# The metrices are topologically but not strongly equivalent

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

Let $X=(0,1)$. For $x,y>0$ we consider the metrices:
1. $d_1(x,y)=|x-y|$
2. $d'(x,y)=\left |\frac{1}{x}-\frac{1}{y}\right |=\frac{|x-y|}{|xy|}$

I want to show that these are topologically equivalent but not strongly equivalent.

$d_1$ and $d'$ are strongly equivalent iff there are constants $k>0$ and $K>0$ such that \begin{equation*}kd_1(x,y)\le d'(x,y)\le \text{ for all $x,y$.}\end{equation*}

From there we get \begin{equation*}k\le \frac{d'(x,y)}{d_1(x,y)}\le K \Rightarrow k\le \frac{1}{|xy|}\le K\end{equation*} for all $x,y$.

It holds that $\frac{1}{|xy|}$ is not bounded for all $x,y$. If $x,y\rightarrow 0$ then $\frac{1}{|xy|}\rightarrow \infty$.

Therefore the metrices $d_1$ and $d'$ are not strongly equivalent.

Is everything correct so far?

Could you give me a hint how to show that these are topologically equivalent?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$d_1$ and $d'$ are strongly equivalent iff there are constants $k>0$ and $K>0$ such that \begin{equation*}kd_1(x,y)\le d'(x,y)\le \text{ for all $x,y$.}\end{equation*}

From there we get \begin{equation*}k\le \frac{d'(x,y)}{d_1(x,y)}\le K \Rightarrow k\le \frac{1}{|xy|}\le K\end{equation*} for all $x,y$.
Hey mathmari !!

We cannot divide by $d_1(x,y)$ since it can be 0. This happens when $x=y$.

It holds that $\frac{1}{|xy|}$ is not bounded for all $x,y$. If $x,y\rightarrow 0$ then $\frac{1}{|xy|}\rightarrow \infty$.

Therefore the metrices $d_1$ and $d'$ are not strongly equivalent.

Is everything correct so far?
The formula still holds if $x\ne y$ so indeed those metrics are not strongly equivalent.

Could you give me a hint how to show that these are topologically equivalent?
From wikipedia:
The two metrics $d_1$ and $d_2$ are said to be topologically equivalent if they generate the same topology on $X$. The adjective "topological" is often dropped.
There are multiple ways of expressing this condition:
• a subset $A \subseteq X$ is $d_1$-open if and only if it is $d_2$-open;
• the open balls "nest": for any point $x \in X$ and any radius $r > 0$, there exist radii $r', r'' > 0$ such that
:: $B_{r'} (x; d_1) \subseteq B_r (x; d_2) \text{ and } B_{r''} (x; d_2) \subseteq B_r (x; d_1).$
• the identity function $I : X \to X$ is both $(d_1, d_2)$-continuous and $(d_2, d_1)$-continuous.
The following are sufficient but not necessary conditions for topological equivalence:
• there exists a strictly increasing, continuous, and subadditive $f:\mathbb{R} \to \mathbb{R}_{+}$ such that $d_2 = f \circ d_1$.
• for each $x \in X$, there exist positive constants $\alpha$ and $\beta$ such that, for every point $y \in X$,
:: $\alpha d_1 (x, y) \leq d_2 (x, y) \leq \beta d_1 (x, y).$

Perhaps we can check one of those 'multiple ways'?

#### mathmari

##### Well-known member
MHB Site Helper
We cannot divide by $d_1(x,y)$ since it can be 0. This happens when $x=y$.
So what has to be done then?

From wikipedia:
The following are sufficient but not necessary conditions for topological equivalence:
• there exists a strictly increasing, continuous, and subadditive $f:\mathbb{R} \to \mathbb{R}_{+}$ such that $d_2 = f \circ d_1$.
• for each $x \in X$, there exist positive constants $\alpha$ and $\beta$ such that, for every point $y \in X$,
:: $\alpha d_1 (x, y) \leq d_2 (x, y) \leq \beta d_1 (x, y).$

Perhaps we can check one of those 'multiple ways'?
We have that $d'=d_1(\frac{1}{x}, \frac{1}{y})$, or not? Can we use that?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So what has to be done then?
We are effectively giving a counter example.
It suffices to just include as part of the counter example that we pick $x\ne y$ so that we won't divide by zero.

We have that $d'=d_1(\frac{1}{x}, \frac{1}{y})$, or not? Can we use that?
We must have that $x$ is in $(0,1)$, but $\frac 1 x$ is not in $(0,1)$ is it?

