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

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