Exercise 6.5.4

Proof. First we note that since $\beta \ne 0$ we have $1\le \beta$. Consider any $\gamma <\alpha +\beta$. If $\gamma <\alpha$ then $\gamma +1\le \alpha <\alpha +1\le \alpha +\beta$. On the other hand if $\gamma \ge \alpha$ then by Lemma 6.5.5 there is an ordinal $\delta$ such that $\alpha +\delta =\gamma$. Then since $\alpha +\delta =\gamma <\alpha +\beta$ it follows from Lemma 6.5.4a that $\delta <\beta$, and so by Lemma ?? we have that $\delta +1<\beta$ since $\beta$ is a limit ordinal. Therefore we have $\gamma +1=(\alpha +\delta )+1=\alpha +(\delta +1)<\alpha +\beta$ by Lemma 6.5.4 parts c and a. Hence in all cases $\gamma +1<\alpha +\beta$ so that $\alpha +\beta$ is a limit ordinal by Lemma ?? since $\gamma$ was arbitrary. □

Proof. We show this by normal (not transfinite) induction on $n$. For $n=0$ we clearly have $\beta +n=\beta +0=\beta <\alpha$. So suppose that $\beta +n<\alpha$ so that we have

by Definition 6.5.1b. The inequality follows from Lemma ?? and the induction hypothesis since $\alpha$ is a limit ordinal. This completes the inductive proof. □

Proof. Existence. First for any ordinal $\alpha$ let

and define $\beta =\sup <!--\; FUNCTION\; APPLICATION\; -->B$.

First we show that $\beta$ is a limit ordinal. To this end consider any $\delta <\beta$. Then since $\beta$ is the least upper bound of $B$ it follows that $\delta$ is not an upper bound of $B$ so that there is a $\gamma \in B$ such that $\delta <\gamma$. Since $\gamma \in B$ it is a limit ordinal so that also $\delta +1<\gamma$ by Lemma ??. Then since $\beta$ is an upper bound of $B$ we have that $\delta +1<\gamma \le \beta$ so that $\beta$ is a limit ordinal by Lemma ?? since $\delta$ was arbitrary.

Now, since each $\gamma \in B$ is in $\alpha +1$ we have that $\gamma <\alpha +1$ so that $\gamma \le \alpha$. Hence $\alpha$ is an upper bound of $B$, and since $\beta$ is the least upper bound of $B$ it follows that $\beta \le \alpha$. Because of this there is a unique ordinal $\xi$ such that $\beta +\xi =\alpha$ by Lemma 6.5.5.

We claim that $\xi <\omega$ so that $\xi$ is a natural number. To the contrary, suppose that $\xi \ge \omega$. It then follows from Lemma 6.5.4 that $\beta +\omega \le \beta +\xi =\alpha$. Also, $\beta +\omega$ is a limit ordinal by Lemma 1 above since $\omega$ is so that $\beta +\omega \in B$ since also $\beta +\omega \in \alpha +1$ (since $\beta +\omega \le \alpha$). Then $\beta +\omega \le \beta$ since $\beta$ is an upper bound of $B$, but this is a contradiction since clearly $\beta =\beta +0<\beta +\omega$ by Lemma 6.5.4a since $0<\omega$. Hence it must be that $\xi <\omega$.

Thus $\alpha =\beta +\xi$ for a limit ordinal $\beta$ and natural number $\xi$, thereby proving existence.

Uniqueness. Suppose that $\alpha ={\beta }_1+n_1={\beta }_2+n_2$ where ${\beta }_1$ and ${\beta }_2$ are limit ordinals and $n_1$ and $n_2$ natural numbers. First suppose that ${\beta }_1\ne {\beta }_2$ so that without loss of generality we can assume that ${\beta }_1<{\beta }_2$. It then follows from Lemma 2 that ${\beta }_1+n_1<{\beta }_2$ as well since ${\beta }_2$ is a limit ordinal and $n_1$ is a natural number. But then we have ${\beta }_1+n_1<{\beta }_2={\beta }_2+0\le {\beta }_2+n_2$, which contradicts the fact that ${\beta }_1+n_1=\alpha ={\beta }_2+n_2$. So it must be that in fact ${\beta }_1={\beta }_2$. But then it follows from Lemma 6.5.4b that $n_1=n_2$ also, which shows uniqueness. □