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What is the advantage of using first-order logic over second-order logic? Second-order logic is more expressive and there is also a way to overcome Russell's paradox...

So what makes first-order logic standard for set theory?

Thanks.

Asaf Karagila
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First order logic has the completeness theorem (and the compactness theorem), while second-order logic does not have such theorems.

This makes first-order logic pretty nice to work with. Set theory is used to transform other sort of mathematical theories into first-order.

Let us take as an example the natural numbers with the Peano Axioms. The second-order theory (replace the induction schema by a second-order axiom) proves that there is only one model, while the first-order theory has models of every cardinality and so on. Given a universe of set theory (e.g. ZFC), we can define a set which is a model of the second-order theory but everything we want to say about it is actually a first-order sentence in set theory, because quantifying over sets is a first-order quantification in set theory.

This makes set theory a sort of interpreter, it takes a second-order theory and says "Okay, I will be a first-order theory and I can prove this second-order theory." and if we have that the set theory is consistent then by completeness it has a model and all the higher-order theories it can prove are also consistent.

To read more:

  1. what is the relationship between ZFC and first-order logic?
  2. First-Order Logic vs. Second-Order Logic
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Asaf Karagila
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    I think some parts of this answer might be slightly misleading. It's only true that second order logic with full semantics lacks a compactness theorem. / It doesn't make sense to say "the second order theory proves that there is only one model". Of course we can prove in the metatheory that there's only one full model of PA2, but that can't even be stated in the second-order theory of $\mathbb{N}$, much less proved. Moreover, it isn't the second order induction axiom that causes categoricity, it's the restriction in the metatheory to only look at full models. – Carl Mummert Feb 21 '14 at 04:12
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    @CarlMummert So if I allow Heniken semantics for second order logic then PA2 wouldn't be categorical any more? – Trismegistos Oct 01 '14 at 19:05
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    Yes, exactly. The thing that makes it categorical is not so much the theory as the declaration that we will only consider full models - although the theory helps via the induction axiom. The theory PA2 (which often goes by the name $Z_2$) has many nonstandard Henkin models. @Trismegistos – Carl Mummert Oct 01 '14 at 19:08
  • @CarlMummert Are $Z_2$ nonstandard models equally exotic as PA1 models? Are they of different cardinality than natural numbers? – Trismegistos Oct 01 '14 at 19:20
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    Yes; the usual Lowenheim-Skolem theorem applies. There are models of every infinite cardinality. Actually, there are no countable full models, since the powerset of the integers in a full model will be uncountable. But in fact for any infinite cardinalities $a,b$ with $a \leq b$ there is a Henkin model of $Z_2$ whose numbers have cardinality $a$ and whose collection of sets has cardinality $b$. @Trismegistos – Carl Mummert Oct 01 '14 at 19:43
  • @Carl: Don't you have to require that $b\leq 2^a$ as well? – Asaf Karagila Oct 01 '14 at 19:45
  • @Carl: I don't understand your comment. You original said that if $a\leq b$ are infinite cardinals, then there is a Henkin model of $Z_2$ whose numbers have $a$ elements and $b$ is the number of subsets of numbers. I was asking for an upper bound, not a lower bound, on $b$. If $2^a<b$, how can you have so many subsets of something whose cardinality was only $a$? – Asaf Karagila Oct 01 '14 at 22:52
  • Ah ha. I did read that inequality in reverse originally. There is no equality sign for sets in the language of $Z_2$. You are correct on the inequality if there is an equality sign for sets that must be interpreted as true equality. @Asaf – Carl Mummert Oct 01 '14 at 23:11
  • @Carl: I see! It seems a bit strange not to add some sort of extensionality axiom. Thanks! – Asaf Karagila Oct 01 '14 at 23:21
  • The convention I'm used to has set equality as a definitional extension rather than an axiom. The motivation is somewhat subtle. One reason is so the truth set of a $\Sigma^0_1$ formula $\phi(n)$ with set parameters $X,Y$ is uniformly r.e. in $X$ and $Y$ (which it morally ought to be, since it's $\Sigma^0_1$). Similarly, it's so the atomic diagram of a countable coded $\omega$-model is computable from the code. A set equality "$X = Y$" is "morally" a $\Pi^0_1$ formula - it hides a universal number quantifier. In elementary settings, it probably makes more sense to include equality. @Asaf – Carl Mummert Oct 02 '14 at 00:06
  • @Carl: Thanks again! – Asaf Karagila Oct 02 '14 at 00:08
  • Set theory is used to transform other sort of mathematical theories into first-order. ZFC axioms can't even be stated in FOL. You need a first-order separation schema (infinitely many separation axioms) – MWB Sep 17 '20 at 19:29
  • @MaxB: I'm confused, the axioms in the schema are not first order? – Asaf Karagila Sep 17 '20 at 19:31
  • You may consist of atoms, but that doesn't make you an atom. – MWB Sep 17 '20 at 19:41
  • @MaxB: The axioms are all first-order logic, therefore the set of axioms is a set of FOL axioms. There's really nothing more to say here. To twist your argument back, if ZFC is not a first-order theory, please point me to an axiom which is not first-order. And do remember, an axiom schema is not an axiom, but rather a schema of many different axioms. – Asaf Karagila Sep 17 '20 at 19:55
  • therefore the set of axioms Aren't you using sets to define sets? I decided to ask a separate question to avoid extended comment discussions: https://math.stackexchange.com/questions/3830319 – MWB Sep 17 '20 at 20:39
  • @AsafKaragila I understand that set theory is a sort of interpreter, but do not understand the importance of completeness theorem. Could you elaborate on it? – Hayatsu May 13 '23 at 23:14