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I have following problem:

Let X be any set and let $c> 0$. We put $d(x,x)=0$, for every $x$ $\in$ X and if $x\neq y$ x,y $\in$ X, we put $d(x,y)=c$. Prove that $(X,d)$ is ultrametric space.

Showing the first condition from the ultrametric space is trivial, since it is basically written above. But can anyone help me, how to prove the second and the third one ?

callculus42
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1 Answers1

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You have $d(x,y) \geq 0$ and $d(x,y) = 0 \Leftrightarrow x=y$ by definition.

Since for every $x,y \in X$, $x \neq y$ we have $d(x,y) = c $ we compute $d(x,y) - d(x,y) = c - c =0$ in other words $d(x,y) = d(y,x)$.

For the ultra metric triangle inequaliy: Say $x \neq y$, $y \neq z$ and $x \neq z$, then we have $d(x,z) = c = \max \{ c, c \} = \max \{d(x,y),d(y,z)\}.$ You can check the other cases for yourself.

Riquelme
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