Show that a subset of $X$ is open in $(X,d)$ if and only if it is open in $(X,\rho)$

Question

I have spent many hours trying to prove this question and only managed to come up with this dodgy "proof". To be honest I have no idea where to start and 99.99% sure that my attempted proof is not good.

I would really appreciate it if you could offer any insight on how to prove this and possibly point out false assumptions that I have made in trying to prove this.

Question

Let $(X, d)$ be a metric space and let $r$ be a real number with $r > 0$. Define $\rho : X \times X \rightarrow \mathbb{R}$ by $\rho(x, y) = rd(x, y).$

Show that a subset of $X$ is open in $(X,d)$ if and only if it is open in $(X,\rho)$.

My proof attempt

forward

Let $A \subseteq X$ where $A$ is open in $(X,d)$.

Suppose by contradiction that $A$ is not open in $(X,\rho)$.

Then $\exists a \in A, \ \forall k > 0, B_\rho(a,k) \cap X \backslash A \neq \emptyset$

Let $x \in B_\rho(a,k) \cap X \backslash A$

Then $\rho(x,a)< k$

Then $\rho(x, y) = rd(x, y)<k, \ \forall k>0$

thus $d(x, y)< \frac{k}{r}$

Let $\frac{k}{r} = \epsilon$

Then $\forall \epsilon > 0, \ B_d(a,\epsilon) \nsubseteq A$

But this is a contradiction as $A$ is open in $(X,d)$, thus $A$ is open in $(X,\rho)$

backward

Let $A$ be open in $(X,\rho)$

then $\forall a \in A, \ B_\rho(a,k) \subseteq A$, where $k= min\{\rho (a,x)| x \in X\backslash A\} >0$

Then $\rho(x, y) = rd(x, y)=k$

Thus let $d(x, y)=\frac{k}{r} > 0$

Let $\epsilon=\frac{k}{r}$

Thus $B_d(a,\epsilon) \subseteq A$

Thus $A$ is open in $(X,d)$

This concludes my attempted proof

Answer 1

Let $\frac{k}{r} = \epsilon$. Then $\forall \epsilon > 0, \ B_d(a,\epsilon) \nsubseteq A$

This last part is a little bit dodgy, but it does take much to make it rigorous. You want to prove a $\forall \epsilon >0, \ldots$, so start by taking some arbitrary $\epsilon>0$. Then define $k$ by $k=\epsilon r$.

Using the fact that $\forall k' >0, B_\rho (a,k)\cap X\backslash A\neq \emptyset$, you get some $x\in B_{\rho}(a,k) \cap X\backslash A$. Now we compute : $$d(x,a) = \frac 1r\rho(x,a)<\frac k r = \epsilon$$ and conclude that $x \in B_{d}(a,\epsilon) \cap X\backslash A$. $\epsilon$ being arbitrary, we have proved that : $$\forall \epsilon >0, B_{d}(a,\epsilon) \cap X\backslash A \neq \emptyset$$ which contradict the fact that $A$ is open in $(X,d)$.

You can adapt this easily to make the second part of the proof rigorous.