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

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.

• I think math.stackexchange.com/questions/3233204/… allow you to conclude. Apr 25 at 1:31
• Yeah, I think it does, thanks a lot. Apr 25 at 1:44
• You're welcome! If you want to, you can answer the question yourself so it is marked as answered. Apr 25 at 1:53
• 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 hols for all positive $n<\omega$. Now, since $\beta>0$ then $\alpha^n\le\alpha^n\cdot\beta=\beta$, so $\alpha^\omega=\sup_n\alpha^n\le\beta$. Apr 25 at 6:44

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 $$\beta$$.
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 $$\alpha^\omega=\sup_n\alpha^n\le\beta$$.
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 $$\alpha^{\omega+1}=2^\omega\cdot 2=\omega\cdot2>\omega=\beta$$.