We see, following the text how to axiomatize the natural numbers and define addition and multiplication from these. We elaborate on only one result, namely one allowing us to make recursive definitions.
Theorem: Let $X$ be a set, let $x_0\in X$ and let $g(n, x)$ be a function $g: \mathbb{N} \times X \to X$. There is a unique function $f: \mathbb{N}\to X$ such that $f(0)= x_0$ and $\forall n \in\mathbb{N}\ f(n++) = g(n, f(n))$.
Proof: The idea is to construct the graph of $f$ as the smallest set containing a specified starting point and closed under a shift.
Let $\mathcal{U} = \{S \subset \mathbb{N}\times X : (0, x_0)\in S,\ (n, x)\in S\Rightarrow (n++, g(n++, x))\in S\}$. Note that $\mathcal{U}$ is non-empty as $\mathbb{N}\times X \in \mathcal{U}$.
Consider the set
It is easy to see that $S_0\in \mathcal{U}$.
Lemma: For every $n\in \mathbb{N}$, there exist a unique $x_n\in X$ such that $(n, x_n) \in S_0$.
Proof: We prove this by induction on $n$. First, let $n=0$. Note that $(0, x_0)\in S$ for all $S\in \mathcal{U}$, and so $(0, x_0) \in S_0$. Now suppose $(0, x_0’)\in S_0$ with $x_0’\neq x_0$. Let $S = \{(n, x) \in S_0: (n, x) \neq (0, x_0’)$. We claim that $S\in \mathcal{U}$.
Firstly, as $(0, x_0) \in S_0$ and $x_0’ \neq x_0$, $(0, x_0)\in S$. Next, suppose $(n, x)\in S \subset S_0$, then $(n++, g(n++, x))\in S_0$. Further, as $n++ \neq 0$, we see that $(n++, g(n++, x)) \neq (0, x_0’)$. Hence $(n++, g(n++, x))\in S$. Thus $S\in \mathcal{U}$.
By definition of $S_0$, as $S\in \mathcal{U}$, $S_0 \subset S$. But $(0, x_0’)$ is in $S_0$ but not $S$, a contradiction. This shows that $x_0\in X$ is the unique point $x \in X$ such that $(0, x)\in S_0$.
The induction step is similar. Assume $(n, x_n)\in S_0$ and this is unique, i.e., $(n, x)\in S_0 \implies x = x_n$. Then $(n, x_n)\in S$ for all $S \in \mathcal{U}$, from which we deduce that $(n++, g(n++, x))\in S_0$. To show uniqueness, suppose $x_{n++}’\in X$ is a point with $(n++, x_{n++}’)\in S_0$ but $x_{n++}’ \neq x_{n++} := g(n++, x)$. Let $S = S_0 \setminus \{(n++, x_{n++}’)\}$. We again show that $S \in \mathcal{U}$ and obtain a contradiction to $S_0 \subset S$.
Namely, as $(0, x_0) \in S_0$ and $n++ \neq 0$, $(0, x_0)\in S$. Next, suppose $(m, x)\in S \subset S_0$, then $(m++, g(m++, x))\in S_0$. If $(m++, g(m++, x)) \notin S$, then we must have $(m++, g(m++, x)) = (n++, x_{n++}’)$. Hence $n = m$, so $x = x_n$. It follows that $x_{n++}’ = g(m++, x) = g(n++, x_n)$, a contradiction. Thus, if $(m, x)\in S$, $(m++, g(m++, x)) \in S$, completing the proof that $S\subset \mathcal{U}$, and contradicting $S_0 \subset S$.
This completes the proof of the lemma.
To prove the theorem, we define the function $f$ by $f(n) = x_n$. As we will see, in terms of set theory the function just corresponds to the set $S_0$. It is straightforward to show by induction than $f$ indeed has the desired properties, and is the unique such function.