Show that if d is a metric, then d'=sqrt(d) is a metric

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In summary: I think you have a good start with the proof. Just try to be more explicit and fill in any gaps in your reasoning. Also, don't be afraid to use symbols and equations to make your argument clearer.
  • #1
docnet
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Homework Statement
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Relevant Equations
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Screen Shot 2022-03-12 at 2.44.10 AM.png


##d'## is a metric on ##X## because it satisfies the axioms of metrics:

Identity of indiscernibles:
##x=y\Longleftrightarrow d(x,y)=0\Longleftrightarrow \sqrt{d(x,y)}=\sqrt{0}##

Symmetry: ##d(x,y)=d(y,x)\Longrightarrow \sqrt{d(x,y)}=\sqrt{d(y,x)}##

Triangle inequality: ##d(x,z)\leq d(x,y)+d(y,z)\Longrightarrow \sqrt{d(x,z)}\leq \sqrt{d(x,y)+d(y,z)}\leq \sqrt{d(x,y)}+\sqrt{d(y,z)}##

Open sets for ##d'##are the same as the open sets for ##d##.

Given ##x## in ##U##, let ##\epsilon>0## such that any ##y## that satisfies ##d(x,y)< \epsilon## is in the open set ##U## for ##d##. Define ##\delta=\sqrt{\epsilon}##. Then,
##d(x,y)<\epsilon\Longrightarrow \sqrt{d(x,y)}<\delta##.
So any ##y## in the open set ##U## for ##d## is in the open set ##U## for ##d'##.

Given ##x## in ##U##, let ##\delta>0## such that any ##y## that satisfies ##\sqrt{d(x,y)}< \delta## is in the open set ##U## for ##'d##. Again, define ##\delta^2=\epsilon##. Then,
##\sqrt{d(x,y)}<\delta \Longrightarrow d(x,y)<\epsilon ##.
So any ##y## in the open set ##U## for ##d'## is in the open set ##U## for ##d##.
 
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  • #2
Looks ok, except for some typos in the last part.
 
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  • #3
All true, but you should probably mention what properties of the square root function you are using in the last two inequalities of the triangle property proof. There are places where you are applying a property of the square root function that you should mention (and maybe prove).
 
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  • #4
I think the last part is best done by noting that for every [itex]x_0 \in X[/itex] and every [itex]r \geq 0[/itex] we have [tex]
\{ x \in X : d(x,x_0) < r \} = \{ x \in X: d'(x,x_0) = \sqrt{d(x,x_0)} < \sqrt{r}\}.[/tex] So if a set contains a [itex]d[/itex]-open ball of each of its points then it contains a [itex]d'[/itex]-open ball of each of its points and vice-versa.
 
  • #5
I would prefer something more explicit. Showing that ##d'## is a metric by almost never writing ##d'## seems not quite right to me. I feel I'm having to fill in too many blanks.

Also, for the open sets question, I'd like to see something more formulaic:

Let ##U## be open under ##d##, then ... ##U## is open under ##d'##.

Let ##U be open under ##d'##, then ... ##U## is open under ##d'##.
 
  • #6
FactChecker said:
All true, but you should probably mention what properties of the square root function you are using in the last two inequalities of the triangle property proof. There are places where you are applying a property of the square root function that you should mention (and maybe prove).
Not sure what name this is called, but it seemed clear from

##\sqrt{d(x,y)+d(y,z)}\leq\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}=\sqrt{d(x,y)}+\sqrt{d(y,z)}##

pasmith said:
I think the last part is best done by noting that for every [itex]x_0 \in X[/itex] and every [itex]r \geq 0[/itex] we have [tex]
\{ x \in X : d(x,x_0) < r \} = \{ x \in X: d'(x,x_0) = \sqrt{d(x,x_0)} < \sqrt{r}\}.[/tex] So if a set contains a [itex]d[/itex]-open ball of each of its points then it contains a [itex]d'[/itex]-open ball of each of its points and vice-versa.

I agree with you. My main difficulty was explaining how ##d(x,y)<r## is the same as ##d'(x,y)<\sqrt{r}##

PeroK said:
I would prefer something more explicit. Showing that ##d'## is a metric by almost never writing ##d'## seems not quite right to me. I feel I'm having to fill in too many blanks.

Also, for the open sets question, I'd like to see something more formulaic:

Let ##U## be open under ##d##, then ... ##U## is open under ##d'##.

