$\beta > 0$ and $\alpha \cdot \beta = \beta \implies \alpha ^{\omega} \leq \beta$

Question

I'm having trouble answering this question of ordinals, let $\alpha, \beta$ be ordinals and suppose $\beta> 0, \alpha \cdot \beta = \beta$ then $\alpha ^ { \omega} \leq \beta$

My try so far goes: suppose $\beta < \alpha ^{\omega}$ then $\beta + \gamma = \alpha ^{ \omega}$ for some ordinal $\gamma$ then $\beta + \gamma=\alpha ^ {\omega} = \alpha \cdot \alpha ^{\omega}=\alpha \cdot ( \beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma = \beta + \alpha \cdot \gamma$

Thus $ \gamma = \alpha \cdot \gamma$ but I don't know if this allows me to conclude something. I would appreciate any suggetions, thanks.

Answer 1

This can be answered fairly straightforward with a contrapositive. Namely, if $\beta<\alpha^\omega$, then $\alpha\cdot\beta\neq\beta$.

Simply note that if $\beta<\alpha^\omega$, then by definition there is some $n$ such that $\alpha^n\leq\beta<\alpha^{n+1}$. Multiply $\alpha$ by these three ordinals, we get $$\alpha\cdot\alpha^n=\alpha^{n+1}\leq\alpha\cdot\beta\leq\alpha\cdot\alpha^{n+1}=\alpha^{n+2}.$$ Even though we allowed a very lax possibility of equality, all three ordinals we get are strictly larger than $\beta$.

Answer 2

By induction on $n>0$, if $\alpha^n\cdot\beta=\beta$, then $\alpha^{n+1}\cdot\beta=\alpha\cdot (\alpha^n\cdot\beta)=\alpha\cdot\beta =\beta$, so the equality holds for all positive $n<\omega$. Now, since $\beta>0$, then $\alpha^n\le\alpha^n\cdot\beta=\beta$, so $\alpha^\omega=\sup_n\alpha^n\le\beta$.

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On the other hand, note that the assumptions are not strong enough to allow us to conclude that $\alpha^{\omega+1}\le\beta$: consider $\alpha=2$, $\beta=\omega$, for which we have $\alpha\cdot\beta=2\cdot\omega=\omega=\beta$, but $\alpha^{\omega+1}=2^\omega\cdot 2=\omega\cdot2>\omega=\beta$.