Show that, for every metric $d$, the metrics $d/(1+d)$ and $\min\{1,d\}$ are equivalent

Question

The actual problem looks like-

Let, $(X,d)$ be a metric space where $X\ne\emptyset$. Define $d_1$ and $d_2$ on $X\times X$ by $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$, $d_2(x,y)=\min\{1, d(x,y)\}\quad\forall(x,y)\in X\times X.$(It is easy to verify $d_1,d_2$ are two metrics on $X$.) Show that, $d_1, d_2$ are equivalent metrics on $X$.

Firstly, I'll use the standard theorem to prove two metrics on a non-empty set are equivalent-
Two metrics $d_1$ and $d_2$ on a non-empty set $X$ are equivalent if and only if for each $x\in X$ and each open ball $B_{d_i} (x,r)$ in $(X,d_i)$(for some $r>0$), there is an $s>0$ such that $B_{d_j} (x,s)\subseteq B_{d_i} (x,r)$, for $i, j=1,2$ and $i\ne j$.
Now, to solve the problem, I break it into to parts- If I can show that both of $d_1,d_2$ are equivalent with the metric $d$, then we are done.
I'm able to show $d$ and $d_2$ are equivalent in the following way-
Let, $x\in X$ and $B_{d_2} (x,r)$ be an open ball in $(X, d_2)$ for some $r>0$. Then $B_{d} (x,r)\subseteq B_{d_2} (x,r)$.
Again, if we take an open ball $B_{d} (x,t)$ in $(X,d)$ for some $t>0$. Then $B_{d_2} (x,s)\subseteq B_{d} (x,t)$, where $s=\min \{1,t\}$.

But how to show $d_1$ and $d$ are equivalent metrics in $X$?
Can anyone assist me to prove $d_1$ and $d$ are equivalent metrics on $X$? Thanks for your help in advance.

Answer 1

Just verify that $B_d(x,r)$ contains $B_{d_1}(x,s)$ where $s= \frac r {1+r}$ and $B_{d_1}(x,r)$ contains $B_d(x,s)$ where $s= \frac r {1-r}$ Note: it is enough to consider the case when $r<1$: if $r\geq 1$ first find $s$ such that $B_{d_1}(x,\frac 1 2)$ contains $B_d(x,s)$ and note that $B_d(x,s)\subset B_{d_1}(x,\frac 1 2) \subset B_{d_1}(x,r)$.

Answer 2

Note that $d_1(x,y)\leq d(x,y)$ for all $x,y\in X$, since $1+d(x,y)\geq 1$. Thus, it follows that $B_{\varepsilon}^d(x)\subseteq B_{\varepsilon}^{d_1}(x)$ for all $x\in X$ and $\varepsilon>0$. Hence, if $U\subseteq X$ is $d_1$-open, it is also $d$-open.

Conversely, suppose that $U$ is $d$-open and let $x\in U$. By definition of an open set in a metric space, there exists $\varepsilon>0$ such that $B_{\varepsilon}^d(x)\subseteq U$. Now let $\delta=\min\{\varepsilon/2,1/2\}$. If $d_1(x,y)<\delta$ then $d(x,y)<\varepsilon$, because: $$d(x,y)=\frac{d_1(x,y)}{1-d_1(x,y)}\leq 2d_1(x,y)< \epsilon$$ Hence, $B_{\delta}^{d_1}(x)\subseteq B_{\varepsilon}^d(x)\subseteq U$, which implies that $U$ is $d_1$-open.

Thus, by the above, $(X,d)$ and $(X,d_1)$ are topologically equivalent.

Since you have shown that $(X,d)$ and $(X,d_2)$ are topologically equivalent, it follows from the fact that topological equivalence is a equivalence relation that $(X,d_1)$ and $(X,d_2)$ are topologically equivalent.

Answer 3

For proving that $d$ and $d_1$ are equivalent metrics it is enough to prove that for any $1>r>0$ there exist $s$ and $t$ positive real numbers s.t. for any $x\in X$ there holds $B_{d_1}(x,s)\subseteq B_d(x,r)\subseteq B_{d_1}(x,t).$

Let suppose that $s=\frac{r}{1+r}$. Let $y\in B_{d_1}(x,s),$ so $d_1(x,y)<s\Rightarrow \frac{d(x,y)}{1+d(x,y)}<s\Rightarrow d(x,y)<\frac{s}{1-s}=r.$ So we proved that $y\in B_{d_1}(x,s) \Rightarrow y\in B_d(x,r)$ or $B_{d_1}(x,s)\subseteq B_d(x,r).$

Now, let $t=\frac{r}{1+r}$ and $y\in B_{d}(x,r),$ so $d(x,y)<r\Rightarrow \frac{1}{d(x,y)}>\frac{1}{r}\Rightarrow 1+\frac{1}{d(x,y)}>1+\frac{1}{r}\Rightarrow \frac{1+d(x,y)}{d(x,y)}>\frac{1+r}{r}\Rightarrow \frac{d(x,y)}{1+d(x,y)}<\frac{r}{1+r}=t.$ So we proved that $y\in B_{d}(x,r) \Rightarrow y\in B_{d_1}(x,t)$ or $B_{d}(x,r)\subseteq B_{d_1}(x,t).$