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I have read in a book:(G. Martínez, G. Piñieiro: "Gödel para todos") that complex axiomatics are Gödel complete, while naturals aren't. How can this be if Naturals are a subset of Complexes, (or at least, the Naturals are isomorphic to a subset of Complexes?

José Carlos Santos
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Perspectiva8
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    Nevertheless, the axioms of $\Bbb C$ do not allow you to formulate a predicate that says something like "$x$ is a natural number" – Hagen von Eitzen Jul 02 '16 at 11:12
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    What does it mean for a system to be Gödel complete? I haven't seen that terminology before, and it's kind of difficult to google. – Dustan Levenstein Jul 02 '16 at 11:18
  • so you should assume also the axiom of choice, I understand. – Perspectiva8 Jul 02 '16 at 11:19
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    Huh? Axiom of Choice would be used set theory, not the theory of either the complex or the natural numbers – Hagen von Eitzen Jul 02 '16 at 11:21
  • what do you mean whit "would be used set theory"? – Perspectiva8 Jul 02 '16 at 11:34
  • I understand that Naturals and Complexes are sets, – Perspectiva8 Jul 02 '16 at 11:42
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    @Perspectiva8: $\mathbb{N}$ and $\mathbb{C}$ are sets only in a much larger theory, usually set theory. In arithmetic there are no such things as sets; the language does not have anything syntactically like set notation. Please look at the Wikipedia page about PA to get a feel for what it is like. The axiom of choice is completely irrelevant to PA and ACF0. – user21820 Jul 02 '16 at 11:52
  • @user21820: The language of logic itself provides us with set-like things. –  Jul 02 '16 at 13:52
  • @Hurkyl: I have completely no idea what you're talking about. Logic has nothing to do with sets. Consider all the various type theories which are extensions of first-order logic but have no underlying notion of sets. And predicates cannot be objects otherwise you get a contradiction. – user21820 Jul 02 '16 at 16:14
  • @user21820: The types are the notion of sets that type theory gives us. –  Jul 02 '16 at 16:37
  • @Hurkyl: Nope I disagree. The types can be built bottom-up in many different ways with many different semantics. In the more constructive semantics, the type $S \to T$ consists of only the expressions for which you can prove is a code for a transformation of a proof of an object of type $S$ to a proof of an object of type $T$. There is no notion of the collection as a single entity at all. One only has such conception of type theory if one wishes to force interpret it in set theory. Types instead correspond to partial indicator functions, not total indicator functions like sets. – user21820 Jul 02 '16 at 17:51
  • @Hurkyl: For another example, consider that we conceive readily of the notion of all objects. In ZFC this notion is necessarily not an object itself. Similarly, the notion of all functions from $S$ to $T$ need not itself be an object, and quantified statements over these functions can be philosophically justifiable without needing there to be a collection of them. An existential statement is proven when you construct a witness. A universal statement $\forall x\ ( P(x) )$ is proven when you construct a procedure to generate a proof of $P(x)$ given an object $x$. – user21820 Jul 02 '16 at 17:59
  • @user21820: But none of this goes against my view. Terms have types. Propositions let us define subtypes and we can ask if a terms of a given type belongs to a subtype. We have functions between types and we can apply them to terms of the appropriate type. We can take intersections and unions of subtypes and depending on the flavor, other type constructors like products or function types. These are the things I want of a simple set theory, and type theory provides them. If I want a richer set theory, higher order logic is up to the task. –  Jul 02 '16 at 18:15
  • And by the elementary theory of the category of sets, we see this is basically enough to give everything you could want of a set theory. It still lacks unbounded quantification, but you usually don't need that. And I think there are even ways you can achieve the same effect without needing anything 'new'. –  Jul 02 '16 at 18:18
  • @Hurkyl: From what you say I don't really know what you mean by "set" then. I had assumed that you meant the usual concept as given by the cumulative hierarchy in ZFC, which is incompatible with constructive viewpoints including many type theories. If you're not saying that then I think I we don't actually disagree; it's just that I interpreted your use of the word "set" to denote more than you did. – user21820 Jul 03 '16 at 05:59
  • And it seems someone doesn't like this question and both our answers... – user21820 Jul 03 '16 at 06:00

2 Answers2

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Perhaps your book didn't explain what it meant by "axiomatics", but you have to learn the precise definitions before it is possible to understand Godel's incompleteness theorems. There is no one formal system for the natural numbers, but there is a widely used one called (first-order) PA. It can prove quite a lot of elementary number theory, but if it is consistent it cannot prove or disprove infinitely many sentences in the language of PA. The language of PA has the signature $(0,1,+,\times)$ at least (one can add the successor function symbol $S$, but it is expressible using the rest by writing "$S(t)$" as "$t+1$"). It turns out that no recursive extension of PA can be complete (there will always be some sentence that can be neither proven nor disproven).

