# What formula of ZFC defines the set of natural numbers?

Let $$\mathsf{ZFC}'$$ be the extension of $$\mathsf{ZFC}$$ containing the constant symbol $$\Bbb N$$, which we take to represent the natural numbers. In order to say that $$\mathsf{ZFC}'$$ is a definitional extension of $$\mathsf{ZFC}$$ we need to find a formula $$\phi$$ in the language of $$\mathsf{ZFC}$$ with a single free variable $$\upsilon$$ such that $$\mathsf{ZFC}\vdash\exists!\upsilon\phi$$ and, for any formula $$\psi$$ containing $$\Bbb N$$, $$\mathsf{ZFC}'\vdash\psi$$ iff $$\mathsf{ZFC}\vdash\exists!\upsilon(\phi\land\psi(\Bbb N\mapsto\upsilon))$$ or equivalently $$\mathsf{ZFC}\vdash \forall\upsilon(\phi\implies\psi(\Bbb N\mapsto\upsilon))$$.

However, since there can be no recursive axiomatization of $$\text{Th}(\Bbb N)$$ - provided that $$\Bbb N$$ is truly the set of natural numbers - there can be no formula $$\phi$$ uniquely characterizing $$\Bbb N$$ (up to isomorphism). So either $$\Bbb N$$ is not the set of natural numbers, $$\mathsf{ZFC}'$$ is not an extension by definitions, or $$\mathsf{ZFC}$$ is inconsistent.

• I disagree with the vote to close; this is absolutely a question about mathematics. That said, I do think it would benefit from a serious presentation change (the dialogue aspect is unnecessary and only makes it harder to read, and the title suggests that the question is less mathematical than it is), so I haven't upvoted yet. Oct 19 at 19:36
• I'm not sure how you would do it in ZFC, but given any Dedekind infinite set, you can a select a subset that satisfies Peano's Axioms. The infinite set in ZFC is, of course, that postulated to exist by the axiom of infinity. IIRC it is Dedekind infinite set. So N is actually already built into ZFC. Oct 19 at 20:14
• @DanChristensen The issue isn't showing that a model of $\mathsf{PA}$ exists, it's showing that $\Bbb N$ specifically exists and is a model of $\mathsf{PA}$. Phrased differently, we need to define $\Bbb N$ in a way that separates it from arbitrary models of $\mathsf{PA}$. Oct 19 at 20:35
• @R.Burton Why first order PA? I don't know if this will help, and I am also not sure how functions work in ZFC. IIUC there are subtle differences between them and the usual functions in most math textbooks, but if we let $I$ be the Dedekind infinite set postulated to exist in AoI, and let $S: I \to I$ be the injective successor function on $I$ and let $0\in I$ have no pre-image under $S$ in $I$, then we can define $N$ as the subset of $I$ such that: $\forall x: [x\in N \iff x \in I \land \forall y\subset I : [0\in y \land \forall z\in y:[z\in y \implies S(z) \in y]\implies x\in y]]$. Oct 20 at 3:16

You write:

Since there can be no recursive axiomatization of $$Th(\mathbb{N})$$ - provided that $$\mathbb{N}$$ is truly the set of natural numbers - there can be no formula $$\varphi$$ uniquely characterizing $$\mathbb{N}$$ (up to isomorphism).

This is incorrect. Very broadly speaking, what we can say is that $$\mathsf{ZFC}$$ (being recursively axiomatizable) must not be able to settle all questions about $$\varphi$$. But this has nothing to do with $$\mathsf{ZFC}$$ proving that exactly one thing satisfying $$\varphi$$ exists or that thing corresponding appropriately to $$\mathbb{N}$$. For example, $$\mathsf{ZFC}$$ also proves "There is exactly one set $$x$$ which is $$\emptyset$$ iff $$\mathsf{CH}$$ holds and is $$\{\emptyset\}$$ iff $$\mathsf{CH}$$ fails," while not settling the question of whether this unique object is empty.

There are various senses in which $$\mathbb{N}$$ is "hard to pin down" and various other senses in which $$\mathbb{N}$$ is "easy to pin down;" you have to be very careful about which sense is being used when applying a given theorem.

