Let's say I have $A=B_{r_{1}}(x)$ and $B=B_{r_{2}}(y)$ and the metric $D(a, c) \leq \max\big[D(a, b) \;,\; D(b, c)\big]$ (ultrametric ). I suppose what I want to prove is: $A \cap B\neq \emptyset \implies A \subset B$. I am not sure where to even start, is it something to do with the triangle inequality? Thanks!
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Suppose that $z \in B_r(x) \cap B_s(y)$ for some $r,s >0$ and $x,y \in X$. I'll denote the ultrametric by $d$. One of the radii $r,s$ must be the smallest, let's assume WLOG that $r \le s$, say.
Let $u \in B_r(x)$. We'll try to show that $u \in B_s(y)$, which will show the inclusion $A \subseteq B$.
So we know that $d(u,x) < r$ and also (from the intersection) that $d(u,z) <r$ and $d(y,z) < s$.
Now we use the strengthened triangle inequality via $z$:
$$d(u,y) \le \max(d(u,z), d(y,z)) < s$$
(if the max is $d(u,z)$, it's $ <r \le s$, and if it's $d(y,z)$ it's also $<s$, so in both cases it is $<s$). This says exactly that $u \in B_s(y)=B$ so we're done. We get $B \subseteq A$ if $s \le r$ happens to be the case.
It's just following the definitions and applying the one distinguishing feature of the ultrametric.
Henno Brandsma
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Here's a potential approach you can take:
First, prove the surprising fact that every point in an ultrametric ball is a center of that ball. In other words, if $B_{r}(z)$ is an arbitrary ultrametric ball in your space, prove that for any $z' \in B_{r}(z)$, we have $B_{r}(z') = B_{r}(z).$
To do this, you can pick an arbitrary point $p \in B_{r}(z)$ and use the ultrametric inequality to show that $p$ is also in $B_{r}(z')$. This gives you one inclusion. Then, swapping the roles of $B_{r}(z)$ and $B_{r}(z')$ gives you the other inclusion, and thus equality.
Once you've shown this fact, use the fact that $A \cap B$ is nonempty to rewrite both $A$ and $B$ as ultrametric balls with the same center. It should then be clear that at least one of the inclusions $A \subseteq B$, $B \subseteq A$ holds.
ckefa
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