Membership is Left Compatible with Ordinal Exponentiation
Theorem
Let $x$, $y$, and $z$ be ordinals.
Suppose $1 < z$.
Then:
- $x < y \iff z^x < z^y$
Proof
Sufficient Condition
The proof shall proceed by Transfinite Induction on $y$.
Basis for the Induction
It is a contradiction that $x < 0$ by the definition of empty set.
This proves the basis for the induction.
Induction Step
The inductive step states that $x < y \implies z^x < z^y$.
Then, if $x < y^+$, then $x = y \lor x < y$.
Lemma
$z^y < z^y \times y$ by Membership is Left Compatible with Ordinal Multiplication.
So $z^y < z^{y^+}$ by the definition of ordinal exponentiation.
$\Box$
Suppose $x = y$. Then $z^x < z^{y^+}$.
Suppose $x < y$. Then:
\(\ds z^x\) | \(<\) | \(\ds z^y\) | Inductive Hypothesis | |||||||||||
\(\ds \) | \(<\) | \(\ds z^{y^+}\) | Lemma above |
In either case, the inductive step holds.
This proves the induction step.
Limit Case
The inductive hypothesis states that:
- $\forall w \in y: z^x < z^w$ for $x < w$
Take any $x < y$.
By Limit Ordinal Equals its Union, it follows that $x < w$ for some $w \in y$.
By the inductive hypothesis, $z^x < z^w$.
Since $w \in y$, it follows that:
\(\ds z^x\) | \(<\) | \(\ds z^w\) | above | |||||||||||
\(\ds \) | \(\le\) | \(\ds \bigcup_{w \in y} z^w\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds z^y\) | Definition of Ordinal Exponentiation |
This proves the limit case.
$\Box$
Necessary Condition
Note that the last part proves that:
- $x \in y \implies z^x \in z^y$
Moreover:
- $x = y \implies z^x = z^y$
Therefore, by the fact that $\in$ is a strong ordering:
- $x \notin y \implies z^x \notin z^y$
By the Rule of Transposition:
- $z^x < z^y \implies x < y$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.33$, $\S 8.34$