Union of Successor Ordinal
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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 |
$\blacksquare$