Vrftjkry

I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.

Below is the verbatim proof by @user642796.

By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.

The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
alpha^{beta + gamma}
&= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
&= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
&= alpha^beta cdot sup { alpha^delta : delta < gamma }
&&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
\
&=alpha ^beta cdot alpha^gamma &&text{(by definition)}
end{align}$$

(The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)

My questions:

1. I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve

$$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
= &alpha^beta cdot sup { alpha^delta : delta < gamma }
left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$

2. The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.

Thank you so much!

Let me provide a few more details. First let's prove the stated

Lemma. Let $1 < alpha$, let $beta$ be a limit ordinal and let $A subseteq beta$ be cofinal. Then $a^beta = sup { a^gamma mid gamma in A }$.

Proof. By definition of ordinal multiplication we have that $alpha^beta = sup { alpha^gamma mid gamma < beta }$. Since ${alpha^gamma mid gamma in A } subseteq { alpha^gamma mid gamma < beta }$ we thus clearly have that $sup { alpha^gamma mid gamma in A } le alpha^beta$.

Conversely let $xi < alpha^beta$. Since $alpha^beta = sup { alpha^gamma mid gamma < beta }$ there is some $gamma < beta$ such that $xi < alpha^gamma$. Now, since $A subseteq beta$ is cofinal, there is some $gamma < gamma^* in A$. Since $gamma mapsto alpha^gamma$ is increasing, we have that $$xi < alpha^gamma < alpha^{gamma^*} le sup { alpha^{delta} mid delta in A }.$$

Thus $alpha^beta = sup { alpha^gamma mid gamma in A }$. Q.E.D.

Now to the main event:

Lemma. Let $alpha > 1$, $beta, gamma$ be ordinals. Then $alpha^{beta + gamma} = alpha^{beta} cdot alpha^{gamma}$.

Proof. We proceed by induction on $gamma$ and only consider the limit case. (As seen in the original post, the case $gamma = delta + 1$ is easy.)

We have
$$
begin{align*}
alpha^{beta + gamma } &= sup{ alpha^ delta mid delta < beta + gamma } && beta + gamma text{ is a limit ordinal} \
&= sup{ alpha^ delta mid beta le delta < beta + gamma } && text{ see lemma above} \
&= sup { alpha^{beta + delta} mid delta < gamma } && text{every } beta le delta < beta + gamma text{ can be written as } delta = beta + delta^* text{ for a unique } delta^* < gamma \
&= sup { alpha^beta cdot alpha^delta mid delta < gamma } && text{ induction hypothesis}
end{align*}
$$
It remains to be seen that $sup { alpha^beta cdot alpha^delta mid delta < gamma } = alpha^beta cdot alpha^gamma$.

Let $delta < gamma$. Then $alpha^delta < alpha^gamma$. Since $(xi, eta) mapsto xi cdot eta$ is increasing in the second coordinate, it follows that $alpha^beta cdot alpha^delta < alpha^beta cdot alpha^gamma$ and thus $sup { alpha^beta cdot alpha^delta mid delta < gamma } le alpha^beta cdot alpha^gamma$.

Conversely, let $xi < alpha^beta cdot alpha^gamma$. Since $cdot$ is increasing in the second coordinate, there is then some $eta < alpha^gamma$ such that $xi < alpha^beta cdot eta$. Since $alpha^gamma = sup { alpha^delta mid delta < gamma }$ there is moreover some $delta < gamma$ such that $xi < alpha^beta cdot alpha^delta$. This shows that $alpha^beta cdot alpha^gamma le sup { alpha^beta cdot alpha^delta mid delta < gamma }$ and hence the desired equality. Q.E.D.