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So ZFC has this property that any consistent theory has a model in ZFC. However I believe you can construct models of ZFC in ZF, so ZF should also have the property that any consistent theory has a model in it. How much weaker can you make ZFC and still have the resulting theory have the property that any consistent theory has a model in it? Which and how many of the seven (non-redundant) axiom/axiom-schema can be removed and the property that consistent theories have models in them be retained?

Pineapple Fish
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1 Answers1

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So ZFC has this property that any consistent theory has a model in ZFC.

The statement as you wrote it is meaningless. The precise statement is that in ZFC, one can prove that every consistent theory has a model (aka the completeness theorem). This does not mean that every theory which actually is consistent can be proved to have a model in ZFC; in fact, this is not true, unless ZFC is inconsistent.

However I believe you can construct models of ZFC in ZF

This is either true or false depending on how you interpret it. ZF cannot prove that there exists a model of ZFC unless both theories are actually inconsistent. However, there is a proper class $L$ which allows ZFC to be interpreted in ZF (and there are many other such proper classes). Since $L$ is a proper class, not a set, it cannot strictly speaking be called a “model in ZF”, but people often use the term “inner model”.

so ZF should also have the property that any consistent theory has a model in it.

Again, this statement is not meaningful as stated. One thing that is known is that ZF cannot prove the completeness theorem (unless ZF is inconsistent), so it is relatively consistent with ZF that some consistent theory has no models. The fact that $L$ is an “inner model” of ZFC means that ZF can prove that every consistent theory in $L$ has a model in $L$; however, it is certainly possible for theories not to lie in $L$.

How much weaker can you make ZFC and still have the resulting theory have the property that any consistent theory has a model in it?

If you are asking how much weaker you can make ZFC such that the completeness theorem still holds, the answer is “pretty weak”. You can prove the completeness theorem in BZ + the Boolean prime ideal theorem, which is vastly weaker than ZFC (much weaker version of choice, no axiom scheme of replacement, and a much weaker version of separation called $\Delta_0$ separation). There is also no need for the axiom of foundation. In fact, the completeness theorem holds in any Boolean topos with natural numbers object where the Boolean prime ideal theorem holds.

Mark Saving
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  • It might be worth noting that ZF does prove the completeness theorem if we restrict our attention to theories in countable (or, more generally, well-orderable) languages. In particular, ZF proves that if ZFC is consistent, then ZFC has a model. – Alex Kruckman Oct 23 '24 at 17:09
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    @AlexKruckman That is indeed relevant, and perhaps I should have stated it. For the special case of a countable recursive theory, this follows immediately from my answer, as all such theories lie in $L$ up to isomorphism. For the general case, you would need to do a bit of work, but no more work than proving the completeness theorem in ZFC. – Mark Saving Oct 24 '24 at 00:19