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For countable transitive models of ZF, is existence of a ZFC extension with the same height a first order property?

In other words, is there a statement $τ$ (in the language of set theory) such that for c.t.m. $M$ of ZF, $(M⊨τ) ⇔∃N⊨\text{ZFC} \, (N⊇M ∧ Ord^N = Ord^M)$?
For the question, we are assuming that a c.t.m. of ZF exists; $Ord$ is the class of all ordinals. The motivation is to understand whether there is a general characterization of when we can add choice to a model of ZF.

Also, if a c.t.m. $M⊨\text{ZF}$ is extendible to a model $N⊨\text{ZFC}$ with the same height, is there such an $N$ such that every set in $N$ is set generic over $M$, and can the forcing be homogeneous?

AC can be added by set forcing iff we have Small Violation of Choice (SVC): $∃X ∀Y ∃α∈Ord \, ∃(\text{onto }f):X×α→Y$ (note that AC proves SVC). However, SVC is not necessary to add choice by class forcing.

Existence of a proper class of Löwenheim-Skolem (LS) cardinals is in a sense a natural generalization (as in weakening) of SVC. A proper class of LS cardinals allows AC to be added by class forcing (A note on Löwenheim-Skolem cardinals by Usuba, 2020). Specifically (if my understanding of forcing theory is correct), we can get AC by iterating with full (alternatively: Easton) support until done: Find the least $κ$ such that $\text{DC}_κ$ fails, and use $\operatorname{Col}(κ,V_λ)$ for the least $λ$ with $\operatorname{cf}(λ)≥κ$ such that $\text{DC}_κ$ holds in the extension.

I suspect (and this is in scope for the question) there are equivalences between certain nice ways of adding choice and having enough LS-like cardinals.

However (without assuming a proper class of LS cardinals), I do not know if AC can sometimes only be forced by other means. Perhaps there is a c.t.m. of ZF where AC can be added by class forcing but DC cannot be added by set forcing. Or a c.t.m. of ZF+DC where AC can be added by class forcing but only by collapsing $ω_1$. Existence of such cases (do they exist?) without a general method of getting AC in them would suggest a negative answer to the top question.

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  • $\begingroup$ Probably stupid question: what happens in your question if we replace “countable transitive model” by just “transitive model”? Is there a statement $\tau$ such that a transitive model $M$ of ZF satisfies $\tau$ iff there is a transitive model $N$ of ZFC containing $M$ with the same ordinal height as $M$? Did you add “countable” because the answer without “countable” is trivial (I don't see it)? $\endgroup$
    – Gro-Tsen
    Commented Oct 8 at 21:07
  • $\begingroup$ @Gro-Tsen Countability is used to ensure that generic sets for M exist in V; otherwise existence of generics would not be a first order property. $\endgroup$ Commented Oct 8 at 21:20
  • $\begingroup$ @Gro-Tsen: See (5) in my answer for that. $\endgroup$
    – Asaf Karagila
    Commented Oct 8 at 21:27

1 Answer 1

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There's no known condition, and this isn't very well researched in the literature.

  1. Monro, in his paper on Dedekind finite sets, constructs a model with a proper class of Dedekind finite sets, so AC and DC cannot be forced by a set forcing, but in that model we can do a class forcing and restore choice.

  2. The Bristol model is locked between $L$ and $L[c]$, so it can be extended to a model of choice. It is not clear if this extension is a class forcing, and if so, is it definable in the model?

  3. If you take Monro's model, or even the Bristol model, and you "do a Feferman-Levy" symmetric extension to make $\omega_1$ singular, then you can still extend to get choice, but you must collapse $\omega_1$.

  4. In the Morris model, we cannot extend the model to a model of $\sf AC$ without adding ordinals because we have countable unions of countable sets that can be mapped on increasingly large sets. So, once we well-order everything, all cardinals must be below the continuum. Similarly in the Gitik model, everything has countable cofinality. This gives you an idea about necessary conditions. You need at least a proper class of regular cardinals. You need at least a bound on iterated power sets of sets that would have some fixed cardinality.

  5. As a side curiosity, Cohen, Solovay, and ultimately Friedman, showed that if $M$ is a countable model of $\sf ZFC$ of height $\alpha$, it can be extended to a model of $\sf ZF$ of the same height which is uncountable. These models cannot be extended back to models of $\sf ZFC$ without collapsing cardinals in the universe itself! Since a countable height implies countable when $\sf ZFC$ holds in the model.


Bibliography.

  1. G. P. Monro, Independence results concerning Dedekind-finite sets. J. Aust. Math. Soc., Ser. A 19, 35-46 (1975) (ZBL0298.02066).

  2. Asaf Karagila, The Bristol model: an abyss called a Cohen real. J. Math. Log. 18 (2), Article ID 1850008, 37 p. (2018) (ZBL1522.03215, arXiv:1704.06939).

  3. Asaf Karagila, The Morris model. Proc. Am. Math. Soc. 148 (3), 1311-1323 (2020) (ZBL1477.03212, arXiv:1811.10977).

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  • $\begingroup$ Sorry, typing on my phone. Will add bibliography later. $\endgroup$
    – Asaf Karagila
    Commented Oct 2 at 7:02

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