# Relationship between Transitive Closure Definitions

## Theorem

Let $x$ be a set.

Let $a$ be the smallest set such that $x \in a$ and $a$ is transitive.

Let $b$ be the smallest set such that $x \subseteq b$ and $b$ is transitive.

Then $a = b \cup \left\{ {x}\right\}$.

## Proof

We have that:

- $x \in a$

and $a$ is transitive.

So:

- $x \subseteq a$

Thus by the definition of $b$ and of smallest set:

- $b \subseteq a$

Since we also have $x \in a$:

- $b \cup \left\{ {x}\right\} \subseteq a$

$x \in \left\{ {x}\right\}$, so:

- $x \in b \cup \left\{ {x}\right\}$

$b \cup \left\{ {x}\right\}$ is transitive:

If $p \in b$ then:

- $p \subseteq b \subseteq b \cup \left\{ {x}\right\}$.

If $p \in \left\{ {x}\right\}$ then:

- $p = x$

So by the definition of $b$:

- $p \subseteq b \subseteq b \cup \left\{ {x}\right\}$

Thus by the definition of $a$:

- $a \subseteq b \cup \left\{ {x}\right\}$

Thus the theorem holds by definition of set equality.

$\blacksquare$