# A problem on equivalent metrics and equivalence classes

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.

• 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... – Gautam Shenoy Jun 22 '13 at 14:12
• Yes I was considering the metric spaces – user83369 Jun 22 '13 at 14:14
• @GautamShenoy $X$ is just a set and any set allows a metric. – Tobias Kildetoft Jun 22 '13 at 18:35
• Which of the many concepts of equivalence of metrics are you using? – Chris Eagle Jun 23 '13 at 9:52
• @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 $d_2 open$} – user83369 Jun 23 '13 at 13:36

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 $X$.
• $X$ 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 $X$. – Tobias Kildetoft Jun 22 '13 at 18:38