You don't have an a priori notion of cardinality, so you cannot really say things like "$0:=|\emptyset|$". In fact, before you can define cardinals you usually define ordinals, and usually the definition of the naturals precedes the definition of the ordinals.
The set-theoretic method begins by using the Axiom of Infinity, which states that there exists at least one inductive set; a set $X$ is inductive if and only if (i) $\emptyset\in X$; and (ii) For all $x$, if $x\in X$, then $s(x) = x\cup\{x\}\in X$.
So, let $X$ be any inductive set. Then we define
$$\mathbb{N} = \cap\\{ A\subseteq X\mid \text{$A$ is inductive}\\}.$$
One can then prove that $\mathbb{N}$ is well defined (it does not depend on the choice of $X$) and satisfies "Peano's Axioms":
- $\emptyset \in\mathbb{N}$.
- If $x\in\mathbb{N}$, then $s(x) \in\mathbb{N}$.
- For all $x\in\mathbb{N}$, $s(x)\neq\emptyset$.
- For all $x,y\in\mathbb{N}$, if $s(x)=s(y)$, then $x=y$.
- If $S\subseteq \mathbb{N}$ is inductive, then $S=\mathbb{N}$.
(In fact, 3 and 4 are just consequences of the definition of $s(x)$). We then define "$0$" to mean $\emptyset$, "$1$" to mean $s(0)$, "$2$" to mean $s(1)=s(s(0))$, etc.
Alternatively, you can begin with Peano's Axioms. Here, there is a primitive notion called "natural number", and a primitive symbol called $0$. We also have a primitive function $s$. The Peano Axioms would be:
- $0$ is a natural number.
- If $n$ is a natural number, then $s(n)$ is a natural number.
- For all natural numbers $n$, $s(n)\neq 0$.
- For all natural numbers $n$ and $m$, if $s(n)=s(m)$, then $n=m$.
- Axiom Schema of Induction. If $\Phi$ is a predicate such that $\Phi(0)$ is true, and for all $n$, $\Phi(n)\Rightarrow \Phi(s(n))$, then for all natural numbers $k$, $\Phi(k)$.
(You can also begin with $1$ instead of $0$; I use $0$ because it then parallels the set-theoretic construction). We then define "$1$" to mean $s(0)$; and "$2$" to mean $s(1)=s(s(0))$, etc.
We then need the Recursion Theorem:
Recursion Theorem. Given a set $X$, an element $a\in X$, and a function $f\colon X\to X$, there exists a unique function function $F\colon\mathbb{N}\to X$ such that $F(0)=a$ and $F(s(n)) = f(F(n))$ for all $n\in\mathbb{N}$.
Once we have these definitions and theorem, we can start defining addition. Fix $n\in\mathbb{N}$. I'm going to define "add $n$", $+_n\colon \mathbb{N}\to\mathbb{N}$ by letting
\begin{align*}
+_n(0) &= n,\\\
+_n(s(m)) &= s(+_n(m)).
\end{align*}
Or, in usual notation,
\begin{align*}
n+0& = n,\\\
n+s(m) &= s(n+m).
\end{align*}
With these definitions, we have:
Theorem. For all $n\in\mathbb{N}$, $n+0=0+n=n$.
Proof. Let $S=\{n\in\mathbb{N}\mid n+0=0+n=n\}$. Note that $0\in S$, since $0+0 = 0$ by the definition of addition. Now assume that $k\in S$; that means that $k+0 = 0+k = k$. Then
$0+s(k) = s(0+k) = s(k)$ (first equality by the definition of addition with $0$, second by the induction hypothesis). And by the definition of addition with $s(k)$, we have $s(k)+0 = s(k)$. Therefore, $k\in S$ implies $s(k)\in S$. Thus, $S=\mathbb{N}$, as desired. QED
Theorem. For all $n\in\mathbb{N}$, $s(n)=n+1$
Proof. Let $S=\{n\in\mathbb{N}\mid s(n)=n+1\}$. First, $0\in S$, since $s(0) = 1 = 0+1$, by the previous theorem. Assume that $k\in S$; that means that $s(k)=k+1$. Then $s(s(k)) = s(s(k)+0) = s(k)+s(0) = s(k)+1$. So $k\in S$ implies $s(k)\in S$, hence $S=\mathbb{N}$. QED
Theorem. For all $\ell,n,m\in\mathbb{N}$, $\ell+(m+n) = (\ell+m)+n$.
Proof. Fix $\ell$ and $m$. Let $S=\{n\in\mathbb{N}\mid \ell+(m+n)=(\ell+m)+n\}$. We have $0\in S$, since
$$\ell + (m+0) = \ell + m = (\ell+m) + 0.$$
Now assume that $k\in S$; that means that $(\ell+m)+k = \ell+(m+k)$. We prove that $s(k)\in S$. We have:
$$(\ell+m)+s(k) = s((\ell+m)+k) = s(\ell+(m+k)) = \ell+s(m+k) = \ell+(m+s(k)).$$
Thus, if $k\in S$ then $s(k)\in S$. Hence, $S=\mathbb{N}$. QED
Lemma. For all $n\in\mathbb{N}$, $1+n = n+1$.
Proof. Let $S=\{n\in\mathbb{N}\mid 1+n=n+1\}$. Then $0\in S$. Suppose that $k\in S$, so that $1+k = k+1 = s(k)$. Then we have:
$$1+s(k) = s(1+k) = s(k+1) = s(k+s(0)) = s(s(k+0)) = s(s(k)) = s(k)+1.$$
Thus, $S=\mathbb{N}$. QED
Theorem. For all $n,m\in\mathbb{N}$, $n+m=m+n$.
Proof. Fix $m$, and let $S=\{n\in\mathbb{N}\mid m+n=n+m\}$. First, $0\in S$, since $m+0=0+m$. Also, $1\in S$ by the previous lemma. Now assume that $k\in S$. Then $m+k=k+m$. To show that $s(k)\in S$, we have:
\begin{align*}
m+s(k) &= s(m+k) = s(k+m) = (k+m)+1 = k+(m+1) = k+(1+m)\\\
&= (k+1)+m = s(k)+m.
\end{align*}
Thus, $S=\mathbb{N}$. QED
And so on. We can then define multiplication similarly, by fixing $n$ and defining
\begin{align*}
n\times 0 &= 0\\\
n\times s(m) &= (n\times m) + n,
\end{align*}
and prove the usual properties of multiplication inductively. Then we can define exponentiation also recursively: fix $n$; then
\begin{align*}
n^0 & = 1\\\
n^{s(m)} &= n^m\times n.
\end{align*}
We later define the order among the natural numbers by
$$a\leq b\Longleftrightarrow \exists n\in\mathbb{N}(a+n=b),$$
and prove the usual properties.
Later, we can construct $\mathbb{Z}$ from $\mathbb{N}$, $\mathbb{Q}$ from $\mathbb{Z}$, $\mathbb{R}$ from $\mathbb{Q}$, $\mathbb{C}$ from $\mathbb{R}$, etc. See for example my answer to this previous question.
