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Let A be a subset of the positive real numbers. It is required to prove that there exists a metric space whose non-zero distances are exactly the set A. Any suggestions and hints on how to proceed will be highly appreciated. Please try not to post the exact solution.

General Grievous
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  • This looks like it can be proven by transfinite induction over $A$. It is enough to show how, having a space with distance set equal to some $A$, build a new space that has a distance space equal to $A \cup {r}$ -- this deals with successor step in induction. For the limit step it's looks like it suffices to consider a union. – xyzzyz Feb 01 '18 at 19:00
  • Is there any other way to solve this thing? – General Grievous Feb 02 '18 at 07:10
  • I’m new to transfinite induction – General Grievous Feb 02 '18 at 07:18
  • If X was a non empty set then, assigning arbitrary values from the set A to pairs of elements won't help since it has to satisfy the triangle inequality. Moreover, the cardinality of the ordered pairs of elements in X has to be equal to the cardinality of A. I simply don't know how to proceed. – General Grievous Feb 02 '18 at 19:10
  • Try thinking about what I wrote in my comment. Let $X$ be a metric space, such that its distances set is equal to $A$. Let $r > 0$ be any real number not already in $A$. Can you, starting out from $X$, build a new metric space $X'$, such that the distances set of $X'$ is equal to $A' = A \cup { r } $? – xyzzyz Feb 02 '18 at 19:44
  • It is hard to give a useful hint yet not the direct solution. Can you solve the problem for the set say $A={1,2}$? – user251257 Feb 03 '18 at 12:15

1 Answers1

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I will try to give a hint without disclose the solution.

Consider the example $A=\{1,3\}$. Let $X=A\cup \{0\}$. We need a function $d:X^2\to A$ with following properties

  1. $d(x,x) = 0$ for every $x\in X$
  2. $d(x,y) = d(y,x)$ for every $x,y\in X$
  3. $d(x,z) \le d(x,y) + d(y,z)$ for every $x,y,z$

Because of 1. and 2. we only need to think about $d(0,1), d(0,3), d(1,3)$. Now assign them to $1$ or $3$ such that $$d(1,3) \le d(0,1) + d(0,3)$$ holds.

Got any idea?

user251257
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