How do I show the existence of a weakly inaccessible cardinal is not provable in ZFC?

Question

So I know that if we assume the existence of inaccessible cardinals than we can show ZFC is consistent. How do I show that the existence of such cardinals is not provable in ZFC?

Answer 1

Pretty much what Alan said in the comment but I will elaborate a little bit.

First in the title of the post you talk about weakly inaccessible cardinals whereas in the body you just talk about inaccessible cardinals. Neither of these can be proven to exist in ZFC since both give you models.

For a strongly inaccessible cardinal $\kappa$, finding the model is particularly easy since $V_\kappa$ is it. This follows in a fairly straightforward manner from strong inaccessibility. Most of the axioms of ZFC will hold even in $V_\beta$ for $\beta$ just a limit ordinal. The problematic ones are Replacement and Power set.

For replacement you need a regular limit (i.e. weakly inaccessible) so you can't get cofinal sequences. For power set you just need pretty much power set, i.e. strongly inaccessible ($\forall \alpha<\kappa(2^\alpha<2^\kappa)$). Notice that assuming GCH, you get "weakly inaccessible" = "strongly inaccessible".

So from what we have seen so far, it seems we need strongly inaccessible to get a model of ZFC ... but there is a trick. Actually there are probably many tricks, but the usual trick I've seen is to look at $L$ (Gödel's constructible universe). In $L$ we get GCH for free and so any weakly inaccessible becomes a strongly inaccessible and $L(\kappa)$ gives us a model for ZFC.

Edit: One of the comments gave me another idea. If for some reason you don't know anything about $L$ yet, but you do know that GCH is consistent (unlikely, but maybe possible), then you can just use that. Since if ZFC proves $\exists \kappa$, $\kappa$ weakly inaccessible, then in a model of ZFC+GCH this provable weakly inaccessible becomes strongly inaccessible.

Note: When I say $X$ is consistent I always mean $Con(ZFC)\implies Con(ZFC+X)$.