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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)$$Cd_2(x, y)$, $d_2(x,y)$$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.

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    show $f$irst 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.2012-11-17

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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.

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    @Dan: Eh, you’re right; I was thinking of tight constants. I’ll have to rewrite that.2012-11-17