A question about the proof of $\sup\{\alpha^\beta\alpha^\delta\mid\delta<\gamma\}=\alpha^\beta\sup\{\alpha^\delta\mid\delta<\gamma\}$ 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 $$\begin{align}
\alpha^{\beta + \gamma} 
&= \sup \{ \alpha^{\beta + \delta} : \delta < \gamma \} &&\text{(}\{ \beta+\delta : \delta < \gamma\}\text{ is cofinal in the limit ord }\beta + \gamma\text{)} \\
&= \sup \{ \alpha^\beta \cdot \alpha^\delta : \delta < \gamma \} &&\text{(by induction hypothesis)} \\
&= \alpha^\beta \cdot \sup \{ \alpha^\delta : \delta < \gamma \} 
&&\text{(}\{ \alpha^\delta : \delta < \gamma \}\text{ is cofinal in the limit ord }\sup \{ \alpha^\delta : \delta < \gamma \}\text{)} 
\\
&=\alpha ^\beta \cdot \alpha^\gamma &&\text{(by definition)}
\end{align}$$
(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:


*

*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



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



*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!
 A: 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 the definition of ordinal exponentiation 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.
