Exponent Not Equal to Zero

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Theorem

Let $x$ and $y$ be ordinals.

Let $x \ne 0$.


Then:

$x^y \ne 0$


Proof

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


Basis for the Induction

$x^0 = 1$ by the definition of ordinal exponentiation.

Therefore, $x^0 \ne 0$.

This proves the basis for the induction.


Induction Step

The inductive hypothesis supposes that $x^y \ne 0$.

\(\displaystyle x^{y^+}\) \(=\) \(\displaystyle x^y \times x\) definition of Definition:Ordinal Exponentiation
\(\displaystyle x^y\) \(\ne\) \(\displaystyle 0\) Inductive Hypothesis
\(\displaystyle x\) \(\ne\) \(\displaystyle 0\) Hypothesis
\(\displaystyle x^y \times x\) \(\ne\) \(\displaystyle 0\) by Ordinals have No Zero Divisors

This proves the induction step.


Limit Case

The inductive hypothesis says that:

$\forall z \in y: x^z \ne 0$


\(\displaystyle \forall z \in y: \ \ \) \(\displaystyle x^z\) \(\ne\) \(\displaystyle 0\) Inductive Hypothesis
\(\displaystyle \implies \ \ \) \(\displaystyle 0\) \(\in\) \(\displaystyle x^z\) by Ordinal Membership is Trichotomy
\(\displaystyle \implies \ \ \) \(\displaystyle 0\) \(\in\) \(\displaystyle \bigcup_{z \in y} x^z\) by the fact that limit ordinals are nonempty
\(\displaystyle \implies \ \ \) \(\displaystyle 0\) \(\in\) \(\displaystyle x^y\) definition of ordinal exponentiation
\(\displaystyle \implies \ \ \) \(\displaystyle x^y\) \(\ne\) \(\displaystyle 0\) definition of empty set

This proves the limit case.

$\blacksquare$


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