The metrices are topologically but not strongly equivalent

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In summary: We have that $d_1(x,y)=|x-y|$ and $d'(x,y)=\left |\frac{1}{x}-\frac{1}{y}\right |$ So $d_1(x,y)=d'(f(x),f(y))$ But $f(x)=\frac{1}{x}$, not $x$, right? (Thinking)In summary, the metrics $d_1$ and $d'$ are not strongly equivalent, as there are no constants $k>0$ and $K>0$ that satisfy the condition $kd_1(x,y)\le d'(x,y)\le Kd_1(x,y)$ for all $x$ and $y
  • #1
mathmari
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Hey! :eek:

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)
 
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  • #2
mathmari said:
$d_1$ 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)

mathmari said:
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)
mathmari said:
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)
 
  • #3
Klaas van Aarsen said:
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)
Klaas van Aarsen said:
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)
 
  • #4
mathmari said:
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)

mathmari said:
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)
 
  • #5
Klaas van Aarsen said:
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!

Klaas van Aarsen said:
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)
 
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  • #6
mathmari said:
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)
 
  • #7
Klaas van Aarsen said:
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)
 
  • #8
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)
 
  • #9
mathmari said:
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)

mathmari said:
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)

mathmari said:
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)
 
  • #10
Klaas van Aarsen said:
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)
 
  • #11
mathmari said:
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)
 

FAQ: The metrices are topologically but not strongly equivalent

1. What does it mean for metrics to be topologically but not strongly equivalent?

Topological equivalence means that two metrics define the same topology on a given set, while strong equivalence means that the two metrics also have the same convergence properties. Therefore, when metrics are topologically but not strongly equivalent, it means that they define the same open sets and neighborhoods, but may have different notions of convergence.

2. How can metrics be topologically but not strongly equivalent?

There are various ways in which metrics can be topologically but not strongly equivalent. One common example is when one metric is complete and the other is not, which affects the convergence of Cauchy sequences. Another example is when one metric is bounded and the other is not, which can lead to different notions of compactness.

3. What implications does topological but not strong equivalence have in analysis?

Topological but not strong equivalence can have significant implications in analysis, as it can affect the properties of functions and sequences defined using the metrics. For example, a function may be continuous with respect to one metric but not with respect to the other, or a sequence may converge to a different limit depending on which metric is used.

4. Is there a way to determine if two metrics are topologically but not strongly equivalent?

Yes, there are several ways to determine if two metrics are topologically but not strongly equivalent. One approach is to look at the properties of the metrics, such as completeness, boundedness, and continuity. Another approach is to compare the open sets and neighborhoods defined by the metrics, as topological equivalence means that they define the same topology on a given set.

5. How does topological but not strong equivalence relate to other mathematical concepts?

Topological but not strong equivalence is closely related to other mathematical concepts, such as topological spaces, continuity, and convergence. It is also related to the concept of equivalence relations, as topological equivalence is a type of equivalence relation between metrics. Understanding topological but not strong equivalence can also provide insights into the properties and behavior of other mathematical structures and spaces.

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