# Limit Ordinals Preserved Under Ordinal Multiplication

## 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.

$x \times y \ne 0$

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 \ \$ $\ds \exists w \in y: \,$ $\ds z^+$ $\in$ $\ds \paren {x \times w} + 1$
$\paren {x \times w} + 1 \subseteq \paren {x \times w} + x$

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
$w^+ \in y$

Therefore:

$z^+ \in x \times y$

contradicting the fact that $z^+ = x \times y$.

Thus:

$z^+ \ne x \times y$

and:

$x \times y \in K_{II}$

$\blacksquare$