Asaf Karagila
I don't have much choice...

Strong chain conditions and preservation of choice principles

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I recently returned from a wonderful week in Italy, where I attended the Young Set Theorist 2013 conference. I met a lot of new people, some old acquaintances, baffled people with oversized pickles, and most importantly shared and learned some great ideas.

One of the nicer things I'd done was to work with Thomas Johnstone on some preservation theorem related to forcing and choice principles (see also this announcement by Victoria Gitman). In order to clean up a bit the proof, I'll introduce a new definition which is going to slightly extend the ideas originally discussed in Italy. So without further jibber jabber, let's talk mathematics.

Canonical names. Given a forcing \(P\in V\), and a \(\{\dot a_i\mid i\in I\}\in V\), we define the canonical name \(\{\dot a_i\mid i\in I\}^\bullet=\{\tup{1_P,\dot a_i}\mid i\in I\}\). For pairs \(\tup{\dot x,\dot y}\) we define \(\tup{\dot x,\dot y}^\bullet\) to be some canonical encoding of the ordered pair, e.g. \(\{\{\dot x\}^\bullet,\{\dot x,\dot y\}^\bullet\}^\bullet\), and similarly for longer tuples.

This notation makes it slightly simpler to talk about names which are generated by collecting some names in the ground model. Next we discuss antichains. Recall that in the lack of choice it is consistent that there is a forcing extension which does not have maximal antichains, or that not all antichains can be extended to maximal antichains. It could also be the case that no antichain is well-ordered. In those cases it is possible to have the usual \(\kappa\).c.c property hold, but in a very formal and useless manner. Sure all antichains are small, but it doesn't help us very much. We therefore introduce the following property.

Definition. We say that a notion of forcing \(P\) has the strong \(\kappa\).c.c property, if every non-empty \(A\subseteq P\) has an \(A\)-maximal antichain of cardinality \(\leq\kappa\).

In particular every strong \(\kappa\).c.c poset has only well-ordered antichains, and has \(\kappa\).c.c, this is because given an antichain \(C\) it must equal to the antichain given from the strong \(\kappa\).c.c-ness. Of course, every well-ordered poset has \(\kappa\).c.c if and only if it has strong \(\kappa\).c.c, so under \(\AC\) it is trivial that the notions of \(\kappa\).c.c and strong \(\kappa\).c.c are equivalent. But as remarked above without choice the situation may be grim.

Theorem. Let $V$ be a model of $\ZF+\AC_\kappa$. If $P\in V$ is a notion of forcing which has strong $\kappa$.c.c, and $G\subseteq P$ is $V$-generic, then $V[G]\models\AC_\kappa$.

Proof. Let \(p\in P\) a condition such that \(p\forces\dot F\text{ is a function from }\kappa\text{ and }\dot F(\check\alpha)\neq\check\varnothing\text{ for all }\alpha\). Then for every \(\alpha\) consider \(D_\alpha\) to be the set \(\{q\leq p\mid\exists\dot x\text{ s.t. }q\forces\dot F(\check\alpha)=\dot x\}\), then this set is non-empty, and therefore contains a maximal antichain \(C_\alpha\) of cardinality at most \(\kappa\). It is not hard to see that \(C_\alpha\) is maximal antichain below \(p\). For every \(q\in C_\alpha\) let consider the collection of names \(\dot x\) such that \(q\forces\dot x=\dot F(\check\alpha)\), and \(\dot x\) has a minimal rank with respect to this property. Then by \(\AC_\kappa\) we can choose \(\dot x_q\) for every \(q\in C_\alpha\), and consider \(\dot A_\alpha=\bigcup_{q\in C_\alpha}\dot x_q\). We have, if so, that \(p\forces\dot F(\check\alpha)=\dot A_\alpha\), for every \(\alpha\lt\kappa\).

Since we have that \(p\forces\exists\tau(\tau\in\dot A_\alpha)\) for every \(\alpha\), we can repeat the above process and end up with names \(\dot a_\alpha\) such that \(p\forces\dot a_\alpha\in\dot A_\alpha\). Now it is easy to see that \(p\forces\{\tup{\check\alpha,\dot a_\alpha}^\bullet\mid\alpha\lt\kappa\}^\bullet\in\prod\dot A_\alpha\) as wanted. \(\square\)

This was an observation made by both myself and Thomas in the case where \(P\) is well-orderable, but it seems that strong \(\kappa\).c.c is a more general notion, and seems to me as the "correct one". The next time I'll post a proof that Thomas showed me:

Theorem. (Johnstone) Let $V$ be a model of $\ZF+\DC_\kappa$. If $P\in V$ is a $\leq\kappa$-strategically closed forcing and $G\subseteq P$ is $V$-generic, then $V[G]\models\DC_\kappa$.

There are 2 comments on this post.

By Victoria Gitman
(Jun 28 2013, 20:59)

Hi Asaf, Question: Is it known that there are posets that have the strong \(\kappa\)-cc but are not well-orderable? I believe that the result about preserving \({\rm AC}_\kappa\) extends to preserving \({\rm DC}_\kappa\) because we use the well-orderability of the poset for mixing, which I guess we can also do with strong \(\kappa\)-cc. Also, I don't think we have the preservation of \({\rm DC}_\kappa\) for \(\leq\kappa\)-Baire posets, but only for \(\leq\kappa\)-strategically closed posets. We actually don't know whether \(\leq\kappa\)-Baire suffices. Do you think it does? I sent Tom the link to your post.

Hi Victoria,

I have to admit that I have nothing of the top of my head, but I also feel that there is no reason for strong \(\kappa\).c.c to imply well-orderability. I am certain that one could concoct something up using all sort of strange pathologies. I'll think about it some more, I'm sure that if I get my brain to overdrive I'll be able to find such example.

As for the second issue, I think that you're right. He did show me the proof for strategically closed forcings, which explains why I had such a difficult time to finish the proof for the Baire case. I feel suddenly very confused about my memory (and unfortunately all I had was the drafts we worked on, which mostly included the Baire case, rather than the strategically closed case). I'll correct this in this post, and perhaps for the next post I'll actually show more than that! (And thank you for sending the link to Tom.)

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