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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.

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    $\begingroup$ 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. $\endgroup$ Oct 19 at 19:36
  • $\begingroup$ 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. $\endgroup$ Oct 19 at 20:14
  • $\begingroup$ @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 $\mathsf{PA}$. $\endgroup$
    – R. Burton
    Oct 19 at 20:35
  • $\begingroup$ @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: $\forall x: [x\in N \iff x \in I \land \forall y\subset I : [0\in y \land \forall z\in y:[z\in y \implies S(z) \in y]\implies x\in y]]$. $\endgroup$ Oct 20 at 3:16
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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.

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  • $\begingroup$ In that case, shouldn't $\varphi$ uniquely characterize $\Bbb N$? $\endgroup$
    – R. Burton
    Oct 19 at 19:58
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    $\begingroup$ @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. $\endgroup$ Oct 19 at 20:07
  • $\begingroup$ And if I don't want to adopt a Platonist perspective? $\endgroup$
    – R. Burton
    Oct 19 at 20:09
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    $\begingroup$ @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. $\endgroup$ Oct 19 at 20:10
  • $\begingroup$ 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 $\mathsf{ZFC}$? $\endgroup$ Oct 19 at 20:11
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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.

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  • $\begingroup$ What is "the language of $\Bbb N$?" $\endgroup$
    – R. Burton
    Oct 19 at 21:01
  • $\begingroup$ @R.Burton Whatever language is being referred to when discussing $Th(\mathbb{N})$ - typically, the language including $0$, $S$, $+$, and $\cdot$. $\endgroup$ Oct 19 at 21:02
  • $\begingroup$ 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? $\endgroup$
    – R. Burton
    Oct 19 at 21:07
  • $\begingroup$ @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. $\endgroup$ Oct 19 at 21:09
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    $\begingroup$ @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? $\endgroup$ Oct 19 at 21:11

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