In Cohen's article, [The Discovery of Forcing](https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-32/issue-4/The-Discovery-of-Forcing/10.1216/rmjm/1181070010.full), 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](https://mathoverflow.net/questions/151234) seems to come close to answering this question, but doesn't directly address it.



Must uncountable standard models of ZFC satisfy CH?

> 
> 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**Remark (1)** In unpublished work, Cohen and Solovay noted that one can use forcing over a countable standard model of ZF to build *uncountable* standard models of $\mathrm{ZF}$ (in which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**](http://dx.doi.org/10.2307/1997333), Trans. Am. Math. Soc. 201, 227-239 (1975). [ZBL0296.02036](https://zbmath.org/?q=an:0296.02036)).


**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](https://en.wikipedia.org/wiki/Zero_sharp) "$0^{\sharp}$ 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 $\mathrm{ZFC + \lnot CH}$ .


