Transitive Closure Always Exists (Set Theory)

Theorem
Let $S$ be a set.

Let $G$ be a mapping such that $G \left({x}\right) = x \cup \bigcup x$.

Let $F$ be defined using the Principle of Recursive Definition:


 * $F \left({0}\right) = S$


 * $F \left({n^+}\right) = G \left({F \left({n}\right)}\right)$

Let $\displaystyle T = \bigcup_{n \mathop \in \omega} F \left({n}\right)$.

Then:


 * $T$ is a set and is transitive


 * $S \subseteq T$


 * If $R$ is transitive and $S \subseteq R$, then $T \subseteq R$.

That is, given any set $S$, there is an explicit construction for its transitive closure.

Proof
$\omega$ is a set by the Axiom of Infinity.

Thus by the Axiom of Replacement, the image of $\omega$ under $F$ is also a set.

Since $T$ is the union of $F \left({\omega}\right)$, it is thus a set by the Axiom of Union.

Furthermore:

By the above equations, $x \in T \land y \in x \implies y \in T$.

Thus by definition $T$ is transitive.

We have that:
 * $S = F \left({0}\right)$

By Set is Subset of Union:


 * $\displaystyle S \subseteq \bigcup_{n \mathop \in \omega} F \left({n}\right)$

So $S \subseteq T$.

Finally, suppose that $S \subseteq R$ and $R$ is transitive.

$T \subseteq R$ follows by finite induction:

For all $n \in \omega$, let $P \left({n}\right)$ be the proposition:
 * $F \left({n}\right) \subseteq R$

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $F \left({0}\right) \subseteq R$

which has been proved above.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $F \left({k}\right) \subseteq R$

Then we need to show:
 * $F \left({k+1}\right) \subseteq R$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $F \left({n}\right) \subseteq R$ for all $n \in \omega$


 * $T \subseteq R$