Timeline for Sigma-complete Lindenbaum algebras? [closed]
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2014 at 21:11 | history | closed |
Andrés E. Caicedo Ricardo Andrade Joseph Van Name Ryan Budney Andrey Rekalo |
Needs details or clarity | |
| S Apr 1, 2014 at 10:06 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Edit: changed countable with sigma-complete (including in title)
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| Apr 1, 2014 at 9:32 | review | Suggested edits | |||
| S Apr 1, 2014 at 10:06 | |||||
| Apr 1, 2014 at 0:24 | comment | added | godelian | @Joel: That sounds interesting! I posted a new more general question here mathoverflow.net/q/162007/12976, but of course you are welcome to answer it for the Boolean case. | |
| Apr 1, 2014 at 0:20 | comment | added | Carl Mummert | The Lindenbaum algebra of a complete theory is the two-element Boolean algebra, of course, which is Ord-complete. It is not hard, given any finite Boolean algebra, to make a first-order theory whose Lindebaum algebra is that Boolean algebra. | |
| Mar 31, 2014 at 23:18 | comment | added | Joel David Hamkins | My current idea is to take the theory of the Boolean algebra $\mathbb{B}$ itself, in the language with a predicate $G$ on $\mathbb{B}$ and the scheme asserting that $G$ is a generic ultrafilter, that is, an ultrafilter meeting every definable dense subset. If $T$ is this theory, then $b\in G$ will have value $b$, and what else is there to say (modulo the theory)? So the corresponding Lindenbaum algebra seems to be very closely related to $\mathbb{B}$, if not $\mathbb{B}$ itself. Perhaps it should be a separate MO question, whether every Boolean algebra arises as a Lindenbaum algebra. | |
| Mar 31, 2014 at 22:33 | comment | added | Joel David Hamkins | I had thought it would be easy, but now I'm not sure whether it is true or not... | |
| Mar 31, 2014 at 22:21 | comment | added | godelian | Yes, that was the assertion I was referring to. Is that true? Sorry, I cannot see it immediately, perhaps it's obvious to you... | |
| Mar 31, 2014 at 21:50 | review | Close votes | |||
| Apr 1, 2014 at 21:11 | |||||
| Mar 31, 2014 at 21:32 | comment | added | Joel David Hamkins | Let's see, I had in mind to build the theory out of the Boolean algebra itself, but I didn't think it all the way through. But I don't understand your comment, since I thought the problem was: given a Boolean algebra $\mathbb{B}$, find a first-order language and a theory $T$ in that language, such that the collection of formulas, modulo provable equivalence in $T$, is a copy of $\mathbb{B}$. Is that true? | |
| Mar 31, 2014 at 21:26 | comment | added | godelian | Joel, do you have a proof of your first assertion? for propositional theories is obvious, but for a first-order theory I'm not even sure it's true... | |
| Mar 31, 2014 at 21:21 | answer | added | Joel David Hamkins | timeline score: 6 | |
| Mar 31, 2014 at 20:50 | comment | added | Joel David Hamkins | Could you clarify a little more exactly what you want? I believe that any Boolean algebra can arise as a Lindenbaum algebra for some suitably chosen first-order theory. And do you really want it to be $\sigma$-complete? In that case, it can't be countably infinite, since there is no countably infinite $\sigma$-complete Boolean algebra. Once the algebra is infinite, you get a countable antichain, and then if it is $\sigma$-complete, you will find continuum many elements arising as joins of subsets of that antichain. | |
| Mar 31, 2014 at 19:04 | review | First posts | |||
| Mar 31, 2014 at 19:34 | |||||
| Mar 31, 2014 at 18:48 | history | asked | John R. | CC BY-SA 3.0 |