Showing two metrics are equivalent.

Question

Let $(X,d)$ be a metric space. Define $$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ (do you know the name of this metric?)

Show that the metrics $d$ and $d_1$ are equivalent.

Edited: Captain Lama pointed out that I was looking for strong equivalence which is not there. So how do I show they are equivalent?

The fact that two metrics are equivalent is not the same as $\alpha d(x,y)\le d_1(x,y)\le \beta d(x,y)$ for some $\alpha,\beta$, it's a weaker condition. That only works for norms. — Captain Lama, Apr 07 '16 at 11:59
Look here https://en.wikipedia.org/wiki/Equivalence_of_metrics#Strong_equivalence — marya, Apr 07 '16 at 12:01
Notice the word "strong". Your metrics are clearly not strongly equivalent because one is bounded and not the other. — Captain Lama, Apr 07 '16 at 12:03
Perhaps useful to show that $d_1$ is indeed a metric - see here. — Dietrich Burde, Apr 07 '16 at 12:03
@DietrichBurde: I didn't got you "they define the same open subsets". — marya, Apr 07 '16 at 12:13
You need to show that a ball centered at $x$ of radius $r$ with respect to $d_1$ contains a ball of radius $r'$ with respect to $d$ centered at $x$ and is contained in a ball of radius $r''$ centered at $x$ with respect to $d$. — Batman, Apr 07 '16 at 12:18

Christian Blatter · Accepted Answer · 2016-04-08T08:11:06.177

3

Consider a point $p\in X$, and let $U^1$ be any $d_1$-neighborhood of $p$. Then there is an $\epsilon>0$ with $U_\epsilon ^1(p)\subset U^1$. Since $d_1(x,y)<d(x,y)$ it follows that $$U_\epsilon(p)=\{x\>|\>d(x,p)<\epsilon\}\subset\{x\>|\>d_1(x,p)<\epsilon\}=U_\epsilon ^1(p)\subset U^1\ .$$ This shows that $U^1$ is a neighborhood of $p$ with respect to $d$ as well.

Conversely: Consider a point $p\in X$, and let $U$ be any $d$-neighborhood of $p$. Then there is a positive $\epsilon<1$ with $U_\epsilon(p)\subset U$. Since $$d(x,y)={d_1(x,y)\over 1-d_1(x,y)}\leq 2d_1(x,y)$$ when $d_1(x,y)<{1\over2}$ it follows that $$U^1_{\epsilon/2}(p)=\{x\>|\>d_1(x,p)<\epsilon/2\}\subset \{x\>|\>d(x,p)<\epsilon\}=U_\epsilon (p)\subset U\ .$$ This shows that $U$ is a neighborhood of $p$ with respect to $d_1$ as well.

Altogether we have proven that $(X,d)$ and $(X,d_1)$ possess the same open sets.

edited Apr 08 '16 at 08:11

answered Apr 07 '16 at 12:57

Christian Blatter

226,825

Can someone give me some intuition why $d_1(x,y)<d(x,y)\Rightarrow {x|d(x,y)<\varepsilon}\subset {x|d_1(x,y)<\varepsilon}$? My expectation was the other direction, means $d_1(x,y)<d(x,y)\Rightarrow {x|d(x,y)<\varepsilon}\supset {x|d_1(x,y)<\varepsilon}$ – Averroes2 Sep 28 '20 at 12:55
1

@Averroes: Its the definition of $\subset$. Each $p\in{x|d(x,y)<\epsilon}$ satisfies $d_1(p,y)<d(p,y)<\epsilon$, hence $p\in{x|d_1(x,y)<\epsilon}$. – Christian Blatter Sep 28 '20 at 13:12

score 2 · Answer 2 · answered Apr 07 '16 at 13:11

In order to show that two spaces, $X,Y$ are topologically equivalent it is enough to prove that the identity function is continuous in both directions $X\to Y$ and $Y\to X$ (so the inverse image of every open set of $Y$ is an open set of $X$ and reciprocally).

A function $f$ from $(X,d)\to (X,d_1)$ is continuous if for all sequence $x_n\to x$ in $(X,d)$ it is verify that

$\forall \epsilon >0$ there exist $\delta>0$ such that $d(x_n-x)<\delta\Rightarrow d_1(f(x_n)-f(x))<\epsilon$

This is clearly verified in both directions when $f(x)=x$ because of the equality $$d_1(x,y)=\frac {d(x,y)}{1+d(x,y)}$$

Thus, the two metrics define the same topology.

Right! It was a lapse because I have considered the same space $X$ with two different distances, (so $(X,d_1)=Y\ne (X,d)). Thank you. — Piquito, Apr 07 '16 at 15:59

Showing two metrics are equivalent.

2 Answers2