Proof verification: $d$ and $\tilde{d}$ are topologically equivalent

Question

Given $X = (0,1]$ a metric space with $\tilde{d}$ defined as $$\tilde{d}(x,y) = \bigl|\frac{1}{x} - \frac{1}{y}\bigr| \ \ \text{for} \ \ x,y \in X$$

I'm trying to prove that $\tilde{d}$ and $d$ the standard metric on $(0,1]$ are topologically equivalent, using the open sets criteria and would like to know if it's correct.

Proof: $U$ is open in $(X,\tilde{d}) \implies U$ is open in $(X,d)$

Let $x \in U$ and $\epsilon > 0$ such that $B_{\epsilon}(x,\tilde{d}) \subseteq U$

If $y \in B_{\epsilon}(x,\tilde{d})$ which means $ \tilde{d}(x,y) =\bigl|\frac{1}{x} - \frac{1}{y}\bigr| = \bigl|\frac{y-x}{xy}\bigr| = \frac{|x-y|}{|xy|} < \epsilon$ and that's equivalent to $d(x,y) = |x-y| < |xy|\epsilon$, then $y \in U$

Therefore, for any $x \in U$ we can find an epsilon $\tilde{\epsilon} = |xy|\epsilon$ such that: $y \in B_{|xy|\epsilon}(x, d) \implies y \in U$

$\Longleftarrow$

Similarly, If $y \in B_{\epsilon}(x,d)$, it follows that $d(x,y) = |x-y| < \epsilon$ which is equivalent to $\tilde{d}(x,y) =\bigl|\frac{1}{x} - \frac{1}{y}\bigr| < \frac{\epsilon}{|xy|}$

Therefore, for any $x \in U$ we can find an epsilon $\tilde{\epsilon} = \frac{\epsilon}{|xy|}$ such that: $y \in B_{\frac{\epsilon}{|xy|}}(x, \tilde{d}) \implies y \in U$

Answer 1

Your argument is not valid. You cannot make $\tilde {\epsilon}$ depend on $y$. Suppose $B_{\epsilon} (x,\tilde {d}) \subset U$ with $0<\epsilon <1$. Choose $ \tilde {\epsilon} >0$ such that $\tilde {\epsilon} < \min \{\frac x 2,\frac {\epsilon x^{2}}2\}$. (Note that $x >0$. This is crucial here). Then $|y-x| < \tilde {\epsilon}$ implies $|\frac 1 y -\frac 1 x|=\frac {|y-x|} {xy} <\epsilon$ because $\epsilon xy >\epsilon x(x-\tilde {\epsilon})=\epsilon x^{2}-\epsilon x{\tilde {\epsilon}} >2\tilde {\epsilon}-\epsilon x{\tilde {\epsilon}} >{\tilde {\epsilon}}$ (since $\epsilon x <1$). Hence $B_{\tilde {\epsilon}} (x,d) \subset B_{\epsilon} (x,\tilde {d}) \subset U$. I leave the other part to you.

Answer 2

There is a fault in your proof. Kindly see the proof below for the other side

\begin{align}\bar{d}(x,y)<r_2 &\iff \left| \dfrac{1}{x}-\dfrac{1}{y} \right| <r_2 \iff \dfrac{1}{x}-r_2<\dfrac{1}{y}<\dfrac{1}{x}+r_2\\& \iff \dfrac{1-x r_2 }{x}<\dfrac{1}{y}<\dfrac{1+xr_2 }{x} \end{align} Choose $r_2<\dfrac{1}{x},$ then \begin{align} \dfrac{1-x r_2 }{x}<\dfrac{1}{y}<\dfrac{1+xr_2 }{x}&\iff \dfrac{x }{1+x r_2}<y<\dfrac{x }{1-xr_2}\\& \iff - \dfrac{x^2 r_2 }{1+x r_2}<y-x<\dfrac{x^2 r_2 }{1-xr_2}. \end{align} As $r_2\to 0,$ then $\dfrac{x^2 r_2 }{1+x r_2}\to 0$ and $\dfrac{x^2 r_2 }{1-xr_2}\to 0.$ Hence, $\forall \, r>0,$ and $\forall \,x\in (0,1],$ there exists $r_2\in (0,1]$ such that $r_2<(r/2x^2).$ So, \begin{align} y-x<\dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <\dfrac{r }{2}<r. \end{align} Also, \begin{align} r_2<\dfrac{r}{2x^2}\iff -r_2>-\dfrac{r}{2x^2}\iff -x^2 r_2>-\dfrac{r}{2}. \end{align} But, $r_2\in (0,1]\implies r_2>-\dfrac{1}{2x},$ implies $x r_2>-\dfrac{1}{2} \iff 1+x r_2>\dfrac{1}{2}\iff \dfrac{1}{1+x r_2}<2.$ Hence, \begin{align} y-x> - \dfrac{x^2 r_2 }{1+x r_2}>-\dfrac{r}{2(1+x r_2)} > -\dfrac{2r}{2}=-r \end{align} Hence, we have that \begin{align} -r=-\dfrac{2r}{2}<-\dfrac{r}{2(1+x r_2)}< - \dfrac{x^2 r_2 }{1+x r_2}<y-x<\dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <\dfrac{r }{2}<r. \end{align} Therefore, \begin{align} |x-y|<r \iff d(x,y)<r \end{align} And you are done. Kindly get back if you have some questions. Remember that they don't necessarily need to have the same radius.