Let $(X,d)$ be a metric space. Let the metric $d'$ be given by $d'(x,y)=\frac{d(x,y)}{1+d(x,y)}$
Show that $d$ and $d'$ are topologically equivalent.
I want to show that $d$ and $d'$ are topologically equivalent.
That means that if we have an open set $U$ with regards to $d$, $U$ is open with regards to $d'$ and vice versa.
So let $U$ be open with regards to $d'$.
Then exists for every $u\in U$ an $\varepsilon > 0$ such that $B_\varepsilon^{d'}(u)\subseteq U$.
Now I have to find for every $u\in U$ an $\delta >0$ such that $B_\delta^{d}(u)\subseteq U$.
I want to give $\delta$ such that holds $B_\delta^d(u)\subseteq B_\varepsilon^{d'}(u)$.
How to give $\delta$?
I have to choose $\delta$ such that $d(v,u)<\delta\Rightarrow d'(v,u)<\varepsilon$.
We choose $\delta=\varepsilon$.
Let $v\in B_\delta^d(u)$. Then we have $d'(v,u)=\frac{d(v,u)}{1+d(v,u)}\stackrel{!}{\leq} d(v,u)<\delta=\varepsilon$.
So we have $v\in B_\varepsilon^{d'}(u)$ and conclude $B_\delta^d(u)\subseteq B_\varepsilon^{d'}(u)\subseteq U$ as desired.
Now:
Let $U$ be open with regards to $d$.
Then exists for every $u\in U$ an $\varepsilon > 0$ such that $B_\varepsilon^d(u)\subseteq U$.
Again I want to choose for every $u\in U$ a $\delta > 0$ such that
$B_\delta^{d'}(u)\subseteq B_\varepsilon^d(u)\subseteq U$.
So $d'(v,u)<\delta\Rightarrow d(u,v)<\varepsilon$.
How can this be done?
I want to find a reasonable choice for $\delta$, so for a sketch we have
$d'(u,v)<\delta\Leftrightarrow \frac{d(u,v)}{1+d(u,v)}<\delta\Leftrightarrow d(u,v)=\delta+\delta d(u,v)\Leftrightarrow d(u,v)(1-\delta)<\delta$.
If $\delta\in (0,1)$ we have:
$d(u,v)<\frac{\delta}{1-\delta}$
If $\delta \geq 1$, then there is nothing to show (can we say so?), since the LHS would be negative, while the RHS is positive.
Is this thought sensible, do a calculation like above, to restrict the choice of $\delta$?
So I thought about choosing $\delta =\min\{1,\varepsilon\}$
I struggle to complete the proof.
Can you help me out?
Thanks in advance.
Best Answer
For your second case (to show $B^{d'}_\delta(u) \subseteq B^d_\varepsilon(u)$) you can choose $$ \delta = \frac{\varepsilon}{1 + \varepsilon} $$ and $\delta > 0$ because $\varepsilon > 0$.
Now you can directly check that this choice works. For if $$d'(u, v) < \delta$$
then $$\frac{d(u,v)}{1 + d(u, v)} < \frac{\varepsilon}{1 + \varepsilon}$$
so since both denominators are positive, cross-multiplying preserves the inequality sign, and we get $$d(u,v) + \varepsilon d(u, v) < \varepsilon + \varepsilon d(u, v)$$
and finally cancelling $\varepsilon d(u, v)$ on both sides gives $$ d(u, v) < \varepsilon $$ as desired.