Equivalence of Definitions of Transitive Closure (Set Theory)

Theorem
Let $x$ and $y$ be sets.

Proof
Let $x^t$ be the transitive closure of $x$ by Definition 2.

Let the mapping $G$ be defined as on that definition page.

$x \in x^t$
$x \in \set x$ by the definition of singleton.

Since $\map G 0 = \set 0$:
 * $\set x \in \map G \N$

Thus $x \in x^t$ by the definition of union.

$x^t$ is a Set
By Denumerable Class is Set, the image of $G$ is a set.

Thus $x^t$ is a set by the Axiom of Unions.

$x^t$ is a Transitive Set
Let $y \in x^t$ and let $z \in y$.

By the definition of $x^t$:
 * $\exists n \in \N: y \in \map G n$

Then by definition of union:
 * $\displaystyle z \in \bigcup \map G n$

But by the definition of $G$:
 * $z \in \map G {n^+}$

Thus by the definition of $x^t$:
 * $z \in x^t$

As this holds for all such $y$ and $z$, $x^t$ is transitive.

$x^t$ is Smallest
Let $m$ be a transitive set such that $x \in m$.

We will show by induction that $\map G n \subseteq m$ for each $n \in \N$.

By Union is Smallest Superset, that will show that $x^t \subseteq m$.

Because $x \in m$:
 * $\map G 0 = \set x \subseteq m$

Suppose that $\map G n \subseteq m$.

Then by Union is Increasing:
 * $\displaystyle \bigcup \map G n \subseteq \bigcup m$

Thus:
 * $\displaystyle \bigcup \map G n \subseteq m$

By Smallest Element is Unique, $x^t$ is the only set satisfying $(2)$.