# Topologically-equivalent/metrically-equivalent metrics and the same topology

Definition: Metrics $d_1$ and $d_2$ on $X$ are topologically equivalent iff $d_1(x,x_n)\to 0 \iff d_2(x,x_n)\to 0$ for every $\{x_n\}\subset X$ and $x\in X$.

Definition: Metrics $d_1$ and $d_2$ on $X$ are metrically equivalent iff $d_1(x,y)\le ad_2(x,y)$ and $d_2(x,y)\le bd_1(x,y)$ for some $a,b\in\mathbb{R}$ and all $x,y\in X$.

I can easily see that metrically equivalent metrics are topologically equivalent.

I'm attempting to prove that $d_1$ and $d_2$ are topologically equivalent iff $(X,d_1)$ and $(X,d_2)$ have the same topology (collection of all open subsets).

Here's my (attempted) proof:

Let $U$ be open in $(X,d_1)$; i.e., $\forall x\in X$, $\exists r_1\in\mathbb{R}^+: d_1(x,y)<r_1 \implies y\in U$.

I'm trying to show that $U$ is open in $(X, d_2)$ by proving that $\exists r_2\in\mathbb{R}^+: d_2(x,y)<r_2\implies d_1(x,y)<r_1\implies y\in U$.

Assume the last statement doesn't hold: $\forall r_2>0, \exists y\in X: d_2(x,y)<r_2$ but $d_1(x,y)\ge r_1$.

Choose $\{x_n\}$ such that $d_2(x,x_n)<1/n$ but $d_1(x,x_n)\ge r_1$. So $d_2(x,x_n)\to 0$ but $d_1(x,x_n)\not\to 0$. Contradiction!

I'm concerned that there might exist $N$ such that $d_2(x,x_N)<1/N\implies x_N = x$. In that case, $d_1(x,x_N) = 0 < r_1$, still contradiction. Is this contradiction valid for my proof?

To prove the converse, that is, if $U$ open in $(X,d_2)$ implies $U$ open in $(X,d_1)$, then $d_1(x,x_n)\to 0$ implies $d_2(x,x_n)\to 0$; I could prove the identity mapping from $(X,d_1)$ to $(X,d_2)$ is continuous, then the preservation of convergence. Is there a direct approach to proving preservation of convergence?

My next question: Are topologically equivalent metrics metrically equivalent? I guess no. Can you give a counter example please?

Thank you!

• You can define a topology in terms of its closure operation. If topologies $T_1$ and $T_2$ on a set $S$ satisfy $Cl_{T_1}A=Cl_{T_2}A$ for every $A\subset S,$ then $T_1 =T_2.$ For a topology generated by a metric, the closure operation is defined by the convergent sequences.... Metric equivalence is also called uniform equivalence. For $x,y \in (0,1)$ let $d_1(x,y)=|x-y|$ and $d_2(x,y)= \tan (\pi |x-y|/2).$ Same topology but not metrically equivalent. – DanielWainfleet Jan 20 '16 at 2:07

Your argument seems kind of off: what is $y$ in your definition of the open set? It should probably start with $\forall x\in U\,\,\exists r_1\,\,\forall y\in X$. I did not carefully analyse the logic of your argument, but there are serious issues with the presentation: you seem to be a little sloppy with the symbolic treatment. If you absolutely need to write those things symbolically, you should take care to do it properly. Moreover, you seem to have fixed $x$ somewhere between the first and the second paragraphs of the proof. You should avoid doing this sort of thing without saying, at least until you become a more seasoned mathematician (and even then you should do it sparingly, if at all, especially in writing!).

The most direct way to prove the fact is probably by noticing that in a metric space, a set is closed if and only if it contains all the limits of (convergent) sequences of its elements. Once you have that, it is easy to see that your definition of topological equivalence makes the closed sets the same.

As for the counterexample, compare the standard Euclidean metric $d_E$ on the real line and $d'(x,y)=\arctan(d_E(x,y))$. They are topologically equivalent (easily), but not metrically equivalent. Even more trivially, you can take $d''(x,y)=\min(d_E(x,y),1)$. A bounded metric will never be metrically equivalent to an unbounded one.

There are two metrics that are commonly considered on the punctured $2$-sphere $S^2\setminus\{N\}$, where $N$ denotes the north pole:

1. The chordal metric. Think of $S^2$ as the unit sphere embedded in $\mathbb{R}^3$. The chordal distance between two points $x$ and $y$ on $S^2$ is defined to be the Euclidean distance between $x$ and $y$ regarded as elements of $\mathbb{R}^3$: $$d_C(x,y) = d_{\mathbb{R}^3}(x,y).$$

2. The induced metric from stereographic projection. Let $F:S^2\setminus\{N\}\to\mathbb{R}^2$ denote stereographic projection from the north pole $N$. For $x,y\in S^2\setminus\{N\}$, define $$d_{sp}(x,y) = d_{\mathbb{R}^2}(F(x),F(y)).$$ That is, the distance between $x$ and $y$ is the Euclidean distance between their projections onto the plane.

It is possible to show that these two metrics on $\mathbb{S}^2\setminus\{N\}$ are topologically equivalent. $d_C$ is just the metric induced from the Euclidean metric on $\mathbb{R}^3$, so the metric topology is the relative topology from $\mathbb{R}^3$. The metric topology induced by $d_{sp}$ can be identified by showing that $F$ is a homeomorphism of $S^2\setminus\{N\}$ onto $\mathbb{R}^2$, a standard exercise.

However, $d_C$ and $d_{sp}$ cannot be equivalent metrics. Note that $d_C$ is bounded by $2$, while $d_{sp}$ is unbounded: if $x\in S^2\setminus\{N\}$ is fixed and we let $y$ approach $N$, then $d_{sp}(x,y)\to\infty$. So there is no uniform constant $a$ that satisfies the inequality $$d_{sp}(x,y) \leq ad_C(x,y)$$ for all $x,y\in\mathbb{S}^2\setminus\{N\}$.

If you wish, you can make the example simpler by moving everything down one dimension: do the same argument, but with $S^1$ replacing $S^2$.