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Are there any significant results in the theory of metric spaces that (are considerably more difficult to reproduce/have not been reproduced) in the theory of uniform spaces?

In particular, I'm interested in any results relating to analysis that might be manifest when viewing $\mathbb{R}$ as a metric space but would be less visible when viewing $\mathbb{R}$ as a uniform space. I'm attempting to define integration in a setting more general than but similar to $\mathbb{R}$, and I'm wondering if I should bother defining a 'nonstandard metric' and exploring that point of view or if I should feel comfortable moving to a uniform point of view that would be induced by such a metric immediately.

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    $\begingroup$ Possibly relevant: mathoverflow.net/questions/212308/… (the Baire category theorem does not hold in complete uniform spaces — I'm not sure if you count this as being "easier" in a metric space, but it seems worth mentioning). $\endgroup$
    – Gro-Tsen
    Commented Aug 16, 2017 at 14:51
  • $\begingroup$ @Gro-Tsen This is relevant to my interests -- I suspect that some counterexamples can be found in the very large, very dense non-Archimedean real closed fields. These are uniform spaces under the canonical entourages induced by their nonstandard metrics, but the fields can get so dense that I suspect they admit countable collections of disjoint dense open subsets. $\endgroup$
    – Alec Rhea
    Commented Aug 16, 2017 at 15:01
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    $\begingroup$ Urysohn's Lemma becomes a lot easier in metric spaces than in its usual setting of normal spaces. If $A$ and $B$ are disjoint closed sets, and $d$ is the metric, then $d(x,A)/(d(x,A)+d(x,B))$ is a continuous function of $x$, identically $0$ on $A$ and identically $1$ on $B$. (I don't know whether a uniform structure would give you an easy proof.) $\endgroup$
    – Andreas Blass
    Commented Aug 17, 2017 at 2:40
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    $\begingroup$ The function Andreas mentioned shows that ZF proves the result for metric spaces. ​ It follows from this answer that ZF does not prove the corresponding result for uniform spaces. ​ ​ ​ ​ ​ ​ ​ (Also, just to make sure: ​ ​ ​ Is the "disjoint" in your comment referring to their overall intersection? ​ In non-empty spaces, pairwise intersections of such subsets are trivially always non-empty, so in non-empty spaces, the only way such collections can be pairwise disjoint is if there is at most one such subset.) ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ $\endgroup$
    – user5810
    Commented Aug 17, 2017 at 8:27
  • $\begingroup$ @RickyDemer Thanks for the link, and my assertion was referencing pairwise disjointness, however I wasn't considering that the sets need be open. I was thinking about the nonstandard analogues of rational and irrational numbers, which are obviously not open or closed under the interval topology/nonstandard metric $\xi$-topology. Thanks for catching the error! $\endgroup$
    – Alec Rhea
    Commented Aug 17, 2017 at 9:10

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