Exercise 6.5.2

Proof. We show this by transfinite induction on $\gamma$. So for $\gamma =0$ we have

Now suppose that $\alpha \cdot (\beta +\gamma )=\alpha \cdot \beta +\alpha \cdot \gamma$ for ordinal $\gamma$. We then have

Lastly, suppose that $\gamma \ne 0$ is a limit ordinal and that $\alpha \cdot (\beta +\delta )=\alpha \cdot \beta +\alpha \cdot \delta$ for all $\delta <\gamma$. If $\alpha =0$ then we have

where we have used Lemma ?? above. So suppose that $\alpha \ne 0$ so that $\alpha \ge 1$. Now, if $\xi <\beta +\gamma$ then $\xi \le \beta +\delta$ for some $\delta <\gamma$. Then by Lemma 6.5.4 we have that $\xi +1\le (\beta +\delta )+1=\beta +(\delta +1)<\beta +\gamma$ since $\gamma$ is a limit ordinal. Hence $\beta +\gamma$ is a limit ordinal so that by Definition 6.5.6c we have $\alpha \cdot (\beta +\gamma )=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\alpha \cdot \xi \mid \xi <\beta +\gamma \right\}$. But then by Definition 6.5.1c we have that $\beta +\gamma =\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\beta +\delta \mid \delta <\gamma \right\}$. It then follows that

by the induction hypothesis. Then by Definition 6.5.6c we have that $\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\alpha \cdot \delta \mid \delta <\gamma \right\}=\alpha \cdot \gamma$ since $\gamma$ is a limit ordinal. It then follows from Lemma ?? that $\alpha \cdot \gamma$ is also a limit ordinal and $\alpha \cdot \gamma \ne 0$ since $\alpha \ne 0$. From this and Definition 6.5.1c we have that

as desired. This completes the inductive proof. □