# Simple counterexample for open sets implying equivalent metrics

Let $\emptyset \neq X$ be a set. We say that two metrics $d_{1}, d_{2}$ on $X$ are equivalent metrics if $\exists \ \alpha_{1}, \alpha_{2} > 0$ such that $$\alpha_{1}d_{2}(x,y) \leq d_{1}(x,y) \leq \alpha_{2}d_{2}(x,y) \ \ \ \forall \ x,y \in X.$$

A recent exercise I came across was to show that if $d_{1}$ and $d_{2}$ are equivalent metrics on $X$, then $(X,d_{1})$ and $(X,d_{2})$ have the same open sets. This was easy enough, but made me question whether the converse were true as well.

I managed to find a counter-example to show that the converse is indeed false by considering $\mathcal{F}(\mathcal{V}) = \left\{F \in \mathcal{L}(\mathcal{V}): F \text{ has finite rank}\right\}$ where $\mathcal{V}$ is an infinite dimensional vector space over $\mathbb{K} = \mathbb{R} \cup \mathbb{C}$ armed with the metric $d(F_{1}, F_{2}) = \text{rank}(F_{1}- F_{2})$.

This wasn't a proof founded in much intuition however, but rather a tedious look at all of the noteworthy metrics that had been mentioned in my notes thus far until realizing that this topology agrees with the discrete topology induced by the discrete metric $\mu$ on $\mathcal{F}(\mathcal{V})$.

My question is whether there are is any more straightforward counter-example to show that metrics having the same open sets does not imply their equivalence. It would be very helpful if someone had some insight as to how one may think of a clever counter-example like this without resorting to ploughing through all of the noteworthy metrics they can think of.

• How about finding two metrics on the real line, both with the usual open sets, but not equivalent? – GEdgar Oct 30 '16 at 20:37
• Thanks @GEdgar, not sure why that wasn't my first thought. – Eric Hansen Oct 30 '16 at 20:51

For $x,y\in\Bbb R$ define $d(x,y)=\min\{|x-y|,1\}$. It’s not hard to check that $d$ defines the same topology on $\Bbb R$ as the usual metric. However, for each $n\in\Bbb Z^+$ we have $d(n,0)=1$ and $|n-0|=n$, so the ratio of $|x-y|$ to $d(x,y)$ can be made as large as you want.