Ordinal Multiplication via Cantor Normal Form/Infinite Exponent
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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
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.45$