Large Cardinals Imply a Model of ZFC 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?
 A: I'd like to put this as a comment, but I don't have sufficient rights: 
It might be interesting to point out that the existence of a transitive model (like the one we get using an inaccessible, mentioned above) does not just imply $Con(ZF)$ but even $Con(ZF+Con(ZF))$.
A: Just one clarification to Guillaume's answer (his answer has been edited by now):  Yes, a model of ZFC is a set
$E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC.
The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded.  $(E,R)$ thinks that $R$ is well-founded, but the real world knows 
an infinite $R$-decreasing chain in $E$.  In this situation, $(E,R)$ is a non-standard model
of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.  
However, an inaccessible cardinal $\kappa$ actually gives you a transitive model of ZFC, 
and this is more than just having any model of ZFC.  (And most approaches to forcing 
would like to have a transitive set that is a model of ZFC (with the binary relation being the usual $\in$-relation).  But there are reasons why we can pretend that we have a transitive set model of ZFC even if we don't really have one.) 
Now, it is not the case that the inaccessibility of $\kappa$ somehow miraculously gives you a new set that is a model of ZFC.  It is more the case that $\kappa$ is so large that if you cut off the universe at $\kappa$, this initial part of the full set-theoretic universe already satisfies ZFC.
I.e., from the point of view of $V_\kappa$, $\kappa$ already looks like the class of all ordinals.
What makes this work is the fact that an inaccessible cardinal has excellent closure properties.
$\forall\lambda<\kappa(2^\lambda<\kappa)$ guarantees that every element of $V_\kappa$ is of size $<\kappa$ and the regularity of $\kappa$ now gives you the axiom of replacement.
The other axioms are rather easily satisfied in this context.  
Note, however, that an inaccesible cardinal can cease to be inaccessible (or even a cardinal) after enlarging the universe (for example by forcing).  The $V_\kappa$ of the ground model (the small original universe) is still a model of ZFC, but the $V_\kappa$ of
the enlarge universe (where $\kappa$ is not inaccessible anymore) need not be a model of ZFC.  
A: 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).
