Exercise 6.5.13

Proof. First we note that $\alpha +\beta$ is a nonzero limit ordinal by Lemma ?? Let $A=\left\{{\gamma }_{\delta }\mid \delta <\alpha +\beta \right\}$ and

We show that $\gamma$ is the least upper bound of $A$, which shows the result by the least upper bound property (Lemma ??). First consider any ${\gamma }_{\delta }$ in $A$ so that $\delta <\alpha +\beta$.

If $\delta <\alpha$ then ${\gamma }_{\delta }\le {\gamma }_{\alpha }={\gamma }_{\alpha +0}\le \gamma$ since the sequence is non-decreasing and $0<\beta$. On the other hand if $\delta \ge \alpha$ then by Lemma 6.5.5 there is a unique ordinal $𝜀$ such that $\alpha +𝜀=\delta$. Then, since $\delta =\alpha +𝜀<\alpha +\beta$ it follows from Lemma 6.5.4a that $𝜀<\beta$. Hence by the definition of $\gamma$ we have that ${\gamma }_{\delta }={\gamma }_{\alpha +𝜀}\le \gamma$. Thus in all cases ${\gamma }_{\delta }\le \gamma$, which shows that $\gamma$ is an upper bound of $A$ since ${\gamma }_{\delta }$ was arbitrary.

Now consider any $𝜀<\gamma$. Then by the definition of $\gamma$ it follows that there is a $\delta <\beta$ such that $𝜀<{\gamma }_{\alpha +\delta }$ since $𝜀$ is not an upper bound $\left\{{\gamma }_{\alpha +\zeta }\mid \zeta <\beta \right\}$. Thus again by Lemma 6.5.4a we have that $\alpha +\delta <\alpha +\beta$ so that ${\gamma }_{\alpha +\delta }\in A$. Hence since $𝜀<{\gamma }_{\alpha +\delta }$ it cannot be that $𝜀$ is an upper bound of $A$. Since $𝜀<\gamma$ was arbitrary this completes the proof that $\gamma =\sup <!--\; FUNCTION\; APPLICATION\; -->A$ as desired. □

Proof. First if $\alpha =0$ then clearly $0^{\alpha }=0^0=1$ by Definition 6.5.9a. Then if $\alpha >0$ we show the result by transfinite induction on $\alpha$. First if $\alpha =1$ then we have

where we have used Definitions 6.5.9b and 6.5.6a. Now suppose that $0^{\alpha }=0$ so that we have

where we again have used the same two definitions. Lastly suppose that $\alpha$ is a nonzero limit ordinal and that $0^{\beta }=0$ for all $\beta <\alpha$. Then we have by Definition 6.5.9c that

where we have used the induction hypothesis. □

Proof. We show this by transfinite induction on $\alpha$. First for $\alpha =0$ we have $1^{\alpha }=1^0=1$ by Definition 6.5.9a. Now assume that $1^{\alpha }=1$ so that $1^{\alpha +1}=1^{\alpha }\cdot 1=1^{\alpha }=1$ by Definition 6.5.9b and the induction hypothesis.

Lastly suppose that $\alpha$ is a nonzero limit ordinal and that $1^{\beta }=1$ for all $\beta <\alpha$. We then clearly have by Definition 6.5.9c that

This completes the inductive proof. □

Proof. Clearly if $\beta =\gamma$ then ${\alpha }^{\beta }={\alpha }^{\gamma }$ so that the conclusion holds. So assume that $\beta <\gamma$.

Case: $\alpha =1$. Then by Lemma 3 we have

Case: $\alpha >1$. Then it follows from Exercise 6.5.14b that ${\alpha }^{\beta }<{\alpha }^{\gamma }$.

Hence in either case ${\alpha }^{\beta }\le {\alpha }^{\gamma }$ is true. □

Proof. We show this by transfinite induction on $\gamma$. First for $\gamma =0$ we have

Now assume that ${\alpha }^{\beta +\gamma }={\alpha }^{\beta }\cdot {\alpha }^{\gamma }$ so that we have

Lastly suppose that $\gamma$ is a nonzero limit ordinal and that ${\alpha }^{\beta +\delta }={\alpha }^{\beta }\cdot {\alpha }^{\delta }$ for all $\delta <\gamma$. First if $\alpha =0$ then ${\alpha }^{\beta +\gamma }=0^{\beta {+}_{\gamma }}=0$ by Lemma 2 since $\beta +\gamma >0$ since $\gamma >0$. Hence we have

where we have used Definition 6.5.6a and the fact that $0^{\gamma }=0$ by Lemma 2 since $\gamma >0$.

On the other hand if $\alpha >0$ then we have by Definition 6.5.9c that

It then follows from Lemma 4 that $\left\{{\alpha }^{\delta }\right\}$ for $\delta <\beta +\gamma$ is a non-decreasing sequence since $\alpha >0$. Thus we can apply Lemma 1 so that

by the induction hypothesis. It then follows from statement (6.6.1) in the text that

by Definition 6.5.9c. This completes the transfinite induction. □

Proof. We show this by transfinite induction on $\gamma$ as well. First for $\gamma =0$ we have

Next suppose that ${\left({\alpha }^{\beta }\right)}^{\gamma }={\alpha }^{\beta \cdot \gamma }$ so that we have

Lastly, suppose that $\gamma$ is a nonzero limit ordinal and that ${\left({\alpha }^{\beta }\right)}^{\delta }={\alpha }^{\beta \cdot \delta }$ for all $\delta <\gamma$.

Case: $\alpha =0$. Then if $\beta =0$ we have

On the other hand if $\beta >0$ then first we note that $0=\beta \cdot 0<\beta \cdot \gamma$ by Definition 6.5.6a and Exercise 6.5.7a since both $\beta \ne 0$ and $0<\gamma$. We then have

Case: $\alpha >0$. Then we have

Since these cases are exhaustive this completes the inductive proof. □