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Because a model of a first order theory is not allowed to use a proper class as its domain, we can't use the universe of the set theory from the "meta-level" directly as a model for a first order theory of its axioms.

This situation alone already appears paradoxical to me, but it gets worse if I look at some different set theories I could use at the "meta-level". If I use NFU, there seems to be no problem to use it directly as a model of its axioms. (However, I'm not very familiar with NFU, so I could be wrong.) If I use ZFC or NBG on the other hand, it looks like they are unusable as a model of their own axioms. Allowing to use a class as the domain for a model also seems problematic, because relations are modeled as sets in ZFC (so for example $\in$ is not a relation in the sense of ZFC). The situation gets even more strange if I use pocket set theory, because then any set theory (including pocket set theory itself) seems to have an inner model (if the axiom of free construction is added) in that set theory (which is necessarily a countable model). However, I have the impression that it would be even possible to allow the use of a class as domain for a model in case pocket set theory is used at the "meta-level". (However, I'm not very familiar with pocket set theory, so I could be wrong.)

Is what I described above considered to be a paradox? Is there a "canonical" resolution of that paradox for ZFC (or NBG). Is it really the case that this paradox only happens to a lesser extent for NFU and pocket set theory, or is it just my unfamiliarity with these theories that prevent me from noticing that the same paradox also applies to these theories?

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Pocket set theory? –  Asaf Karagila Jan 3 '13 at 15:05
@Asaf: Wikipedia to the rescue! –  Arthur Fischer Jan 3 '13 at 15:10
I learned about this from SEP in the article about alternative set theories, which was written by Randall Holmes. The wikipedia article about pocket set theory cites Rudy Rucker as the inventor of that theory, but everything I have read about it so far was written by Randall Holmes. –  Thomas Klimpel Jan 3 '13 at 15:11
Any sufficiently powerful set theory cannot possibly contain a model of itself as a set, because then that set theory would prove its own consistency. So, for example, in NF, even though there is a universal set, there is no set that implements the $\in$ relation. (I suspect something even worse is true: for "most" sets $X$ there is no set $R$ such that $R = [\in] \cap (X \times \mathscr{P} X)$.) On the flip side, in ZF there is no universal set, but for every set $X$ there is a set $R$ such that $R = [\in] \cap (X \times \mathscr{P} X)$. So either way you have to give up something. –  Zhen Lin Jan 3 '13 at 15:32
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1 Answer

up vote 1 down vote accepted

"Because a model of a first order theory is not allowed to use a proper class as its domain, we can't use the universe of the set theory from the "meta-level" directly as a model for a first order theory of its axioms."

Proper classes are used as models in the metatheory especially in relative consistency proofs. Proper class models provide finitistic proofs in the metatheory.

For example, the proper class model $V = \{x : x = x\}$ gives you the finistic proof in the metatheory that $\text{Con}(ZFC) \Rightarrow \text{Con}(ZFC)$, which is absolutely trival.

Some less trivial example of class models, let $ZF^-$ denote $ZFC - \text{foundation}$. The proper class $WF$ of well-founded sets is a proper class model satisfying the axiom of foundation. Hence $\text{Con}(ZF^-) \Rightarrow \text{Con}(ZF)$.

Also letting $L$ be Godel's Constructible universe, $L$ is proper class model of $ZF$, $V = L$, $AC$, and $GCH$. This proper class model yield the finitistic proof of relative consistency of these axioms from consistency of $ZF$.

So at the "meta-level", proper class models are useful for providing consistency proof in the metatheory. The formal precise logic of class models involve relativization.

However, sufficiently strong set theory is not capable of proving that there is a model of itself. The existence of a model of $ZFC$ proves consistency of $ZFC$. Hence $ZFC \vdash \text{Con}(ZFC)$. By the incompleteness theorem, this can only occur if $ZFC$ is inconsistent.

$ZFC \vdash Con(X)$ is stronger than the finitistic $\text{Con}(ZFC) \Rightarrow Con(X)$, where $X$ is some theory.

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I don't understand the downvote. If I understand you and Zhen Lin correctly, you are claiming that if we could prove that there is a set model in $ZFC$ for itself without assuming $Con(ZFC)$, then we would have proved $ZFC \vdash Con(ZFC)$. This may be true, but it's not obvious to me why. –  Thomas Klimpel Jan 4 '13 at 0:42
@ThomasKlimpel This is a basic model theory fact. ZFC can formalize the notion of first order logic, models, $\models$, $\vdash$, and other logical notions. ZFC (choice may be necessary) can then prove the Godel's Completeness Theorem that states that a any first order theory is consistent if and only if it has a model. Therefore, if you can produce a set that is a model of $ZFC$, you have proved in ZFC that $ZFC$ has a model. By the completeness theorem, which ZFC can prove, $ZFC$ is then consistent. –  William Jan 4 '13 at 4:39
@ThomasKlimpel But note that intuitively $V = \{x : x = x\}$, the universe, "should be a model" of $ZFC$ since all the axioms of $ZFC$ holds in $V$. However, $ZFC$ can prove $V$ exist. However the first part of my answer is point out that proper class model like $L$ and $WF$ are useful for proving relative consistency result but these proof are not proofs in $ZFC$ just in the finite metatheory. –  William Jan 4 '13 at 4:43
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