Timeline for Profinite completion of a partial order
Current License: CC BY-SA 3.0
20 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Aug 7, 2014 at 11:21 | vote | accept | Yann Pequignot | ||
| Jul 21, 2014 at 14:04 | comment | added | Yann Pequignot | @JosephVanName I did post my proof as an answer. The profinite completion of a discrete partial order indeed coincides with its Stone Cech compactification. Thanks for the link about the Nachbin compactification. | |
| Jul 19, 2014 at 15:31 | answer | added | Yann Pequignot | timeline score: 3 | |
| Jul 19, 2014 at 15:30 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
brought my answer to the proper place
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| Jul 18, 2014 at 18:44 | comment | added | Joseph Van Name | Also, it seems as if the compactification that we are talking coincides with the Nachbin compactification. I gave an answer here mathoverflow.net/a/140625/22277 sketching basic facts about ordered topological spaces and I talked a little about the Nachbin compactification there. Also, Guram Bezhanishvili and Patrick Morandi both from New Mexico State University have written several relevant papers on partially ordered spaces and their ordered compactifications. The paper sierra.nmsu.edu/gbezhani/tos.pdf describes the Nachbin compactification and other ordered compactifications. | |
| Jul 18, 2014 at 18:41 | comment | added | Joseph Van Name | Yann Pequignot. You should post your proof as an answer to this question. Users are encouraged to answer their own questions if they find an answer after they ask the question. | |
| Jul 18, 2014 at 16:34 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
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| Jul 18, 2014 at 14:50 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
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| Jul 18, 2014 at 14:03 | comment | added | Yann Pequignot | @JosephVanName the description you give of the profinite completion of $P$ is simply the Priestley dual of the lattice of downsets of $P$. I wrote a sketch of a proof that this indeed the case. Your comments are welcome! | |
| Jul 18, 2014 at 13:59 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
Gave a sketch of the proof.
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| Jul 18, 2014 at 7:55 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
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| Jul 18, 2014 at 6:49 | comment | added | Yann Pequignot | @JosephVanName Thanks for your guess. I need to think about it. I gave another more pedestrian try. I don't know yet if your construction is equivalent. | |
| Jul 18, 2014 at 6:39 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
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| Jul 18, 2014 at 6:33 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
Gave it a pedestrian try.
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| Jul 17, 2014 at 19:03 | comment | added | Joseph Van Name | I would guess that the profinite completion of a partial order is a sort of generalized Stone Cech compactification of the partially ordered set $P$. More specifically, if $P$ is a poset, then let $\mathcal{L}(P)$ be the collection of all downwards closed subsets of $P$. Then let $\iota:P\rightarrow 2^{\mathcal{L}(P)}$ be the mapping where $\iota(x)(L)=1$ iff $x\in L$. Then let $X=\overline{\iota[P]}$. Then $X$ becomes a Priestley space with its natural partial ordering and it seems like $X$ is the profinite completion of $P$. | |
| Jul 17, 2014 at 16:28 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
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| Jul 17, 2014 at 15:50 | history | edited | Yann Pequignot | CC BY-SA 3.0 |
I gave it a try.
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| Jul 17, 2014 at 11:13 | review | First posts | |||
| Jul 17, 2014 at 11:45 | |||||
| Jul 17, 2014 at 11:11 | history | asked | Yann Pequignot | CC BY-SA 3.0 |