# How do I proceed to prove $\tau<\alpha\cdot\beta\iff \tau=\alpha\cdot\eta+\zeta$ for a unique $\eta<\beta,\zeta<\alpha$?

Let $$\alpha,\beta,\gamma,\tau$$ be ordinals. Then

1. $$\tau<\alpha\cdot\beta\iff \tau=\alpha\cdot\eta+\zeta$$ for a unique $$\eta<\beta,\zeta<\alpha$$

2. $$\tau<\alpha^\beta\alpha^\gamma\implies \tau=\alpha^\beta\cdot\eta+\zeta$$ for some $$\eta,\zeta\le\alpha^\delta$$ and $$\delta<\gamma$$

My attempt:

1a. $$\tau<\alpha\cdot\beta\implies \tau=\alpha\cdot\eta+\zeta$$ for a unique $$\eta<\beta,\zeta<\alpha$$

Uniqueness

Assume that $$\alpha\cdot\eta+\zeta=\alpha\cdot\eta'+\zeta'$$.

Assume $$\eta\neq\eta'$$. WLOG, we assume $$\eta<\eta'$$ and thus $$\eta+1\le\eta'$$. It follows that $$\alpha\cdot(\eta+1)\le\alpha\cdot\eta'$$ and thus $$\alpha\cdot\eta+\alpha\le\alpha\cdot\eta'$$. Then $$(\alpha\cdot\eta+\alpha)+\zeta'\le\alpha\cdot\eta'+\zeta'$$ and thus $$\alpha\cdot\eta+(\alpha+\zeta')\le\alpha\cdot\eta'+\zeta'$$. Moreover, $$\zeta<\alpha$$, then $$\zeta<\alpha+\zeta'$$. Hence $$\alpha\cdot\eta+\zeta<\alpha\cdot\eta+(\alpha+\zeta')\le\alpha\cdot\eta'+\zeta'$$ and thus $$\alpha\cdot\eta+\zeta<\alpha\cdot\eta'+\zeta'$$. This is a contradiction and thus $$\eta=\eta'$$.

It follows that $$\zeta=\zeta'$$ and thus $$(\eta,\zeta)=(\eta',\zeta')$$.

Existence

We prove by induction on $$\tau$$.

• If $$\tau=0$$ then $$\tau=\alpha\cdot 0+0$$.

• If $$\tau=\tau'+1$$, then by IH $$\tau'=\alpha\cdot\eta'+\zeta'$$ and thus $$\tau'+1=(\alpha\cdot\eta'+\zeta')+1=$$ $$\alpha\cdot\eta'+(\zeta'+1)$$.

• If $$\tau$$ is a limit ordinal, then by IH $$\forall \tau'<\tau:\tau'=\alpha\cdot\eta'+\zeta'$$

I'm stuck at showing $$\tau=\alpha\cdot\eta+\zeta$$

1b. $$\tau<\alpha\cdot\beta\Longleftarrow \tau=\alpha\cdot\eta+\zeta$$ for a unique $$\eta<\beta,\zeta<\alpha$$

$$\eta<\beta\implies\eta+1\le\beta\implies\alpha\cdot(\eta+1)\le\alpha\cdot\beta\implies\alpha\cdot\eta+\alpha\le\alpha\cdot\beta$$. Moreover, $$\zeta<\alpha\implies\alpha\cdot\eta+\zeta<\alpha\cdot\eta+\alpha$$. Thus $$\alpha\cdot\eta+\zeta=\tau<\alpha\cdot\beta$$.

Pleas help me prove $$\tau=\alpha\cdot\eta+\zeta$$. Thank you so much!

I have figured out the proof and posted here.

1a. $$\tau<\alpha\cdot\beta\implies \tau=\alpha\cdot\eta+\zeta$$ for a unique $$\eta<\beta,\zeta<\alpha$$

Uniqueness

Assume that $$\alpha\cdot\eta+\zeta=\alpha\cdot\eta'+\zeta'$$.

