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 Finite Recursion:


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


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

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
$T$ is a countable union of sets and is thus a set.

Furthermore:

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

$T$ is transitive.

Furthermore, note that $S = F\left({ 0 }\right)$

By Subset of Union, it follows that:


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

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:
 * $F\left({n}\right) \subseteq R$

$P \left({0}\right)$ is true, as this just says $S \subseteq R$.

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 $n \in \omega$


 * $T \subseteq R$