Timeline for Antichains of Cardinals in ZF Without Choice...
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2011 at 23:28 | comment | added | Asher M. Kach | Joel: My motivation for requiring infinite $S$ was the initial context: How complicated can the theory of cardinals in a model of $ZF$ (in the language of posets) be? That said, now my interest goes beyond the initial context, so I am curious about sets such as $\{1,2\}$. Thanks for elaborating! <p> For comparison, the theory of the cardinals of a model of $ZFC$ (as a poset or under addition) is decidable. <p> Note that, so the underlying universe is a set rather than a proper class, I am restricting the "model" of set theory to any downward closed subset of cardinals within the "model". | |
| Dec 9, 2011 at 6:35 | history | edited | Asaf Karagila♦ |
edited tags
|
|
| Dec 9, 2011 at 5:33 | comment | added | Asaf Karagila♦ | Asher, suppose that $\{\a\}$ was a maximal antichain (again, I'm ignoring finite sets) then $a$ is comparable with its Hartog number and thus well orderable. So everyone below $a$ can be well ordered as well. It's clear that $W_{\aleph_0}$ implies $1\in S$; on the other hand, if $W_{\aleph_0}$ does not hold, pick $a$ to be a D-finite set, every cardinal above it cannot be well ordered and thus incomparable with its Hartog number. Therefore no one above $a$ can be a maximal antichain of size $1$. | |
| Dec 9, 2011 at 0:25 | comment | added | Joel David Hamkins | Asher, the connection is that I believe that whenever there is an antichain of size 2, then one can make larger finite antichains, but it wasn't clear to me how to make larger finite maximal antichains. But if the answer to my question was that it always held, then one could not attain $S=\{1,2}$, for example. Meanwhile, you had asked about infinite $S$, but it isn't clear to me why you insist on that. | |
| Dec 9, 2011 at 0:17 | comment | added | Asher M. Kach | Joel: Naively, I would expect models of ZF with finite antichains having no finite maximal antichain extension. Is there a connection between this phenomena and the primary question? | |
| Dec 9, 2011 at 0:15 | comment | added | Asher M. Kach | Trevor: Thanks, I was not even certain that (nontrivial) finite maximal antichains were consistent with ZF. Asaf and Joel: Indeed, the only reason I required $0 \not \in S$ and $1 \in S$ was to avoid such trivialities with the finite cardinals. Excepting them, why is $1 \in S$ if and only if $W_{\aleph_0}$ holds? Naively, it seems plausible to have a set an infinite set $S$ with no countable subset, with some other cardinal (with countable subsets and size $S$ subsets) comparable to all cardinals. | |
| Dec 8, 2011 at 23:19 | comment | added | Joel David Hamkins | Asaf, I don't think your embedding fact settles the question I raise in my comment, since perhaps the finite antichain can be extended to a maximal antichain that does not lie in the copy of the poset in the cardinals. | |
| Dec 8, 2011 at 22:40 | comment | added | Asaf Karagila♦ | @Joel: Sure, if you allow finite cardinals then $1\in S$ is a requirement. I just don't think it is very interesting that way. Considering only infinite cardinals, on the other hand... :-) As for your second question, I'd not think so. Jech has a theorem [Axiom of Choice, Thm 11.1] which embeds every poset into the cardinals, take a poset with only infinite maximal antichains, then its embedding will produce a model in which there are finite antichains that cannot be extended into finite maximal antichains. | |
| Dec 8, 2011 at 22:29 | comment | added | Joel David Hamkins | Do we have any reason to hope that every finite antichain of cardinals can be extended to a finite maximal antichain? | |
| Dec 8, 2011 at 22:27 | comment | added | Joel David Hamkins | Asaf, I don't think what you say is correct. We need $1\in S$ always because, for example, {7} is a maximal antichain, since every set either has fewer than 7 elements or at least 7. Similarly for {0} or {k} for any finite k. | |
| Dec 8, 2011 at 21:53 | comment | added | Asaf Karagila♦ | To answer a very small part too, $1\in S$ if and only if $W_{\aleph_0}$ holds, that is every infinite set has a countable subset. | |
| Dec 8, 2011 at 18:48 | comment | added | Trevor Wilson | To answer a very small part of your question, $\lbrace \omega_1, \mathbb{R} \rbrace$ is a maximal antichain in all known natural models of the Axiom of Determinacy. See A.E. Caicedo and R. Ketchersid, \emph{A trichotomy theorem in natural models of AD}, available online at sites.google.com/site/richardketchersid/home/research/adcf-final.pdf. | |
| Dec 8, 2011 at 15:09 | history | asked | Asher M. Kach | CC BY-SA 3.0 |