Exercise 6.5.6

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

So for $\beta =0$ we clearly have $\alpha \cdot \beta =\alpha \cdot 0=0$, which is a limit ordinal.

Now suppose that $\alpha \cdot \beta$ is a limit ordinal. Then we have $\alpha \cdot (\beta +1)=\alpha \cdot \beta +\alpha$ by Definition 6.5.6b. If $\alpha =0$ then $\alpha \cdot (\beta +1)=\alpha \cdot \beta +\alpha =\alpha \cdot \beta +0=\alpha \cdot \beta$, which is a limit ordinal by the induction hypothesis. On the other hand if $\alpha \ne 0$ then by Lemma ?? $\alpha \cdot (\beta +1)=\alpha \cdot \beta +\alpha$ is a limit ordinal since $\alpha$ is.

Lastly suppose that $\beta$ is a limit ordinal and that $\alpha \cdot \gamma$ is a limit ordinal for all $\gamma <\beta$. Let $A=\left\{\alpha \cdot \gamma \mid \gamma <\beta \right\}$ so that by Definition 6.5.6b we have $\alpha \cdot \beta =\sup <!--\; FUNCTION\; APPLICATION\; -->A$. Consider any $\delta <\alpha \cdot \beta$ so that $\delta$ is not an upper bound of $A$. Hence there is a $\gamma <\beta$ such that $\delta <\alpha \cdot \gamma$. Then by the induction hypothesis $\alpha \cdot \gamma$ is a limit ordinal so that $\delta +1<\alpha \cdot \gamma$ by Lemma ??. But then we have $\delta +1<\alpha \cdot \gamma \le \sup <!--\; FUNCTION\; APPLICATION\; -->A=\alpha \cdot \beta$. Thus by Lemma ?? this shows that $\alpha \cdot \beta$ is also a limit ordinal, which completes the inductive proof. □

Proof. ( $\to$) Suppose that $\alpha =\omega \cdot n$ for natural number $n$. Then by Lemma 1 $\alpha$ is a limit ordinal since $\omega$ is. Also clearly by Exercise 6.5.7a $\alpha =\omega \cdot n<\omega \cdot (n+1)\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\omega \cdot k\mid k<\omega \right\}=\omega \cdot \omega ={\omega }^2$.

( $\leftarrow$) Now suppose that $\alpha$ is a limit ordinal and $\alpha <{\omega }^2$. We have $a<{\omega }^2=\omega \cdot \omega =\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\omega \cdot n\mid n<\omega \right\}$ so that $\alpha$ is not an upper bound of $\left\{\omega \cdot n\mid n<\omega \right\}$. Hence there is a $k<\omega$ such that $\alpha <\omega \cdot k$. It therefore follows that the set $A=\left\{n\in \omega \mid \alpha \le \omega \cdot n\right\}$ is nonempty. Since $A$ is nonempty set of natural numbers (which are well-ordered) it follows that $A$ has a least element $n$. If $n=0$ it follows that $\alpha =0=\omega \cdot 0$ since $\alpha \le \omega \cdot n=\omega \cdot 0=0$ and $0$ is the only ordinal for which this is true. Then if $n>0$ we have that $n-1$ is a natural number and moreover it follows that $\omega \cdot (n-1)<\alpha$ since otherwise $n-1$ would have been the least element of $A$.

Thus we have that $\omega \cdot (n-1)<\alpha \le \omega \cdot n$. But since $\omega \cdot n=\omega \cdot \left[(n-1)+1\right]=\omega \cdot (n-1)+\omega$ clearly any ordinal $\gamma$ where $\omega \cdot (n-1)+0=\omega \cdot (n-1)<\gamma <\omega \cdot n=\omega \cdot (n-1)+\omega$ must have the form $\omega \cdot (n-1)+m$ for a natural number $m>0$. From this it clearly follows that $\gamma$ is a successor ordinal. Since $\alpha$ is a limit ordinal it can thus not be such a $\gamma$ so that it cannot be that $\alpha <\omega \cdot n$. But since we have established that $\alpha \le \omega \cdot n$ (since $n\in A$) it has to be that $\alpha =\omega \cdot n$. □

Proof. ( $\to$) Suppose that $\alpha =\omega \cdot n+k$ for natural numbers $n$ and $k$. We then have

as desired.

( $\leftarrow$) Now suppose that $\alpha <{\omega }^2$. By Exercise 6.5.4 we have that $\alpha =\beta +k$ for a unique limit ordinal $\beta$ and natural number $k$. We also have by Lemma 6.5.4 that $\beta +0\le \beta +k=\alpha <{\omega }^2$ since obviously $0\le k$. Hence $\beta$ is a limit ordinal such that $\beta <{\omega }^2$ so that by Lemma 2 there is a natural number $n$ such that $\beta =\omega \cdot n$, thereby proving the result since this means that $\alpha =\beta +k=\omega \cdot n+k$. □

Proof. To see this we first show that ${\omega }^2$ is a limit ordinal. So consider any $\alpha <{\omega }^2$ so that by Lemma 3 there are natural numbers $n$ and $k$ such that $\alpha =\omega \cdot n+k$. We then have $\alpha +1=(\omega \cdot n+k)+1=\omega \cdot n+(k+1)<{\omega }^2$ again by Lemma 3 since $k+1$ is a natural number. Hence ${\omega }^2$ is a limit ordinal by Lemma ??.

Next we show that ${\omega }^2$ is an additive ordinal. So again consider $\alpha <{\omega }^2$ so that by Lemma 3 there are natural numbers $n$ and $k$ such that $\alpha =\omega \cdot n+k$. We then have

since $k+{\omega }^2={\omega }^2$ and $n+\omega =\omega$ by Lemma ??.

Lastly we show that if $\omega <\alpha <{\omega }^2$ then $\alpha$ is not an additive ordinal. Clearly by Lemma 3 there are natural numbers $n$ and $k$ such that $\alpha =\omega \cdot n+k$. Now let $\beta =\omega$ so that clearly $\beta <\alpha$. We then have that

We also clearly have

so that $\beta +\alpha \ne \alpha$. Since $\beta <\alpha$ this shows that $\alpha$ is not an additive ordinal. □