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:

 $\displaystyle x$ $\in$ $\displaystyle T$ Assumption $\displaystyle \leadsto \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle x$ $\in$ $\displaystyle \map F n$ Definition of Union of Family $\displaystyle \leadsto \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle x$ $\subseteq$ $\displaystyle \bigcup \map F n$ Set is Subset of Union $\displaystyle y$ $\in$ $\displaystyle x$ Assumption $\displaystyle \leadsto \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle y$ $\in$ $\displaystyle \bigcup \map F n$ $\displaystyle \leadsto \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle y$ $\in$ $\displaystyle \map F {n + 1}$ Definition of $F$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $\in$ $\displaystyle T$ Definition of $T$

By the above equations:

$x \in T \land y \in x \implies y \in T$

Thus by definition $T$ is transitive.

$\Box$

We have that:

$S = \map F 0$
$\displaystyle S \subseteq \bigcup_{n \mathop \in \omega} \map F n$

So $S \subseteq T$.

$\Box$

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:

 $(1):\quad$ $\displaystyle \map F k$ $\subseteq$ $\displaystyle R$ Hypothesis $\displaystyle \leadsto \ \$ $\displaystyle \bigcup \map F k$ $\subseteq$ $\displaystyle \bigcup R$ Set Union Preserves Subsets $(2):\quad$ $\displaystyle$ $\subseteq$ $\displaystyle R$ Class is Transitive iff Union is Subset $\displaystyle \leadsto \ \$ $\displaystyle \map F k \cup \bigcup \map F k$ $\subseteq$ $\displaystyle R$ by $(1)$, $(2)$, Set Union Preserves Subsets $\displaystyle \leadsto \ \$ $\displaystyle \map F {k + 1}$ $\subseteq$ $\displaystyle R$ Definition of $\map F {k + 1}$

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 $n \in \omega$
$T \subseteq R$

$\blacksquare$