Topologically equivalent metric space Is the metric $d'(x,y) = |e^x - e^y|$ topologically equivalent to the usual absolute value metric $d(x,y) = |x-y|$ on the reals?
EDIT: How would I show that the open sets in $( \mathbb R,d)$ and the open sets in $(\mathbb R,d')$ are the same using the fact that the function $e^x: \mathbb R \to (0, \infty)$ is a homeomorphism.
 A: Yes. The induced topologies are the same. This can most easiliy be seen my noting that $\mathbb{R}\to\mathbb{R}_+: x\mapsto e^x$ is a homeomorphism which transforms these metrics to one another.
EDIT: Note however that the uniform structures are not the same. This means that there are sequences (example: $a_n= -n$) which is cauchy in $d'$ but not in the standard metric. (Thanks Mike for noting)
Here is the proof to get from homeomorphism to same topologies: Let $f(x)=e^x$ and note that a homeomorphism maps open sets to open sets.
Let $O\subset\mathbb{R}$ be an open set (w.r.t. $d$). Then $f(O)\subset\mathbb{R}_+$ is open because $f$ is a homeomorphism. This means $f(O) = \bigcup_i B_{\epsilon_i}(x_i)$ is the union of (possibly infinitly many) unit balls (and all $x_i\in\mathbb{R}_+$). Then
$\begin{align}
O = f^{-1}(f(O)) &= f^{-1}(\bigcup_i B_{\epsilon_i}(x_i)) \\
&= \bigcup_i f^{-1}(B_{\epsilon_i}(x_i)) \\
&= \bigcup_i \{f^{-1}(y) \ |\ d(x_i,y)<\epsilon_i \} \\
&= \bigcup_i \{y \ |\ d(x_i,f(y))<\epsilon_i \} \\
&= \bigcup_i \{y \ |\ d'(f^{-1}(x_i),y)<\epsilon_i \} \\
&= \bigcup_i \{y \ |\ d'(f^{-1}(x_i),y)<\epsilon_i \} \\
&= \bigcup_i B'_{\epsilon_i}(f^{-1}(x_i))
\end{align}$
is the union of some $d'$-open balls, i.e. $O$ is $d'$-open as well.
Conversely, let $O'\subset\mathbb{R}$ be an $d'$-open set. Then it is the union of some $d'$-open balls. i.e.
$\begin{align}
O' &= \bigcup_i\{y\ |\ d'(x_i,y)<\epsilon_i\} \\
   &= \bigcup_i\{y\ |\ d(f(x_i),f(y))<\epsilon_i\} \\
   &= \bigcup_i\{f^{-1}(y)\ |\ d(f(x_i),y)<\epsilon_i\} \\
   &= \bigcup_i f^{-1}(B_{\epsilon_i}(f(x_i)). \\
\end{align}$
Each of these $f^{-1}(B(f(x_i))$ is open (w.r.t. d) because $f^-1$ is a homeomorphism as well.
