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, $V_{\kappa}\models ZFC$


1 Answer 1


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 $\kappa$).

and your third

$$ ZFC\vdash \forall\kappa \mbox{ inaccessible},\forall \phi\in \ulcorner ZFC\urcorner,\;\phi^{V_\kappa}$$

using Quine corners to denote when there's some coding of formulas from the metatheory involved.

  • $\begingroup$ 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. $\endgroup$
    – Lorenzo
    Commented May 18 at 16:53
  • $\begingroup$ @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 $\kappa$ ? $\endgroup$ Commented May 18 at 17:03
  • $\begingroup$ @spaceisdarkgreen Could you prove it ? $\endgroup$ Commented May 18 at 17:09
  • $\begingroup$ 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 $ZFC$ ? $\endgroup$ Commented May 18 at 17:15
  • $\begingroup$ @Lorenzo yes, you’re right. $\endgroup$ Commented May 18 at 18:12

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