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.
$\blacksquare$