It's well-known that not all choice principles are preserved under forcing, e.g. in this answer https://mathoverflow.net/a/77002/109573 Asaf shows the ordering principle can hold in $V$ and fail in a generic extension. Indeed, the standard proof for preservation of AC is based on the fact that well-orderability is preserved under surjection, a fact that doesn't seem to have any nice generalization for weaker choice principles at all. So I wonder if we can get any results in the opposite direction.

Are there any known results of the form "If all generic extensions satisfy [some weak choice principle] then [some stronger choice principle] holds in $V$"?

I take choice principles to include e.g. AC, DC, AC$_{\omega}$, the selection principle, "all infinite sets are Dedekind-infinite," and "(strongly) amorphous sets don't exist." Two conjectures I want to focus on are:

Plausible conjecture: AC$_{\omega}$ in all generic extensions implies AC in $V$ (the idea here is that if there's a set in $V$ without a choice function, maybe there's a way to collapse its cardinality to $\omega$ without adding a choice function),


Ridiculous conjecture: If every generic extension has no strongly amorphous sets, then AC holds in $V$ (I can't believe this is true, but I also have no idea what property $V$ can have to prevent forcing amorphous sets).

  • 3
    $\begingroup$ I am including trivial forcing. I don't see how that gives a positive answer. I'm only assuming the generic extensions (including $V$) satisfy weak choice principles. $\endgroup$ – Elliot Glazer Jul 31 '18 at 18:08
  • 1
    $\begingroup$ This is a tough question. Looking back after seven years, I can now say that I feel that the main reason that AC is preserved under generic extensions is that AC is stating that surjections admits inverses. But in reality this amounts to injections rather than inverses. So the Partition Principle should be enough for that. The problem, however, is that the relation between full choice and the Partition Principle is one of the toughest nuts to crack in choiceless set theory. [...] $\endgroup$ – Asaf Karagila Jul 31 '18 at 23:28
  • 1
    $\begingroup$ While your questions and conjectures are very interesting, I feel that somehow people are often too focused on these families of choice principles, and that's somehow limiting. PP and KWP (Kinna–Wagner Principles) seem to be much more pertinent to the generic multiverse. While they are not as famous, PP does imply DC (for example), so if it is indeed the case that PP is preserved in generic extensions, then countable choice would be preserved too and it would serve as a counterexample. Or a proof that PP implies full choice. In either case, it's a hard problem to tackle. [...] $\endgroup$ – Asaf Karagila Jul 31 '18 at 23:31
  • 1
    $\begingroup$ Unless, of course, the obvious solution works. But for that one has to sit down and think about it. And it's mighty late here for doing just that now. $\endgroup$ – Asaf Karagila Jul 31 '18 at 23:32
  • 1
    $\begingroup$ I am pleased to see such a question, which can be expressed explicitly in modal terms as $\Box\text{AC}^-\to\text{AC}^+$, using the forcing modality, where $\text{AC}^-$ is the weak choice principle and $\text{AC}^+$ is the stronger one. $\endgroup$ – Joel David Hamkins Jul 31 '18 at 23:33

Here is a partial answer.

Theorem. Suppose that $\mathcal X=\{X_i\mid i\in I\}$ is a family of pairwise disjoint sets which does not admit a choice function, and let $\Bbb P$ be a forcing which is well-orderable. Then $\Bbb P$ cannot force a choice function from $\cal X$.

Proof. Suppose that $\dot f$ is a name such that $p\Vdash\dot f\text{ is a choice function from }\check{\mathcal X}$. Enumerate $\Bbb P$, and let $F(i)=x$ if and only if the least condition $q\leq p$ in the enumeration, such that $q$ decides the value of $\dot f(\check i)$, forced $\dot f(\check i)=\check x$. $\quad\square$

Corollary. Suppose that $\sf AC_\kappa$ fails, then there is a generic extension where $\sf AC_\omega$ fails.

Proof. Note that $\kappa^{<\omega}$, or $\operatorname{Col}(\omega,\kappa)$ is a well-orderable forcing. $\quad\square$

Corollary. If $\sf AC_\omega$ holds in every generic extension then $\sf AC_{\rm WO}$, and therefore $\sf DC$ hold in every generic extension.

Proof. Otherwise, collapse a suitably large $\kappa$ to be countable. Additionally, note that a generic extension of a generic extension is itself a generic extension. $\quad\square$

This means that it is enough to verify that if $\mathsf{AC}_{\rm WO}$ holds, i.e. $\forall\alpha\in\mathrm{Ord},\sf AC_{\aleph_\alpha}$, then there is a generic extension where it fails. This principle is weaker than $\sf AC$, but it does imply $\sf DC$ (not $\sf DC_{\aleph_1}$, though).

Unfortunately, the above method hits a roadblock since $X^{<\omega}$ is not well-orderable for an arbitrary set, and indeed forcing with $X^{<\omega}$ can easily add choice functions to the universe, and in fact on occasion also the axiom of choice in its full glory.

| cite | improve this answer | |
  • $\begingroup$ Very nice! I like it very much. $\endgroup$ – Joel David Hamkins Aug 2 '18 at 14:48
  • $\begingroup$ To be fair, this is something I have noticed a few years ago. I recall telling Uri Abraham this theorem on a car trip back from Jerusalem to Beersheba when I was a student. I might have even written this in some answer on MO before, because I have a strong feeling of deja vu about it at the moment. $\endgroup$ – Asaf Karagila Aug 2 '18 at 14:49
  • $\begingroup$ (I am also pretty sure that I am not the first person to have noticed that. If I were a gambling man, I would have bet that Hugh had known this since the 1980s at least. Probably the same can be said about a few other logicians too.) $\endgroup$ – Asaf Karagila Aug 2 '18 at 14:51
  • 2
    $\begingroup$ I guess you also get DC, since AC_WO implies DC. So if countable choice holds in all forcing extensios, then DC, which is an attractive way to say it. $\endgroup$ – Joel David Hamkins Aug 2 '18 at 14:54
  • $\begingroup$ Oh, that's a good observation. I will edit it in once I'm on the train. $\endgroup$ – Asaf Karagila Aug 2 '18 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.