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https://en.wikipedia.org/wiki/Skolem%27s_paradox

In reading about Skolem's paradox on wikipedia, I encountered the claim that every consistent countable axiomatisation of set theory in first order logic has a countable model. That got me thinking that it's strange that set theory could have more than one axiomatisation at all, countably modeled or otherwise. The only set theory I typically hear about is ZFC, and then I recalled NBG set theory, but that is an extension of ZFC, not a separate axiomatisation. Or is NBG considered a separate axiomatisation?

Moreover, I had thought that set theory required second order logic, since it involves quantifying over both sets and relations between sets. How could it have an axiomatisation in first-order logic at all, let alone more than one?

Finally, if indeed set theory (and n.b. that they didn't clarify which set theory, so I would take it to mean ZFC) has more than one axiomatisation in first-order logic, does it have finitely many, denumerably many, or perhaps even non-denumerably many?

Thank you for any advice.

Hank Igoe
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    Are you asking about the motivation for the different set theories that exist or simply how many there are? You can make variants of ZFC such as ZF, ZF + countable choice, ZF - axiom of infinity that may or may not be different enough to meaningfully count as distinct set theories. – Greg Nisbet Jan 05 '21 at 01:06
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    @GregoryNisbet Oh I wonder if I misintrepreted the statement, because I took it to mean that you could get the same set theory, say ZFC, but with a different set of axioms. In other words, you could have ZFC1 and ZFC2 which had the exact same theorems, even though the axioms themselves were different. That is to say, someone might want to swap out axioms from ZFC to some different axioms, and then show that the two axiom systems led to exactly the same set of theorems. Any theorem in the first set theory is derivable in the second theory, and conversely. Is such a thing possible? – Hank Igoe Jan 05 '21 at 01:11
  • Your question touches on a few different subjects, but I think what you're after is a categorical set theory (exactly one model), which necessarily must be non-first-order. You may want to check out this answer, which defines one version of ZFC2 and search for ZFC2 more generally (I can't find many resources on it). Also check out this question. – Greg Nisbet Jan 05 '21 at 01:44

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You can indeed get the same theory from a different set of axioms. A bit more precisely, for a set of sentences $\Gamma$, let the deductive closure of $\Gamma$ be the set $$Ded(\Gamma):=\{\varphi:\Gamma\vdash\varphi\}$$ of sentences which $\Gamma$ proves (I'm tacitly fixing a particular language to work in here, e.g. the language of set theory). Then we can have $Ded(\Gamma_1)=Ded(\Gamma_2)$ even if $\Gamma_1\not=\Gamma_2$.

For example, given $\Gamma$ consider $$\Gamma_{double}:=\{\varphi\wedge\varphi:\varphi\in\Gamma\}.$$ Obviously $Ded(\Gamma)=Ded(\Gamma_{double})$, but in general $\Gamma$ will be a different set of sentences than $\Gamma_{double}$. More interestingly, we can show the following:

  • Every finitely axiomatizable theory is axiomatizable by a single sentence. That is, if $\Gamma$ is finite, then there is some single sentence $\gamma$ such that $Ded(\Gamma)=Ded(\{\gamma\})$. There's actually less to this than meets the eye: just let $\gamma$ be the conjunction of the finitely many sentences in $\Gamma$.

  • Every computably enumerably axiomatizable theory is in fact computably axiomatizable, and indeed primitive recursively axiomatizable and more - this is Craig's trick, and is often useful in the context of incompleteness problems.

And getting back to $\mathsf{ZFC}$ in particular, it's easy to show that the Separation scheme is in fact redundant: we can prove each instance of Separation from the remaining axioms of $\mathsf{ZFC}$. Similarly, we can replace Choice with Zorn's Lemma, the Well-Ordering Principle, or various other statements. So there are lots of natural alternative axiomatizations of $\mathsf{ZFC}$.

(As to how many axiomatizations $\mathsf{ZFC}$ has, it's a good exercise to show that there are in fact continuum-many ways to axiomatize $\mathsf{ZFC}$. In fact, there are continuum-many ways to do this non-redundantly, but that takes more work. Even constructing a single non-redundant axiomatization is nontrivial - see here.)

Noah Schweber
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