We all know that the collection of all natural numbers $\mathbb{N}$ is a set. But before that how do we prove it really is a set, by using only Zermelo-Fraenkel (ZF) axioms.
I tried using the axiom of infinity, but could not come up with a satisfactory proof ( or you can say construction). Im now wondering can we actually prove that $\mathbb{N}$ is a set using only ZF axioms? If yes how?. Any help is appreciated.
Edit: Here the collection of natural numbers is already assume to exist by Peano's axiom. The collection of natural numbers is define to be a collection $\mathbb{N}$, that satisfies the following:
Peano's Axioms
(i) $1\in \mathbb{N}$.
(ii) For each $a\in \mathbb{N}$, there exists a unique $a'\in \mathbb{N}$, called the successor of $a$.
(iii) $a\neq a'$ for any $a\in \mathbb{N}$.
(iv) For any $a,b \in \mathbb{N}$, $a'=b' \implies a=b$.
(v) If $S$ is a subset $\big($I should say $S$ is a subcollection, because we didn't know yet whether $\mathbb{N}$ is a set$\big)$of $\mathbb{N}$ such that : (a) $1\in S$, and (b) if $n\in S $ then $ n'\in S$. Then $S= \mathbb{N}$.
Now I want to prove that the above collection $\mathbb{N}$ is indeed a set using the following ZF axioms.
ZF Axioms
(i)(Extension Axiom.) Any two sets are equal if and only if they have the same elements.
(ii)(Empty Set Axiom.) There exist a set with no elements.(It is called the empty set and denoted by $\emptyset$.)
(ii)(Pairing Axiom) For any two sets $A$ and $B$, there exist a set $\{A,B\}$ consisting of exactly $A$ and $B$ as elements.
(iv)(Union Axiom.) let $\mathcal{F}$ be a set. Then there exist a set $S$ whose elements belong to atleast one element of $\mathcal{F}$ .
(v)(Subset Axiom.) Let $\phi(x)$ be a formula. If $A$ is a set then the collection of all $x\in A$ such that $\phi(x)$ is a set. ($S$ is called a subset of $A$.)
(Vi)(Power Set Axiom.) Let $A$ be a set then there exist a set $P(A)$ consisting of all subsets of $A$.($P(A)$ is called the power set of $A$)
(vii)(Replacement Axiom). Let $\phi(x,y)$ be a formula in two variables. Let $A$ be a set. If for each $x\in A$ there exist a unique $y$ such that $\phi(x,y)$, then the collection of all $y$ in which there exist atleast one $x\in A$ such that $\phi(x,y)$ is a set.
(viii)(Infinity Axiom.) There exist a set $I$ such that $\emptyset\in I$ and for each $x\in I$, $x\cup\{x\}\in I.$
(ix)(Regularity Axiom.) For every non-empty set $A$ there exist atleast one element $x\in A$ that $x$ is disjoint from $A$.(i.e., there exist $x\in A $ such that $x\cap A = \emptyset$.)
Best Answer
There are various ways to prove the existence of $\mathbb{N}$. I assume that $\mathbb{N}$ is the least inductive set (i.e., a set $I$ such that $0\in I$ and $x\in I\to S(x)\in I$, where $S(x)=x\cup\{x\}$ is a successor of $x$) and the axiom of infinity is the statement there is an inductive set.
In your case (i.e., working with $\mathsf{ZF}$), relying on the axiom of separation suffices. Observe that $\mathbb{N}$ is the intersection of all inductive sets. Hence the following equality holds: $$\mathbb{N} = \{x\in I \mid \forall J (\text{$J$ is inductive}\to x\in J)\}.$$
Some textbooks state the axiom of infinity as there is an infinite set, and do not mention anything about inductive sets. Even in that case, we can define what finite sets and infinite sets are. (For example, we may define natural numbers as ordinals whose elements are 0 or successor ordinals, and finite sets as an image of a natural number.)
Moreover, we can still prove $\mathbb{N}$ exists: let $I$ be an infinite set. For each natural number $n$, we can find $x\subseteq I$ whose cardinality is $n$. Hence
$$\{x\in\mathcal{P}(I): |x|=n\}$$
is a non-empty set. (where $\mathcal{P}(X)$ is the power set of $X$.) Now consider the set
$$A = \{x\in\mathcal{P}(I) \mid \text{$x$ is finite}\}$$ and a map $F$ of domain $A$, defined by $F(x)=|x|$. By the axiom of replacement, the image of $F$ exists, and it is $\mathbb{N}$.