Unable to solve this metric space question

Question

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.

Answer 1

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 $A = d(X^2) \setminus \{0\}$.