Ordinal Multiplication is Left Distributive
Theorem
Let $x$, $y$, and $z$ be ordinals.
Let $\times$ denote ordinal multiplication.
Let $+$ denote ordinal addition.
Then:
- $x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
Proof
The proof shall proceed by Transfinite Induction, as follows:
Basis for the Induction
Let $0$ denote the ordinal zero.
\(\ds x \times \paren {y + 0}\) | \(=\) | \(\ds x \times y\) | Definition of Ordinal Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + 0\) | Definition of Ordinal Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times 0}\) | Definition of Ordinal Multiplication |
This proves the basis for the induction.
Induction Step
\(\ds x \times \paren {y + z}\) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times z}\) | Inductive Hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \times \paren {y + z^+}\) | \(=\) | \(\ds x \times \paren {y + z}^+\) | Definition of Ordinal Addition | ||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times \paren {y + z} } + x\) | Definition of Ordinal Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {x \times y} + \paren {x \times z} } + x\) | Inductive Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + \paren {\paren {x \times z} + x}\) | Ordinal Addition is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times z^+}\) | Definition of Ordinal Multiplication |
This proves the induction step.
Limit Case
The inductive hypothesis for the limit case states that:
- $x \times \paren {y + w} = \paren {x \times y} + \paren {x \times w}$ for all $w \in z$ and $z$ is a limit ordinal.
The proof shall proceed by cases:
Case 1
Suppose $x = 0$.
\(\ds x \times \paren {y + z}\) | \(=\) | \(\ds 0\) | Ordinal Multiplication by Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 + 0\) | Definition of Ordinal Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times z}\) | Ordinal Multiplication by Zero |
Case 2
Suppose that $x \ne 0$.
Since $w$ is a limit ordinal, $y + w$ and $x \times w$ are limit ordinals by Limit Ordinals Preserved Under Ordinal Addition and Limit Ordinals Preserved Under Ordinal Multiplication.
\(\ds x \times \paren {y + z}\) | \(=\) | \(\ds \bigcup_{w \mathop \in \paren {y + z} } \paren {x \times w}\) | Definition of Ordinal Multiplication | |||||||||||
\(\ds \paren {x \times y} + \paren {x \times z}\) | \(=\) | \(\ds \bigcup_{v \mathop \in \paren {x \times z} } \paren {x \times y} + v\) | Definition of Ordinal Addition |
Take any $w \in y + z$.
It follows that $w \in y \lor \paren {y \subseteq w \land w \in y + z}$ by Relation between Two Ordinals and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
Thus, $w \in y \lor w = y + u$ for some $u \in z$ by Ordinal Subtraction when Possible is Unique.
If $w < y$, then:
\(\ds x \times w\) | \(\in\) | \(\ds x \times y\) | Membership is Left Compatible with Ordinal Multiplication | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \paren {x \times y} + v\) | Ordinal is Less than Sum |
If $w = y + u$, then:
\(\ds x \times w\) | \(=\) | \(\ds x \times \paren {y + u}\) | Definition of $w$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times u}\) | Inductive Hypothesis | |||||||||||
\(\ds x \times u\) | \(\in\) | \(\ds x \times z\) | Membership is Left Compatible with Ordinal Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + v\) | setting $v$ to $x \times u$ |
By Supremum Inequality for Ordinals:
- $x \times \paren {y + z} \subseteq \paren {x \times y} + \paren {x \times z}$
Conversely, if $v \in x \times z$, then:
\(\ds \exists w \in z: \, \) | \(\ds v\) | \(\in\) | \(\ds x \times w\) | Ordinal is Less than Ordinal times Limit | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \times y} + v\) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times w}\) | Substitutivity of Class Equality | ||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {y + w}\) | Inductive Hypothesis |
By Supremum Inequality for Ordinals:
- $\paren {x \times y} + \paren {x \times z} \subseteq x \times \paren {y + z}$
By definition of set equality:
- $x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
This proves the limit case.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.25$