Exponent Base of One

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Theorem

Let $x$ be an ordinal.


Then:

$1^x = 1$


Proof

The proof shall proceed by Transfinite Induction on $x$.


Basis for the Induction

\(\ds 1^0\) \(=\) \(\ds 1\) Definition of Ordinal Exponentiation

This proves the basis for the induction.


Induction Step

The inductive hypothesis supposes that $1^x = 1$ for some $x$.

\(\ds 1^{x^+}\) \(=\) \(\ds 1^x \times 1\) Definition of Ordinal Exponentiation
\(\ds \) \(=\) \(\ds 1 \times 1\) Inductive Hypothesis
\(\ds \) \(=\) \(\ds 1\) Ordinal Multiplication by One

This proves the induction step.


Limit Case

The inductive hypothesis supposes that $\forall y \in x: 1^y = 1$.

It follows that:

\(\ds \forall y \in x: \, \) \(\ds 1^y\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \bigcup_{y \mathop \in x} 1^y\) \(=\) \(\ds \bigcup_{y \mathop \in x} 1\) Indexed Union Equality
\(\ds \leadsto \ \ \) \(\ds 1^x\) \(=\) \(\ds 1\) Definition of Ordinal Exponentiation

This proves the limit case.

$\blacksquare$


Sources