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

Let $\sequence {b_i}$ be a sequence of natural numbers.


Then:

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


Proof

By Upper Bound of Ordinal Sum:

$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$


Furthermore:

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


It follows that:

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


It follows that:

\(\ds x^{a_1 \mathop + y}\) \(\le\) \(\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} b_i}^y\) Ordinal Power of Power
\(\ds \) \(\le\) \(\ds x^{a_1 \mathop + y}\) proven above
\(\ds \leadsto \ \ \) \(\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} b_i}^y\) \(=\) \(\ds x^{a_1 \mathop + y}\) Definition 2 of Set Equality

$\blacksquare$


Also see


Sources

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