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If I understand its formulation correctly, the completeness theorem states that any consistent theory has set-sized models, as opposed to models of any size. On the other hand, I've seen a lot of answers phrased something like this:

It's consistent with ZFC that there are no set models of ZFC, but there would still be class models such as the von Neumann Universe.

Which seems a bit strange, because it seems to suggest that ZFC could have only proper classes for models, despite being consistent.

I realize this is partially about which things are sets in which models: if a smallest inaccessible cardinal $\kappa$ exists, then the model $V_\kappa$ is a model of ZFC and none of the sets in that model are themselves models of ZFC (right?). But it still seems like we should be able to talk about whether set models "actually" exist, in terms of something like: "There exists a model of ZFC one of whose elements is an inner model of ZFC."

So the question is: what am I missing in the assertion that ZFC is relatively consistent with the absence of set models? Does this just mean that some models of ZFC have no elements which are inner models? Or is it saying something stronger along the lines of "it's possible that although ZFC is consistent, none of its models can be formalized as sets in any other model of ZFC?"

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So the question is: what am I missing in the assertion that ZFC is relatively consistent with the absence of set models? Does this just mean that some models of ZFC have no elements which are inner models?

Yes (though "inner model" has a technical meaning which I'm pretty sure is not what you intend here). To be precise, it means that ZFC (assuming it is consistent) cannot prove there exists a model of ZFC (here by model I will always mean set model). Therefore, by the completeness theorem, assuming ZFC is consistent, there exists a model $(M,\epsilon)$ of ZFC such that there does not exist any $m\in M$ such that $(M,\epsilon)\vDash\text{"$m$ is a model of ZFC"}$, where "$m$ is a model of ZFC" is the usual statement in the language of set theory encoding the statement that $m$ is a model of ZFC.

(Note that such a model $(M,\epsilon)$ cannot think that ZFC is consistent, since if it did, then by the completeness theorem internalized to $(M,\epsilon)$ there would be some element $m\in M$ which $(M,\epsilon)$ thinks is a model of ZFC. So in this setup, ZFC really is consistent, but $(M,\epsilon)$ believes it is inconsistent, and has a nonstandard natural number that encodes a "proof" of a contradiction from ZFC.)

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  • $\begingroup$ Thanks -- and you're right, I was using inner model wrong. That said, I still sort of wonder whether there's a definitive verdict on the second framing I posed. In other words: could it be the case that ZFC, although consistent, has no model which we could view (in some other model) as a set? $\endgroup$
    – Spencer Dembner
    Jul 14, 2018 at 3:56
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    $\begingroup$ Yes: it could be that ZFC is consistent but ZFC+Con(ZFC) is inconsistent. Then every model of ZFC must believe that ZFC is inconsistent and thus has no models. $\endgroup$
    – Eric Wofsey
    Jul 14, 2018 at 4:02
  • $\begingroup$ Well, that's strange. Thanks for clearing it up! $\endgroup$
    – Spencer Dembner
    Jul 14, 2018 at 4:04
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    $\begingroup$ Note though that in this scenario, ZFC would fail to be sound about arithmetic statements. Specifically, ZFC would prove that there exists a proof of a contradiction from ZFC, even though no such proof actually exists. $\endgroup$
    – Eric Wofsey
    Jul 14, 2018 at 4:05
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    $\begingroup$ Continuing by induction: if iterating $T\mapsto T+\mathrm{Con}(T)$ any number of times starting from ZFC produces an inconsistent theory, then we'll need to discard ZFC as a foundation of general mathematics. $\endgroup$
    – hmakholm left over Monica
    Jul 15, 2018 at 1:19

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