# Upper Bound of Ordinal Sum

## Theorem

Let $x$ and $y$ be ordinals.

Suppose $x > 1$.

Let $\sequence {a_n}$ be a finite sequence of ordinals such that:

- $a_n < x$ for all $n$

Let $\sequence {b_n}$ be a strictly decreasing finite sequence of ordinals such that:

- $b_n < y$ for all $n$

Then:

- $\ds \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$

## Proof

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

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

- $\ds \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$

### Basis for the Induction

- $\map P 0$ says that:

- $\ds \sum_{i \mathop = 1}^0 x^{b_i} a_i < x^y$

\(\ds \sum_{i \mathop = 1}^0 x^{b_i} a_i\) | \(=\) | \(\ds 0\) | Definition of Ordinal Sum | |||||||||||

\(\ds \) | \(<\) | \(\ds x^y\) | Exponent Not Equal to Zero |

This is our basis for the induction.

$\Box$

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

- $\ds \sum_{i \mathop = 1}^k x^{b_i} a_i < x^y$

Then we need to show:

- $\ds \sum_{i \mathop = 1}^{k \mathop + 1} x^{b_i} a_i < x^y$

### Induction Step

This is our induction step:

By the inductive hypothesis:

- $\ds \sum_{i \mathop = 1}^k x^{b_{i + 1}} a_{i + 1} < x^{b_1}$ since $b_1$ is greater than all $b_i$ for $i > 1$.

Then:

\(\ds \sum_{i \mathop = 1}^{k + 1} x^{b_i} a_i\) | \(=\) | \(\ds x^{b_1} a_1 + \sum_{i \mathop = 1}^k x^{b_{i + 1} } a_{i + 1}\) | General Associative Law for Ordinal Sum | |||||||||||

\(\ds \) | \(<\) | \(\ds x^{b_1} a_1 + x^{b_1}\) | Membership is Left Compatible with Ordinal Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{b_1} a_1^+\) | Ordinal Multiplication is Left Distributive | |||||||||||

\(\ds \) | \(\le\) | \(\ds x^{b_1^+}\) | Membership is Left Compatible with Ordinal Multiplication and Successor of Element of Ordinal is Subset | |||||||||||

\(\ds \) | \(\le\) | \(\ds x^y\) | Membership is Left Compatible with Ordinal Exponentiation and Successor of Element of Ordinal is Subset |

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So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

- $\ds \sum_{i \mathop = 1}^n a_i x^{b_i} < x^y$

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 8.43$