# Proving, for every three ordinals $\alpha,\beta,$ and $\gamma$, that $\alpha^{\beta+\gamma}=\alpha^{\beta}\alpha^{\gamma}$

I am trying to prove, for every three ordinals $\alpha,\beta,$ and $\gamma$, that $\alpha^{\beta+\gamma}=\alpha^{\beta}\alpha^{\gamma}$.

Proof is by induction on $\gamma$.

Basis: If $\gamma = 0$, then $\alpha^{\beta+0}=\alpha^{\beta}=\alpha^{\beta}*1=\alpha^{\beta}*\alpha^0.$

Assumption: Suppose true for every $\delta < \gamma$.

Show for $\gamma$:

If $\gamma$ is a successor ordinal, then there exist $\delta$ such that $\gamma=\delta+1$, and $\alpha^{\beta+\gamma}=\alpha^{\beta+\delta+1}=\alpha^{(\beta+\delta)+1}= \alpha^{\beta+\delta}\alpha^1=_{\mathrm{assumption}}\alpha^{\beta}\alpha^{\delta}\alpha= \alpha^{\beta}\alpha^{\delta+1}.$

If $\alpha$ is a limit ordinal, then $\alpha^{\beta+\gamma}=\sup\{\alpha^{\beta+\delta};\delta < \gamma\}= \sup\{\alpha^{\beta}\alpha^{\delta};\delta < \gamma\}=_{***}\alpha^\beta\alpha^\gamma.$

I am mainly intrested in the last part of the proof signed with $***$. Can I say that? And if not, how do I continue from here?

Thank you! Shir

The proof is generally okay. As for that last part, if you already know that exponentiation and multiplication are continuous, then the last equality is true and trivial enough to stand on its own. But if you don't, then you need to prove it - or at least this equality.

Since $\alpha^\beta$ is fixed here, we can show that $\gamma_1<\gamma_2\implies\alpha^\beta\alpha^{\gamma_1}\leq\alpha^\beta\alpha^{\gamma_2}$, this is either by induction or by using the fact that exponentiation is preserving $\leq$.

Now we can finish the proof. Clearly $\delta<\gamma$ implies $\alpha^\beta\alpha^\delta\leq\alpha^\beta\alpha^\gamma$. On the other hand if $\tau<\alpha^\beta\alpha^\gamma$, then we can write it as $\alpha^\beta\cdot\tau_1+\tau_2$. Show that $\tau_2$ and $\tau_1$ are both $\leq\alpha^\delta$ for some $\delta<\gamma$, and you're practically done.

• Yes, I am actually using the continuity of this action. I will check what I do know about continuous actions on ordinals. Thank you! – topsi Nov 11 '13 at 12:41

As Asaf mentions in his answer, the limit case requires looking at the continuity of the ordinal functions. I'll look at it a bit differently, but it's all the same in the end.

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.)

This is part of a more general theory of normal functions. In short, a "function" $F : \mathbf{Ord} \to \mathbf{Ord}$ is normal if

1. $\alpha < \beta$ implies $F(\alpha) < F(\beta)$ ($F$ is strictly increasing); and
2. for limit ordinals $\beta$, $F(\beta) = \sup_{\gamma < \beta} F(\gamma)$.

The basic fact about normal functions is that we can interchange them with sups, meaning that $${\textstyle \sup_{\alpha \in S} F(\alpha)} = F ( \sup S )$$ for any set $S$ of ordinals.

It can be shown that, given any ordinal $\alpha$ the following mappings are normal:

• $\xi \mapsto \alpha + \xi$.
• $\xi \mapsto \alpha \cdot \xi$ (assuming $\alpha > 0$).
• $\xi \mapsto \alpha^\xi$ (assuming $\alpha > 1$).
• I see. Very intersting, the term, normal function.. Thank you for your proof and explanation! – topsi Nov 11 '13 at 12:39
• For the topologically inclined it may be useful to say explicitly that the normal functions are the strictly increasing continuous functions. – Brian M. Scott Nov 11 '13 at 14:13