Existence of inaccessible cardinals implies the consistency of ZFC I wonder if any can sketch for me in very broad lines the proof of the fact that the existence of inaccessible cardinals implies the consistency of ZFC? I don´t know much about set theory, but I find it extremely interesting that this should be the case.
 A: In $\sf ZFC$ one can define a "rank" function on sets. This means, in rough words, how many times we need to iterate the power set function before we can generate a set (taking union at limit stages).
If $\kappa$ is inaccessible then the sets whose rank is smaller than $\kappa$ form a model of $\sf ZFC$. Therefore by the completeness theorem one has the $\sf ZFC$ is consistent, if there exists an inaccessible cardinal.
I'm not getting into details of what exactly are the sets of rank less than $\kappa$, or how to show that all the axioms of $\sf ZFC$ hold in that set. But if one is familiar with these basics definitions then one can easily show that it is the case.
Finally, even if we assume that $\sf ZFC$ is consistent, and therefore has a model, it is still far from implying that there is an inaccessible cardinals. There is a long and curious hierarchy of stronger and stronger assertions regarding the consistency of $\sf ZFC$ (we can require just the consistency of $\sf ZFC$, or the consistency of the theory $\sf ZFC+\rm Con(\sf ZFC)$, and so on; we can require the there are "nice" models of $\sf ZFC$; and more and more).
