When I am reading through higher Set Theory books I am frequently met with statements such as '$V$ is a model of ZFC' or '$L$ is a model of ZFC' where $V$ is the Von Neumann Universe, and $L$ the Constructible Universe. For instance, in Jech's 'Set Theory' pg 176, in order to prove the consistency of the Axiom of Choice with ZF, he constructs $L$ and shows that it models the ZF axioms plus AC.
However isn't this strictly inaccurate as $V$ and $L$ are proper classes? For instance, by this very method we might as well take it as a $Theorem$ in ZFC that ZFC is consistent since $V$ models ZFC. However this is obviously impossible as ZFC cannot prove its own consistency. I highly doubt that Jech would make a mistake in such classic textbook, so I must be missing something.
How could we, for instance, show Con(ZF) $\implies$ Con(ZF + AC) without invoking the use of proper classes? I imagine, for instance, that we would start with some (set sized) model $M$ of ZFC and apply some sort of 'constructible universe' construction to $M$.
Best Answer
What is shown in the cases you mention is not that the model is a model of ZFC, made as a single statement, but rather the scheme of statements that the model satisfies every individual axiom of ZFC, as a separate statement for each axiom.
The difference is between asserting "$L$ is a model of ZFC" and the scheme of statements "$L$ satisfies $\phi$" for every axiom $\phi$ of ZFC.
This difference means that from the scheme, you cannot deduce Con(ZFC).
For the proof that Con(ZF) implies Con(ZFC), one assumes Con(ZF), and so there is a set model $M$ of ZF. The $L$ of this model, which is a class in $M$ but a set for us in the meta-theory, is a model of ZFC, since it satisfies every individual axiom of ZFC. So we've got a model of ZFC, and thus Con(ZFC).