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I want to show that with p-adic ultrametric : $|.|_{p}=p^{-ord(.)}$ where . is a integer and p is a prime , with $|0|_{p}=0$ if we have balloons $B(a,R)=\{z\in \mathbb{Z}||z-a|_{p}\le R\}$ $B(b,R)=\{z\in \mathbb{Z}||z-b|_{p}\le R\}$

then they can never overlap.

Proposition: They can never overlap without being the same.

Assume we have two different points y and y' in one ball, if $y\in B(a,R)$ with the strong triangle inequality $|a+b|_{p}\le sup\{|a|_{p} ,|b|_{p}\}$

now set: a:=z-y' and b:= y'-y so we get : |z-y|_{p}\le sup\{|z-y'|_{p}. |y'-y|_{p} \}

So that means |z-y|_{p} \le R \Leftrightarrow |z-y'|_{p}\le R e.g. every point of the ball is a center point so the balloons can never overlap without being the same.

Is this a proof for the proposition? (I never made use of the second balloon...)

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Your proof is correct and usually the point where most people would stop, but you can complete the proof explicitly to make it clearer how the second balloon is related:

You have shown that for all y, y' \in B(a,R), you have B(a,R) = B(y,R)=B(y',R). Now suppose there is some intersection between the two balloons, say at $z$. Then since $z$ is in the first ball, by the previous result $ B(a,R) = B(z,R).$ Similarly, since $z$ is in the second balloon, $B(b,R) = B(z,R)$, and thus $B(a,R) = B(b,R)$ so the balloons are the same.