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.
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.
Let $S_1$ be the open sets for $d_1$-metric and the same goes for $S_2$. We WTS $S_1 = S_2$, and we just show that $S_1 \subset S_2$ because the other direction has an analogous proof. Pick an arbitrary open set $S \in S_1$. To show that $S \in S_2$, we want to show that for every $y \in S$, there exists some $r$ such that $B_r(y,d_2) \subset S$. We know that $d_1(x,y) \leq c_2d_2(x,y)$ and there exists some $r'$ such that $B_{r'}(y,d_1) \subset S$ by definition of open set. Now, consider $B_r(y,d_2) = \{x : d_2(x,y) < r = r'/c_2\}$, and grab an arbitrary $x \in B_r(y,d_2)$. We know that $$d_1(x,y) \leq c_2d_2(x,y) \leq c_2r = c_2\frac{r'}{c_2} = r'$$ which means that $x \in B_{r'}(y,d_1)$, and because $B_r(y,d_2) \subset S$, we know that $x \in S$, and so $B_r(y,d_2) \subset S$.
