Is set theory used in proofs of theorems in logic ? For example, is set theory used in Gödel's incompleteness theorem ? If yes, than does it mean that if we use other from ZFC set theory or even other theory (I mean not set-theory) as foundation than this theorem will not hold ?
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Goedels incompleteness theorem is not only about ZFC. It states that every sufficient strong system (PA is enough) is either inconsistent or incomplete. – Peter Jul 28 '18 at 11:31
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There are statements provable in ZFC, but not in PA , the most famous example is the statement that every Goodstein-sequence eventually terminates. – Peter Jul 28 '18 at 11:33
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3@Peter I think the question is asking about the theory in which, not about which, these results are proved. – Noah Schweber Jul 28 '18 at 12:47
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Yes, set theory is used in some theorems (e.g. the completeness theorem), but not in the incompleteness theorem.
When Gödel proved his incompleteness theorem he worked very hard on proving it using only Peano axioms of the natural numbers, since those were "indisputable" compared to the set theoretic axioms which were still being scrutinized by some people. This was later improved by various mathematician and we know now that you only need a much weaker theory for incompleteness to kick in.
On the other hand, the completeness theorem speaks about models, which are inherently sets. So of course that set theory comes into play there. In fact, the general statement of the completeness theorem for first-order logic is equivalent to the Boolean Prime Ideal theorem, and is therefore not provable without the axiom of choice. It should be remarked that if the language is countable, then choice is not needed.
We can talk about completeness in theories like Peano arithmetic, which provides us with weaker results, though. There the notion of a model becomes slightly less natural than the one in set theory.
Asaf Karagila
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2+1. Two comments for the OP: first of all, the completeness theorem holds without choice as long as the language is well-orderable; even countability isn't necessary. Moreover, ZF is (provably in a very weak base theory) equiconsistent with ZFC, so we shouldn't be too skeptical of adding choice, and indeed the "concrete" consequences of ZF and ZFC are (again, provably in a weak theory) the same. (cont'd) – Noah Schweber Jul 28 '18 at 12:40
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1Second, and more significantly I think, doing model theory in PA (or any theory of first-order arithmetic - that is, numbers only) is inherently difficult since PA doesn't seem to have a way to talk directly about infinite structures (whereas finite structures, in a finite language at least, can be coded by natural numbers). There are ways around this, and in particular you may be interested in the arithmetized completeness theorem, but the difficulty is significant enough in my opinion that it's worth explicitly mentioning. – Noah Schweber Jul 28 '18 at 12:44
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3Good comments, Noah. I think you missed a link there at the second comment. :) – Asaf Karagila Jul 28 '18 at 13:03
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Yes, changing the "background theory" can lead to alternate proof/model theories. However, in order to get really meaningful differences (that is, differences at the "basic level" like the incompleteness theorem) you need to do significant damage to the background, to the point that it doesn't seem to hold much interest. I believe Hajek and Pudlak's book goes into proving basic proof theoretic results in weak theories of arithmetic, and I recommend it highly.
That said, violence is certainly entertaining! Visser's lovely paper Oracle bites theory shows how bad things can get once we kill arithmetic thoroughly enough (in particular, he shows that over a truly feeble base theory the set of consequences of a consistent theory need not be a consistent theory!), and Visser's other papers are probably also of interest to you.
Basically, in order to be skeptical of the incompleteness theorem we need to already be skeptical as such basic arithmetical principles as the totality of exponentiation. This is why from a foundational point of view the possibility has not received serious attention, though it may be of technical interest (I wouldn't know, I'm not an expert in the relevant area). This is similar to the situation with Cantor's diagonal argument, where the amount of set theory you have to break to make it non-applicable is so great that it's hardly worth doing, and certainly not foundationally compelling (so far as I know).
Noah Schweber
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1I seem to remember that Ed Nelson was sceptical about totality of exponentiation. – Angina Seng Jul 28 '18 at 14:28
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The main point about Gödel's Theorem is the method of proof. This shows that if the model or theory being used is rich enough, then it contains within it the resources to encode its own limitations in that it will either be inconsistent or incomplete (in the sense that there will be true statements which are unprovable within the system or model).
There are systems which escape Gödel's method, but they do so by being deliberately limited, rather than being too strong to fail.
Mark Bennet
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2I don't think this really answers the question - I think the OP is asking about the theory in which, not about which, Godel's theorem is proved. – Noah Schweber Jul 28 '18 at 12:47
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@NoahSchweber I put this in as an answer because there seemed to me to be an implicit question "can I find a theory which avoids Gödel's Theorem?" and it seemed to be worth more than a comment. – Mark Bennet Jul 28 '18 at 15:32
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1"can I find a theory which avoids Gödel's Theorem?" can mean two things: "is there a theory to which GT doesn't apply" or "is there a (reasonable) theory which thinks that GT isn't true?" (e.g. which is consistent and thinks that PA is complete). I think it's clear from the question that the latter is meant. – Noah Schweber Jul 28 '18 at 15:56