# Let $\alpha,\gamma$ be ordinals such that $0<\alpha\le\gamma$. Then there is a greatest ordinal $\beta$ such that $\alpha\cdot\beta\le\gamma$

Let $$\alpha,\gamma$$ be ordinals such that $$0<\alpha\le\gamma$$. Then there is a greatest ordinal $$\beta$$ such that $$\alpha\cdot\beta\le\gamma$$.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

Let $$A=\{\delta \in {\rm Ord} \mid \alpha \cdot \delta > \gamma\}$$. Since $$\alpha\cdot(\gamma+1)=\alpha\cdot\gamma+\alpha>\alpha\cdot\gamma\ge\gamma$$, $$\gamma+1\in A$$ and thus $$A\neq\emptyset$$. Let $$\xi=\min A$$.

We next prove that $$\xi$$ is a successor ordinal. Assume the contrary that $$\xi$$ is a limit ordinal, then $$\alpha\cdot\xi=\sup_{\eta<\xi}(\alpha\cdot\eta)>\gamma$$. Then $$\alpha\cdot\eta>\gamma$$ for some $$\eta<\xi$$. Thus $$\eta\in A$$ and $$\eta<\xi$$. This contradicts the minimality of $$\xi$$. Hence $$\xi$$ is a successor ordinal and $$\xi=\beta+1$$. Then $$\beta=\max\{\delta \in {\rm Ord} \mid \alpha \cdot \delta \le \gamma\}$$.

Update: I add the proof of $$\beta=\max\{\delta \in {\rm Ord} \mid \alpha \cdot \delta \le \gamma\}$$.

For $$\delta>\beta$$: $$\delta\ge\beta+1=\xi\implies\alpha \cdot\delta\ge\alpha \cdot\xi>\gamma\implies\alpha \cdot\delta>\gamma\implies$$ $$\delta\notin\{\delta \in {\rm Ord} \mid \alpha \cdot \delta \le \gamma\}$$. Moreover, $$\beta\in\{\delta \in {\rm Ord} \mid \alpha \cdot \delta \le \gamma\}$$. Hence $$\beta=\max\{\delta \in {\rm Ord} \mid \alpha \cdot \delta \le \gamma\}$$.

The argument for $$\xi$$ not being a limit is more clearly written as follows, I think:

If $$\xi$$ is a limit ordinal, then by minimality of $$\xi$$, $$\alpha \cdot \delta \le \gamma$$ for all $$\delta < \xi$$, as $$\delta \notin A$$, and so $$\alpha \cdot \xi = \sup\{\alpha \cdot \delta : \delta < \xi\}\le \gamma$$, which contradicts $$\alpha \cdot \xi > \gamma$$.

So $$\xi = \beta+1$$ I agree with, but you have not yet shown that $$\beta$$ is then as required, you just claim so, without an argument.

Well, $$\beta < \xi$$ already gives $$\alpha \cdot \beta \le \gamma$$, by minimality, so $$\beta \in \{\delta \in \text{Ord}: \alpha \cdot \delta \le \gamma\}$$.

And if $$\beta' > \beta$$ we know $$\beta' \ge \beta+1= \xi$$ so we need to have the lemma that

$$\beta \ge \beta'$$ implies $$\alpha \cdot \beta \ge \alpha \cdot \beta'$$ for any fixed $$\alpha$$,

and this can quite easily be shown by transfinite induction. (It might be in your text already). Having this as a lemma, we can say $$\beta' > \beta$$ then $$\beta' \ge \xi$$ and $$\alpha \cdot \beta' \ge \alpha \cdot \xi > \gamma$$ and so $$\beta' \notin \{\delta \in \text{Ord}: \alpha \cdot \delta \le \gamma\}$$ and $$\beta$$ is indeed maximal.