I could not find a way to prove the following. Could you please help me?
Regards
Let $X$ be a set, and let $d_1(x, y)$ and $d_2(x, y)$ be two metrics in $X$. Suppose that the metrics $d_1(x, y)$ and $d_2(x, y)$ are equivalent in the sense that there is a constant $C ≥ 1$ such that
$d_1(x,y) ≤ Cd_2(x, y)$, $d_2(x,y) ≤ Cd_1 (x, y)$, $x,y ∈ X$.
Prove that the metrics $X$, $d_1(x,y)$ and $X$, $d_2(x,y)$ have the same open sets.
Best Answer
Let $\tau_1$ be the collection of sets that are open in the $d_1$ metric, and let $\tau_2$ be the collection of sets that are open in the $d_2$ metric; you want to show that $\tau_1=\tau_2$. The most straightforward way is to show that $\tau_1\subseteq\tau_2$ and $\tau_2\subseteq\tau_1$. The two proofs are virtually identical, so let’s see how to prove $\tau_1\subseteq\tau_2$.
There’s only one reasonable way to start: let $U$ be an arbitrary element of $\tau_1$. To show that $U\in\tau_2$, we must show that for each $x\in U$ there is an $\epsilon>0$ such that $B_{d_2}(x,\epsilon)\subseteq U$. Since $U\in\tau_1$ we know that there is a $\delta>0$ such that $B_{d_1}(x,\delta)\subseteq U$. Can you now use the fact that $d_1(x,y)\le Cd_2(x,y)$ for all $x,y\in X$ to find the desired $\epsilon$ in terms of $\delta$? You may find this question and its answer helpful.
Note, by the way, that the set of constants $C$ such that $$d_1(x,y)\le Cd_2(x,y)$$ for all $x,y\in X$ is in general not the same as the set such that $$d_2(x,y)\le Cd_1(x,y)$$ for all $x,y\in X$. However, it’s true that if $C_1$ is in the first set and $C_2$ in the second, then $\max\{C_1,C_2\}$ is in both, so you can in fact assume that use a single constant for both inequalities. You can also assume that it’s greater than or equal to $1$, because if some constant $C$ works, so does any larger constant.