Show that the collection of open balls in two metric spaces are identical I am having trouble trying to prove the following statement. I can see why it would be true intuitively, however, I am having trouble formalising the proof as the notation is quite confusing.
Show that the collection of open balls in the metric space $(\Bbb{R},d_{st})$ coincides with that for the metric space $(\Bbb{R},d)$, where $d_{st}$ denotes the standard metric on $\Bbb{R}$ and:
$d(x,y) = \ln{(1 + \lvert x - y\rvert)}$.
We have defined a subset $U \subset X$ to be open [in $(X,d)$] if:
$\forall$ $u \in U$ $\exists$ $r_u \gt 0$ such that $B_{r_u}^{d}(u) \subset U$.
If someone could help to formalise the proof for this, it would help greatly. There is a follow on question which asks to show that $f: (\Bbb{R},d_{st}) \to (\Bbb{R},d_{st})$ is continuous if and only if $f: (\Bbb{R},d) \to (\Bbb{R},d)$ is continuous. Would I be correct in saying that if we induce the topology on $\Bbb{R}$ with respect to the two distance functions, by the question above, we have that $T_d$ = $T_{d_{st}}$ and thus, if $f$ is continuous in one space it must follow that $f$ is continuous in the other space because the spaces are identical as topological spaces?
Thanks!
 A: Let $U$ be open in the metric space $(\Bbb{R},d_{st})$. We prove that it is open in $(\Bbb{R},d)$ as well.
Take $u\in U$. Since $U$ is open in the metric $d_{st}$, there is an $r_{st}>0$, such that $B^{d_{st}}_{r_{st}}(u)\subseteq U$.
You know that $d(x,y) = \ln(1+d_{st}(x,y))$, so let $r = \ln(1+r_{st})>0$. Then $B_r^d(u) = B_{r_{st}}^{d_{st}}(u)$ (as I show below), and thus $B_r^d(u)\subseteq U$, meaning that $U$ is also open in $(\Bbb{R},d)$. You can do the other direction similarly.
Argument why $B_r^d(u) = B_{r_{st}}^{d_{st}}(u)$:
Take $x\in B_r^d(u)$. Then $d(x,u) < r$, or $\ln(1+d_{st}(x,u))<r$, and thus $d_{st}(x,u) < e^r-1=r_{st}$. This means that $x\in B_{r_{st}}^{d_{st}}(u)$.
For the other direction, take $x\in B_{r_{st}}^{d_{st}}(u)$. Then $d_{st}(u,x) < r_{st}$, and since $d(x,y) = \ln(1+d_{st}(x,y))$ and $r = \ln(1+r_{st})$, we get $d(u,x)<r$, which proves $x\in B_r^d(u)$.
A: If $u$ and $r$ are two non-negative numbers such that $\ln(1+u) < r$ then $u \leq e^{r}-1$, and vice versa. This means that $$d_{\text{log}}(x,y) < r\quad \Rightarrow \quad d_{\text{standard}}(x,y) < e^{r}-1,$$ and $$d_{\text{standard}}(x,y) < r\quad \Rightarrow \quad d_{\text{log}}(x,y) < \ln(1+r).$$  
Thus, if point $x$ lies in an open "logarithmic ball" centered at $y$ and with radius $r$ then point $x$ also lies in an open "standard ball" centered at $y$ and with radius $e^{r}-1\ .$ If point $x$ lies in a open standard ball centered at $y$ with radius $r$ then point $x$ also lies in an open "logarithmic ball" centered at $y$ with radius $\ln(1+r)\ .$
A: These are my thoughts on your second question.
A function $f:(\mathbb{R},\mathcal{T})\to(\mathbb{R},\mathcal{T})$ is continuous with respect to the topology $\mathcal{T}$ if $f^{-1}(\mathcal{T})\subseteq \mathcal{T}\ .$ Since the topology in question is generated by the open balls, the standard topology is equivalent to the "logarithmic topology". If function $f$ is continuous in the standard topology then $$f^{-1}(\mathcal{T}_{\text{log}}) = f^{-1}(\mathcal{T}_{\text{standard}}) \subseteq \mathcal{T}_{\text{standard}} = \mathcal{T}_{\text{log}}\ ,$$ so function f is continuous in the "logarithmic topology" as well. Conversely, if function $f$ is continuous in the "logarithmic topology" then $$f^{-1}(\mathcal{T}_{\text{standard}}) = f^{-1}(\mathcal{T}_{\text{log}}) \subseteq \mathcal{T}_{\text{log}} = \mathcal{T}_{\text{standard}}\ ,$$ so function $f$ is continuous in the standard topology also. 
