We say that a model $M$ of $\mathsf{ZF}$ satisfies Small Violations of Choice ($\mathsf{SVC}$) if all (any) of the following apply:

  1. There is a model $V\subseteq M$ such that $V\vDash\mathsf{ZFC}$, and $M$ is a symmetric extension of $V$.
  2. There is a forcing $\mathbb{P}\in M$ such that $\mathbb{P}\mathrel{\Vdash}\mathsf{AC}$.
  3. There is $A\in M$ such that for all $X\in M$, there is an ordinal $\eta$ and a surjection $f\colon A\times\eta\to X$ in $M$.
  4. There is $A\in M$ such that for all $X\in M$, there is an ordinal $\eta$ and an injection $f\colon X\to A\times\eta$ in $M$.

The equivalence is true, standard, and very difficult. [1] introduces $\mathsf{SVC}$ and has some work towards proving their equivalence. For the purposes of this question, (1), (3), and (4) are not strictly required, but may be a helpful tool. What I am really interested in is "If (2) then can we force $\mathsf{AC}$ without identifying the cardinalities of distinct ordinals?".

I am particularly interested in condition (2), and in forcing $\mathsf{AC}$ in such a way that if $\lambda\neq\kappa\in M$ are well-ordered cardinals, $\mathbb{P}\mathrel{\Vdash}\check{\lambda}\neq\check{\kappa}$.

Question 1. Is this always possible? No! As Asaf Karagila has pointed out, we certainly require that all successors are regular. Any model of $\text{cf}(\omega_1)=\omega$ would not be able to be extended to a model of $\mathsf{AC}$ without collapsing $\omega_1$ to $\omega$.

Question 2. How much can we save? If $M\vDash\mathsf{DC}_{{<}\lambda}$, can we always keep cardinals at least the size of $\lambda$ distinct?

[1] Blass, Andreas, Injectivity, projectivity, and the axiom of choice, Trans. Am. Math. Soc. 255, 31-59 (1979). ZBL0426.03053.

  • 2
    $\begingroup$ Is the equivalence of (1)–(4) obvious? Is it a standard fact? $\endgroup$
    – Gro-Tsen
    Jun 12 at 15:16
  • $\begingroup$ You obviously want to require that all successors are regular. Otherwise just consider a model where $\omega_1$ is singular. $\endgroup$
    – Asaf Karagila
    Jun 12 at 16:20
  • $\begingroup$ What is the question? (I think you are asking whether (2) (and hence (1), (3) or (4)) implies that there is a set forcing $\mathbb{P}$ which forces AC and preserves the distinction of well-ordered cardinals; is that right?) $\endgroup$
    – Farmer S
    Jun 12 at 16:21
  • $\begingroup$ Also, (1) should say that $V$ satisfies ZFC. $\endgroup$
    – Asaf Karagila
    Jun 12 at 16:21
  • $\begingroup$ @Gro-Tsen: Standard? Yes. Obvious? Not at all. $\endgroup$
    – Asaf Karagila
    Jun 12 at 16:22

1 Answer 1


The answer is no.

Firstly, let's get the obvious out of the way. The Feferman–Levy model is a symmetric extension in which $\omega_1$ is singular. If we force the axiom of choice, we must collapse the ordinal to be countable.

Right. So perhaps we want to require that all successor cardinals are regular. This may be enough? But it is not going to be enough.

Gitik, Moti; Koepke, Peter, Violating the singular cardinals hypothesis without large cardinals, Isr. J. Math. 191, Part B, 901-922 (2012). ZBL1291.03094.

In that paper, the authors start with a model of $\sf GCH$, for any $\lambda>\aleph_{\omega+1}$, they build a symmetric extension which does not change cardinals and cofinalities of ordinals, where $\aleph^*(\mathcal P(\omega_\omega))>\lambda$. This is a violation of the Singular Cardinals Hypothesis which will require the existence of large cardinals, had it taken place in $\sf ZFC$.

Therefore, constructing the Gitik–Koepke model, where there is no inner model with the necessary large cardinal hypotheses, will result in a model where any extension by forcing to restore choice must collapse cardinals.

To your second question, Fernengel and Koepke had extended the original Gitik–Koepke model, first to a proper class situation, then for any set-many cardinals with $\sf DC_{<\lambda}$, under some additional constraints, which we can even show must violate Silver's theorem, and therefore collapse cardinals.

But, if $\sf DC_{<\lambda}$ holds, then at the very least we can force $\sf AC$ with a $\lambda$-closed forcing and therefore not add any bounded subsets to $\lambda$, thus preserving all cardinals up to $\lambda$ itself.

To see that, note that if $A$ is the seed for $\sf SVC$, and $\sf DC_{<\lambda}$ holds but $\sf DC_\lambda$ fails, then $\aleph(A)\geq\lambda$. Simply force with $A^{<\lambda}$, collapsing $A$ to have cardinality $\lambda$.


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