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Let $A$ be any subset of the set of positive real numbers $\mathbb{R}_+$ ; then does there exist a metric space $(X,d)$ such that $d\colon X \times X \to A\cup\{0\}$ is a surjection ?

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  • $\begingroup$ What are your thoughts on this problem? $\endgroup$ – Math1000 Jan 27 '15 at 15:49
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Let's call $B=A\cup \{0\}$ and we define $d:B\times B\to B$ as $d(a,b)=0$ if $a=b$ and $d(a,b)=\max\{a,b\}$ if $a\neq b$. Note that $d(0,b)=b$ for each $b\in B$ so $d$ is indeed a surjection. The other axioms for a metric are easily verified (there are several cases though).

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