Proving/disproving that weakly inaccessible cardinals imply $V_{\kappa}$ models ZFC

Question

I'm trying a problem which asks which axioms of ZFC hold - under the assumption that there exists a weakly inaccessible cardinal $\kappa$ - inside $V_{\kappa}$ ($\kappa$ being such a cardinal).

I am relatively sure I have argued that everything holds except replacement which I am struggling with. I don't seem to have used the weakly inaccessible cardinal part with the others, so I am guessing this is where it comes in, but I'm not even sure what I'm meant to be considering. If I take $x\in V_{\kappa}$ and consider some function defined on $x$, am I considering functions into $V_{\kappa}$ only? Even if this is the case, how do I use regularity of $\kappa$ to argue $f(x)\in V_{\kappa}?$

Answer 1

If $\kappa$ is strongly inaccessible, then Replacement holds in $V_\kappa$; but if $\kappa$ is only weakly inaccessible, then there is some $\lambda<\kappa$ such that $2^\lambda\geq\kappa$.

Now, ZF proves that every well-ordered set is isomorphic to an ordinal, but we can now find a well-ordered set inside $V_\kappa$ whose order type is $\kappa$, and therefore is not isomorphic to any ordinal inside $V_\kappa$.

Interestingly, the regularity of $\kappa$ is not what gives you Replacement. In fact, the least $\kappa$ such that $V_\kappa$ is a model of ZF has cofinality $\omega$ (if such $\kappa$ exists, of course). However, regularity does provide you with an guarantee that if all sets inside $V_\kappa$ have cardinality less than $\kappa$, then any function applied to them will also have cardinality less than $\kappa$.