Equivalent Metrics

**FLT** · February 14th 2010, 09:31 AM

Hi, I'd be greatful for any help on the following problem; I have to show that two metrics $\rho$ and $\sigma$ are equivalent if and only if every open ball $B_s^{\sigma}(x)$ contains an open ball $B_r^{\rho}(x)$ and every open ball $B_r^{\rho}(x)$ contains an open ball $B_s^{\sigma}(x)$ .

Here is my proof thus far:

If $\rho$ and $\sigma$ are equivalent, then we have that $c\rho(x,y) \leq \sigma(x,y) \leq C\rho(x,y)$ . Now, taking $\sigma(x,y) \leq C\rho(x,y)$ ...

Here is where I get stuck - I want to say the following:

...for large enough s, we can say that $\sigma(x,y) < C\rho(x,y) < s$ , and so we have that $\sigma(x,y) < s$ is indeed our set $B_s^{\sigma}(x)$ . Taking $C\rho(x,y) < s$ , and setting $r = \frac{s}{C}$ , we also have our set $B_r^{\rho}(x)$ , and so we conclude that $B_s^{\sigma}(x) \subset B_r^{\rho}(x)$ .

I am not allowed to use any arguments using cauchy sequences - only the equivalence inequality relation used above. I'd be so grateful for any help provided!

**Nyrox** · February 14th 2010, 10:46 AM

Two metrics are said to be equivalent if they generate the same topology, and you know that a basis for a metric topology consists of the set of open balls. So, given $B_r^\sigma (x)$ , you know that this is an open set in both topologies generated by $\sigma$ and $\rho$ , thus $B_r^\sigma (x)$ can be written as a union of balls $B_{s_i}^\rho (x_i)$ , where $x_i \in B_r^\sigma (x)$ . One of these must contain $x$ , and consequently must contain a ball of radius $s$ centered at $x$ .

Do you also need help with the other direction?

**FLT** · February 14th 2010, 06:35 PM

Originally Posted by Nyrox

Two metrics are said to be equivalent if they generate the same topology, and you know that a basis for a metric topology consists of the set of open balls. So, given $B_r^\sigma (x)$ , you know that this is an open set in both topologies generated by $\sigma$ and $\rho$ , thus $B_r^\sigma (x)$ can be written as a union of balls $B_{s_i}^\rho (x_i)$ , where $x_i \in B_r^\sigma (x)$ . One of these must contain $x$ , and consequently must contain a ball of radius $s$ centered at $x$ .

Do you also need help with the other direction?

Thanks for your response. As I said previously, we are only allowed to use the definition of equivalence given by the inequality in the proof. I understand that there are Topological and Cauchy sequence treatments of the proof, but I'd like to see how this can be proved using only the inequality.

**Nyrox** · February 15th 2010, 02:30 AM

However the inequality above is NOT the definition of equivalent metrics. It is a sufficient condition, but not necessary. Consider for instance $\mathbb R$ with the standard metric $\rho$ and a second metric $\sigma$ given by: $\sigma (x,y)=\min \{1,\rho (x,y)\}$ . These two are equivalent since

$\begin{cases}B_{\frac{r}{r+1}}^\sigma (x) \subseteq B_r^\rho (x)\\B_r^\rho (x)\subseteq B_r^\sigma (x)\end{cases}$ .

The inequality, though, is not satisfied.

The notion of equivalent metrics is purely topological, so you must use topological arguments

**FLT** · February 15th 2010, 04:56 AM

Originally Posted by Nyrox

However the inequality above is NOT the definition of equivalent metrics. It is a sufficient condition, but not necessary. Consider for instance $\mathbb R$ with the standard metric $\rho$ and a second metric $\sigma$ given by: $\sigma (x,y)=\min \{1,\rho (x,y)\}$ . These two are equivalent since

$\begin{cases}B_{\frac{r}{r+1}}^\sigma (x) \subseteq B_r^\rho (x)\\B_r^\rho (x)\subseteq B_r^\sigma (x)\end{cases}$ .

The inequality, though, is not satisfied.

The notion of equivalent metrics is purely topological, so you must use topological arguments

Why not? I can find a C and a c such that $C\rho(x,y) \leq \sigma(x,y) \leq c\rho(x,y)$ . If $\sigma(x,y)$ is the minimum of $\rho(x,y)$ and 1, then clearly $\sigma(x,y) \leq \rho(x,y)$ for any x,y. Likewise, for the upper bound, I can pick some C such that $\sigma(x,y) > C\rho(x,y)$ , with that C satisfying 0 < C < 1. Where's the flaw in my argument?

