# Transitive Closure Always Exists (Set Theory)

## Contents

## 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 $n \in \omega$

- $T \subseteq R$

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 9.1$