How can I show these two metrics give the same topology?

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.

Solution 1:

Let $(X,d)$ be a metric space and $f: \mathbb R_+ \rightarrow \mathbb R_+$ a function as you've described. Given $r>0$ and $x \in X$ we want to find an $\varepsilon>0$ such that $B_{f\circ d}(x,\varepsilon)\subset B_d(x,r)$. Set $\varepsilon=f(r)$ and take $w \in B_{f\circ d}(x,\varepsilon)$ then $f(d(x,w))<f(r)$ and since $f$ is non-decreasing it must be the case that $d(x,w)<r$ and therefore $w \in B_d(x,r)$.

To show that we can find $\varepsilon>0$ such that $B_d(x,\varepsilon) \subset B_{f\circ d}(x,r)$ is a bit trickier. Let $y \in (0,r)$ and pick $\varepsilon \in f^{-1}(y)$. Suppose $w \in B_d(x,\varepsilon)$ then $d(x,w)<\varepsilon$ and $f(d(x,w))\leq f(\epsilon)=y<r$ so $w \in B_{f\circ d}(x,r)$ as desired.

I think it's interesting to note that to get the balls to be contained in each other you don't really need the other two properties triangle inequality for $f$. But that it is needed in order to ensure $f\circ d$ is a metric.

Solution 2:

Here is a Theorem from my book Topology, Sts. Cyril and Methodius University, Skopje, 2002 (in Macedonian) . The proof consists of five rows.

Theorem. X is a set with two topologies T1 and T2 , B1 is a basis for topology T1 and B2 is a basis for topology T2 . If for every B1 in B1 and every point x in B1 there exists B2 in B2 , such that x is in B2 and B2 is a subset of B1, then T1 is a subset of T2 .

Nikita Shekutkovski