My question arised while I was reading Kechris' "Classical Descriptive Set Theory", pp. $58$.
At row $14$ of that page, the author states that
If $d$ is a left-invariant complete metric on a metrizable topological group $G$, consider the new metric
$$\rho(x,y)=d(x,y)+d(x^{-1},y^{-1}).$$
It is easy to see that it is also compatible (but not necessarily left-invariant).
Question: Why is $\rho$ compatible with $d$?
Clearly $d\le \rho$, but I stuck when I try to prove the opposite implication.
If I add the extra hypothesis that $d$ is both left and right invariant, then the result follows by the fact that $d(x,y)=d(xy^{-1},1)=d(1,xy^{-1})=d(x^{-1},y^{-1})$, but this is not the case in general.
Thank you in advance for your help.
Best Answer
Compatibility of two metrics merely means that the two metrics induce the same topologies. That is, a sequence $x_n$ converges in one if and only if it converges in the other, and this is clear here, because if $d(x_n,x)\to 0$ then by continuity of the inverse operation with respect to the metric $d$, you also have $d(x_n^{-1},x^{-1})\to 0$, so $\rho(x_n,x)\to 0$. Note that $G$ is a topological group with respect to the metric $d$, which guarantees the continuity of the group-inverse operation. Conversely, if $\rho(x_n,x)\to 0$ then clearly the non-larger nonnegative sequence also tends to zero, that is, $d(x_n,x)\to 0$.