prove two metrics determine the same topology I want to prove that the euclidean topology $d_E$ and $d(x,y) = \left| \frac{1}{2^x}- \frac{1}{2^y}\right|$ determine the same topology. 
My work: I assume you want to show that for every $x \in \mathbb{R}$, and every $\epsilon > 0$, where  $B^{d_E}_{\epsilon}(x)$ is some open ball in the euclidean metric, there exists some $\delta > 0$ such that there is some $B_{\delta}^{d}(x) \subseteq B^{d_E}_{\epsilon}(x)$. The problem is I cannot find a relation between $\left| \frac{1}{2^x}- \frac{1}{2^y}\right|$ and $|x-y|$ to use to find $\delta$. 
Any help on this would be really appreciated! Thank you so much
 A: You actually need to go both ways. I’ll do the other direction as a sort of model.
Draw the graph of $f(x)=\dfrac1{2^x}$. Note that for any $x,y\in\Bbb R$ you have $d(x,y)=|f(x)-f(y)|$. In other words, the $d$-distance between $x$ and $y$ is the vertical distance between the points $\langle x,f(x)\rangle$ and $\langle y,f(y)\rangle$, and the Euclidean distance is the horizontal distance between those points. 
Take the $\epsilon$-ball $B_\epsilon^d(x)$, where $\epsilon<f(x)$. Examining the graph, you should see that 
$$\begin{align*}
B_\epsilon^d(x)&=\{y\in\Bbb R:f(x)-\epsilon<f(y)<f(x)+\epsilon\}\\
&=\left(f^{-1}\big(f(x)+\epsilon\big),f^{-1}\big(f(x)-\epsilon\big)\right)\;,
\end{align*}$$
which is an open interval around $x$. If you let 
$$\delta=\min\left\{x-f^{-1}\big(f(x)+\epsilon\big),f^{-1}\big(f(x)-\epsilon\big)-x\right\}\;,$$
you’ll have $B_\delta^{d_E}(x)\subseteq B_\epsilon^d(x)$.
Now take $B_\epsilon^{d_E}(x)$; $y\in B_\epsilon^{d_E}(x)$ if and only if $f(y)$ satisfies a certain inequality condition. Use that condition to find a $\delta$ such that $B_\delta^d(x)\subseteq B_\epsilon^{d_E}(x)$.
A: If you know, that the standard topology on $\bf R$ is determined by all open intervals, the solution is simple: function $1/2^x$ transforms $\bf R$ onto $(0,+\infty)$ and for every $(c,d)$ in $(0,+\infty)$ the ends $c$, $d$ are determined by their logarithms (base $1/2$) in $\bf R$.