Let ##U## be open under ##d'##, then ... ##U## is open under ##d'##.
##d'## is a metric on ##X## because it satisfies the axioms of metrics:

Identity of indiscernibles:
##x=y\Longleftrightarrow d(x,y)=0\Longleftrightarrow \sqrt{d(x,y)}=\sqrt{0}\Longleftrightarrow d'(x,y)=0##

Symmetry: ##d(x,y)=d(y,x)\Longrightarrow \sqrt{d(x,y)}=\sqrt{d(y,x)}\Longleftrightarrow d'(x,y)=d'(y,x)##

Triangle inequality: ##d(x,z)\leq d(x,y)+d(y,z)\Longrightarrow \sqrt{d(x,z)}\leq \sqrt{d(x,y)+d(y,z)}\leq \sqrt{d(x,y)}+\sqrt{d(y,z)}\Longleftrightarrow d'(x,z)\leq d'(x,y)+d'(y,z)## (because the square root of the sum is less than equal to the sum of the square roots)Let ##U## be open under ##d##, then for all ##x## in ##U##, there is an ##\epsilon>0## such that any ##y## that satisfies ##d(x,y)< \epsilon## is in the open set ##U## for ##d##. Define ##\delta## as ##\sqrt{\epsilon}##, then any ##y## that satisfies ##d(x,y)<\epsilon## also satisfies ##d'(x,y)< \delta##, and is in the open set ##U## for ##d##.

Let ##U## be open under ##d'##, then for all ##x## in ##U##, there is a ##\delta>0## such that any ##y## that satisfies ##d(x,y)< \delta## is in the open set ##U## for ##d'##. Define ##\epsilon## as ##\delta^2##, then any ##y## satisfies ##d'(x,y)< \delta## also satisfies ##d(x,y)<\epsilon##, and is in the open set ##U## for ##d##.
 
  • #7
docnet said:
Let ##U## be open under ##d##, then for all ##x## in ##U##, there is an ##\epsilon>0## such that any ##y## that satisfies ##d(x,y)< \epsilon## is in the open set ##U## for ##d##. Define ##\delta## as ##\sqrt{\epsilon}##, then any ##y## that satisfies ##d(x,y)<\epsilon## also satisfies ##d'(x,y)< \delta##, and is in the open set ##U## for ##d##.
It's more about technique that correctness, but here's how I would do this:

Let ##U## be open under ##d## and ##x \in U##. There exists ##\epsilon>0## such that ##d(x, y) < \epsilon \ \Rightarrow \ y \in U##.

Now ##d'(x, y) < \sqrt{\epsilon} \Rightarrow \ d(x, y) < \epsilon \ \Rightarrow \ y \in U##. And, as ##x## was arbitrary, we see that ##U## is open under ##d'##.
 
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  • #8
#1 It's good practice to briefly mention what you are going to do before you start chasing the epsilons around. We take a subset of ##X## and show it is open with respect to ##d## if and only if it is open with respect to ##\sqrt{d}##. Mentioning such things explicitly makes things easier for yourself and also helps convince your grader more.
 
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  • #9
Here are some steps where I think you could/should have mentioned some properties of the square root function and included some more detail.
docnet said:
Not sure what name this is called, but it seemed clear from

##\sqrt{d(x,y)+d(y,z)}\leq\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}=\sqrt{d(x,y)}+\sqrt{d(y,z)}##
##\sqrt{d(x,y)+d(y,z)}\leq\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}## because the square root function is increasing and ##d(x,y)+d(y,z) \leq d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}##.

##\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}= \sqrt{(\sqrt{d(x,y)}+\sqrt{d(y,z)})^2} = \sqrt{d(x,y)}+\sqrt{d(y,z)}##

Another step could use some more explanation:
docnet said:
##d(x,y)=0\Longleftrightarrow \sqrt{d(x,y)}=\sqrt{0}##
Because the square root function is one-to-one.
 
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FAQ: Show that if d is a metric, then d'=sqrt(d) is a metric

What is a metric?

A metric is a mathematical function that measures the distance between two points in a given space. It is used to define the concept of distance in mathematics and is essential in many fields such as geometry, topology, and analysis.

How is a metric defined?

A metric is defined as a function that satisfies four properties: non-negativity, identity of indiscernibles, symmetry, and the triangle inequality. These properties ensure that the metric function accurately measures the distance between two points in a space.

What is the relationship between d and d' in the statement?

The statement shows that if d is a metric, then d' is also a metric. The function d' is defined as the square root of d, which means it is derived from d. This relationship allows us to prove that d' also satisfies the four properties of a metric.

Why is it important to show that d' is a metric?

It is important to show that d' is a metric because it allows us to use the properties of a metric to analyze and understand the space defined by d'. This can help us make new discoveries and solve problems in various fields of mathematics and science.

Are there any other examples of deriving a metric from an existing one?

Yes, there are many other examples of deriving a metric from an existing one. For example, the Euclidean metric can be derived from the Manhattan metric by taking the square root of the sum of squared distances. This process is known as metric transformation and is commonly used in mathematics and physics.

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