0

I've asked this question before on stackexchange for any hints or suggestions. However, there weren't sufficient replies and I was not able to solve this question. So now I'm looking for a solution.

Let $A$ be a subset of the set of all non negative real numbers. It is required to show the existence of a metric space $X$, such that the set of all non-zero distances of $X$ equal the set $A$. All solutions will be highly appreciated.

General Grievous
  • 2,121
  • "'I've asked this question before on stackexchange for any hints or suggestions. However, there weren't sufficient replies and I was not able to solve this question. "... maybe you should take a hint. – fleablood Feb 02 '18 at 18:42
  • 1
    Since you're new: the proper way to deal with a question that hasn't had responses is to edit it to clarify / expand it (which bumps the post), or offer a bounty to raise attention. Also, please give more than a day for a response. –  Feb 02 '18 at 18:45
  • A = {0} is a counter example. – William Elliot Feb 02 '18 at 19:36
  • 1
    @WilliamElliot ${0} \not \subset \mathbb R^+$. – fleablood Feb 02 '18 at 20:49
  • @fleablood. Even though 0 is not a positive number 0 is a non-negative number. – William Elliot Feb 03 '18 at 02:44

1 Answers1

2

Take $X = \{0\} \cup A$ and define $d:X^2 \to X$ by $$ d(a,b) = \begin{cases} 0 &\text{if } a=b,\\ \max\{a,b\} &\text{otherwise.} \end{cases} $$ This defines a distance because:

  • $d(x,y)=0$ iff $x=y$ follows from, if $x \neq y$, then suppose wlog $x < y$, and it follows that $\max\{x,y\}=y \neq 0$.
  • If $x=y$, then $d(x,y)=0=d(y,x)$; if $x < y$, then $d(x,y)=y=d(y,x)$.
  • $d(x,z) = \max\{x,z\} \leq \max\{x,y\}+\max\{y,z\}$, because each of $x$ and $z$ is less than $\max\{x,y\}+\max\{y,z\}$; so $d(x,z) \leq d(x,y) + d(y,z)$.

It is clear that for $a \in A$, we have $d(0,a)=a$, so $A \subseteq d(X^2)$.
If $a,b \in X$ are such that $a \neq b$, say, $a < b$, then $d(a,b) = b \in A$ because if $a < b$ then $b \neq 0$; so $d(X^2) \setminus \{0\} \subseteq A$. Hence $A = d(X^2) \setminus \{0\}$.

amrsa
  • 12,917
  • What if $\max (a,b) \not \in A$? – fleablood Feb 02 '18 at 20:51
  • @fleablood that can only happen if $a=b=0$, since $X = {0}\cup A$. Am I missing something? – amrsa Feb 02 '18 at 20:52
  • no, you are not missing anything. I was missing something in that I didn't get that we can choose $X$ and .... I don't know what I was thinking. I racked my brain but for some reason I had convinced myself that X couldn't be specific and we had to account for all possible X. ... You are absolutely right and I was absolutely wrong. – fleablood Feb 02 '18 at 20:55
  • @fleablood Ok, I was trying now to tell you that. By the way, do you confirm the example is right? – amrsa Feb 02 '18 at 20:56
  • Oh, yes. Your answer is perfect. I had considered it and the like but for some brain-farty lapse of sensibility I thought had to allow for all possible $X$ up to and including the entirety of $\mathbb R^+$... which I don't think would be possible. – fleablood Feb 02 '18 at 20:59
  • @fleablood Yes I suppose it's like we misread a logical formula swapping the quantifiers. It happens. Thank you for checking it! – amrsa Feb 02 '18 at 21:01
  • 1
    I will nitpick that $X$ need not include $0$. ... nitpick..:) (Speaking of swapping quantifiers. $d(x,y)$ must be in A, not X, and only if x \ne y. So there is no need to include $0$ in X. ... There's no harm in include $0$ in X... but there's no need. – fleablood Feb 02 '18 at 21:03
  • @fleablood After changing the answer I realised that we actually need $0$, for suppose $A$ has a minimum, $a_0$. Then, $d(a_0,a_0)=0$ and for all other $x, y \in A$ we have $d(x,y) = \max{x,y} > a_0$. Thanks anyway! – amrsa Feb 02 '18 at 22:02
  • Yes, but the question was only about a set of non-zero real numbers and non-zero distances. – fleablood Feb 02 '18 at 23:49