Timeline for Topological spaces that resemble the space of irrationals
Current License: CC BY-SA 4.0
7 events
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| Oct 17, 2021 at 23:25 | history | edited | LSpice | CC BY-SA 4.0 |
TeX while this is on the front page
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| Apr 5, 2019 at 9:07 | comment | added | Tony Huynh | @TimothyChow Yes, that's what I meant when I said ''necessary and sufficient''. Thanks for the reference! | |
| Mar 21, 2019 at 22:58 | comment | added | Timothy Chow | @TonyHuynh : When you say "necessary and sufficient" are you looking for a theorem that the above list of conditions cannot be weakened? I recently learned of a nice result by Mel Currie ("A Metric Characterization of the Irrationals Using a Group Operation", Topology and Its Applications 21 (1985), 223-236) that if the word "completely" is dropped, then there are uncountably many non-homeomorphic examples. In particular we can take any metric space $(S,d)$ satisfying $\forall x\in S \forall r\in\mathbb{R}^+ \exists ! y\in S : d(x,y) = r$. Currie calls these "spyc spaces". | |
| Sep 14, 2017 at 10:07 | history | edited | Tony Huynh | CC BY-SA 3.0 |
added 10 characters in body
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| Jul 31, 2010 at 1:46 | comment | added | Tony Huynh | @Pete: Yes, it would be nice to rephrase to get necessary and sufficient conditions. Indeed, nowhere compact means what you think it means. | |
| Jul 31, 2010 at 1:09 | comment | added | Pete L. Clark | Since $J$ is not complete in its given metric, maybe a better way to phrase the result is "completely metrizable". Also, does "nowhere compact" mean that no nonempty open set has compact closure? (If not, what?) | |
| Jul 30, 2010 at 23:09 | history | answered | Tony Huynh | CC BY-SA 2.5 |