Subset is Right Compatible with Ordinal Multiplication

Theorem

Let $x, y, z$ be ordinals.

Then:

$x \le y \implies \paren {x \cdot z} \le \paren {y \cdot z}$

Proof

The proof shall proceed by Transfinite Induction on $z$.

Basis for the Induction

By definition of ordinal multiplication:

$x \cdot 0 = 0$

so the statement simply reduces to:

$0 \le 0$

This proves the basis for the induction.

Induction Step

\(\ds x\)

\(\le\)

\(\ds y\)

\(\ds \leadsto \ \ \)

\(\ds x \cdot z\)

\(\le\)

\(\ds y \cdot z\)

Inductive Hypothesis

\(\ds \leadsto \ \ \)

\(\ds \paren {x \cdot z} + x\)

\(\le\)

\(\ds \paren {y \cdot z} + y\)

Membership is Left Compatible with Ordinal Addition and Subset is Right Compatible with Ordinal Addition

\(\ds \leadsto \ \ \)

\(\ds x \cdot z^+\)

\(\le\)

\(\ds y \cdot z^+\)

Definition of Ordinal Multiplication

This proves the induction step.

Limit Case

\(\ds x\)

\(\le\)

\(\ds y\)

\(\ds \leadsto \ \ \)

\(\ds \forall w \in z: \, \)

\(\ds x \cdot w\)

\(\le\)

\(\ds y \cdot w\)

Inductive Hypothesis

\(\ds \leadsto \ \ \)

\(\ds \bigcup_{w \mathop \in z} \paren {x \cdot w}\)

\(\le\)

\(\ds \bigcup_{w \mathop \in z} \paren {y \cdot w}\)

Indexed Union Subset

\(\ds \leadsto \ \ \)

\(\ds x \cdot z\)

\(\le\)

\(\ds y \cdot z\)

Definition of Ordinal Multiplication

This proves the limit case.

$\blacksquare$

Also see

• Subset is Right Compatible with Ordinal Addition

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.21$