Using the reflection theorem we can prove in $\mathsf{ZFC}$ that for any finite $\Lambda\subseteq\mathsf{ZFC}$ there is some limit ordinal $\gamma$ such that $R(\gamma)\models\Lambda$. Then by the completeness and compactness theorem we get that there is some set $M$ and some binary relation $E$ on $M$ such that $\mathsf{ZFC}\vdash[(M,E)\models \mathsf{ZFC}]$. But then as $\mathsf{ZFC}$ cannot prove the existence of a transitive model of $\mathsf{ZFC}$, $E$ cannot be well-founded on $M$; for otherwise the transitive collapse of $(M,E)$ would be a transitive model of $\mathsf{ZFC}$.

Can we use this to prove that $(M,E)$ does not satisfy the axiom of foundation?, and if so, why this does not contradict $\mathsf{Con(ZFC)}$?


  • 3
    $\begingroup$ I think that everyone coming across the Levy reflection theorem ends up with this question at some point. $\endgroup$
    – Asaf Karagila
    Aug 28, 2013 at 13:27
  • 5
    $\begingroup$ Can we really prove in ZFC that "for any finite $\Lambda\subseteq ZFC$ there is some [model of] $\Lambda$" If so, compactness would immediately give us a proof that ZFC itself has a model, which Gödel wouldn't like much. Isn't the situation rather than for any finite $\Lambda$ there is a proof that $\Lambda$ has a model (but the proof will in general depend on $\Lambda$, so it cannot be prefixed with a quantification over $\Lambda$). $\endgroup$ Aug 28, 2013 at 13:32
  • $\begingroup$ Remember that ZFC is not a finitely axiomatised theory, so the compactness theorem only guarantees that there exists $(M, E)$ such that, for each axiom $\phi$, $(M, E)$ satisfies $\phi$. But that does not mean that $(M, E)$ is internally a model of ZFC! Whether or not $(M, E)$ is well-founded is besides the point. $\endgroup$
    – Zhen Lin
    Aug 28, 2013 at 14:24
  • $\begingroup$ @HenningMakholm, you're saying that we know that in the metatheory we can prove that for any finite subset of $\mathsf{ZFC}$, $\mathsf{ZFC}$ proves it has a model, but that we cannot see that each finite subsets of $\mathsf{ZFC}$ have a model simultaneously, and thus we cannot even use the compactness theorem? $\endgroup$ Aug 28, 2013 at 21:32
  • $\begingroup$ @CamiloArosemena: Yes, that's what I'm saying. $\endgroup$ Aug 28, 2013 at 21:52

1 Answer 1


Let $\mathsf{ZFC}$ denote the recursively-axiomatised first-order theory of ZFC, formalised in a metatheory (say, second-order arithmetic) sufficiently strong to prove (the syntactic version of) compactness. Lévy's reflection principle says the following:

For each formula $\phi$ in the language of set theory, $\mathsf{ZFC} \vdash \exists \alpha . \phi^{V_\alpha} \leftrightarrow \phi$, where $\phi^M$ denotes the relativisation of $\phi$ to $M$.

In particular, this is a metatheorem: it involves a quantification in the metatheory. Nonetheless, using the compactness theorem, we can carry out a version of your argument. Let $\mathsf{F}$ be the first-order theory obtained by adding to the language of set theory a constant $\alpha$, with the axioms of $\mathsf{F}$ being the axioms of $\mathsf{ZFC}$ plus an axiom $\phi^{V_\alpha} \leftrightarrow \phi$ for each formula $\phi$ in the language of set theory. ($\mathsf{F}$ stands for Feferman.) Then:

$\mathsf{F}$ is consistent if and only if $\mathsf{ZFC}$ is consistent.

Now, let $\mathtt{ZFC}$ denote the (recursively-axiomatised first-order) theory of ZFC formalised in $\mathsf{ZFC}$. Your claim is essentially that "$\mathsf{F} \vdash (V_\alpha \vDash \mathtt{ZFC})$", but this is not correct.

Consider the theory $\mathsf{F} + (\mathtt{ZFC} \vdash \bot)$: this theory is consistent if and only if $\mathsf{F}$ is, by the construction of $\mathsf{F}$. Clearly, in $\mathsf{F} + (\mathtt{ZFC} \vdash \bot)$, we cannot have $V_\alpha \vDash \mathtt{ZFC}$. Thus it must be that $\mathsf{F} \not\vdash (V_\alpha \vDash \mathtt{ZFC})$ (if $\mathsf{F}$ is consistent). This sounds confusing but it is actually quite simple: if $\mathsf{F}$ (hence $\mathsf{ZFC}$) is consistent, then any model of $\mathsf{F} + (\mathtt{ZFC} \vdash \bot)$ must contain non-standard numbers, and these non-standard numbers imply that $\mathtt{ZFC}$ contains non-standard axioms. So while it is true that $V_\alpha$ (in any model of $\mathsf{F}$) is (externally) a model of $\mathsf{ZFC}$, it need not be (internally) a model of $\mathtt{ZFC}$.


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