Set Contained in Smallest Transitive Set

Theorem
Let $S$ be a set.

Then there is a transitive set $G$ such that $S \subseteq G$ and if $Q$ is any transitive set such that $S \subseteq Q$ then $G \subseteq Q$.

Construction of $G$
Let $U$ be the class of all sets.

Define the mapping $F: \N \to U$ recursively:


 * $F \left({0}\right) = S$
 * $F \left({n + 1}\right) = \bigcup F \left({n}\right)$

Applying the axiom of union inductively, $F \left({n}\right)$ is a set for each $n \in \mathbb N$.

Let $\displaystyle G = \bigcup_{i \mathop = 0}^\infty F \left({i}\right)$.

By the axiom of union, $G$ is a set.

Transitivity
$G$ is transitive. That is, if $a \in b$ and $b \in G$, then $a \in G$.

Proof
Suppose $a \in b$ and $b \in G$.

Then by the definition of $G$, there exists an $n\in \mathbb N$, $b \in F(n)$.

By the definition of $F$, $F(n+1) = \bigcup F(n)$. Then by the definition of union, $a \in F(n+1)$.

Thus by the definition of $G$, $a \in G$.

Minimality
If $Q$ is a transitive set and $S \subseteq Q$ then $G \subseteq Q$.

Proof
Suppose $Q$ is transitive and $S \subseteq Q$.

Define $F$ as above.

Prove by induction that $F(n) \subseteq Q$ for each $n$.