In an answer and comment to
it was suggested that given a forcing argument using a c.t.m., one could always translate the same argument into a setting without c.t.m.'s. If this is the case, then what is one to make of the following argument by Paul Cohen in his paper "The Discovery of Forcing" (pp. 1090-91):
There was another negative result, equally simple, that remained unnoticed until after my proof was completed. This says one cannot prove the existence of any uncountable standard model in which AC holds, and CH is false (this does not mean that in the universe CH is true, merely that one cannot prove the existence of such a model even granting the existence of standard models, or even any of the higher axioms of inﬁnity). The proof is as follows: 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. I only saw this after I was asked at a lecture why I only worked with countable models, whereupon the above proof occurred to me.
The same proof can be used to show that one cannot prove the existence of a uncountable standard model in which AC holds, and there exists a nonconstructible real.
If one was to use Boolean-valued models or a Boolean ultrapower approach to 'constructing' models in which CH was false or there existed a nonconstructible real, does this mean that one cannot prove that the models so constructed (assuming the models so constructed were standard models) are uncountable, even if the proof using these two methods make no mention of the models' 'countability'?