Must uncountable standard models of ZFC satisfy CH?

Question

In Cohen's article, The Discovery of Forcing, he says that "one cannot prove the existence of any uncountable standard model in which AC holds, and CH is false," and offers the following proof.

If $M$ is an uncountable standard model in which AC holds, it is easy to see that $M$ contains all countable ordinals. If the axiom of constructibility is assumed, this means that all the real numbers are in $M$ and constructible in $M$. Hence CH holds.

But this argument, on the surface of it, invokes $V = L$. Can we eliminate the use of $V = L$? The discussion in a related MO question seems to come close to answering this question, but doesn't directly address it.

Answer 1

Remarks (2) and (3) are added in this edit.

What Cohen's quoted proof outline is leaving implicit is the following statement in which $\mathrm{Con}(T)$ means "$T$ is consistent".

$(*)$ Assuming $\mathrm{Con(ZF + SM)}$, $\mathrm{V} \neq \mathrm {L}$ is not provable from $\mathrm{ZF + SM}$, where $\mathrm{SM}$ stands for the statement "there is standard (i.e., well-founded) model of ZF".

$(*)$ is an immediate consequence of the the well-known fact that $\mathrm{Con(ZF + SM)}$ implies $\mathrm{Con(ZF + SM + V = L)}$. This well-known fact, in turn, follows from absoluteness considerations: if $\mathcal{M}\models \mathrm{ZF + SM}$, then $\mathrm{L}^{\mathcal{M}} \models \mathrm{ZF + SM+V=L}$, where $\mathrm{L}^{\mathcal{M}}$ is the constructible universe as computed in $\mathcal{M}$.

By the way: The quoted statement of Cohen in his article is phrased as the theorem below on pages 108-109 of his book "Set Theory and the Continuum Hypothesis". In Cohen's terminology SM stands for the statement "there is standard (i.e., well-founded) model of $\mathrm{ZF}$".

Theorem. From $\mathrm{ZF + SM}$ or indeed from any axiom system containing $\mathrm{ZF}$ which is consistent with $\mathrm{V = L}$, one cannot prove the existence of an uncountable standard model in which $\mathrm{AC}$ is true and $\mathrm{CH}$ is false, nor even one in which AC holds and which contains nonconstructible real numbers.

Three remarks are in order:

Remark (1) In unpublished work, Cohen and Solovay noted that one can use forcing over a countable standard model of ZF to build uncountable standard models of $\mathrm{ZF}$ (in which AC fails by Cohen's aforementioned result). Later, Harvey Friedman extended their result by showing that every countable standard model of $\mathrm{ZF}$ of (ordinal) height $\alpha$ can be generically extended to a model with the same height but whose cardinality is $\beth_{\alpha}$ (Friedman, Harvey, Large models of countable height, Trans. Am. Math. Soc. 201, 227-239 (1975). ZBL0296.02036).

Remark (2) It is easy to see (using the reflection theorem and relativizing to the constructible universe) that, assuming the consistency of $\mathrm{ZF + SM}$, the theory $\mathrm{ZF + SM}$ + "there is no uncountable standard model of $\mathrm{ZFC}$" is also consistent.

Remark (3) Within $\mathrm{ZF}$ + "there is an uncountable standard model $\mathcal{M} \models \mathrm{ZFC+V=L}$ such that $\omega_3^{\mathcal{M}}$ is countable", one can use forcing to build a generic extension $\mathcal{N}$ of $\mathcal{M}$ that violates $\mathrm{CH}$; thus $\mathcal{N}$ is an uncountable standard model of $\mathrm{ZFC + \lnot CH}$. More specifically, the assumption of countability of $\omega_3^{\mathcal{M}}$, and the fact that GCH holds in $\mathcal{M}$, assures us that there exists a $\mathbb{P}$-generic filter over $\mathcal{M}$, where $\mathbb{P}$ is the usual notion of forcing in $\mathcal{M}$ for adding $\omega_2$ Cohen reals. Thus, in the presence of the principle "$0^{\sharp}$ exists" (which is implied by sufficiently large cardinals, and implies that every definable object in the constructible universe is countable) there are lots of uncountable standard models of $\mathrm{ZFC + \lnot CH}$ .

Answer 2

Here is a dichotomy which is close to what you want.

Given a model $V$ of $\mathrm{ZFC} + \mathrm{CH}$.

1. If $\omega_1^V$ is inaccessible to reals then every real number is contained in an uncountable standard model of $\mathrm{ZFC}+\lnot\mathrm{CH}$.

2. If $\omega_1^V$ is not inaccessible to reals then there is a real $x$ such that every uncountable standard model of $\mathrm{ZFC}$ containing $x$ satisfies $\mathrm{CH}$ (perhaps vacuously).

For 1, if $x$ is any real then $\omega^V_1$ is inaccessible in $L[x]$. Then $L_{\omega^V_1}[x]$ is an uncountable standard model of $\mathrm{ZFC}$ and we can force to add $\kappa$ Cohen reals over $L_{\omega^V_1}[x]$ for any $\omega_2^{L[x]} \leq \kappa < \omega_1^V$ to obtain an uncountable standard model of $\mathrm{ZFC}+\lnot\mathrm{CH}$.

For 2, there must be a real $x$ such that $\omega^V_1 = \omega_1^{L[x]}$. Then we can mimic Cohen's argument to see that every uncountable standard model $M$ of $\mathrm{ZFC}$ that contains $x$ in must satisfy $\mathrm{CH}$ since such a model must have $\omega_1^M = \omega_1^V$.