TITLE: Must uncountable standard models of ZFC satisfy CH?
QUESTION [17 upvotes]: 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.

REPLY [5 votes]: Here is a dichotomy which is close to what you want.
Given a model $V$ of $\mathrm{ZFC} + \mathrm{CH}$.

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

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 $\omega_1^M = \omega_1^V$.