#### mathmari

##### Well-known member
MHB Site Helper
We are effectively giving a counter example.
It suffices to just include as part of the counter example that we pick $x\ne y$ so that we won't divide by zero.
Ahh ok!!

We must have that $x$ is in $(0,1)$, but $\frac 1 x$ is not in $(0,1)$ is it?
Why is $x$ in $(0,1)$ ? Isn't it in $(0,\infty)$ ? I got stuck right now.

Last edited:

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#### mathmari

##### Well-known member
MHB Site Helper

I saw now that there is a typo, it should be $X=(0,\infty)$.

What can we do in this case? Can we use the criterion with the function of compoisition?

#### mathmari

##### Well-known member
MHB Site Helper
Is the following true?

Two metrices $d,d′$ on a set $X$ are topologically equivalent iff the map $(X,d)\rightarrow (X,d'), x\mapsto x$ is continuous and the map $(X,d')\rightarrow (X,d), x\mapsto x$ is continuous.

From that we get that each convergent sequence in respect to $d$ converges also in respect to $d'$ with the same limit.

Or is this not correct?

So, we have that $d'(x,y)=d_1(f(x), f(y))$ with $f(x)=\frac{1}{x}$ and the function $f$ is contiuous on $X$.

We also have that $d_1(x,y)=d'(f(x), f(y))$ with $f(x)=\frac{1}{x}$ and the function $f$ is contiuous on $X$.

Does it follow therefore that the two metrices are topologically equivalent?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Is the following true?

Two metrices $d,d′$ on a set $X$ are topologically equivalent iff the map $(X,d)\rightarrow (X,d'), x\mapsto x$ is continuous and the map $(X,d')\rightarrow (X,d), x\mapsto x$ is continuous.
That looks as if it is the same as:
The two metrics $d_1$ and $d_2$ are said to be topologically equivalent if they generate the same topology on $X$.
There are multiple ways of expressing this condition:
• the identity function $I : X \to X$ is both $(d_1, d_2)$-continuous and $(d_2, d_1)$-continuous.

So yes, that is true.

From that we get that each convergent sequence in respect to $d$ converges also in respect to $d'$ with the same limit.

Or is this not correct?
Maybe. How is that relevant?

So, we have that $d'(x,y)=d_1(f(x), f(y))$ with $f(x)=\frac{1}{x}$ and the function $f$ is contiuous on $X$.

We also have that $d_1(x,y)=d'(f(x), f(y))$ with $f(x)=\frac{1}{x}$ and the function $f$ is contiuous on $X$.

Does it follow therefore that the two metrices are topologically equivalent?
I guess you are trying to apply:
The following are sufficient but not necessary conditions for topological equivalence:
• there exists a strictly increasing, continuous, and subadditive $f:\mathbb{R} \to \mathbb{R}_{+}$ such that $d_2 = f \circ d_1$.
Is that the case?

If so, what you have is a $d_1\circ f$ instead of $f\circ d_1$, don't you?
I don't think we can apply it that way.

#### mathmari

##### Well-known member
MHB Site Helper
I guess you are trying to apply:
The following are sufficient but not necessary conditions for topological equivalence:
• there exists a strictly increasing, continuous, and subadditive $f:\mathbb{R} \to \mathbb{R}_{+}$ such that $d_2 = f \circ d_1$.
Is that the case?

If so, what you have is a $d_1\circ f$ instead of $f\circ d_1$, don't you?
I don't think we can apply it that way.
Ah so we cannot use the fact that $d'(x,y)=d_1(f(x), f(y))$ and $d_1(x,y)=d'(f(x), f(y))$ with $f(x)=\frac{1}{x}$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ah so we cannot use the fact that $d'(x,y)=d_1(f(x), f(y))$ and $d_1(x,y)=d'(f(x), f(y))$ with $f(x)=\frac{1}{x}$ ?
I don't see how.

Suppose we try to apply:
• the open balls "nest": for any point $x \in X$ and any radius $r > 0$, there exist radii $r', r'' > 0$ such that
:: $B_{r'} (x; d_1) \subseteq B_r (x; d_2) \text{ and } B_{r''} (x; d_2) \subseteq B_r (x; d_1).$

Whatever we pick for $x$ and $r$, the ball $B_r (x; d')$ will be an open interval that is a subset of $(0,\infty)$ won't it?
Say that it is the interval $(a,b)$ and $x\in(a,b)$.
Then we can always find an $r'$ such that $B_{r'} (x; d_1) \subseteq (a,b)$ can't we?
We can pick for instance $r'=\frac 12\min(x-a,b-x)$.