Krdytkyu

Let $alpha,gamma$ be ordinals such that $0<alphalegamma$. Then there is a greatest ordinal $beta$ such that $alphacdotbetalegamma$.

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 $alphacdot(gamma+1)=alphacdotgamma+alpha>alphacdotgammagegamma$, $gamma+1in A$ and thus $Aneqemptyset$. Let $xi=min A$.

We next prove that $xi$ is a successor ordinal. Assume the contrary that $xi$ is a limit ordinal, then $alphacdotxi=sup_{eta<xi}(alphacdoteta)>gamma$. Then $alphacdoteta>gamma$ for some $eta<xi$. Thus $etain A$ 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 $delta>beta$: $deltagebeta+1=xiimpliesalpha cdotdeltagealpha cdotxi>gammaimpliesalpha cdotdelta>gammaimplies$ $deltanotin{delta in {rm Ord} mid alpha cdot delta le gamma}$. Moreover, $betain{delta in {rm Ord} mid alpha cdot delta le gamma}$. 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.