Exercise 6.5.5

Proof. ( $\to$) We show this by induction on $n$. So for $n=0$ and any ordinals $\alpha$ and $\beta$ where $\alpha <\beta$ we clearly have

Now suppose that $\alpha +n<\beta +n$ for any ordinals $\alpha$ and $\beta$ where $\alpha <\beta$. Suppose that $\alpha$ and $\beta$ are such ordinals so that by Lemma ?? we have $\alpha +1\le \beta <\beta +1$. Hence $\alpha +1$ and $\beta +1$ are ordinals such that $\alpha +1<\beta +1$. It then follows from the induction hypothesis that

noting that clearly natural numbers commute with respect to addition. This completes the proof by induction.

( $\leftarrow$) Suppose that $\alpha +n<\beta +n$ for ordinals $\alpha$ and $\beta$ and natural number $n$. It cannot be that $\beta <\alpha$ for then it would follow that $\beta +n<\alpha +n$ by what was just shown. Nor can it be that $\alpha =\beta$ since then clearly $\alpha +n=\beta +n$. Hence by the linearity of the ordinal ordering it follows that $\alpha <\beta$. □

Proof. This follows directly from Lemma 1 in the same way as the proof of Lemma 6.5.4b. □

Proof. For any natural number $n$ we show this by transfinite induction on $\alpha$. So if $\alpha =\omega$ then as explained in the text we have $n+\alpha =n+\omega =\omega =\alpha$. Now suppose that $n+\alpha =\alpha$ so that we have $n+(\alpha +1)=(n+\alpha )+1=\alpha +1$. Lastly, suppose that $\alpha >\omega$ is a limit ordinal and that $n+\gamma =\gamma$ for all $\gamma <\alpha$. We then have by Definition 6.5.1c that

by the induction hypothesis and comments in the text after Theorem 6.2.10. This completes the inductive proof. □

Proof. In what follows suppose generally that $\alpha =\gamma +n$ and $\beta =\delta +m$ where $\gamma$ and $\delta$ are limit ordinals and $n$ and $m$ are natural numbers. Note that $\alpha$ and $\beta$ can be expressed in this way uniquely by Exercise 6.5.4.

Case: $\beta =0$. If $\alpha =0$ then clearly $\xi =0$ is the only solution since for any other $\xi \ne 0$ we have $\xi +\alpha =\xi +0=\xi \ne 0=\beta$. On the other hand if $\alpha \ne 0$ then there is no solution since for any $\xi$ we have by Lemma 6.5.4 that $\xi +\alpha >\xi +0=\xi \ge 0=\beta$ so that $\xi +\alpha \ne \beta$.

Case: $\beta$ is a successor. From this it follows that clearly $m\ne 0$ since otherwise $\beta =\delta +m=\delta +0=\delta$ would be a limit ordinal.

Now, if $\alpha =0$ then clearly $\xi =\beta$ is the only solution so that $\xi +\alpha =\xi +0=\xi =\beta$.

If $\alpha$ is a successor then similarly $n\ne 0$. Suppose that $\gamma =0$ so that since $\alpha \le \beta$ we have $0+n=n\le \delta +m=\beta$. Then if $n\le m$ then $\xi =\delta +(m-n)$ is a solution since then

Moreover clearly this is the only solution since if $\xi \ne \delta +(m-n)$ then

by Corollary 2. On the other hand if $n>m$ then for any ordinal $\xi =𝜀+k$ where $𝜀$ is a limit ordinal and $k$ is a natural number, we have that $k+n>0+n=n>m$ so that clearly $k+n\ne m$. From this it follows that

since both the limit ordinal and the natural number part of ordinals must be equal for the overall ordinals to be equal. Hence this case has no solutions. Now suppose that $\gamma \ne 0$. Then if $n\ne m$ then there are no solutions since for any ordinal $\xi$ we have

since $n\ne m$ and $\xi +\gamma$ is a limit ordinal by Lemma ??. On the other hand if $n=m$ then since $\gamma +n=\alpha \le \beta =\delta +m=\delta +n$ it follows from Lemma 1 and Corollary 2 that $\gamma \le \delta$, and since $\gamma \ne 0$ we have that $0<\gamma \le \delta$.

Then we have that $\xi$ is a solution if and only if $\xi +\gamma =\delta$. For, supposing that $\xi +\gamma =\delta$, we have

and if it is not the case then

by Corollary 2. Since $\gamma$ and $\delta$ are both nonzero limit ordinals it then follows from the final case below (where both $\alpha$ and $\beta$ are nonzero limit ordinals) there are either zero or infinite such $\xi$.

If $\alpha$ is a nonzero limit ordinal then clearly there are no solutions since, for any ordinal $\xi$, Lemma ?? tells us that $\xi +\alpha$ is a limit ordinal whereas $\beta$ is a successor so that it has to be that $\xi +\alpha \ne \beta$.

Case: $\beta$ is a nonzero limit ordinal.

If $\alpha =0$ then clearly $\xi =\beta$ is the only solution since $\xi +\alpha =\xi +0=\xi$.

If $\alpha$ is a successor ordinal then $n\ne 0$ and for any ordinal $\xi$ we have that $\xi +\alpha =\xi +(\gamma +n)=(\xi +\gamma )+n$, which is a successor ordinal as well since $\xi +\gamma$ is a limit ordinal again by Lemma ??. Hence there are no solutions since $\beta$ is a limit ordinal and $\xi$ was arbitrary.

Lastly, suppose that $\alpha$ is also a nonzero limit ordinal. Suppose that there at least one solution $\xi$ such that $\xi +\alpha =\beta$. Now consider any natural number $k$ so that we have

by Lemma 3 since $\alpha$ is a nonzero limit ordinal and therefore clearly $\alpha \ge \omega$. Hence $\xi +k$ is also a solution and so there are an infinite number of solutions since $k$ was arbitrary. Thus there are either zero solutions or an infinite number of solutions (since a nonzero number of solutions implies an infinite number of solutions). This completes the case structure.

These results are summarized in the following table:

Since this case structure is exhaustive the result follows. □