Questions about this metric

Question

Let $(A,a)$ be a metric space and define $a':A\times A\rightarrow\mathbb{R}$ with $a'(x,y)=\frac{a(x,y)}{1+a(x,y)}$.
I know that this is indeed a metric, but:

1. How do I show that a subset $X$ is open in $(A,a)\iff$ $X$ is open in $(A,a')$?
2. Is there an isometry $(\mathbb{R},a)\rightarrow (\mathbb{R},a')$?
3. Is there an isometry $(\mathbb{R},a')\rightarrow (\mathbb{R},a)$?
   For the last two questions $a$ is Euclidian.

1. $\Rightarrow$ There is an $\epsilon$ s.t. $B_a(x,\epsilon)\subset X$. I want to find an $\epsilon'$ s.t. $B_{a'}(x,\epsilon')\subset X$. Since $\frac{a(x,y)}{1+a(x,y)}\leq a(x,y)<\epsilon$, I can take $\epsilon=\epsilon'$.
   $\Leftarrow$ Which $\epsilon$ can we take here?

2. and 3.
   We need to find two maps $f,g$ such that $$|x-y|=\frac{|f(x)-f(y)|}{1+|f(x)-f(y)|}$$ and $$\frac{|x-y|}{1+|x-y|}=|g(x)-g(y)|$$ Which ones can we take?

Answer 1

Since the map $t \mapsto \dfrac{t}{1+t}$ is strictly increasing on $[0,+\infty)$, not only does $a'$ generate the same topology as $a$, the two metrics even have the same open balls (including $A$ as the ball with infinite radius and arbitrary centre). As $a(x,y) < r \iff a'(x,y) < \dfrac{r}{1+r}$ for $r \in (0,+\infty]$ (where we set $\frac{+\infty}{1+\infty} = 1$), we have

$$B_r^{(a)}(x) = B_{\frac{r}{1+r}}^{(a')}(x)$$

for all $x\in A$ and $r \in (0,+\infty]$.

Regarding the existence of isometries $f\colon (\mathbb{R},a) \to (\mathbb{R},a')$ and/or $g \colon (\mathbb{R},a') \to (\mathbb{R},a)$, look at for example

$$a'\bigl(f(0), f(2)\bigr),$$

and for the other direction consider the points $g(-2), g(0), g(2)$. If $\lvert g(0) - g(-2)\rvert = a'(0,-2)$ and $\lvert g(2) - g(0)\rvert = a'(2,0)$, can you have $\lvert g(2) - g(-2)\rvert = a'(2,-2)$?