17
$\begingroup$

With the Axiom of Choice, the cardinals form a nice linearly ordered "set". In the absence of the Axiom of Choice, the cardinals form a partially ordered "set". Broadly, I am wondering what properties these posets can have.

A specific question I am interested in is the following.

Is there, for each (infinite) subset $S \subseteq \mathbb{N}$ containing $1$ and not containing $0$, a model of ZF in which there is a maximal antichain of cardinals of size $n$ if and only if $n \in S$?

Though I have a mild interest in knowing how such a model would be constructed (assuming a positive answer), my primary interest is in knowing that it is (is not) possible. Hence, if such a result exists in the literature, a citation would be all that I ask for.

Improve this question
$\endgroup$
11
  • 3
    $\begingroup$ 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. $\endgroup$
    – Trevor Wilson
    Commented Dec 8, 2011 at 18:48
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 8, 2011 at 21:53
  • 1
    $\begingroup$ 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. $\endgroup$
    – Joel David Hamkins
    Commented Dec 8, 2011 at 22:27
  • $\begingroup$ Do we have any reason to hope that every finite antichain of cardinals can be extended to a finite maximal antichain? $\endgroup$
    – Joel David Hamkins
    Commented Dec 8, 2011 at 22:29
  • $\begingroup$ @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. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 8, 2011 at 22:40

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .