# metrics, open sets

i am studying for my exam and trying to solve some questions. I have got a problem about proving the following.

Let $X$ be a set, and let $d_1$ and $d_2$ be two metrics on $X$. Suppose that $d_1$ and $d_2$ are equivalent in the sense that there is a constant $C \ge 1$ such that $d_1(x,y) \le Cd_2(x,y)$; $d_2(x, y)\le Cd_1(x, y)$; and $x, y$ are elements of $X$. Show that the metric spaces $(X,d_1)$ and $(X,d_2)$ have the same open sets.

I would appreciate if someone can help. Thanks!

-
Do you mean that those inequalities are supposed to hold for all elements $x$ and $y$ of $X$, or what? – Chris Eagle Nov 24 '12 at 14:24
what i understand from the question is, not for all x and y, these are just two random metrics. Am i right? – bigO Nov 24 '12 at 14:25
$x$ and $y$ are certainly not random metrics. You said yourself they are elements of $X$. – Chris Eagle Nov 24 '12 at 14:26

## 2 Answers

You have to prove that every open set for $d_1$ is an open set for $d_2$ and conversely (however here you do not even need to prove the converse considering the symmetry in the definition). So take an open set $O$ for $d_1$, and prove that it is an open set for $d_2$. You probably have a definition of open sets in a metric space that says "A subset $O$ of $X$ is open iff for every $x \in O$, there is an open ball centered in $x$ within $O$". The rest should be easy.

-

Let $\tau_j$ be the topology induced by $d_j$ for $j=1,2$, and assume $B(x;r)$ be an open ball in $\tau_1$. It suffices to show there is an open ball $B(x;\delta)$ in $\tau_2$ such that $B(x;\delta)\subset B(x;r)$. (In this way we show that every open set in $\tau_1$ is also open in $\tau_2$.)

But since the two metrics are equivalent, this is not difficult to do this once you pick $\delta$ small enough.

-