# Transitive Closure Always Exists (Set Theory)

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## 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:

 $\displaystyle x$ $\in$ $\displaystyle T$ Assumption $\displaystyle \implies \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle x$ $\in$ $\displaystyle F \left({n}\right)$ Definition of Indexed Union $\displaystyle \implies \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle x$ $\subseteq$ $\displaystyle \bigcup F\left({n}\right)$ Set is Subset of Union $\displaystyle y$ $\in$ $\displaystyle x$ Assumption $\displaystyle \implies \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle y$ $\in$ $\displaystyle \bigcup F \left({n}\right)$ $\displaystyle \implies \ \$ $\, \displaystyle \exists n: \,$ $\displaystyle y$ $\in$ $\displaystyle F \left({n+1}\right)$ definition of $F$ $\displaystyle \implies \ \$ $\displaystyle y$ $\in$ $\displaystyle T$ 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:

$S = F \left({0}\right)$
$\displaystyle S \subseteq \bigcup_{n \mathop \in \omega} F \left({n}\right)$

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 $P \left({n}\right)$ 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:

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

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

$\blacksquare$