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A distance space is a pair $(X,d)$ where $X$ is a set and $d:X \times X \rightarrow \mathbb{R}$ is a symmetric, non-negative map such that $d(x,x)=0$ for all $x \in X$. These are sometimes called semi-metrics or pseudo-metrics, but these terms are used for a variety of different concepts in the literature.

Does anyone know of a characterization of distance spaces for which there exists a metric space $(X,\rho)$ such that $d(x,y) \leq \rho(x,y)$ for all $x,y \in X$? This is straight-forward when $X$ is finite or bounded, but seems more challenging for general $(X,d)$.

Two observations to start:

  • The existence of such a metric $\rho$ is equivalent to the existence of a function $f:X \rightarrow \mathbb{R}$ such that $f(x) + f(y) \geq d(x,y)$ for all $x,y \in X$.
  • An example where there is no such metric $\rho$ is given by $X = [0,1]$ and $d(x,y) = \frac{1}{|x-y|}$ for all (distinct) $x,y$.
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    $\begingroup$ If $X$ is countable this is easy by induction. Namely $X=\{x_i:i\ge 0\}$, $x_i$ distinct, and define $\rho(x_i,x_j)=u_j$ for $i<j$ with $u_j$ increasing, and large enough (namely $\ge \max_{i:i<j} d(x_i,x_j)$). $\endgroup$
    – YCor
    Commented Feb 8 at 16:18
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    $\begingroup$ Your notation $\Re$ is very unusual (it sometimes means real part, but not the set of reals). You probably mean $\mathbf{R}$ or $\mathbb{R}$. $\endgroup$
    – YCor
    Commented Feb 8 at 17:38
  • $\begingroup$ @YCor: Agreed, though it’s not absolutely nonexistent — I remember seeing it in some old-ish books (~60s or earlier), I think possibly in set theory? $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Feb 8 at 17:44
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    $\begingroup$ Just a comment on the last example ($(X,\delta)$ any infinite compact metric space with $d(x,y)=1/\delta(x,y)$ for $x\neq y$): every uncountable metric space has an finite (and even uncountable) bounded subspace. In this example, every infinite subset has an accumulation point, which means has pair of points for which the function $d$ tends to infinity, so cannot be bounded for any distance $\ge d$. $\endgroup$
    – YCor
    Commented Feb 8 at 19:12

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