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Let $\phi$ be a property that's independent of $ZFC$, so that there are strcutures ${\mathfrak A}=(A,{\in}_A)$ (where $A$ is a set or class and ${\in}_A$ is a binary relation on $A$) that are models of $ZFC+\phi$.

I am wondering if for any such structure, we can always find a larger structure ${\mathfrak B}=(B,{\in}_B)$ (so $A \subseteq B$ and ${\in}_B$ coincides with ${\in}_A$ on $A \times A$) that models $ZFC+\lnot{\phi}$ ?

A special case : it would seem that if $\phi$ is the statement "an inaccessible cardinal exists", then inaccessible cardinals in $\mathfrak A$ will stay inaccessible in $\mathfrak B$. But since independence results like Easton's show the arithmetical operations can be counter-intuitive somehow on cardinals, I am not even sure about this. Can the experts help me on this ?

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  • $\begingroup$ The problem makes the most sense for transitive models, or a similar restriction, rather than just when $A$ is a substructure of $B$. Otherwise, there's no reason that the set which is the empty set in $A$ has to be empty in $B$, since the condition only says that no elements of $A$ can appear in the set, not that no new elements of $B$ can appear in the set. $\endgroup$
    – Carl Mummert
    May 1, 2012 at 12:04

3 Answers 3

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If you take a transitive $\omega$-model $\mathcal{A}$ of ZFC, it will satisfy Con(ZFC). Any transitive extension $\mathcal{B}$ will still be an $\omega$-model, and so every transitive extension will also satisfy Con(ZFC). (To see that $\mathcal{B}$ will still be an $\omega$-model, note that $\mathcal{B}$ will still contain the original $\omega^\mathcal{A}$, and $\mathcal{B}$ will recognize $\omega^\mathcal{A}$ as an inductive set containing $\emptyset$, so $\omega^\mathcal{B}$ will be an inductive subset of $\omega^\mathcal{A}$. But since $\omega^\mathcal{A}$ is $\omega$, it has no proper inductive subset, so $\omega^\mathcal{B} = \omega$. )

More generally, you should look into the paper "The Modal Logic of Forcing" by Hamkins and Löwe. That paper only considers forcing extensions, not arbitrary extensions, but it has interesting results about turning independent statements on and off by taking extensions.

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  • $\begingroup$ You don't need the models to be transitive, you just need $B$ to not add any new elements to sets that were already in $A$. In other words, if $B \models x \in y$ and $y \in A$ then $x \in A$ and $A \models x \in y$. $\endgroup$
    – Carl Mummert
    May 1, 2012 at 12:02
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I will give an answer to your special case:

Suppose that $\frak A$ is a transitive model of $ZFC+I$, where $I$ is "there exists a single inaccessible cardinal.". In $\frak A$ we define the notion of forcing $(P,\le)$ where $p\in P$ is a function with a finite domain from $\omega$ to the inaccessible cardinal, and $p\leq q\iff q\subseteq p$ (in this notation, $p$ is stronger).

Let $G$ be a $P$-generic filter over $\frak A$, note that $G$ defines a surjective function from $\omega$ onto the previously-inaccessible cardinal. Denote by $\frak B$ is the generic extension we generate by adding $G$, that is ${\frak A}[G]$. In $\frak B$ there is no inaccessible cardinals. We therefore added a single set $G$ and now $\mathfrak B\models ZFC+\lnot I$.

Note that $\frak A$ thinks that the cardinality of $P$ is the cardinality of the inaccessible cardinal that it knows. However if we also add the requirement that $P$ itself has cardinality of less than the inaccessible then we cannot change its inaccessibility when we go to $\frak B$.

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$\newcommand\ZFC{\text{ZFC}}$Although it is from some time ago, this interesting question has just come to my attention. I must say that you ask some extremely interesting questions.

For this one, despite the indications of the comments and answers, one may actually achieve a positive answer when the initial model is countable. Surprising as it may be, the situation for countable models $\mathfrak{A}$ is that the answer is yes!

Theorem. If $\phi$ is independent of $\ZFC$ and $\mathfrak{A}=\langle A,\in^A\rangle$ is a countable model of $\ZFC+\phi$, then there is a model $\mathfrak{B}=\langle B,\in^B\rangle\models\ZFC+\neg\phi$, where $A\subset B$ and $\in^A$ agrees with $\in^B$ on $A\times A$.

Proof. This follows from the embedding result in my paper

This is the paper that was directly inspired by (and that answers) your previous interesting question. The main result of the paper shows that a countable model of set theory $M$ is isomorphic to a submodel of another model $N$ just in case the ordinals of $M$ embed order-preservingly into the ordinals of $N$. In particular, every nonstandard model of set theory is universal for all countable models of set theory. Since $\phi$ is independent, there is a nonstandard countable model $\mathfrak{B}\models\ZFC+\neg\phi$, and since it is nonstandard, we may find an isomorphic copy of $\mathfrak{A}$ as a submodel of $\mathfrak{B}$, as desired. QED

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