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 $x^y \ne 0$.
\(\ds x^{y^+}\) | \(=\) | \(\ds x^y \times x\) | Definition of Ordinal Exponentiation | |||||||||||
\(\ds x^y\) | \(\ne\) | \(\ds 0\) | Inductive Hypothesis | |||||||||||
\(\ds x\) | \(\ne\) | \(\ds 0\) | by hypothesis | |||||||||||
\(\ds x^y \times x\) | \(\ne\) | \(\ds 0\) | Ordinals have No Zero Divisors |
This proves the induction step.
Limit Case
The inductive hypothesis says that:
- $\forall z \in y: x^z \ne 0$
\(\ds \forall z \in y: \, \) | \(\ds x^z\) | \(\ne\) | \(\ds 0\) | Inductive Hypothesis | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(\in\) | \(\ds x^z\) | Ordinal Membership is Trichotomy | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(\in\) | \(\ds \bigcup_{z \mathop \in y} x^z\) | Definition:Limit Ordinal | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(\in\) | \(\ds x^y\) | Definition of Ordinal Exponentiation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^y\) | \(\ne\) | \(\ds 0\) | Definition of Empty Set |
This proves the limit case.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.32$