Ordinal Exponentiation via Cantor Normal Form/Limit Exponents

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Theorem

Let $x$ and $y$ be ordinals.

Let $x$ and $y$ be limit ordinals.

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

Let $\left\langle{b_i}\right\rangle$ be a sequence of natural numbers.


Then:

$\displaystyle \left({\sum_{i \mathop = 1}^n x^{a_i} \times b_i}\right)^y = x^{a_1 \mathop \times y}$


Proof

By Upper Bound of Ordinal Sum:

$\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_i} \times b_i }\right) \le x^{a_1} \times \left({b_1 + 1}\right)$


Furthermore:

$\displaystyle x^{a_1} \le \sum_{i \mathop = 1}^n \left({ x^{a_i} \times b_i }\right)$


It follows that:

\(\displaystyle \left({ x^{a_1} }\right)^y\) \(\le\) \(\displaystyle \left({\sum_{i \mathop = 1}^n x^{a_i} \times b_i}\right)^y\) Subset is Right Compatible with Ordinal Exponentiation
\(\displaystyle \) \(\le\) \(\displaystyle \left({x^{a_1} \times \left({b_1 + 1}\right)}\right)^y\) Subset is Right Compatible with Ordinal Exponentiation
\(\displaystyle \) \(=\) \(\displaystyle x^{a_1 \times y}\) Ordinal Exponentiation of Terms


It follows that:

\(\displaystyle x^{a_1 \mathop + y}\) \(\le\) \(\displaystyle \left({ \sum_{i \mathop = 1}^n x^{a_i} b_i }\right)^y\) by Ordinal Power of Power
\(\displaystyle \) \(\le\) \(\displaystyle x^{a_1 \mathop + y}\) Proven above
\(\displaystyle \implies \ \ \) \(\displaystyle \left({\sum_{i \mathop = 1}^n x^{a_i} b_i}\right)^y\) \(=\) \(\displaystyle x^{a_1 \mathop + y}\) Definition of Set Equality

$\blacksquare$


Also see


Sources