Exercise 6.5.1

Proof. We show this by transfinite induction on $\alpha$. For $\alpha =0$ we have $0\cdot \alpha =0\cdot 0=0$ by Definition 6.5.6a. Now suppose that $0\cdot \alpha =0$ so that we have

by the induction hypothesis, Definition 6.5.6b, and Definition 6.5.1a. Lastly, suppose that $\alpha \ne 0$ is a limit ordinal and that $0\cdot \beta =0$ for all $\beta <\alpha$. Then by Definition 6.5.6c we have that

by the induction hypothesis. This completes the proof. □

Proof. ( $\to$) We show this by contrapositive. So suppose that there is a $\beta <\alpha$ such that $\beta +1\ge \alpha$. Then by Lemma ?? we have also that $\beta +1\le \alpha$, from which it follows that $\alpha =\beta +1$ so that $\alpha$ is a successor ordinal and not a limit ordinal.

( $\leftarrow$) Suppose that $\beta +1<\alpha$ for every $\beta <\alpha$. Suppose also that $\alpha =\gamma +1$ is a successor ordinal. Then clearly $\gamma <\gamma +1=\alpha$ so that also $\gamma +1<\alpha$, which is an immediate contradiction. Hence it must be that $\alpha$ is a limit ordinal.

Note that the bi-conditional is vacuously true for $\alpha =0$. □

Proof. Since $\alpha \ne 0$ we have that $\alpha \ge 1$. Now consider any $\gamma <\alpha \cdot \beta$. We claim that $\gamma \le \alpha \cdot \delta$ for some $\delta <\beta$. Suppose to the contrary that $\gamma >\alpha \cdot \delta$ for all $\delta <\beta$. Then $\gamma$ is an upper bound of the set $\left\{\alpha \cdot \delta \mid \delta <\beta \right\}$. But from this it follows that

by Definition 6.5.6c, which contradicts the definition of $\gamma$. Hence the claim is true so that $\gamma \le \alpha \cdot \delta$ for some $\delta <\beta$. Then we have by Lemma 6.5.4a and the fact that $1\le \alpha$ that

by Exercise 6.5.7a since $\delta +1<\beta$ since $\beta$ is a limit ordinal. Note that we also used Definition 6.5.6b. Thus since $\gamma +1<\alpha \cdot \beta$ and $\gamma$ was arbitrary it follows from Lemma 2 that $\alpha \cdot \beta$ is a limit ordinal.

We also have that $\alpha \ne 0$ and $0<\beta$ (since $0\ne \beta$) so that by Exercise 6.5.7a above and Definition 6.5.6a

Thus $\alpha \cdot \beta \ne 0$. □

Proof. We show this by transfinite induction on $\gamma$.

First for $\gamma =0$ we have

where we have used Definition 6.5.6a repeatedly. Now suppose that $(\alpha \cdot \beta )\cdot \gamma =\alpha \cdot (\beta \cdot \gamma )$ for ordinal $\gamma$. Then

Now suppose that $\gamma \ne 0$ is a limit ordinal and $(\alpha \cdot \beta )\cdot \delta =\alpha \cdot (\beta \cdot \delta )$ for all $\delta <\gamma$. First if $\beta =0$ then

where we have used Definition 6.5.6a and Lemma 1. So assume that $\beta \ne 0$. Then we have $(\alpha \cdot \beta )\cdot \gamma =\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{(\alpha \cdot \beta )\cdot \delta \mid \delta <\gamma \right\}=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\alpha \cdot (\beta \cdot \delta )\mid \delta <\gamma \right\}$ by Definition 6.5.6c and the induction hypothesis. Then, since $\beta \ne 0$, by Lemma 3 we have that $\beta \cdot \gamma$ is a limit ordinal and $\beta \cdot \gamma \ne 0$. From this and Definition 6.5.6c we have that

as desired. This completes the inductive proof. □