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I have the following question in my book:

Show that any subset $A$ of the set of non negative reals with $0 \in A$, is the set of all distances between points of some metric spaces.

And the solution is given to be:

For a given set $A$ of non negative reals with $0 \in A$ , we define a function $ d: A \times A \rightarrow \mathbb R$ by $$ d(x,y) = \begin{cases} \text{max}\{x,y\}, & \text{if} x \neq y \\ 0 , & \text {if} x=y \end {cases} $$

Then $d$ is clearly a Metric on $A$ and for any $x \in A$ , $d(0,x) = x$ ; also for any $x , y \in A $ , $d(x,y) = \text{max}(x,y)$ or $0$ so that $d(x,y) \in A$. Thus $(A,d)$ is the required metric space.

Now my question is: Is the metric taken in the solution fixed? Can we take any other metric e.g the usual metric or any other metric on $A$?

My next question is there any special significance of the 'max' metric used in the solution on any space? By special I mean like the usual metric is the usual distance between points in the co-ordinate plane, such as if this metric has any significance.

hiren_garai
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  • The d you have defined is not a metric as the $\Delta$ inequality fails. Let $x=0, y=10, z=10$. The$d(x,y)+d(y,z)=9+0 < 10=d(x,z)$ – Mathemagical Oct 01 '17 at 02:26
  • How $d(x,y) = 9 $ ? $\text{max} {0,10 } = 10 $ I think. – hiren_garai Oct 01 '17 at 02:55
  • You're right. I mixed up two examples I was trying. – Mathemagical Oct 01 '17 at 03:06
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    It is easy to furnish a metric on $A$: any metric on $\mathbb{R}_+$ can be restricted appropriately as needed. It's also not that hard to furnish a metric whose range contains $A$, for example the standard metric does this, as would any other metric satisfying $d(0,x)=x$. But generally the range of the standard metric will contain other elements too, for example when $A={ 0,1,3 }$ the standard metric has $2$ in its range. So I'm not sure how you would make the range of $d$ be exactly equal to $A$ except this way. – Ian Oct 01 '17 at 03:14
  • @Mathemagical . You probably have noticed by now that for non-negative $x,y,z,$ if $x\ne y\ne z \ne x$ then $d(x,z)=\max (x,z)\leq x+z\leq \max (x,y)+\max (y,z)=d(x,y)+d(y,z).$ – DanielWainfleet Oct 01 '17 at 07:07
  • @DanielWainfleet Yes. But thanks for writing it out. This d is actually quite cool. Bends my intuition a bit to see this metric which ignores the size of the smaller of the two numbers altogether. – Mathemagical Oct 01 '17 at 07:18
  • @Mathemagical .With this metric every $p\in A$ \ ${0}$ is an isolated point for if $0\ne p\in A$ then the open ball of radius $p/2$ centered at $p$ contains only the point $p$. And if $0\in A$ then $0$ is also isolated $unless$ $ \inf (A\setminus {0})=0.$ – DanielWainfleet Oct 01 '17 at 07:21
  • @DanielWainfleet Cool. Is see that what you say is true. Metric spaces are amongst the least strange topological spaces and even amongst them there are some strange ones. – Mathemagical Oct 01 '17 at 10:05

1 Answers1

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Is the metric taken in the solution fixed?

If you mean "is it unique", the answer is no. For example, take any bijection $\varphi:A\to A$ and let $d_\varphi(x,y) = \max(\varphi(x), \varphi(y)) $ for $x\ne y$. More generally, let $B$ be any set of the same cardinality as $A$ and define $d_\varphi$ on $B$ as above, using a bijection $\varphi:B\to A$. This is described as "pulling a metric back by $\varphi$".

Also, the underlying set need not have the same cardinality as $A$. Given any set $X$, we can define a metric on $A\times X$ by $d((a,x),(a',x'))=\max (a,a')$ unless $(a,x)=(a',x')$. In particular, we can put such a metric on $A\times \mathbb{R}$ and then pull it back to $\mathbb{R}$ by some bijection $\mathbb{R}\mapsto A\times \mathbb{R}$.

any special significance of the 'max' metric

It has a reasonable interpretation as the required fuel capacity of a plane that has to fly from $x$ to $y$ with a required stopover at $0$, where it is refueled.

It's also an example of a ultrametric, meaning $d(x,y)\le \max(d(x,z), d(y,z))$. Absent any assumptions on the structure of $A$, we pretty much have to use an ultrametric: given that we must have a triangle with two side length $a,b\in A$, what should the third sidelength of the triangle be? We don't know if any other elements of $A$ (if they even exist) can fit the triangle inequality. Using the smaller of $a,b$ risks violating the triangle inequality, if it happens to be less than half of the larger. So the only safe choice is to have the third side be $\max(a,b)$. Hence, an ultrametric. Notice that $\max$ function comes up naturally here.

  • My first doubt is cleared, while the significance of the 'max' metric is not totally cleared ! Hope to get it in future. Thank you for your answer. @Michelle. – hiren_garai Oct 02 '17 at 02:12
  • If by significance you mean "is it used anywhere except to solve this problem", the answer is probably no. –  Oct 02 '17 at 02:14