Let $X=(0,\infty)$ be a space, $d_{1}(x,y)=\mid{x-y} \mid$ and $d_{2}=\mid lnx-lny\mid$ be two metric spaces. Show that these metrics generates the same topology.
My attempt:
$\tau_{1} \subset \tau_{2} \iff (\forall B_{d_{1}}(x,\epsilon)\in\beta_{d}) (\exists\delta>0):B_{d_{2}}(x,\delta)\subset B_{d_{1}}(x,\epsilon)$
$B_{d_{1}}(x,\epsilon)=(y\in X : d_{1}(x,y)<\epsilon)=(x-\epsilon,x+\epsilon)$
$B_{d_{2}}(x,\epsilon)=(y\in X : d_{2}(x,y)<\epsilon)=(xe^{-\epsilon},xe^{\epsilon})$
EDİT:
I want to show that these metrics are Lipschitz equivalent.
We need to find $\alpha,\beta>0$ st $\alpha .\mid x-y \mid< \mid lnx - lny \mid<\beta . \mid x-y \mid$
Since the derivative of $ln$ is bounded in such space, we can say that it is Lipschitz, so there exists a positive number $\beta$ st. $\mid ln(x)-ln(y)\mid <\beta \mid x-y\mid$.
On the other hand, does there exist any $\alpha$ st. $\alpha .\mid x-y \mid< \mid lnx - lny \mid$ ?
You need to prove that for each $x \in X$ and each $\varepsilon \geq 0$, there exists $\delta \geq 0$ such that $B_2(x,\delta) \subseteq B_1(x,\varepsilon)$, and the converse.
So fix $\varepsilon$, and take $\delta = \min\{\ln(\frac{x+\varepsilon}{x}),\ln(\frac{x}{x-\varepsilon})\}$.
It follows that $$xe^{-\delta}\geq xe^{-\ln(x/(x-\varepsilon))}=xe^{\ln((x-\varepsilon)/x)}=x\frac{x-\varepsilon}{x}=x-\varepsilon,$$ and $$xe^{\delta}\leq xe^{\ln((x+\varepsilon)/x)}=x\frac{x+\varepsilon}{x}=x+\varepsilon,$$ whence $$(xe^{-\delta},xe^{\delta}) \subseteq (x-\varepsilon,x+\varepsilon).$$
Conversely, if $\delta=\min\{x(1-e^{-\varepsilon}),x(e^{\varepsilon}-1)\}$, you get that $$(x-\delta,x+\delta) \subseteq (xe^{-\varepsilon},xe^{\varepsilon}).$$