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

Solution 1:

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.

Solution 2:

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 $B_d(x,s)\subset B_{d_1}(x,\frac 1 2) \subset B_{d_1}(x,r)$.