Timeline for How is a MacNeille completion "universal" like a beta-compactification is "universal"?
Current License: CC BY-SA 4.0
9 events
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Aug 2, 2022 at 9:35 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
fix TeX etc.
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Feb 21, 2015 at 12:42 | vote | accept | James Brewer | ||
Feb 19, 2015 at 2:51 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
added 414 characters in body
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Feb 19, 2015 at 0:39 | comment | added | Joseph Van Name | Qiaochu Yuan. I agree that for different purposes there are other reasonable notions of a completion of a poset. For instance, anyone interested in forcing would be interested in the Boolean completion of a partially ordered set where the Boolean completion of a separative poset is the unique complete Boolean algebra that contains the poset as a join-dense subset. | |
Feb 18, 2015 at 23:19 | comment | added | Joseph Van Name | Qiaochu Yuan. I was not being clear about what I meant saying that the Dedekind-MacNeille completion is the only reasonable completion of a poset. I apologize for that. I only meant to say that the Dedekind-MacNeille completion is only completion in which the original poset is both meet dense and join dense and if one wants the original poset to be meet-dense and join-dense in the complete lattice, then the completion is the Dedekind-MacNielle completion. This is the sense in which the Dedekind-MacNielle completion is the only completion of a poset. | |
Feb 18, 2015 at 18:02 | comment | added | Qiaochu Yuan | What? I can think of at least two other reasonable completions, namely the free cocompletion $P^{op} \Rightarrow 2$, and the free completion $(P \Rightarrow 2)^{op}$. These correspond to looking at downward-closed and upward-closed sets respectively. | |
Feb 18, 2015 at 13:40 | vote | accept | James Brewer | ||
Feb 18, 2015 at 13:40 | |||||
Feb 18, 2015 at 0:04 | history | answered | Joseph Van Name | CC BY-SA 3.0 |