Finite Ordinal Times Ordinal/Lemma

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Lemma

Let $m$ be a finite ordinal.

Let $m \ne 0$, where $0$ is the zero ordinal.


Then:

$m \times \omega = \omega$

where $\omega$ denotes the minimal infinite successor set.


Proof

\(\displaystyle \forall n \in \omega \ \ \) \(\displaystyle m \times n\) \(\in\) \(\displaystyle \omega\) Natural Number Multiplication is Closed
\(\displaystyle \bigcup_{n \mathop \in \omega} \left({ m \times n }\right)\) \(\le\) \(\displaystyle \omega\) Supremum Inequality for Ordinals
\(\displaystyle \implies \ \ \) \(\displaystyle m \times \omega\) \(\le\) \(\displaystyle \omega\) Definition of Ordinal Multiplication

Also, $\omega \le \left({ m \times \omega }\right)$ by Subset is Right Compatible with Ordinal Multiplication.

The lemma follows from the definition of equality.

$\Box$