X$ is a set and $d:X\times X\rightarrow\mathbb{R}_{\geq 0}$ is a distance function satisfying the usual axioms for a distance function, together with the strict triangle inequality $d(x,z)\leq\mathrm{max}(d(x,y),d(y,z))$, is called an ultrametric space.

A ball in an ultrametric space $(X,d)$ is a subset of $X$ of the form $\{x\in X\mid d(x,a)<\epsilon\}$ or of the form $\{x\in X\mid d(x,a)\leq\epsilon\}$, for a fixed $a\in X$ and a fixed positive real number $\epsilon>0$.

It is claimed in Peter Schneider's "$p$-Adic Lie Groups" in Part I, Chapter 1, on page 6, that it is possible to have a descending sequence $B_{1}\supseteq B_{2}\supseteq \ldots \supseteq B_{n}\supseteq \ldots$ of balls in a complete ultrametric space with empty intersection. (If the limit of the diameters is zero then this forces the intersection to be nonempty.) I was wondering if anyone could give an example where the intersection is empty.

Rupert
  • 928

2 Answers2

2

The idea for this example was partly suggested by Abstraction’s answer.

Define a function $d:\Bbb N\times\Bbb N\to\Bbb R$ by

$$d(m,n)=\begin{cases} 0,&\text{if }m=n\\ 1+2^{-\min\{m,n\}},&\text{if }m\ne n\;; \end{cases}$$

it’s straightforward to verify that $d$ is a discrete ultrametric on $\Bbb N$. (E.g., if $\ell<m<n$, then $d(\ell,m)=d(\ell,m)=1+2^{-\ell}>1+2^{-m}=d(m,n)$, so any one of the three distances is at most the larger of the other two.) The only Cauchy sequences are the eventually constant sequences, which certainly converge, so $d$ is complete.

If $k,\ell,n\in\Bbb N$ with $k\ne\ell$, then $d(k,\ell)=1+2^{-\min\{k,\ell\}}<1+2^{-n}$ iff $\min\{k,\ell\}>n$, so for each $n\in\Bbb N$ we have

$$B\left(n+1,1+2^{-n}\right)=\{k\in\Bbb N:k\ge n+1\}\;.$$

For $n\in\Bbb N$ let $B_n=B\left(n+1,2^{-n}\right)$; clearly $B_n\supseteq B_{n+1}$ for each $n\in\Bbb N$, and $\bigcap_{n\in\Bbb N}B_n=\varnothing$.

Community
  • 1
Brian M. Scott
  • 616,228
1

Here is an idea about how this example can be built (it relies on specific space properties which seem to be possible, but I can't provide an example of such).

Let $(X,d)$ be separable complete ultrametric space, with $\{x_\imath\}$ being dense subset of $X$. Let $\{r_\imath\}$ be a descending real sequence with non-zero limit for which holds the condition $$\forall \imath \in \mathbb{N}, \forall a \in X, \exists b \in X : r_{\imath+1} < d(a,b) \le r_\imath$$ Note that most well-known ultrametric spaces don't have such sequence. When they don't even have distance values other than values of an infinitely-small sequence (such as with $p$-adic metric on $\mathbb{Q}$ where you only have values $1 \over p^n$), the claim is obviously false.

Now let $B_\imath$ be $B(c_\imath,r_\imath)$ where $c_{\imath+1} \in X$ is chosen so that for $x_{\imath+1} \notin B_\imath$, $c_{\imath+1}=c_\imath$ and for $x_{\imath+1} \in B_\imath$, $c_{\imath+1} : r_{\imath+1} < d(c_{\imath+1},x_{\imath+1}) \le r_\imath$.

1) $\{B_\imath\}$ is a descending sequence of balls (since $d(c_\imath,c_{\imath+1}) \le r_\imath$, $B_{\imath+1} \subseteq B_\imath$).

2) Intersection of $\{B_\imath\}$ is empty: $B_\imath \cap B(x_\imath,\lim_{n \to \infty} r_n) = \emptyset$ (since $d(x_\imath,c_\imath) > r_\imath$ and $r_\imath > \lim_{n \to \infty} r_n$), and $\forall a \in X, \exists \imath \in \mathbb{N} : a \in B(x_\imath,\lim_{n \to \infty} r_n) \implies a \notin B_\imath$.

Abstraction
  • 2,482