Proving that if $\alpha \neq 0$ then $\alpha \cdot \beta$ is limit when $\beta$ is limit ordinal

Question

This is my problem. I have to prove that if $\alpha \neq 0$ is any ordinal then $\alpha \cdot \beta$ is a limit ordinal when $\beta$ is limit ordinal. The definition of limit ordinal I have is that $A$ is a limit ordinal if for each $\alpha \in A$, it is true that: $\alpha+1 \in A$ So I tried to use the definition but can't see a clear answer.

¿There are another equivalences of limit ordinals I can use? ¿Any hint?

Please and thank you !

Answer 1

Prove the following fact first:

$A$ is a limit ordinal if and only if $\sup\{\alpha\in A\mid\alpha+1\in A\}=A$.

In other words, you don't have to check that for every $\alpha\in A$, $\alpha+1$ is also in $A$. Just unboundedly many of them.

Next, $\alpha\cdot\beta=\sup\{\alpha\cdot\gamma\mid\gamma<\beta\}$, and since $\alpha>0$, you can check that $(\alpha\cdot\gamma)+1\leq(\alpha\cdot\gamma)+\alpha=\alpha\cdot(\gamma+1)$. Finally, apply the fact.

Answer 2

If you're using the recursive definition of product and $\delta<\alpha·\beta$, pick $\gamma<\beta$ with $\delta<\alpha·\gamma$. Then $$ \delta+1 < \alpha·\gamma +1 \le \alpha·\gamma + \alpha =\alpha ·(\gamma +1) \le \alpha\cdot\beta $$ because $\gamma+1<\beta$.

If you're using the definition $\alpha\cdot\beta =\text{otp}(\beta\times \alpha)$ (ordered lexicographically), then any pair $(\xi,\zeta)$ is lexicographically below $(\xi+1,\zeta)$, so that $\beta\times \alpha$ has no largest element, and so neither does $\alpha\cdot\beta$ (under $\in$). But then $\alpha\cdot \beta$ is a limit.