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Let $ X $ be a non empty set and $ \tau= \{d\mid d$ is a metric on $X\}$ Define the relation $\sim $ on $\tau$ by $ d \sim d' $ iff $ d $ and $ d'$ are equivalent metrics on $X$. Show that $\sim $ is an equivalence relation on $\tau $. Identify equivalence classes.

I could prove that it is an equivalence relation but I couldn't understand how to identify equivalence classes? Any help would be very much appreciated.

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  • $\begingroup$ What if X is a space that is not metrizable? Then $\tau$ would be empty. Perhaps you are assuming X to be a metric space... $\endgroup$ Jun 22, 2013 at 14:12
  • $\begingroup$ Yes I was considering the metric spaces $\endgroup$
    – user83369
    Jun 22, 2013 at 14:14
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    $\begingroup$ @GautamShenoy $X$ is just a set and any set allows a metric. $\endgroup$ Jun 22, 2013 at 18:35
  • $\begingroup$ Which of the many concepts of equivalence of metrics are you using? $\endgroup$ Jun 23, 2013 at 9:52
  • $\begingroup$ @ChrisEagle The definition given was two $d_1 d_2 $ metrics are equivalent if $\tau_1 $= $\tau_2$ where $\tau_1$= {$ G \subseteq X | G $is $d_1$ open} and $\tau_2 $={$ G \subseteq X $| G is $d_2 open$} $\endgroup$
    – user83369
    Jun 23, 2013 at 13:36

1 Answer 1

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If $X$ is a metric space,you can determine equivalent classes of this equivalent relation with metrizable topologies on $X$ Because,equivalent metrics generate same topology on $X$.

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  • $\begingroup$ Can you please show a way of finding an equivalence class? $\endgroup$
    – user83369
    Jun 22, 2013 at 16:43
  • $\begingroup$ $X$ is just a set, neither a metric space nor any other sort of topological space. But yes, one can identify the equivalence classes with the metrizable topologies on $X$. $\endgroup$ Jun 22, 2013 at 18:38
  • $\begingroup$ @TobiasKildetoft Can you tell me how to find the equivalence classes in this please? $\endgroup$
    – user83369
    Jul 2, 2013 at 12:12

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