I'm studiyng set theory in college and I found this theorem:
If $V_\kappa$ is strongly inaccesible, then $V_\kappa \models ZFC$.
The proof of other axioms I understood but there is this statement that I can't understand why.
Since there is $g:a \rightarrow \kappa$ and $\bar{\bar{a}}<\kappa$ and $\kappa$ is regular we have $sup \{ g(c) :c \in a\}<\kappa$
Can someone explain why $sup\{g(c):C\in a\}<\kappa$ ?
We only need the regularity of $\kappa$ here:
Fix a bijection $b$ between $a$ and an ordinal $\alpha$ of the same cardinality ($\alpha=\vert a\vert$ would be a natural choice). Then $g\circ b$ is a map from $\alpha$ to $\kappa$. Since $\alpha<\kappa$ and $\kappa$ is regular, there is no cofinal map from $\alpha$ to $\kappa$ (this is basically the definition of a regular cardinal).
So $g\circ b$ is not cofinal - that is, $\sup\{(g\circ b)(\beta): \beta\in\alpha\}<\kappa$. But since $\{(g\circ b)(\beta): \beta<\alpha\}=\{g(b): b\in a\}$ (why? think about what you know about $g$), we've proved the statement we wanted to.