Assume $$\eta\neq\eta'$$. WLOG, we assume $$\eta<\eta'$$ and thus $$\eta+1\le\eta'$$. It follows that $$\alpha\cdot(\eta+1)\le\alpha\cdot\eta'$$ and thus $$\alpha\cdot\eta+\alpha\le\alpha\cdot\eta'$$. Then $$(\alpha\cdot\eta+\alpha)+\zeta'\le\alpha\cdot\eta'+\zeta'$$ and thus $$\alpha\cdot\eta+(\alpha+\zeta')\le$$ $$\alpha\cdot\eta'+\zeta'$$. Moreover, $$\zeta<\alpha$$, then $$\zeta<\alpha+\zeta'$$. Hence $$\alpha\cdot\eta+\zeta<\alpha\cdot\eta+(\alpha+\zeta')\le$$ $$\alpha\cdot\eta'+\zeta'$$ and thus $$\alpha\cdot\eta+\zeta<\alpha\cdot\eta'+\zeta'$$. This is a contradiction and thus $$\eta=\eta'$$.

It follows that $$\zeta=\zeta'$$ and thus $$(\eta,\zeta)=(\eta',\zeta')$$.

Existence

For $$\tau<\alpha\cdot\beta$$, let $$X=\{\gamma\mid\alpha\cdot\gamma\le\tau\}$$ and $$\eta=\sup X$$. Since $$\tau<\alpha\cdot\beta$$, $$\forall\gamma\in X:\gamma<\beta$$ and thus $$\eta\le\beta$$.

First, we prove that $$\eta<\beta$$.

• If $$\beta=\beta'+1$$, then $$\forall\gamma\in X:\gamma\le\beta'$$ and thus $$\eta=\sup X\le\beta'<\beta$$.

• If $$\beta$$ is a limit ordinal, we assume the contrary that $$\eta=\beta$$. Then $$\gamma<\beta\implies\gamma<\eta=\sup X$$ $$\implies\gamma<\gamma'$$ for some $$\gamma'\in X$$ $$\implies\alpha\cdot\gamma<\alpha\cdot\gamma'\le\tau$$ for some $$\gamma'\in X$$. Thus $$\gamma<\beta\implies$$ $$\alpha\cdot\gamma<\tau$$. We have $$\alpha\cdot\beta=\sup\{\alpha\cdot\gamma\mid\gamma<\beta\}\le\sup\{\tau\mid\gamma<\beta\}=\tau$$, which is a contradiction. It follows that $$\eta\neq\beta$$ and thus $$\eta<\beta$$.

Second, we prove $$\alpha\cdot\eta\le\tau$$.

• If $$\eta\in X$$, then $$\eta=\gamma$$ for some $$\gamma\in X$$. It follows that $$\alpha\cdot\eta=\alpha\cdot\gamma\le\tau$$.

• If $$\eta\notin X$$, then $$\eta$$ is clearly a limit ordinal. We have $$\gamma<\eta\implies\gamma<\sup X\implies\gamma<\gamma'$$ for some $$\gamma'\in X$$ $$\implies\alpha\cdot\gamma<\alpha\cdot\gamma'\le\tau$$ for some $$\gamma'\in X$$. It follows that $$\gamma<\eta\implies\alpha\cdot\gamma<\tau$$. Then $$\alpha\cdot\eta=\sup\{\alpha\cdot\gamma\mid\gamma<\eta\}\le\sup\{\tau\mid\gamma<\eta\}=\tau$$. Thus $$\alpha\cdot\eta\le\tau$$ and hence $$\eta\in X$$, which contradicts to our very first assumption that $$\eta\notin X$$. As a result, this case does not exist.

As a result, there is a unique $$\zeta$$ such that $$\tau=\alpha\cdot\eta+\zeta$$.

Finally, we prove $$\zeta<\alpha$$. Assume the contrary that $$\alpha\le\zeta$$, then $$\alpha+\delta=\zeta$$ for some $$\delta$$. Then $$\tau=\alpha\cdot\eta+\zeta=\alpha\cdot\eta+(\alpha+\delta)=(\alpha\cdot\eta+\alpha)+\delta=\alpha\cdot(\eta+1)+\delta$$. This contradicts the fact that $$\eta=\sup X$$.

1b. $$\tau<\alpha\cdot\beta\Longleftarrow \tau=\alpha\cdot\eta+\zeta$$ for a unique $$\eta<\beta,\zeta<\alpha$$

$$\eta<\beta\implies\eta+1\le\beta\implies\alpha\cdot(\eta+1)\le\alpha\cdot\beta\implies\alpha\cdot\eta+\alpha\le\alpha\cdot\beta$$. Moreover, $$\zeta<\alpha\implies\alpha\cdot\eta+\zeta<\alpha\cdot\eta+\alpha$$. Thus $$\alpha\cdot\eta+\zeta=\tau<\alpha\cdot\beta$$.