This question came up while showing the composition of a metric with a certain other function gives another metric. Suppose I have some metric space $(X,d)$ and a continuous, non-decreasing function $f$ on the nonnegative reals. Moreover, suppose that $f(x)=0$ iff $x=0$, and $f$ also satisfies the triangle inequality in that $f (x+y)\leq f(x)+f(y)$.
Using these properties, it is not difficult to show that $f\circ d$ is yet another metric on $X$, so $(X,f\circ d)$ is also a metric space.
I notice the (open) balls given by the metrics are of form $ B_d(x,r)=\{y\in X\mid d(x,y)\lt r\} $ and $ B_{f\circ d}(x,r)=\{y\in X\mid (f\circ d)(x,y)\lt r\}, $ so it seems that the topologies generated by the base of open balls in each case would probably be the same. I would like to see how one would go about showing the topologies given by these two metrics are indeed the same. Thanks for any insight.