Show that $d(x,y)$ and ${d(x,y)\over d(x,y)+1}$ are topologically equivalent

Question

Let $(X,d)$ be a metric space and let $q(x,y)={d(x,y)\over 1+d(x,y)}$ be another metric(which we know it is). Show that $d(x,y)$ and ${d(x,y)\over d(x,y)+1}$ are topologically equivalent.

I used the definition that metrics $f,g$ on metric spaces $(X,d/ q)$ are called equivalent if $U\subset X$ is open with respect to $f$ if and only if it is open with respect to $g$.

I managed one direction: $q(x,y)\le d(x,y)$. Let $U\subset X$ be open with respect to $q$ and let $x\in U$. So there exists $r>0$ such that $B_q(x,r)\subset U$. For any $y\in X$, such that $d(x,y)<r$, $q(x,y)\le d(x,y)<r$. That implies that $y\in B_d(x,r)\Rightarrow y\in B_q(x,r)\Rightarrow B_{d}(x,r)\subset B_q(x,r)\subset U$ (So I hope. All those transitions seem a little counterintuitive.). With that uncertainty in mind, I am a little lost in the other direction.

I would appreciate any help on the issue.

Answer 1

Hint: use

$0 \le \frac{x}{2} \le \frac{x}{1+x} \le x$ for $0\le x \le 1$. The last inequality holds for all $x\ge0$

So $\frac{1}{2}d(x,y)\le q(x,y)\le d(x,y)$

More detail:

Take $B_d(x,\epsilon)$, take $\epsilon_1=\min\{\epsilon,1\}$, take $r=\frac{\epsilon_1}{2}$

Then $B_q(x,r)\subseteq B_d(x,\epsilon_1)\subseteq B_d(x,\epsilon)$

Conversely, take $B_q(x,\epsilon)$ and $r=\epsilon$

Then $B_d(x,r)\subseteq B_q(x,r)=B_q(x,\epsilon)$

Answer 2

For one direction: we have that for any $x\in X$ and any $\epsilon>0$ exists $r>0$ such that

$$y\in\Bbb B_d(x,r)\implies y\in\Bbb B_q(x,\epsilon)$$

this mean that

$$d_d(x,y)<r\implies d_q(x,y)=\frac{d_d(x,y)}{1+d_d(x,y)}<\epsilon$$

If $\frac{r}{1+r}<\epsilon$ and when $r$ decreases then the LHS decreases (i.e. the function $\frac{r}{1+r}$ is increasing) we are done. Then we can check that choosing $r<\frac{\epsilon}{1-\epsilon}$ works whenever $\epsilon<1$.

If $\epsilon\ge 1$ then using any $\epsilon^*<1$ for the bound $r<\frac{\epsilon^*}{1-\epsilon^*}$ would work (in particular $r=1$ works whenever $\epsilon\ge1$).

In general $r=\epsilon$ is an easy solution too.