Timeline for Results that are easier in a metric space
Current License: CC BY-SA 3.0
6 events
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| Aug 17, 2017 at 9:10 | comment | added | Alec Rhea | @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! | |
| Aug 17, 2017 at 8:27 | comment | added | user5810 | 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.) | |
| Aug 17, 2017 at 2:40 | comment | added | Andreas Blass | 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.) | |
| Aug 16, 2017 at 15:01 | comment | added | Alec Rhea | @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. | |
| Aug 16, 2017 at 14:51 | comment | added | Gro-Tsen | 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). | |
| Aug 16, 2017 at 14:42 | history | asked | Alec Rhea | CC BY-SA 3.0 |