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).

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    $\begingroup$ 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. $\endgroup$ Jul 31, 2018 at 18:08
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    $\begingroup$ 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, 2018 at 23:28
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    $\begingroup$ 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, 2018 at 23:31
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    $\begingroup$ 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, 2018 at 23:32
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    $\begingroup$ 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. $\endgroup$ Jul 31, 2018 at 23:33

2 Answers 2


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.

  • $\begingroup$ Very nice! I like it very much. $\endgroup$ Aug 2, 2018 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, 2018 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, 2018 at 14:51
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    $\begingroup$ 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. $\endgroup$ Aug 2, 2018 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, 2018 at 14:58

Working on a related topic for a new paper with Jonathan Schilhan I've stumbled into a disproof of your ridiculous conjecture.

Definition. The class $W$ is the closure of $\{\{a\}\mid a\in V\}$ under well-ordered unions. Sets in $W$ are called "almost well-orderable sets".

The axiom $V=W$ is consistent with $\lnot\sf AC$, for example in Gitik's model, and in a large cardinals-free model, Blass' model where all ultrafilters are principal.

Theorem. If $V=W$, then there are no amorphous sets.

Proof. We define a rank function based on how many steps it requires to generate a set using well-ordered unions of singletons, call this the $W$-rank of a set. If $A$ is amorphous of minimal rank, then it is a well-ordered union of proper subsets. Since $A$ is minimally ranked, those must be finite sets, as an infinite subset of an amorphous set is amorphous. So the union must be infinite, and therefore $A$ can be mapped onto $\omega$, at the very least, which is impossible. $\square$

Theorem. If $V=W$ and $G$ is $V$-generic, then $V[G]=W^{V[G]}$.

Proof. Note that there is a surjection from $V$ onto $V[G]$, definable in $V[G]$. Or, formally speaking, for every set in $V[G]$ there is a set in $V$ which maps onto it. Moreover, the closure of $\{\{a\}\mid a\in V[G]\}$ under well-ordered unions, in $V[G]$, contains $V$.

So if we show that this closure is itself closed under surjective images, we are done. This too is done by induction on the $W$-rank. If $x=\bigcup_{\alpha<\delta}x_\alpha$ and $f\colon x\to y$ is a surjection, then by the induction hypothesis $f``x_\alpha=y_\alpha$ is in the closure, and of course $y=\bigcup_{\alpha<\delta}y_\alpha$. $\square$

Finally, the wanted corollary.

Corollary. If $V=W$, then no generic extension of $V$ contains amorphous sets.

On the other hand, over the Cohen model, where $W$ is the class of well-orderable sets, there is a forcing adding an amorphous set (as was shown by Monro). So while the above shows that the ridiculous conjecture is false, it is not entirely clear how ridiculous it was to begin with. Namely, can we refine it to something "more or less" in the spirit of $V\neq W$?

  • $\begingroup$ "Almost well-orderable" is a nice notion! The Bernstein set result I told you about can be phrased as "if $\mathbb{R}$ is almost well-orderable, then there is a Bernstein set." $\endgroup$ Dec 23, 2021 at 21:01
  • $\begingroup$ It is a nice notion. Normally you just start with the class of well-orderable sets as the "ground level", but I originally adapted this from work related to the Gitik model, so I started with countable sets and countable unions. And I got too lazy to rewrite everything from scratch with better details. $\endgroup$
    – Asaf Karagila
    Dec 23, 2021 at 21:31
  • $\begingroup$ Re the last paragraph: this sounds like an approach to resolving the plausible conjecture. If $V \neq W$ implies that countable choice fails in some extension, that combined with your other answer would prove the conjecture. $\endgroup$ Dec 24, 2021 at 23:25
  • $\begingroup$ @ElliotGlazer: Well, the plausible one anyway. Keep an eye out in the next week or so, the first draft is complete, and I expect that by the end of next week we'll finish proofing it and put it online. $\endgroup$
    – Asaf Karagila
    Dec 25, 2021 at 0:34

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