Ordinal Multiplication via Cantor Normal Form/Infinite Exponent

Theorem

Let $x$ and $y$ be ordinals.

Let $x > 1$.

Let $y \ge \omega$ where $\omega$ denotes the minimally inductive set.

Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.

Let $\sequence {b_i}$ be a sequence of ordinals such that $0 < b_i < x$ for all $1 \le i \le n$.

Then:

$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$

Proof

It follows that:

\(\ds x^{a_1}\)

\(\le\)

\(\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i}\)

Ordinal is Less than Sum

\(\ds \)

\(<\)

\(\ds x^{a_1 + 1}\)

Upper Bound of Ordinal Sum

By multiplying the inequalities by $x^y$ on the left:

\(\ds x^{a_1} \times x^y\)

\(\le\)

\(\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y\)

Subset is Right Compatible with Ordinal Multiplication

\(\ds \)

\(\le\)

\(\ds x^{a_1 \mathop + 1} \times x^y\)

Solving for both sides of the inequality:

\(\ds x^{a_1} \times x^y\)

\(=\)

\(\ds x^{a_1 \mathop + y}\)

Ordinal Sum of Powers

\(\ds x^{a_1 \mathop + 1} \times x^y\)

\(=\)

\(\ds x^{\paren {a_1 \mathop + 1} \mathop + y}\)

Ordinal Sum of Powers

\(\ds \)

\(=\)

\(\ds x^{a_1 \mathop + \paren {1 \mathop + y} }\)

Ordinal Addition is Associative

\(\ds \)

\(=\)

\(\ds x^{a_1 \mathop + y}\)

Finite Ordinal Plus Transfinite Ordinal

Therefore:

\(\ds x^{a_1 \mathop + y}\)

\(\le\)

\(\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y\)

Substitutivity of Class Equality

\(\ds \)

\(\le\)

\(\ds x^{a_1 \mathop + y}\)

Substitutivity of Class Equality

\(\ds \leadsto \ \ \)

\(\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y\)

\(=\)

\(\ds x^{a_1 \mathop + y}\)

Definition 2 of Set Equality

$\blacksquare$

Also see

• Definition:Cantor Normal Form

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.45$