Equivalence of Definitions of Transitive Closure (Set Theory)

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

Then the following are equivalent:


 * $(1): \quad y$ is the transitive closure of $x$ by the Definition 2.
 * $(2): \quad y$ is the smallest transitive set such that $x \in y$.

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

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

$x \in x^t$
$x \in \left\{{x}\right\}$ by the definition of singleton.

Since $G \left({0}\right) = \left\{{0}\right\}$:
 * $\left\{{x}\right\} \in G \left({\N}\right)$

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 Union.

$x^t$ is a transitive set
Let $y \in x^t$ and let $z \in y$.

By the definition of $x^t$:
 * $\exists n \in \N: y \in G \left({n}\right)$

Then by definition of union:
 * $\displaystyle z \in \bigcup G \left({n}\right)$

But by the definition of $G$:
 * $z \in G \left({n^+}\right)$

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 $G \left({n}\right) \subseteq m$ for each $n \in \N$.

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

Because $x \in m$:
 * $G \left({0}\right) = \left\{{x}\right\} \subseteq m$

Suppose that $G \left({n}\right) \subseteq m$.

Then:
 * $\displaystyle \bigcup G \left({n}\right) \subseteq \bigcup m$

Thus:
 * $\displaystyle \bigcup G \left({n}\right) \subseteq m$

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