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Given a metric space $(M,d)$ and $\sim$ an equivalence relation on $M$.

I wonder if the function $\widetilde{d}:M/_\sim \times M/_\sim \rightarrow \mathbb{R}_{\geq 0}$ defined by

$\widetilde{d}([x]_{\sim},[y]_{\sim}) = inf_{x \in [x]_{\sim}, y \in [y]_{\sim}} d(x,y)$

Is it a metric on $M/\sim$?

5xum
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jaogye
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  • Not in general. It works if you replace $\inf$ with $\sup$ though, at least if the space is bounded (so that $\tilde d$ does not take value $+\infty$) – Del Dec 05 '16 at 09:24

1 Answers1

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No, the triangle inequality typically fails. For instance, let $M=\mathbb{R}$ with the usual metric and let $x\sim y$ iff either $x=y$ or $\{x,y\}=\{0,1\}$. Then $\widetilde{d}([-1],[2])=3$ but $\widetilde{d}([-1],[0])=1$ and $\widetilde{d}([0],[2])=1$ since $[0]=[1]$.

It is also possible to have $\widetilde{d}([x],[y])=0$ even if $[x]\neq [y]$. For instance, just take any equivalence relation with an equivalence class $[x]$ that is not closed, and let $y$ be some point in the closure of $[x]$ but not in $[x]$.

Eric Wofsey
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