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

$\mathrm{\exists }I\left(\mathrm{\exists }e\right(e\in I\wedge \mathrm{\neg }\mathrm{\exists }t(t\in e))\wedge \mathrm{\forall }y(y\in I\to \mathrm{\exists }s(s\in I\wedge \mathrm{\forall }t(t\in s\leftrightarrow (t\in y\vee t=y)))\left)\right)$

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 $\mathrm{\varnothing }\in I$ such that $y\in I$ implies $y\cup \{y\}\in I$. The first conjunct is $\mathrm{\varnothing }\in I$ written out, and the inner clause $\mathrm{\forall }t(t\in s\leftrightarrow (t\in y\vee t=y))$ 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 $y\mathrm{\notin }y$ (Under Foundation, $x\mathrm{\notin }x$ 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.