Union of Successor Ordinal

Theorem

Let $x$ be an ordinal.

Let $x^+$ denote the successor of $x$.

Then:

$\map \bigcup {x^+} = x$

Proof

\(\ds \map \bigcup {x^+}\)

\(=\)

\(\ds \map \bigcup {x \cup \set x}\)

Definition of Successor Set

\(\ds \)

\(=\)

\(\ds \paren {\bigcup x \cup \bigcup \set x}\)

Set Union is Self-Distributive/Sets of Sets

\(\ds \)

\(=\)

\(\ds \paren {\bigcup x \cup x}\)

Union of Singleton

\(\ds \)

\(=\)

\(\ds x\)

Class is Transitive iff Union is Subclass

$\blacksquare$