# Inaccessible cardinals and consistency of ZFC

I don't see how to state and prove rigorously the following result. Which of the following statement makes sense and is true ?

For all $$\phi$$ in $$ZFC$$, we have : $$ZFC+\exists \kappa'(\kappa' \text{ inaccessible})\vdash \forall\kappa(\kappa \text{ inaccessible}\rightarrow \phi^{V_{\kappa}})$$

For all $$\phi$$ in ZFC and for all inaccessible cardinal $$\kappa$$, we have : $$ZFC+\exists \kappa'(\kappa' \text{ inaccessible})\vdash \phi^{V_{\kappa}}$$

For $$\kappa$$ an inaccessible cardinal, $$V_{\kappa}\models ZFC$$

For all $$\phi$$ in $$ZFC$$, we have : $$ZFC+\exists \kappa'(\kappa' inaccessible)\vdash \forall\kappa(\kappa inaccessible\rightarrow \phi^{V_{\kappa}})$$

This makes sense and is true, provided we interpret $$\phi^{V_\kappa}$$ to be the statement internal to set theory that the code of the formula $$\phi$$ is true in $$V_\kappa.$$ And you don't need the extra assumption that there exists an inaccessible cardinal... it would hold just fine (vacuously) if there were none.

EDIT: As Lorenzo notes in the comments, there's not actually an issue with interpreting the notation $$\phi^{V_\kappa}$$ as just the relativization of $$\phi$$... not sure what I was thinking.

For all $$\phi$$ in ZFC and for all inaccessible cardinal $$\kappa$$, we have : $$ZFC+\exists \kappa'(\kappa' inaccessible)\vdash \phi^{V_{\kappa}}$$

This one does not make sense, since it doesn't make sense to talk about provability in ZFC (+inaccessibles) of a formula with parameters, which $$\phi^{V_\kappa}$$ is.

For $$\kappa$$ an inaccessible cardinal, $$V_{\kappa}\models ZFC$$

This makes sense as a statement of set theory and is provable in ZFC. Your first statement is almost the statement that this last one is provable in ZFC (+inaccessibles, but as I mentioned, that's irrelevant). Only the difference is there you quantified over formulas in the metatheory, whereas $$V_\kappa\models ZFC$$ means we're quantifying over formulas internally in the set theory . i.e., using similar notation as your first, we have $$ZFC\vdash \forall\kappa \mbox{ inaccessible},\forall \phi\in ZFC,\;\phi^{V_\kappa}$$

It can be useful to abuse notation less to keep things straight. If I were being careful, I'd write your first statement

For every $$\phi$$ in ZFC, we have $$ZFC\vdash \forall \kappa\mbox{ inaccesible,}\; V_\kappa\models \ulcorner \phi\urcorner$$

although (see edit above) the original phrasing does work as is when we take $$\phi^{V_\kappa}$$ as an abbreviation of the relativization of $$\phi$$ (which is itself a first order formula in $$\kappa$$).

$$ZFC\vdash \forall\kappa \mbox{ inaccessible},\forall \phi\in \ulcorner ZFC\urcorner,\;\phi^{V_\kappa}$$
• We don't need codes for formulae to make sense of the first statement. The sentence $\phi^{V_\kappa}$ is the the relativization of the formula $\phi$ to $V_\kappa$ (which is a $\Delta_1^{\mathsf{ZF}}$-definable set with parameter $\kappa$). Codes of formulae are needed only to make formal sense of the third statement. Commented May 18 at 16:53
• @Lorenzo What do you mean by $V_{\kappa}$ is a $\Delta_1^{ZF}$-definable set with parameter $\kappa$ ? Could you write explicitly a formula characterizing the set $V_{\kappa}$ with a parameter $\kappa$ ? Commented May 18 at 17:03
• For the third statement, is it enough to simply show $V_{\kappa}\models \phi$ as if we were in basic logic,model theory ? I mean just by making a basic metatheoric proof of the fact that $V_{\kappa}$ satisfies each axiom of $ZFC$ ? Commented May 18 at 17:15