Transitive Closure Always Exists (Set Theory)

Theorem
Let $S$ be a set.

Let $G$ be a mapping such that $\map G x = x \cup \bigcup x$.

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


 * $\map F 0 = S$


 * $\map F {n^+} = \map G {\map F n}$

Let $\displaystyle T = \bigcup_{n \mathop \in \omega} \map F n$.

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 $\map F \omega$, it is thus a set by the axiom of unions.

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:
 * $S = \map F 0$

By Set is Subset of Union:


 * $\displaystyle S \subseteq \bigcup_{n \mathop \in \omega} \map F n$

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 $\map P n$ be the proposition:
 * $\map F n \subseteq R$

Basis for the Induction
$\map P 0$ is the case:
 * $\map F 0 \subseteq R$

which has been proved above.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\map F k \subseteq R$

Then we need to show:
 * $\map F {k + 1} \subseteq R$

Induction Step
This is our induction step:

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

Therefore:
 * $\map F n \subseteq R$ for all $n \in \omega$


 * $T \subseteq R$