First, let
0 denote the empty set. Now let
S(n)=n∪{n}. Then intuitively the natural numbers are all sets of the form
S(S(S(⋯S(S(0))⋯))) for some
finite number of applications of
S.
But how do we capture this intuition?
Call a set
I “inductive” if
0∈I and for
∀a(a∈I→S(a)∈I).
Note existence of the predicate
nat(n)=∀I:(I inductive→a∈I). This says that a set “is a natural number” if it’s in every inductive set.
Now invoke the axiom of infinity to produce an inductive set
I0, and define
N:={n∈I0:nat(n)}. Intuitively we are thinking of
N as the intersection of all inductive sets, and using
I0 to guarantee that at
least on such set exists (and hence the intersection is nonempty).
Observe the following lemma (the “principle of induction”). If
A is an inductive subset of
N, then
A=N.