# Proving ZFC is consistent

I've heard from a friend that we can actually prove the consistency of ZFC if we assume at least one inaccessible cardinal exists. How is this carried out, precisely? Googling doesn't help and my friend just know this neat fact, that's all.

This post on the other math forum goes into more detail, as regards the metamathematical theory (it's subtle!). The mentioned rank initial segment $V_\kappa$ are all sets with rank $< \kappa$ (the inaccessible), where every set has a rank by the axiom of regularity in ZFC. See wikipedia with its link to the cumulative hierarchy.

If you think of all sets as built up from the empty set from using the axioms (so forming pairs, unions, power sets etc.), with these axioms we can never cross (in rank) the rank $\kappa$ (this uses that $\lambda < \kappa \rightarrow 2^\lambda < \kappa$, for the power set axiom, e.g.) So if some model (a class) exists, the small sets (i.e. of rank $<\kappa$) also form a model, and is then a set (not a class). And we can prove this within ZFC.

• Specifically, ZFC proves "If $\kappa$ is inaccessible, then $V_\kappa$ is a model of ZFC." (Just to avoid talking about class models.) – Noah Schweber Apr 15 '16 at 17:04