I've run across the statement, "The existence of a strongly inaccessible cardinal implies the consistency of ZFC" in several places (Cohen's Set Theory and the Continuum Hypothesis p. 80, for one). His argument is that the set of all sets of rank less than that large cardinal form a model. It seems to me that since we have access to all of those sets without the large cardinal, we could make a model of ZFC without the large cardinal axiom, and thus prove ZFC consistent. This, of course, is false. So, how does the large cardinal provide a model of ZFC that doesn't exist without the large cardinal?



Large Cardinals Imply a Model of ZFC
A model of ZFC is a *set* $E$ together with a binary relation $R$ satisfying the axioms of ZFC (and we can always **if we can** suppose that $E$ is transitive and that $R=\in|\_{E\times E}$ **then the model is said *standard***).
So we not only need to find a class of sets satisfying the axioms of ZFC, but this class must be itself a set.


If you have a strongly inaccessible cardinal $\kappa$, then there is a *set* $V\_\kappa$ which is a **(standard)** model of ZFC, hence ZFC is consistent.
But without the existence of $\kappa$, even if you could consider the class of all sets of rank “less than $\kappa$”, this would not be a set and the theorem ”there is a model implies the consistency of ZFC” would not be applicable (because there is no model).


