Limit Ordinals Preserved Under Ordinal Multiplication
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Theorem
Let $x$ and $y$ be ordinals.
Let $x$ be non-empty.
Let $y$ be a limit ordinal.
It follows that the ordinal product $\left({x \times y}\right)$ is a limit ordinal.
Proof
$y$ is a limit ordinal and thus is nonzero, by definition.
$x$ and $y$ are both nonzero.
So by Ordinals have No Zero Divisors:
- $x \times y \ne 0$
So by definition of limit ordinal:
- $x \times y \in K_{II} \lor \exists z \in \On: x \times y = z^+$
Suppose that $x \times y = z^+$ for some ordinal $z$.
\(\ds x \times y\) | \(=\) | \(\ds \bigcup_{w \mathop \in y} \paren {x \times w}\) | Definition of Ordinal Multiplication |
It follows that:
\(\ds z\) | \(\in\) | \(\ds \bigcup_{w \mathop \in y} \tuple {x \times w}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists w \in y: \, \) | \(\ds z\) | \(\in\) | \(\ds x \times w\) | Definition of Set Union | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists w \in y: \, \) | \(\ds z^+\) | \(\in\) | \(\ds \paren {x \times w}^+\) | Successor is Less than Successor | |||||||||
\(\ds \paren {x \times w}^+\) | \(=\) | \(\ds \paren {x \times w} + 1\) | Ordinal Addition by One | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists w \in y: \, \) | \(\ds z^+\) | \(\in\) | \(\ds \paren {x \times w} + 1\) |
But by Subset is Right Compatible with Ordinal Addition:
- $\paren {x \times w} + 1 \subseteq \paren {x \times w} + x$
Therefore:
\(\ds \exists w \in y: \, \) | \(\ds z^+\) | \(\in\) | \(\ds \paren {x \times w} + x\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times w^+}\) | Definition of Ordinal Multiplication |
But by Successor of Ordinal Smaller than Limit Ordinal is also Smaller:
- $w^+ \in y$
Therefore:
- $z^+ \in x \times y$
contradicting the fact that $z^+ = x \times y$.
Thus:
- $z^+ \ne x \times y$
and:
- $x \times y \in K_{II}$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.23$