• In that case, shouldn't $\varphi$ uniquely characterize $\Bbb N$? Oct 19 at 19:58
• @R.Burton In a sense, but I suspect not the sense you have in mind. In any model $M$ of $\mathsf{ZFC}$ there will be a unique object picked out by $M$ via $\varphi$, denoted $\varphi^M$, which we can think of as "$M$'s version of" $\mathbb{N}$. Moreover, adopting a Platonist perspective for the moment we'll have $\varphi^V=\mathbb{N}$ where $V$ is the "actual" set-theoretic universe (or something close enough anyways). But different models of $\mathsf{ZFC}$ may disagree about the behavior of their $\varphi$-versions. Oct 19 at 20:07
• And if I don't want to adopt a Platonist perspective? Oct 19 at 20:09
• @R.Burton Then the question of whether a formula successfully defines $\mathbb{N}$ in $\mathsf{ZFC}$ is, without further elaboration, vague. However, this is inessential to the point at issue: we'll still have a $\varphi$ which $\mathsf{ZFC}$ proves defines a unique structure subject to an appropriate "unaxiomatizability" property, and that's really the only way $\mathbb{N}$ as such is appearing here. Oct 19 at 20:10
• For now I would focus on the following: does it make sense that you can have a formula which $\mathsf{ZFC}$ proves defines a unique object, but which defines differently-behaving objects across different models of $\mathsf{ZFC}$? Oct 19 at 20:11

I will argue that both are wrong.

First, the statement

Since there can be no recursive axiomatization of $$Th(\mathbb{N})$$ ... thre can be no formula $$\phi$$ uniquely characterizing $$\mathbb{N}$$ (up to isomorphism)

This is subtly wrong. There can be no first-order formula in the language of $$\mathbb{N}$$ where all quantifiers range over $$\mathbb{N}$$ which uniquely characterises $$\mathbb{N}$$.

But there can be a formula in the language of set theory which uniquely characterises $$\mathbb{N}$$. Let $$\psi(v) := (\exists e . e \in v \land \forall y . \neg (y \in e)) \land \forall e . e \in v \to \exists k . k \in v \land \forall u . u \in k \iff (u = e \lor u \in e)$$. Then let $$\phi(v) := \psi(v) \land \forall u . \psi(u) \to \forall x . x \in u \to x \in v$$. The axiom of infinity allows us to prove $$\exists! v . \phi(v)$$. This is what allows us to define $$\mathbb{N}$$ as a definitional extension.

Therefore, B is wrong. ZFC does have a notion of the "set of natural numbers" and can be definitionally extended to include such a set, as seen above.

However, A is potentially wrong in that A assumes that the set $$\mathbb{N}$$, defined in ZFC, has anything to do with the "actual natural numbers". If ZFC were inconsistent, then ZFC would prove statements like "$$0 = 1$$" (interpreted suitably in $$\mathbb{N}$$). Even if ZFC is consistent, how do we know that all statements it proves about $$\mathbb{N}$$ are actually true about the actual natural numbers? We would know that all $$\Pi_1$$ statements it proves about $$\mathbb{N}$$ actually hold, but we wouldn't necessarily know that all first-order statements proved about $$\mathbb{N}$$ are actually true.

If we adopt constructive logic and accept principles like "all functions are recursive", then we can actually come up with specific statements about $$\mathbb{N}$$ that ZFC proves (such as "for every general-recursive unary function $$f$$, either $$f(0)$$ exists or $$f(0)$$ does not exist") but which are not actually true about the natural numbers.

• What is "the language of $\Bbb N$?" Oct 19 at 21:01
• @R.Burton Whatever language is being referred to when discussing $Th(\mathbb{N})$ - typically, the language including $0$, $S$, $+$, and $\cdot$. Oct 19 at 21:02
• I was assuming that we stay within $\mathsf {ZFC}$, so $\text{Th}(\Bbb N)$ would consist of the fragment of $\mathsf {ZFC}$ defining the naturals. Now that I look at it closer, doesn't you $\psi$ just say that there is an inductive set, as opposed to defining $\Bbb N$ uniquely? Oct 19 at 21:07
• @R.Burton You are correct that this is what $\psi$ says. But $\phi$ is the relevant predicate for doing a definitional extension. I just thought it would take too much space to directly state $\phi$ without stating $\psi$ first. Oct 19 at 21:09
• @R.Burton I'm not sure what you mean by "the fragment of ZFC defining the naturals". What is the formal language you're referring to here? Oct 19 at 21:11