Ordinal Multiplication via Cantor Normal Form/Limit Base

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Theorem

Let $x$ and $y$ be ordinals.

Let $x$ be a limit ordinal.

Let $y > 0$.

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 finite ordinals.


Then:

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




Proof

The proof shall proceed by finite induction on $n$:

For all $n \in \N_{\le 0}$, let $\map P n$ be the proposition:

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


Since $x$ is a limit ordinal, it follows that $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiation.


Basis for the Induction

$\map P 1$ is the statement:

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


\(\ds \sum_{i \mathop = 1}^1 \paren {x^{a_i} b_i} \times x^y\) \(=\) \(\ds \paren {x^{a_i} \times b_i} \times x^y\) Definition of Ordinal Sum
\(\ds \) \(=\) \(\ds x^{a_i} \times \paren {b_i \times x^y}\) Ordinal Multiplication is Associative
\(\ds \) \(=\) \(\ds x^{a_i} \times x^y\) Finite Ordinal Times Ordinal
\(\ds \) \(=\) \(\ds x^{a_i \mathop + y}\) Ordinal Sum of Powers

This is our basis for the induction.

$\Box$


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

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


Then we need to show:

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


Induction Step

This is our induction step:

By Upper Bound of Ordinal Sum, it follows that:

$\ds \sum_{i \mathop = 1}^n \paren {x^{a_{i \mathop + 1} } b_{i \mathop + 1} } < x^{a_1}$


By Membership is Left Compatible with Ordinal Multiplication:

\(\ds x^{a_1} b_1\) \(\le\) \(\ds x^{a_1} b_1 + \sum_{i \mathop = 1}^k \paren {x^{a_{i \mathop + 1} } b_{i \mathop + 1} }\) Ordinal is Less than Sum
\(\ds \) \(\le\) \(\ds x^{a_1} b_1 + x^{a_1}\) Membership is Left Compatible with Ordinal Addition
\(\ds \leadsto \ \ \) \(\ds x^{a_1} \times b_1 \times x^y\) \(\le\) \(\ds \sum_{i \mathop = 1}^{k \mathop + 1} \paren {x^{a_i} b_i} \times x^y\) Subset is Right Compatible with Ordinal Multiplication and General Associative Law for Ordinal Sum
\(\ds \) \(\le\) \(\ds x^{a_1} \times \paren {b_1 + 1} \times x^y\) Subset is Right Compatible with Ordinal Multiplication


But:

\(\ds x^{a_1} \times b_1 \times x^y\) \(=\) \(\ds x^{a_1} \times x^y\) Finite Ordinal Times Ordinal
\(\ds \) \(=\) \(\ds x^{a_1 \mathop + y}\) Ordinal Sum of Powers
\(\ds x^{a_1} \times \paren {b_1 + 1} \times x^y\) \(=\) \(\ds x^{a_1} \times x^y\) Finite Ordinal Times Ordinal
\(\ds \) \(=\) \(\ds x^{a_1 \mathop + y}\) Ordinal Sum of Powers


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