Proving the existence of a metric space

Question

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.

Answer 1

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?