Timeline for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?)
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2013 at 12:06 | comment | added | Harry Gindi | @Todd: I guess you need compactness! | |
| Jul 29, 2013 at 19:01 | comment | added | Todd Trimble | @HarryGindi: Sorry, but there is no freaking way that Hausdorff spaces are monadic over sets. Because this would imply that the forgetful functor $U: Haus \to Set$ (being monadic) reflects isomorphisms, i.e., if $f: X \to Y$ is a bijective continuous maps between Hausdorff spaces, then $f$ is a homeomorphism. Which is false of course. | |
| Feb 10, 2010 at 6:26 | history | made wiki | Post Made Community Wiki by Harry Gindi | ||
| Feb 9, 2010 at 0:00 | history | edited | Harry Gindi | CC BY-SA 2.5 |
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| Feb 8, 2010 at 23:13 | comment | added | Joel David Hamkins | Harry, you say in "a Hausdorff space...every ultrafilter converges uniquely to a point," but this isn't true. The real ine has ultrafilters going off to infinity. I think you either want to add compactness or to make a statement only about convergent ultrafilters. | |
| Feb 8, 2010 at 21:00 | comment | added | Harry Gindi | Neat! But I mean, can't you just describe all quantale-looking things as monoidal categories? | |
| Feb 8, 2010 at 18:41 | comment | added | Tim Porter | There are even quantaloids and quantaloid enriched categories! Fun and relevant to discussions on the café perhaps. | |
| Feb 8, 2010 at 17:37 | comment | added | Harry Gindi | @Tim, I didn't know that was a thing. I just looked it up, and it seems to be what I was talking about. | |
| Feb 8, 2010 at 17:25 | comment | added | Tim Porter | @Harry : are by any chance you working your way towards a quantale? | |
| Feb 8, 2010 at 12:42 | comment | added | Clark Barwick | More generally, for any sober space (and thus for $\mathrm{Spec}(R)$ for any ring $R$), the homeomorphism type is specified by the lattice of opens. The points of a sober space are in bijective correspondence with the completely prime filters on the lattice of open sets. | |
| Feb 8, 2010 at 12:23 | history | edited | Harry Gindi | CC BY-SA 2.5 |
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| Feb 8, 2010 at 12:15 | history | edited | Harry Gindi | CC BY-SA 2.5 |
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| Feb 8, 2010 at 11:47 | history | answered | Harry Gindi | CC BY-SA 2.5 |