I am quite confused about models of ZF(C) set theory; in particular the wording that is used frequently. I have seen many cases of statements such as "assume $V$ satisfies ZFC" where $V$ is the class $\{ x | x = x \}$, i.e. the class of all sets. But, of course, $V$ is an inner model of ZF(C), as ZF(C) $\vdash \phi^V$ vacuously holds for all $\phi \in$ ZF(C). Thus, I thought, there cannot be $V$ which is not an inner model of ZF(C). (Indeed, ZF(C) proves the existence of $V_{\alpha}$ for all $\alpha \in$ ON.)

My question: if "assume $V$ satisfies ZF(C)" is stated, does that mean we are assuming Con(ZF(C)) and are alluding to the Completeness Theorem (i.e. are we working externally with a model, $(V,\in)$, say)? If so, why the clashing nomenclature?

PS: If possible, please edit the title so that it is more appropriate -- I am not quite sure what a specific title for this question would look like.


1 Answer 1


Yes, when we say that $V$ is assumed to satisfy $\sf ZF(C)$, we mean that externally, we assume that $\sf ZFC$ is consistent, and that $V$ is a model of $\sf ZFC$.

An inner model is a technical term in the set theoretic jargon. It means that we work internally to $V$, and we have a proper class $M$, which contains all the ordinals, and $(M,\in)$ satisfies each of the axioms of $\sf ZF$ or $\sf ZFC$ (depending on the context).

So yes, every model of $\sf ZF(C)$ is an inner model of itself. And indeed, every inner model is a model in the meta-theory. But there is a key difference. Inner models are used to calibrate consistency strength.

If we start with a model of $\sf ZF+DC+LM$ (all sets of reals are Lebesgue measurable), then we can find an inner model of that model where there is an inaccessible cardinal (specifically $L$ in this case). We didn't need to create a whole "new universe" or change the ordinals, we only needed to remove a few of the sets.

In that sense, inner models are generally "close enough" to the universe. They cannot be used to prove consistency statement in the theory, but rather in the meta-theory.

To some extent, the opposite of inner models are generic extensions, where we use forcing to carefully add sets into our model so that we do not add ordinals, or "change the model too much" in a technical sense. And indeed, if $V[G]$ is a generic extension of $V$, then $V$ is an inner model of $V[G]$.

  • $\begingroup$ This has been illuminating, thank you very much! A quick follow-up: does that mean we use inner models only as vehicles to prove relative consistency results? $\endgroup$
    – MacRance
    Nov 11, 2018 at 12:22
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    $\begingroup$ @MacRance No. Inner models can be useful in all sorts of situations. Suppose for example that you work in a model without choice. Sometimes it's useful to consider the inner model $L[a]$ for some set $a$ and leverage the fact that $L[a]$ satisfies choice to obtain some statement about $V$. $\endgroup$ Nov 11, 2018 at 13:03
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    $\begingroup$ @MacRance: Indeed, as Stefan points out, there are other uses for inner models such as appealing to absoluteness arguments and so on. But for the most part, inner models are "famous" for their importance in relative consistency results. $\endgroup$
    – Asaf Karagila
    Nov 11, 2018 at 13:10
  • $\begingroup$ @Asaf A follow-up question has arisen, I hope you don't mind: Is it true that any class is a set in the meta-theory? (I am referring to your sentence any inner model is a model in the meta-theory, i.e. a set in the meta-theory but a class in the theory, right?) $\endgroup$
    – MacRance
    Nov 14, 2018 at 9:54
  • $\begingroup$ @MacRance: In "standard setting" where the universe is a set in the meta-universe, which satisfies some basic set theory, yes. It's an object of the meta-universe. $\endgroup$
    – Asaf Karagila
    Nov 14, 2018 at 9:57

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