Timeline for Partial orders arising from $l$-spaces
Current License: CC BY-SA 4.0
9 events
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| Feb 3, 2022 at 11:03 | history | edited | Glorfindel | CC BY-SA 4.0 |
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| Nov 4, 2010 at 22:39 | comment | added | Andrej Bauer | @Henrik: I don't quite understand what it is you don't believe. Are you doubting the Stone representation theorem for Boolean algebras? The powerset $P(\mathbb{N})$ of the natural numbers is a Boolean algebra. The corresponding Stone space is the space of ultrafilters on $P(\mathbb{N})$, whereas you seem to think that the corresponding Stone space is $\mathbb{N}$, which is false. | |
| Nov 4, 2010 at 16:24 | comment | added | HenrikRüping | I still can't believe that the powerset of the naturals is obtained in this way. $\mathbb{N}=\bigcup_{n\in \NN}\\{n\\}$ and there is no finite subcovering. | |
| Nov 3, 2010 at 23:25 | comment | added | Martin Brandenburg | Ah, of course! I thought that generalized boolean algebras have more structural data than partial orders, but they don't. | |
| Nov 3, 2010 at 23:05 | comment | added | Andrej Bauer | I thought I answered that: a poset comes from a Boolean ring precisely when it is a generalized Boolean algebra. | |
| Nov 3, 2010 at 19:48 | comment | added | Martin Brandenburg | Hm, I know this duality (maddin.110mb.com/pdf/boolean.pdf; of course, Stone's terminology is older). But can we describe boolean rings as special partial ordered sets? Of course the boolean structure induces the partial order, but how can we determine if a partial order comes from a boolean ring? | |
| Nov 3, 2010 at 17:59 | history | edited | Andrej Bauer | CC BY-SA 2.5 |
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| Nov 3, 2010 at 17:31 | history | edited | Andrej Bauer | CC BY-SA 2.5 |
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| Nov 3, 2010 at 16:24 | history | answered | Andrej Bauer | CC BY-SA 2.5 |