# Two metrics have the same open sets (proof help needed)

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

$d_1(x,y)$ ≤ $Cd_2(x, y)$, $d_2(x,y)$ ≤ $Cd1 (x, y)$, x,y ∈ $X$.

Prove that the metrics $X$, $d_1(x,y)$ and $X$, $d_2(x,y)$ have the same open sets.

• show first that each open ball with respect to one metric contains an open ball, of perhaps different radius, with respect to the other metric. Note also that your condition that C>=1 is not needed, C>0 suffices. – Ittay Weiss Nov 17 '12 at 9:16

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.