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

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 $f(x)\in V_{\kappa}?$

• Funny! I recently asked a question in the same spirit. Jan 23, 2017 at 13:46

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 $\kappa$.
• Thank you. This is all quite new to me so I'm a little (or very!) confused. 1. what material are we using to show $\exists \lambda<\kappa$ with $\kappa\leqslant 2^{\lambda}$ if $\kappa$ weakly inacc.? 2. I am having trouble relating such a $\lambda$ to stuff in $V_{\kappa}$. We cannot have $x\in V_{\kappa}$ with $|x|=\lambda$ else power-set also fails (?); so is it that there is an $x\subseteq V_{\kappa}$ with that size? [feel free to "explain like I'm five", feeling quite lost] Jan 23, 2017 at 15:28
• (1) Oh right - so, is the assumption/axiom "there is a weakly inacc. cardinal" enough to justify the existence such a $\lambda$? I thought it was consistent with ZFC that all weak inaccessibles are strongly inaccessible (through independence of GHC). Sorry if I'm slow. (2) so what is confusing me is that power set is fine, so it's not clear to me why $2^{\lambda}\geq \kappa$ breaks replacement. What do $\lambda$ and $2^{\lambda}$ correspond to in terms of $V_{\kappa}$ (or $V$ I suppose)? Jan 23, 2017 at 16:55
• As I wrote. if $\kappa$ is strongly inaccessible, there is no such $\lambda$. If $\kappa$ is weakly inaccessible AND NOT strongly inaccessible, then there is such $\lambda$ by definition. Jan 23, 2017 at 17:18
• Ok yes, so in the case of this particular problem (where I am only told $\kappa$ is weakly inaccessible and nothing more), is it true that I cannot give a definitive answer as to whether $V_{\kappa}$ models replacement? Jan 23, 2017 at 17:23