Can the truth value of an independent property be changed at will by enlarging the model?

Question

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 ?

Answer 1

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.

Answer 2

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 $\frak B$.