In contrast, the theory of the complex numbers (the first-order theory that consists of all the true sentences over the language of fields), denoted by $Th(\mathbb{C})$, has a recursive axiomatization. The language of fields has the same signature as what I gave above for $PA$. It turns out that one possible recursive axiomatization for $Th(\mathbb{C})$ is the axiomatization ACF0 of algebraically closed fields with characteristic zero, which consists of axioms of fields, plus axioms stating the existence of a root for every non-constant $1$-variable polynomial, plus axioms of the form "$1+\cdots+1 \ne 0$" for all possible number of ones. In fact, one of the easiest way to show that $Th(\mathbb{C})$ has a recursive axiomatization is simply to prove that ACF0 gives a complete theory, by first proving quantifier elimination for ACF0 and then showing that quantifier-free sentences are either proven or disproven by ACF0.

The key point is that PA and ACF0 have different incompatible axioms. PA proves the sentence "$\neg \exists n\ ( n + n = 1 )$" while ACF0 proves its negation "$\exists n\ ( n + n = 1 )$". So it is not at all surprising that they are completely different. Don't forget that PA is supposed to capture the properties of the natural numbers as a whole, and just because we can embed natural numbers into the complex numbers says nothing about the relation between the properties of $\mathbb{N}$ and those of $\mathbb{C}$.

If we attempt to extend the language with a predicate symbol $N$ so that we can axiomatize both $\mathbb{C}$ and its subset $\mathbb{N}$ by including the axioms of ACF0, and the axioms of PA modified such that all quantifiers involved are restricted to $N$, then the resulting theory will no longer be complete because it correctly proves and disproves the same sentences as PA (after they are translated accordingly).

If you are unfamiliar with logic, please see here for resources that I recommend.

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  • I am familiar with Peano axioms, but I don't know similar axiomatization for the complex numbers – Perspectiva8 Jul 02 '16 at 12:03
  • @Perspectiva8: The link I included has references that describe ACF0, specifically Stephen Simpson's as well as Wolfgang Rautenberg's. – user21820 Jul 02 '16 at 12:04
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    @Perspectiva8: If you already know PA, then the key point is probably to understand that isomorphism is an external notion in some meta-theory such as ZFC. Both PA and ACF0 have no notion of sets and cannot say anything like "function", not to say "embedding" or "structure-preserving". – user21820 Jul 02 '16 at 12:07
  • ACF0 ought to have axioms stating that every polynomial with coefficients in the field has a root, rather than just every integer polynomial. If you only speak of integer polynomials, then the theory is not complete since $\exists y.\forall x.x^2\ne y$ is independent. It is false in $\mathbb C$ but true in $\mathbb A(\pi)$ (where $y=\pi$ is an example). – hmakholm left over Monica Jul 02 '16 at 14:19
  • @HenningMakholm: Thanks for that; I at first wrote what you said and then thought integers were enough and didn't check lol! – user21820 Jul 02 '16 at 16:10
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The short answer is that the (first-order) arithmetic of the complex numbers is insufficient to even propose the question

  • Is this complex number a natural number?

which means it cannot even speak about the kinds of questions that cause problems with integer arithmetic. For example, the problem of Diophantine equations — the theory of complex arithmetic is capable of asking whether systems of equations have real solutions, but cannot go on to discuss the more refined question of whether any of those solutions are integer solutions.

The simplest formulation of a theory of complex arithmetic is probably the theory of real closed fields which, by a theorem of Tarski, is complete.

Note that I am considering complex conjugation as part of the arithmetic of the complexes; the theory of complex arithmetic is able to use complex conjugation to define the real numbers by $z = \bar{z}$, so the theory of complex arithmetic can talk about real arithmetic. Conversely, the theory of real arithmetic can talk about complex arithmetic by mimicking the usual construction of complexes as pairs of real numbers.

The only subsets of the real line you can define the theory of real closed fields are finite unions of intervals whose endpoints are algebraic numbers. (infinite 'endpoints' are allowed; e.g. $(0,\infty)$ is allowed, and is defined by the proposition $x > 0$)

In particular, the natural numbers can only be given as an infinite union of intervals:

$$ \mathbb{N} = [0,0] \cup [1,1] \cup [2,2] \cup \ldots $$

and thus cannot be described in the theory of real closed fields.