Exercise 6.5.10

Proof. Consider any ordinal $\alpha$. First we note that clearly $\alpha =\alpha +0<\alpha +\omega$ by Lemma 6.5.6a since $0<\omega$. It also clearly follows from Lemma ?? that $\alpha +\omega$ is a limit ordinal.

Now suppose that $\beta$ is any ordinal such that $\alpha <\beta <\alpha +\omega$. Since $\alpha <\beta$ there is a unique ordinal $\gamma$ such that $\alpha +\gamma =\beta$ by Lemma 6.5.5. Now, since $\alpha +0=\alpha <\beta =\alpha +\gamma$ it follows again from Lemma 6.5.6a that $0<\gamma$. Similarly we have $\alpha +\gamma =\beta <\alpha +\omega$ so that by the same lemma $\gamma <\omega$. Hence $0<\gamma <\omega$ so that $\gamma$ is a nonzero natural number. In particular $\gamma \ge 1$ so that $n=\gamma -1$ is also a natural number and $\gamma =n+1$. We then have that $\beta =\alpha +\gamma =\alpha +(n+1)=(\alpha +n)+1$ so that clearly $\beta$ is a successor ordinal. □

Proof. ( $\to$) Suppose to the contrary that not every limit ordinal is equal to $\omega \cdot \beta$ for some $\beta$. Then let $\alpha$ be a limit ordinal such that $\alpha \ne \omega \cdot \gamma$ for every ordinal $\gamma$. Now let $\beta$ be the set of ordinals $\delta$ such that $\omega \cdot \delta <\alpha$, noting that it could potentially be the empty set. We claim that $\beta$ is an ordinal number and that moreover it is a limit ordinal.

Clearly if $\beta =\varnothing =0$ then it is a limit ordinal, so assume that $\beta \ne \varnothing$. Then consider any $\delta \in \beta$ so that $\omega \cdot \delta <\alpha$. Also consider any $x\in \delta$ so that $x$ is also an ordinal number by Lemma 6.2.8 and moreover that $x<\delta$. It then follows from Exercise 6.5.7a that $\omega \cdot x<\omega \cdot \delta <\alpha$. Hence we have that $x\in \beta$ so that $\delta \subseteq \beta$ since $x$ was arbitrary. Since $\delta$ was also arbitrary this shows that $\beta$ is transitive. Also since $\beta$ is nonempty set of ordinals it follows from Theorem 6.2.6d that $\beta$ is well-ordered. Thus by definition $\beta$ is an ordinal number.

Now consider any $\delta <\beta$ so that $\delta \in \beta$ so that $\omega \cdot \delta <\alpha$. By Lemma 1 we have that $\omega \cdot \delta +\omega =\omega \cdot (\delta +1)$ is the next limit ordinal after $\omega \cdot \delta$ (i.e. there are no limit ordinals between them) so it has to be that $\omega \cdot (\delta +1)\le \alpha$ since otherwise $\alpha$ would be a limit ordinal between $\omega \cdot \delta$ and $\omega \cdot (\delta +1)$. But since $\alpha \ne \omega \cdot \gamma$ for all ordinals $\gamma$ it cannot be that $\alpha =\omega \cdot (\delta +1)$. Thus it must be that $\omega \cdot (\delta +1)<\alpha$ so that $\delta +1\in \beta$ so that $\delta +1<\beta$. This shows that $\beta$ is a limit ordinal by Lemma ??.

Now we claim that $\omega \cdot \beta =\alpha$, which is of course is a contradiction that proves the desired result. To see this let $A=\left\{\omega \cdot \delta \mid \delta <\beta \right\}$ so that by Definition 6.5.6c we have that $\omega \cdot \beta =\sup <!--\; FUNCTION\; APPLICATION\; -->A$ since $\beta$ is a limit ordinal. Now since $\delta <\beta$ means that $\delta \in \beta$ so that $\omega \cdot \delta <\alpha$ by the definition of $\beta$, clearly $\alpha$ is an upper bound of $A$ so that $\omega \cdot \beta =\sup <!--\; FUNCTION\; APPLICATION\; -->A\le \alpha$. However, if it were the case that $\omega \cdot \beta <\alpha$ then we would have by definition that $\beta \in \beta$ which contradicts Lemma 6.2.7 since we have shown that $\beta$ is an ordinal. Thus the only possibility is that $\omega \cdot \beta =\alpha$, which gives rise to the contradiction.

Moreover, we can show that $\beta$ is unique. To see this, consider ${\beta }_1$ and ${\beta }_2$ where $\alpha =\omega \cdot {\beta }_1$ and $\alpha =\omega \cdot {\beta }_2$. Then clearly $\omega \cdot {\beta }_1=\omega \cdot {\beta }_2$ so that ${\beta }_1={\beta }_2$ by Exercise 6.5.7b since clearly $\omega \ne 0$.

( $\leftarrow$) Suppose that $\alpha =\omega \cdot \beta$ for some ordinal $\beta$. Then the result that $\alpha$ is a limit ordinal follows immediately from Lemma ?? since $\omega$ is a limit ordinal. □