Timeline for Maximal chains and antichains of statements weaker than AC
Current License: CC BY-SA 3.0
14 events
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| Dec 28, 2015 at 16:05 | comment | added | Joel David Hamkins | @EmilJeřábek Ah, of course, that's great! I'd suggest that you post an answer will a fuller explanation. I had been thinking there might be atoms, but we know in fact it is dense. | |
| Dec 28, 2015 at 15:09 | comment | added | Keith Kearnes | @Emil: that answers every order-theoretic question here. How about a specific countable sequence of sentences representing free generators? | |
| Dec 28, 2015 at 12:43 | comment | added | Emil Jeřábek | The Lindenbaum-Tarski algebra of any countable theory interpreting a modicum of arithmetic is the unique countable atomless Boolean algebra. This follows from the incompleteness theorem. | |
| Dec 28, 2015 at 12:37 | comment | added | Joel David Hamkins | @Bartek, Given that you're basically talking about the Lindenbaum algebra, which tells the whole story about your order, I would think the natural questions here are about the structure of it as a Boolean algebra, rather than specifically about maximal pairwise incomparable sets. You should want to know if there are atoms, and how many, whether it is an atomic Boolean algebra. Since it is countable and infinite, it cannot be complete. Does it have the Cohen algebra as a subalgebra? | |
| Dec 28, 2015 at 11:10 | comment | added | Bartek | Yes, I meant the "incomparable" sense. Thank you @Keith and Joel; I was confused by the answers because I didn't know the other meaning of the word. I think I have to edit the question then, unless that makes it uninteresting. | |
| Dec 28, 2015 at 4:34 | comment | added | Joel David Hamkins | @KeithKearnes I was thinking of the sense of antichain in a Boolean algebra that is commonly used, where one refers to pairwise incompatible (disjoint) elements rather than merely incomparable elements (that is, they pairwise meet to $0$). In this case, any element and its complement do form a maximal antichain, and there needn't be atoms. But perhaps the OP intends to refer to antichain in the "incomparable" sense, in which case I would agree with you. | |
| Dec 28, 2015 at 4:20 | comment | added | Keith Kearnes | I think a Boolean algebra with a finite maximal antichain of more than one element must have an atom. Here by "antichain" I mean a set of pairwise incomparable elements. It is nontrivial if it has more than one element. Simply choosing an element and its complement will not not produce a maximal antichain, usually. | |
| Dec 28, 2015 at 2:51 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Dec 28, 2015 at 2:49 | comment | added | Joel David Hamkins | Yes, I was thinking of the lower cone, but I should have meant the upper cone. But this is still a Boolean algebra. I have edited. | |
| Dec 28, 2015 at 2:12 | comment | added | Bartek | I think I got my least and greatest elements wrong in the question. AC is what proves everything weaker than itself so it is the least element, or 0, or "the falsest" element. Or, $\text{AC}\vee\phi\sim\phi$. It seems to me that if $\varphi$ is weaker than AC, then $\text{AC}\wedge\neg\varphi$ is impossible (and also not weaker than AC). | |
| Dec 28, 2015 at 1:01 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Dec 28, 2015 at 0:29 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Dec 28, 2015 at 0:27 | comment | added | Asaf Karagila♦ | I recall giving the example of long chains with bounded AC or DC on math.SE, but it's always 's relevant answer! | |
| Dec 28, 2015 at 0:23 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |