Timeline for Locales as spaces of ideal/imaginary points
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Mar 10, 2019 at 4:00 | answer | added | Qiaochu Yuan | timeline score: 5 | |
| Mar 9, 2019 at 22:43 | answer | added | Joseph Van Name | timeline score: 3 | |
| Aug 21, 2018 at 15:36 | vote | accept | Maxime Ramzi | ||
| Jul 22, 2018 at 13:44 | comment | added | Simon Henry | @Max : yes and know. Essentially all the toposes in algebraic geometry are "coherent toposes" (it is a kind of compactness property) and it is a well known theorem of Deligne that (assuming AC) any such topos always have enough points. So their points are not really imaginaries. Though often these points corresponds to things that one cannot really construct in practice: prime ideal, algebraic closure etc... so working with the topos is a way to avoid to actually have to consider its points (but, as these points always exists, it is only for convenience, unless you want to avoid AC). | |
| Jul 22, 2018 at 8:28 | comment | added | Maxime Ramzi | @Hurkyl : But they're not used to refer to spaces of imaginary points, are they ? | |
| Jul 22, 2018 at 6:27 | comment | added | user13113 | Even in classical mathematics, one of the main approach to algebraic geometry is not about spaces, but about toposes, which are already defined in this "point free" manner. | |
| Jul 21, 2018 at 15:18 | answer | added | Simon Henry | timeline score: 12 | |
| Jul 21, 2018 at 14:30 | answer | added | Ingo Blechschmidt | timeline score: 16 | |
| Jul 21, 2018 at 8:29 | comment | added | Maxime Ramzi | @Qfwfq : $\mathbb{Q}$ and any dense, linear order without endpoints. It's known that if this second order is countable, then it is in fact isomorphic to $\mathbb{Q}$. If it's not countable though, then by collapsing its cardinal to $\aleph_0$ in a forcing extension makes it isomorphic to $\mathbb{Q}$ | |
| Jul 20, 2018 at 22:32 | comment | added | Qfwfq | I see. I had never heard of this notion before (not a logician speaking...). I know I'll be asking a slightly off topic thing but could you make a concrete example of such a pair of structures being "not isomorphic just for stupid reasons"? | |
| Jul 20, 2018 at 21:24 | comment | added | Maxime Ramzi | I wasn't sure whether it was standard notation. It means they're finitely partially isomorphic, i.e. there are many finite partial isomorphisms between them (many in the sense that you can always enlarge a finite domain or a finite codomain of a finite partial isomorphism to any larger finite domain or codomain). Some authors describe it as "$A$ is isomorphic to $B$, except if they're not; but then it's for stupid reasons, such as cardinality". In particular it's easy to see that this is the same as "there exists a forcing extension where they're isomorphic" @Qfwfq | |
| Jul 20, 2018 at 20:30 | comment | added | Qfwfq | What does $A\cong_p B$ mean? Does it denote elementary equivalence of structures? | |
| Jul 20, 2018 at 17:08 | answer | added | Harry Gindi | timeline score: 1 | |
| Jul 20, 2018 at 16:54 | history | asked | Maxime Ramzi | CC BY-SA 4.0 |