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Let $(X,d)$ be a metric space and define an equivalence relation $\sim$ on $X$. Then $$ d'([x],[y]):= \inf\{d(x',y'): x' \in [x],\, y' \in [y]\}, $$ may fail the triangle inequality, where $[x]$ is the equivalence class of $x\in X$ under $\sim$ (and similarly for $[y]$). However, does $d'$ define an ultrametric?

AB_IM
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1 Answers1

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Well, it can also fail to be a metric at all, as two distinct classes can have $d'$-distance $0$.

Also, it need not be an ultrametric (trivial example: start with a non-ultrametric $d$, and have trivial classes..).

Henno Brandsma
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