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I am trying to proove that if $(X,d)$ is a metric space with this propery: $$y\in B(x,r)\Rightarrow B(x,r) = B(y,r)$$ then the metric is an ultra metric.

This is my proof:

Let $x,y,z\in X$ be some arbitrary elements in the space. Let $r=\max\{d(x,y),d(y,z)\}$ and also let $\varepsilon >0$ be some positive number. notice that $y\in B(x,r+\varepsilon)$ and also $z\in B(y,r+\varepsilon)$. By the given property we can conclude that $$B(x,r+\varepsilon)=B(y,r+\varepsilon)$$ Then $z\in B(x, r+\varepsilon )$, so: $$d(x,z)<r+\varepsilon=max\{d(x,y),d(y,z)\}+\varepsilon$$

and this is true for all $\varepsilon >0$, so when $\varepsilon \rightarrow 0$ we get: $$d(x,z)\leq max\{d(x,y),d(y,z)\}$$ So $d$ is an ultra metric. $\blacksquare$

Is this a legal proof? Is the last transition correct? I am 99% precent sure but still I just want to get some clarification.

Tair Galili
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1 Answers1

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Yep, looks good. The last transition is valid, and is a standard trick in analysis. If you want to prove it, then prove it indirectly. Suppose that $$d(x, z) > \max\{d(x, y), d(y, z)\},$$ and consider the particular $\varepsilon$, $$\varepsilon = d(x, z) - \max\{d(x, y), d(y, z)\} > 0.$$

user790072
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