Do choice principles in all generic extensions imply AC in $V$?

Question

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),

and

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).

Answer 1

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. $\quad\square$

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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.

Answer 2

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.

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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 $V\neq W$?