Ordinal Addition is Closed
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Theorem
Let $\On$ be the class of all ordinals.
Then:
- $\forall x, y \in \On: x + y \in \On$
That is: the sum $x+y$ is an ordinal.
Proof
Using Transfinite Induction on $y$:
Base Case
| \(\ds x\) | \(\in\) | \(\ds \On\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x + \O\) | \(\in\) | \(\ds \On\) | Definition of Ordinal Addition |
Inductive Case
| \(\ds x + y\) | \(\in\) | \(\ds \On\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {x + y}^+\) | \(\in\) | \(\ds \On\) | Successor Set of Ordinal is Ordinal | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x + y^+\) | \(\in\) | \(\ds \On\) | Definition of Ordinal Addition |
Limit Case
| \(\ds \forall z \in y: \, \) | \(\ds x + z\) | \(\in\) | \(\ds \On\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \bigcup_{z \mathop \in y} \paren {x + z}\) | \(\in\) | \(\ds \On\) | Union of Set of Ordinals is Ordinal: Corollary | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x + z\) | \(\in\) | \(\ds \On\) | Definition of Ordinal Addition |
$\blacksquare$
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Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.2$