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Let $d$ be a metric on a group $G$ and define $d^{-1}$ by $d^{-1}(x,y)=d(x^{-1},y^{-1})$. Why do $d$ and $d^{-1}$ generate the same metric topology on $G$?

Let $g \in G$ and $\epsilon >0$. Let $B_d=B_d(g,\epsilon)=\{h \in G : d(g,h) < \epsilon\}$. I need to find a $d^{-1}$ ball inside of $B_d$ but am having trouble. Did I set the problem up correctly?

Any hints/solutions are appreciated here.

user41728
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    I suppose $G$ is a topological group in the topology induced by $d$? Otherwise, the conclusion would not necessarily hold. If $d$ makes $G$ a topological group, the inversion $x \mapsto x^{-1}$ is by definition of a topological group a homeomorphism. – Daniel Fischer Aug 30 '13 at 23:30
  • It must be a typo then, I will close the question soon. Thank you. – user41728 Aug 30 '13 at 23:35

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I believe you are missing an assumption. Your metric shouldn't be any old metric on the set of elements of $G$, it needs to respect the group structure.

For example, taking the inverse of an element should be continuous function with regards to $d$. How could you use that continuity to find a $\epsilon_2$ such that $B_{d^{-1}}(g,\epsilon_2)\subset B_d(g,\epsilon)$?

amueller
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