The Axiom of Infinity: there is a set containing a set with no elements and closed under $y \mapsto y \cup \{y\}$

The Axiom of Infinity is the sentence

$\exists I\,\Bigl(\exists e\,\bigl(e \in I \wedge \neg\exists t\,(t \in e)\bigr) \wedge \forall y\,\bigl(y \in I \to \exists s\,(s \in I \wedge \forall t\,(t \in s \leftrightarrow (t \in y \vee t = y)))\bigr)\Bigr)$

of the language of set theory (The first-order language of set theory: $\in$, $=$, formulas with parameters, and class abbreviations).

It is written here in $\in$ and $=$ alone, with no abbreviations, because the notation it is usually stated in is introduced later on this page. Once that notation is available, the sentence reads: there is a set $I$ with $\varnothing \in I$ such that $y \in I$ implies $y \cup \{y\} \in I$. The first conjunct is $\varnothing \in I$ written out, and the inner clause $\forall t\,(t \in s \leftrightarrow (t \in y \vee t = y))$ says exactly that $s$ is $y \cup \{y\}$.

• The only unconditional existence assertion on this page. Extensionality and Foundation produce no sets at all, and every other axiom produces new sets from sets already given; this one asserts outright that a set $I$ with the two stated properties exists. Two results below say what its closure clause does: $y \cup \{y\}$ is never $y$, since $y$ is one of its elements while $y \notin y$ (Under Foundation, $x \notin x$ for every set $x$, there are no sets with $x \in y \in x$, and there are no sets with $x \in y \in z \in x$), and $y \mapsto y \cup \{y\}$ is injective (If $x \cup \{x\} = y \cup \{y\}$ then $x = y$).

• Inductive sets and $\mathbb{N}$. A set satisfying the two conjuncts above is what Inductive set ↗ calls inductive, and the natural numbers are built at The natural numbers $\mathbb{N}$ (von Neumann) ↗ as the smallest inductive set. Those items state the axiom in the abbreviated form; the sentence displayed above is the same assertion with the abbreviations expanded.