Given that $d_1$ and $d_2$ are metrics at $X$ and there are $m, M> 0$ so that $md_1 (x, y) ≤ d_2 (x, y) ≤ Md_1(x, y)$ for each $x, y \in X$

Question

Given that $d_1$ and $d_2$ are metrics on $X$ and there are $m, M> 0$ so that $md_1 (x, y) \le d_2 (x, y) \le Md_1 (x, y)$ for each $x, y \in X$. Prove that $A \subseteq X$ is open in $( X, d_1)$ if and only if $A \subseteq X$ is open in $(X, d_2)$.

Answer 1

$A$ is open in $(X,d_1)$ implies that, for each $a\in A$ there exists $r>0$, such that $B_{d_1}(a,r)\subseteq A$, where $B_{d_1}(a,r)=\{x|d_1(x,a)<r\}$. Now $d_1(x,a)<r\implies d_2(x,a)\le Md_1(x,a)<Mr$. Hence $B_{d_2}(a,Mr)\subseteq B_{d_1}(a,r)\subseteq A$. Hence $A$ is open in $(X,d_2)$.

$A$ is open in $(X,d_2)$ implies that, for each $a\in A$ there exists $r>0$, such that $B_{d_2}(a,r)\subseteq A$. Now $d_2(x,a)<r\implies d_1(x,a)\le d_1(x,a)/m<r/m$. Hence $B_{d_1}(a,r/m)\subseteq B_{d_2}(a,r)\subseteq A$. Hence $A$ is open in $(X,d_1)$.