**Nyrox** · February 15th 2010, 06:03 AM

Originally Posted by FLT

Likewise, for the upper bound, I can pick some C such that $\sigma(x,y) > C\rho(x,y)$ , with that C satisfying 0 < C < 1.

This is false! To see why, observe that

$\rho (0,10^n)=10^n$ and $\sigma (0,10^n)=1$ for all $n\in \mathbb N$

thus, the constant $C$ satisfies $C\leq 10^{-n}$ , for all $n\in \mathbb N$ , so $C\leq 0$ , resulting in a contradiction.

**FLT** · February 15th 2010, 06:31 AM

Originally Posted by Nyrox

This is false! To see why, observe that

$\rho (0,10^n)=10^n$ and $\sigma (0,10^n)=1$ for all $n\in \mathbb N$

thus, the constant $C$ satisfies $C\leq 10^{-n}$ , for all $n\in \mathbb N$ , so $C\leq 0$ , resulting in a contradiction.

No, for any $n\in \mathbb N$ , C will satisfy 0<C<1! But this is besides the point; I can still pick such a C! It doesn't matter what C I pick, I can always pick a small enough one, ad infinitum.

**Nyrox** · February 15th 2010, 10:20 AM

C is a constant: it cannot depend on $n$ . The inequality condition doesn't state that for each pair $(x,y)$ there exist a $C_{(x,y)}$ and a $c_{(x,y)}$ such that

$C_{(x,y)}\rho(x,y)\leq\sigma(x,y)\leq c_{(x,y)}\rho(x,y)$ .

All metrics satisfy this!

What the condition states is that, there exist some fixed $C$ and $c$ , such that, for all $x,y$

$C\rho(x,y)\leq\sigma(x,y)\leq c\rho(x,y)$

But there isn't any number $C>0$ such that $C<10^{-n}$ for all $n$ . For each $n$ there exists a $0<C<10^{-n}$ , but not for all $n$ at the same time.

**Drexel28** · February 15th 2010, 07:03 PM

Originally Posted by FLT

Hi, I'd be greatful for any help on the following problem; I have to show that two metrics $\rho$ and $\sigma$ are equivalent if and only if every open ball $B_s^{\sigma}(x)$ contains an open ball $B_r^{\rho}(x)$ and every open ball $B_r^{\rho}(x)$ contains an open ball $B_s^{\sigma}(x)$ .

Here is my proof thus far:

If $\rho$ and $\sigma$ are equivalent, then we have that $c\rho(x,y) \leq \sigma(x,y) \leq C\rho(x,y)$ . Now, taking $\sigma(x,y) \leq C\rho(x,y)$ ...

Here is where I get stuck - I want to say the following:

...for large enough s, we can say that $\sigma(x,y) < C\rho(x,y) < s$ , and so we have that $\sigma(x,y) < s$ is indeed our set $B_s^{\sigma}(x)$ . Taking $C\rho(x,y) < s$ , and setting $r = \frac{s}{C}$ , we also have our set $B_r^{\rho}(x)$ , and so we conclude that $B_s^{\sigma}(x) \subset B_r^{\rho}(x)$ .

I am not allowed to use any arguments using cauchy sequences - only the equivalence inequality relation used above. I'd be so grateful for any help provided!

I agree with Nyrox. What definition of "equivalent metrics" are you using. I also think that two metrics are equivalent if they generate the same topology. Otherwise, the only other natural definition would be that they are literally the same metric space.

**FLT** · February 15th 2010, 08:36 PM

Originally Posted by Drexel28

I agree with Nyrox. What definition of "equivalent metrics" are you using. I also think that two metrics are equivalent if they generate the same topology. Otherwise, the only other natural definition would be that they are literally the same metric space.

OK, the definition I'm using is the following:

Two metrics $\rho$ and $\sigma$ on the same space $X$ are said to be equivalent if there exists $c,C > 0$ and independent of $x,y$ such that

$<br /> c\rho(x,y) \leq \sigma(x,y) \leq C\rho(x,y)<br />$

**Nyrox** · February 16th 2010, 03:24 AM

Well, it's obviously legitimate to use that inequality as a definition, but in such case you cannot prove that every open ball in one metric containing an open ball in the other metric implies the equivalence of metrics

At least not without some extra condition.

Thread: Equivalent Metrics

LinkBack

Thread Tools

Display

Equivalent Metrics

Similar Math Help Forum Discussions

Fun with metrics

Equivalent Metrics

Are equivalent metrics comparable on compact sets?

Challenging Problem!!! (Equivalent Metrics)

Metrics

Search Tags