The definition of the set of natural numbers in ZFC

Question

I am reading about the axiom of infinity. One thing I learned is that only with this axiom one is able to define the set of natural numbers. Here is my step-by-step try and I wonder if it is working because the standard definition is such a short-cut that it confuses beginners like me of what it actually means.

First let $x$ be the set whose existence is postulated by the axiom of infinity. By the way: I use the old version of this axiom as it was given by Zermelo, but that should only be a very formal difference to what nowadays is done.

First I define more loosely

$\mathbb N := \{y \mid \text{$y \in x$ & $y$ is in all inductive sets}\}$.

This defined $\mathbb N$ exists because $x$ exists and the axiom schema of separation postulates the existence of a set that is created by subsetting from an existing set.

Now, I’d define a set $I$ of all inductive sets $a$,

$I := \{a \mid \text{$\emptyset \in a$ and if $z \in a$ then $\{z\} \in a$}\}$.

$I$ exists in ZFC, because $x$ exists and one can apply the pairing axiom on $x$ to get $I$ with $x$ and the other inductive sets are just assumptions which does not harm $I$ in any way.

Now we can further formalize:

$\mathbb N :=\left\{y \mid \text{$y \in x$ & $y \in \bigcap_{a \in I} a$}\right\}$

and because $x$ is definitely in $I$ we can waive off the weaker first part of the conjunction and get eventually

$\mathbb N := \left\{y \mid y \in \bigcap_{a \in I} a\right\} = \big\{\emptyset, \{\emptyset\}, \{\{\emptyset\}\},\dots\big\}$.

Answer 1

The common way of constructing $\mathbb{N}$ is as follows:

(I will use your definition of inductive which has "the successor" of the set $x$ to be $\{x\}$; note that the usual definition is that $x^+=x\cup\{x\}$, rather than just $\{x\}$).

Theorem. If $\{X_j\}_{j\in J}$ is a family of inductive sets, then $\bigcap_{j\in J}X_j$ is an inductive set.

Proof. Let $X=\cap X_j$. Each $X_j$ satisfies that $\varnothing\in X_j$ and if $a\in X_j$, then $\{a\}\in X_j$. Therefore, $\varnothing\in X$. If $a\in X$, then $a\in X_j$ for all $j\in J$, hence $\{a\}\in X_j$ for all $j\in J$, hence $\{a\}\in X$. Thus, $X$ is inductive. $\Box$

Let $S$ be an inductive set; we know such a thing exists, by the Axiom of Infinity. Now let $$\mathbb{N}_S = \cap\{T\subseteq S\mid T\text{ is inductive.}\}$$ Note that this makes sense, since the collection of inductive subsets of $S$ is nonempty, since it certainly includes $S$. So this makes sense. In addition, by the Theorem, $\mathbb{N}_S$ is inductive.

This is "the natural numbers of $S$". Obviously it depends on $S$. But it turns out that it actually does not depend on $S$:

Proposition. Let $T$ be any inductive set. Then $\mathbb{N}_S=\mathbb{N}_T$.

Proof. SInce $\mathbb{N}_S$ is inductive, $\mathbb{N}_S\cap T$ is inductive by the Theorem. Thus, $\mathbb{N}_S\cap T$ is an inductive subset of $T$. Therefore, $\mathbb{N}_T\subseteq \mathbb{N}_S\cap T\subseteq\mathbb{N}_S$. A symmetric argument gives $\mathbb{N}_S\subseteq \mathbb{N}_T$. $\Box$

Having proven this, we just define $\mathbb{N}$ to be the set $\mathbb{N}_S$ for any inductive set $S$, since the Proposition shows that we will obtain the same set regardless of which inductive set $T$ we may have started with. This set has the property that it is contained in any inductive set, and so it is the "smallest" inductive set.

This avoids the issue of quantifying over "all inductive sets" or trying to construct the class of all inductive sets (which is not a set).

Answer 2

Maybe this is the non-obvious detail: Even though the collection of all inductive sets, $I$, is not a set itself, as long as it is non-empty, the intersection of all the sets in it (that is, the intersection of all inductive sets) is a set. By definition, any intersection, even of a class-worth of sets, is a subset of any of the sets in the class. So by the axiom of separation, it is a set, and, in this case, an inductive set. This intersection of all inductive sets, $\omega$, is therefore the smallest inductive set. The axiom of infinity tells you that $I$ is, indeed, non-empty, that at least one inductive set exists.

Another potentially confusing this is, you can actually build any natural number you want even without infinity, all you need is repeated application of pairing. Define $0 = \emptyset$ and then for any $m$ you want, build it up recursively as $n+1 = n \cup \lbrace n\rbrace$. By construction, if $n < m$ it is $n \in m$, so $m$ is the set of all numbers that are smaller than $m$. You only need infinity to show that there is a set that contains all natural numbers.