The metrices are topologically but not strongly equivalent

[mathmari]

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

 

[I like Serena]

$d_1$ 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 $x=y$. (Worried)

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. (Nod)

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'? (Thinking)

 

[mathmari]

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

So what has to be done then? (Wondering)

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'? (Thinking)

We have that $d'=d_1(\frac{1}{x}, \frac{1}{y})$, or not? Can we use that? (Wondering)

 

[I like Serena]

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. (Nerd)

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? (Worried)

 

[mathmari]

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. (Nerd)

Ahh ok!

We must have that $x$ is in $(0,1)$, but $\frac 1 x$ is not in $(0,1)$ is it? (Worried)

Why is $x$ in $(0,1)$ ? Isn't it in $(0,\infty)$ ? I got stuck right now. (Wondering)

 

[I like Serena]

Why is $x$ in $(0,1)$ ? Isn't it in $(0,\infty)$ ? I got stuck right now.

Because your problem statement says:

Let X=(0,1).

(Thinking)

 

[mathmari]

Because your problem statement says:

(Thinking)

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? (Wondering)

 

[mathmari]

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? (Wondering)

 

[I like Serena]

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. (Smile)

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? (Wondering)

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? (Wondering)

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. (Shake)

 

[mathmari]

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? (Wondering)

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. (Shake)

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 $f(x)=\frac{1}{x}$ ? (Wondering)

 

[I like Serena]

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. (Worried)

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 $r'=\frac 12\min(x-a,b-x)$. (Thinking)