Condition for Membership is Right Compatible with Ordinal Exponentiation
Jump to navigation
Jump to search
Theorem
Let $x, y, z$ be ordinals.
Let $z$ be the successor of some ordinal $w$.
Then:
- $x < y \iff x^z < y^z$
Proof
Sufficient Condition
Suppose $x < y$.
By Subset is Right Compatible with Ordinal Exponentiation:
- $x^w \le y^w$
Then:
\(\ds x^z\) | \(=\) | \(\ds x^w \times x\) | Definition of Ordinal Exponentiation | |||||||||||
\(\ds y^z\) | \(=\) | \(\ds y^w \times y\) | Definition of Ordinal Exponentiation | |||||||||||
\(\ds x^w \times x\) | \(\le\) | \(\ds y^w \times x\) | Subset is Right Compatible with Ordinal Exponentiation | |||||||||||
\(\ds \) | \(<\) | \(\ds y^w \times y\) | Membership is Left Compatible with Ordinal Exponentiation |
Thus the sufficient condition is satisfied.
$\Box$
Necessary Condition
Suppose $x^z < y^z$.
From Subset is Right Compatible with Ordinal Exponentiation:
- $y \le x \implies y^z \le x^z$
This contradicts:
- $x^z < y^z$
so:
- $y \nleq x$
By Ordinal Membership is Trichotomy:
- $x < y$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.36$