Transitive Closure Always Exists (Set Theory)

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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:

$S = \map F 0$

By Set is Subset of Union:

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

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:


\(\text {(1)}: \quad\) \(\displaystyle \map F k\) \(\subseteq\) \(\displaystyle R\) Hypothesis
\(\displaystyle \leadsto \ \ \) \(\displaystyle \bigcup \map F k\) \(\subseteq\) \(\displaystyle \bigcup R\) Set Union Preserves Subsets
\(\text {(2)}: \quad\) \(\displaystyle \) \(\subseteq\) \(\displaystyle R\) 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 $n \in \omega$
$T \subseteq R$

$\blacksquare$


Sources