Union of Transitive Class is Transitive
Theorem
Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let $A$ be transitive.
Then $\bigcup A$ is also transitive.
Proof 1
Let $A$ be transitive.
By Class is Transitive iff Union is Subclass:
- $\bigcup A \subseteq A$
By Union of Subclass is Subclass of Union of Class:
- $\map \bigcup {\bigcup A} \subseteq \bigcup A$
Then by Class is Transitive iff Union is Subclass:
- $\bigcup A$ is transitive.
$\blacksquare$
Proof 2
Let $A$ be transitive.
Let $x \in \bigcup A$.
By Class is Transitive iff Union is Subclass we have that:
- $\bigcup A \subseteq A$
Thus by definition of subclass:
- $x \in A$
As $A$ is transitive:
- $x \subseteq A$
Let $z \in x$.
As $x \subseteq A$, it follows by definition of subclass that:
- $z \in A$
Thus we have that:
- $\exists x \in A: z \in x$
and so by definition of union of class:
- $z \in \bigcup A$
Thus we have that:
- $z \in x \implies z \in \bigcup A$
and so by definition of subclass:
- $x \subseteq \bigcup A$
Thus we have that:
- $x \in \bigcup A \implies x \subseteq \bigcup A$
Hence $\bigcup A$ is a transitive class by definition.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.3. \ \text {(c)}$