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 minimally inductive set.


Proof

\(\ds \forall n \in \omega: \, \) \(\ds m \times n\) \(\in\) \(\ds \omega\) Natural Number Multiplication is Closed
\(\ds \bigcup_{n \mathop \in \omega} \paren {m \times n}\) \(\le\) \(\ds \omega\) Supremum Inequality for Ordinals
\(\ds \leadsto \ \ \) \(\ds m \times \omega\) \(\le\) \(\ds \omega\) Definition of Ordinal Multiplication

Also, by Subset is Right Compatible with Ordinal Multiplication:

$\omega \le \paren {m \times \omega}$

The lemma follows from the definition of equality.

$\blacksquare$