A continuous, bijective, function such that two different metrics have the same open sets.

Question

Let $ f: \Re \rightarrow \Re$ be a continuous and strictly increasing function such that $f(0)=0$. Suppose $d_1$ and $d_2$ are two different metrics on the nonempty set $X$ and that $d_1(x,y)=f(d_2(x,y))$. Show that $(X,d_1)$ and $(X,d_2)$ have the same open sets.

$\textbf{My Attempt:}$

Let $r_1$ and $r_2$ be the collection of open subsets in the $d_1$ and $d_2$ metric. We want to show that $r_1=r_2$. So let $U \subseteq r_1$. I have seen, from an answer to a similar question on this website, that I must show for that for any $x \in U$ there exist $\epsilon >0$ such that the open ball $B_{d_2} (x,\epsilon) \subseteq U$. Also since $x \in U$ there exist $\delta > 0$ such that $B_{d_1} (x,\epsilon) \subseteq U$. I am quite lost as what to do from here on using that $f$ is a continuous bijection. We have not yet covered homeomorphisms in class so I will not be able to use any such theorems.

Also, sorry for not directly linking to the similar question I mention. I still need to look up how to do this.

Answer 1

Note that since $f$ is increasing and continuous with $f(0)=0,$ then the range of $f$ is some open interval containing $0$ (possibly a ray, or all of the real line). (Why?) Moreover, letting $I$ be the range of $f,$ we have that $f$ is invertible and that $f^{-1}:I\to\mathfrak R$ is increasing and continuous with $f^{-1}(0)=0.$ (Why?)

To prove your claim, then, it suffices to prove that for every $\epsilon>0$ and every $x\in X,$ there exist $\delta,\delta'$ such that $B_{d_1}(x,\epsilon)=B_{d_2}(x,\delta)$ and $B_{d_2}(x,\epsilon)=B_{d_1}(x,\delta').$ (Why?)

As a hint for how to show this, you should use the definition $$B_d(x,\epsilon)=\{y\in X:d(x,y)<\epsilon\}$$ together with what you know about $f$ and $f^{-1}$.