Inductive set

Working in ZFC, for a set $x$ write $x^{+} := x \cup \{x\}$ for its successor (a set, by the axioms of Pairing and Union, The Axiom of Pairing: $\forall x\,\forall y\,\exists z\,\forall t\,(t \in z \leftrightarrow (t = x \vee t = y))$ and The Axiom of Union: $\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow \exists s\,(s \in x \wedge z \in s))$). A set $I$ is inductive when

$\varnothing \in I \quad\text{and}\quad \bigl(x \in I \implies x^{+} \in I\bigr)\ \text{for every } x.$

That is, $I$ contains the empty set and is closed under the successor operation.

The Axiom of Infinity (The Axiom of Infinity: there is a set containing a set with no elements and closed under $y \mapsto y \cup \{y\}$) is precisely the assertion that an inductive set exists. Inductive sets can be large and are far from unique (if $I$ is inductive so is $I \cup \{\text{anything}\}$ once closed off), so an inductive set is not yet a good definition of $\mathbb{N}$. The natural numbers are carved out as the smallest inductive set, the intersection of all of them (The natural numbers exist: a smallest inductive set, The natural numbers $\mathbb{N}$ (von Neumann)); minimality is what delivers the induction principle.

Here $0 := \varnothing$ and the successor of $n$ is $\sigma(n) := n^{+} = n \cup \{n\}$, so $1 = \{0\}$, $2 = \{0,1\}$, $3 = \{0,1,2\}$: each natural number is the set of all smaller natural numbers.