Ordinal Multiplication by Zero
Jump to navigation
Jump to search
Theorem
Let $x$ be an ordinal.
\(\ds x \cdot \O\) | \(=\) | \(\ds \O\) | ||||||||||||
\(\ds \O \cdot x\) | \(=\) | \(\ds \O\) |
Proof
\(\ds x \cdot \O\) | \(=\) | \(\ds \O\) | Definition of Ordinal Multiplication |
For $\O \cdot x = \O$, the proof shall proceed by Transfinite Induction on $x$.
Basis for the Induction
\(\ds \O \cdot \O\) | \(=\) | \(\ds \O\) | Definition of Ordinal Multiplication |
This proves the basis for the induction.
Induction Step
\(\ds \O \cdot x\) | \(=\) | \(\ds \O\) | Inductive Hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\O \cdot x} + \O\) | \(=\) | \(\ds \O\) | Definition of Ordinal Addition | ||||||||||
\(\ds \O \cdot x^+\) | \(=\) | \(\ds \paren {\O \cdot x} + \O\) | Definition of Ordinal Multiplication | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \O \cdot x^+\) | \(=\) | \(\ds \O\) | Equality is Transitive |
This proves the induction step.
Limit Case
\(\ds \forall y \in x: \, \) | \(\ds \O \cdot y\) | \(=\) | \(\ds \O\) | Hypothesis | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bigcup_{y \mathop \in x} \paren {\O \cdot y}\) | \(=\) | \(\ds \O\) | Indexed Union Equality | ||||||||||
\(\ds \O \cdot x\) | \(=\) | \(\ds \bigcup_{y \mathop \in x} \paren {\O \cdot y}\) | Definition of Ordinal Multiplication | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \O \cdot x\) | \(=\) | \(\ds \O\) | Equality is Transitive |
This proves the limit case.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.18$