Is Cohen's model of $\mathsf{ZF}+\neg \mathsf{AC}$ transitive?

Question

How many models did Cohen provide for $\mathsf{ZF}+\neg \mathsf{AC}$? Did he find a inner/transitive model?

Answer 1

There is a confusion here about the terms.

An inner model refers to a proper class which is a transitive class and a model of $\sf ZF$ (or $\sf ZFC$ depending on the context).

Transitive model refers usually to a set $M$ such that:

1. $M$ is a transitive set.
2. $(M,\in)$ is a model of set theory.

The mechanism of forcing is usually presented by starting with a countable transitive model. Forcing is nice that way in that it preserves transitivity, and transitive models are nice to begin with.

But forcing has a very good internal definition so that in practice we don't think about that transitive model, instead we work internally to it, and so the forcing is "over the universe". So in a way, one can think of this as though the real universe is the generic extension we're interested in, and the ground model is an inner model over which we do the forcing.

In that sense, Cohen's model is an inner model.

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TL;DR Cohen's model is an inner model of a generic extension of the universe. If the universe is a transitive model in a larger universe, then Cohen's model is also a transitive model in that universe.

Answer 2

Let me add that Cohen's argument of "going to a bigger model and then shrinking to an inner model of the outer model" is, in a sense, inevitable. More precisely, there is no hope (unless ZFC is inconsistent) to always be able to find an inner model of $\sf{ZF}+\neg \sf{AC}$, starting from an arbitrary model of $\sf ZF$.

Suppose that $\sf ZF$ defined a transitive class model of $\sf {ZF}+\neg \sf AC$. In other words, there is some formula $\phi(v)$ in the language $\{\in\}$ with one free variable such that, if $M:=\{x:\phi(x)\}$, then $\sf ZF$ proves that $M$ is a transitive proper class and, for every axiom $\sigma$ of $\sf ZF$, $$ \sf{ZF}\vdash \sigma ^M $$ and $$ \sf{ZF}\vdash\neg \sf{AC}^M $$ Here, $\psi^M$ for a formula $\psi$ denotes the relativization of $\psi$ to $M$, i.e. you replace every $\forall x$ in $\psi$ by $\forall x(\phi(x)\to\dots)$ and every $\exists x$ by $\exists x(\phi(x)\wedge \dots)$.

But wait, the above procedure can be carried out within any model of $\sf ZF$. In particular, it can be done in $L$, to obtain the relativized $M^L$. But $M^L$ is a transitive class model of $\sf ZF$, which means that $L\subset M^L$. But obviously $M^L\subset L$, so $M^L=L$. This is a problem, because $M^L$ thinks choice is false, while $L$ thinks choice is true.