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

Question

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

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

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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 $\beta=\max\{\delta \in {\rm Ord} \mid \alpha \cdot \delta \le \gamma\}$.

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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 $\beta=\max\{\delta \in {\rm Ord} \mid \alpha \cdot \delta \le \gamma\}$.

Answer 1

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.