How can I prove the existence of the set of natural numbers $\mathbb{N}$ in ZFC Set Theory?

Question

How can I prove the existence of the set of natural numbers $\mathbb{N}$ in ZFC Set Theory?

We have the following axioms and theorems:

1. $\forall X\exists! Y\forall x(x\in Y\iff x\in X\land \phi(x))$
   Theorem from the Axiom Schema of Specification, from which the functional predicate "set builder notation" $\{x\in X|\phi(x)\}$ is introduced.
2. $\exists X(\emptyset\in X\land\forall x\in X(x\cup\{x\}\in X))$
   Axiom of Infinity

$\mathbb{N}$ is defined here (https://proofwiki.org/wiki/Definition:Natural_Numbers) as the minimally inductive set, which is defined as:

$\mathbb{N}=\omega=\bigcap\{X\in\mathcal{P}(I)|\emptyset\in X\land\forall x\in X(x\cup\{x\}\in X)\}$ where $I$ is any inductive set.

The problem is $\{X\in\mathcal{P}(I)|\emptyset\in X\land\forall x\in X(x\cup\{x\}\in X)\}$ cannot be proven to exist because $I$ cannot appear at the conclusion of any proof by the restriction on existential instantiation.

The restrictions of existential instantiation are:

1. the constant must be new to the proof
2. the constant cannot be in the conclusion of the proof
3. the constant must replace all instances of the bound variable

See here:

1. $\exists X(\emptyset\in X\land\forall x\in X(x\cup\{x\}\in X))$
   axiom of infinity
2. $\emptyset\in I\land\forall x\in I(x\cup\{x\}\in I)$
   existential instantiation $I$/$X$ 1
3. $\forall X\exists! Y\forall x(x\in Y\iff x\in X\land \phi(x))$
   theorem 1
4. $\forall X\exists! Y\forall x(x\in Y\iff x\in X\land \emptyset\in X\land\forall x\in X(x\cup\{x\}\in X))$
   schema elimination $\emptyset\in X\land\forall x\in X(x\cup\{x\}\in X)$/$\phi(x)$
5. $\exists! Y\forall x(x\in Y\iff x\in \mathcal{P}(I)\land \phi(x))$
   universal instantiation $\mathcal{P}(I)$/$X$ 4

From here I would be able to assign $Y$ a constant symbol ($\mathbb{N}$,$\omega$,$\aleph_0$, etc.),

but I cannot include the constant $I$ in the conclusion of my proof. Therefore I cannot conclude that such a $Y$ exists.

How can I prove the existence of the set of natural numbers $\mathbb{N}$?

Answer 1

Here's a sketch

Let $\phi(x)$ be any formula in which $x$ appears free, and neither $y, z$ nor $t$ appear free. Prove the theorem schema

$$\exists x \phi(x) \implies \exists ! y \forall z (z\in y \iff \forall t( \phi(t)\implies z\in t)) $$

Now apply this to $$\phi(x) := \text{Inductive}(x) := \emptyset\in x \land \forall p (p\in x \implies p\cup\{p\}\in x)$$

So letting $$\text{Omega}(u) := \forall z (z\in u \iff \forall t (\text{Inductive}(t)\implies z\in t))$$

You get $$\exists x \text{Inductive}(x) \implies \exists !y \text{Omega}(y)$$

You know $$\exists x \text{Inductive}(x)$$ is true by the axiom infinity. So you can deduce that $$\exists !y \text{Omega}(y)$$ and this set is defined to be the set of natural numbers.

Answer 2

I'll give a constructive approach.

Definition 1

Successor of a set x is $ x\cup\{x\} $

S(x) := $ x\cup\{x\} $

Definition 2

A set x is a successor if ∃y$\in$x($ x = y\cup\{y\} $)

Succ(x) ↔ ∃y$\in$x($ x = y\cup\{y\} $)

That's all we need alongside the ZF axioms to construct the Natural Numbers.

By the Infinity Axiom, there exists an Inductive set.

Let I be an inductive set

Let

$\omega$ = {x$\in$I | x = ∅ ∨ ( Succ(x) ∧ ∀y$\in$x(y = ∅ ∨ Succ(y))}

$\omega$ exists by the axiom schema of specification

I leave to you to show that it is the smallest inductive set.

First show that it is inductive.

Then Let X be an inductive set. For the purpose of contradiction assume X is a proper subset of $\omega$

Note: $\omega$ is not dependent on a particular inductive set I, since it is the smallest inductive set. If choosing a different inductive set I' , resulted in an inductive set ω', as both ω' and $\omega$ are the smallest inductive set. They must be equal.