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 alternative definition.
 * $(2)\quad$ $y$ is the smallest transitive set such that $x \in y$.

Proof
Let $x^t$ be the transitive closure of $x$ by the alternative definition.

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

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

Since $G(0) = \{ 0 \}$, $\{ x \} \in G(\N)$.

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 Union.

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

By the definition of $x^t$, $y \in G(n)$ for some $n \in \N$.

Then $z \in \bigcup G(n)$ by the definition of union.

But by the definition of $G$, $z \in G(n^+)$.

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 $G(n) \subseteq m$ for each $n \in \N$.

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

$G(0) = \{ x \} \subseteq m$ because $x \in m$.

Suppose that $G(n) \subseteq m$.

Then $\bigcup G(n) \subseteq \bigcup m$.

Thus $\bigcup G(n) \subseteq m$.

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