I want to prove that $d(x,y)=|x-y|$ and $d'(x,y)=|\ln(x)-\ln(y)|$ are topologically equivalent.
Proof: $\tau_d \subset \tau_{d'} \iff \forall x \in X, \forall \epsilon >0, \exists r>0$ such that $B_{d'}(x,r) \subset B_d(x,\varepsilon)$ $ \iff \forall y, d'(x,y)<r \implies d(x,y)<\varepsilon \iff |\ln(x)-\ln(y)| < r \implies |x-y|<\varepsilon$
Now I don't know what to do next in this case because I cannot find a relation between the two distances, I'm familiar for proving the distances that are related to be topologically equivalent, like $d(x,y)$ and $d'(x,y)=\ln(1+d(x,y))$ for example, were I can say that $\ln(1+d(x,y))<r \implies d(x,y)<\varepsilon$$ \iff d(x,y)<e^r -1 \implies d(x,y)<\varepsilon \iff e^r -1 =\varepsilon$ then calculate $r$ and do the same for the converse $\supset$.
But how can I prove that $d$ and $d'$ are topologically equivalent using this method for the case of:
- $d(x,y)=|x-y|$ and $d'(x,y)=|\ln(x)-\ln(y)|$
- or for $d(x,y)=|x-y|$ and $d'(x,y)=|\arctan(x)-\arctan(y)|$
- or for $d(x,y)=|x-y|$ and $d'(x,y)=|e^{-x}-e^{-y}|$
when I can't seem to find any relation between the two distances in these cases.