Metric space

Set

 context $X$ … set definiendum $\langle X,d\rangle\in\mathrm{it}$ postulate $d$ … metric $X$

We can reconstruct the set underlying a metric via $\text{dom}(\text{dom}(d))=\text{dom}(X\times X)=X$, so the set of metrics and the set of metric spaces over $X$ are in bijection.

Reference

Wikipedia: Metric space