# Equivalent characterizations of ordinals of the form $\omega^\delta$

Let $\alpha$ be a limit ordinal. Show the following are equivalent:

1. $\forall \beta, \gamma<\alpha (\beta+\gamma<\alpha)$
2. $\forall \beta<\alpha(\beta+\alpha=\alpha)$
3. $\forall X\subset \alpha(\text{type}(X)=\alpha, \text{ or,} \text{ type}(\alpha-X)=\alpha)$
4. $\exists \delta(\alpha=\omega^\delta )$.

What I've tried: I've shown that: $1\rightarrow2$ and $1\rightarrow3$.

Could anybody help me? Thanks ahead:)

-
You should: explain what you have tried for the remaining parts, and what relevant facts you know; add the homework tag if applicable; and give this question a more specific title. –  Nate Eldredge Aug 6 '12 at 2:51
Perhaps expressing $\alpha$, $\beta$, $\gamma$ in Cantor normal form might help in some parts of the proof. –  Martin Sleziak Aug 6 '12 at 2:53
@Nate Okey. Thanks for your kindful reminding. –  Paul Aug 6 '12 at 2:58
Thanks Martin. I will read Cantor normal form carefully. –  Paul Aug 6 '12 at 2:59

I would recommend showing the implications 1 $\Rightarrow$ 3 $\Rightarrow$ 2 $\Rightarrow$ 1, and 1 $\Rightarrow$ 4 $\Rightarrow$ 1.

The following are hints of those you haven't completed.

(3 $\Rightarrow$ 2.) Given $\beta < \alpha$ what does (3) say about $\mathrm{type} ( \beta )$ and $\mathrm{type} ( \alpha \setminus \beta )$?

(2 $\Rightarrow$ 1.) Use the monotonicity of ordinal addition.

(1 $\Rightarrow 4$.) Given a limit ordinal $\alpha$ stisfying (1), define $$\delta = \min \{ \delta \in \mathbf{ON} : \alpha < \omega^{\delta+1} \}.$$ Note that $\omega^\delta \leq \alpha$, and so there are unique ordinals $\gamma , \zeta$ such that $\zeta < \omega^\delta$ and $\omega^\delta \cdot \gamma + \zeta = \alpha$. Show that $\zeta = 0$ and $\gamma = 1$.

(4 $\Rightarrow$ 1.) Prove by transfinite induction on $\delta > 0$ that whenever $\beta , \gamma < \omega^\delta$, then $\beta + \gamma < \omega^\delta$. You will have to handle the base case $\delta = 1$ separately, but it shouldn't be difficult. For the inductive case, you will have to worry about whether $\delta$ is a successor or a limit ordinal, as $\beta , \gamma < \omega^\delta$ will imply something slightly different in each case.

-
Arthur, I thought it might be good to ping you that Paul posted a new answer as a reaction to yours. –  Martin Sleziak Aug 6 '12 at 8:06

(@Arthur: Thanks very much. For the part $4 \Rightarrow 1$, I've tried to prove. But I'm not sure I'm right. So I write here as an answer, for the comment cannot contain the long proof.)

The part $(4 \Rightarrow 1)$

For $\delta=1$, it is obviously right, because the sum of any two finite ordinals is less then $\omega$.

Now assuming that for any $x<\delta$ the case is always right.

Suppose that $\delta$ is a successor. Let $\delta=y+1$. Then $\omega^\delta=\omega^y \times \omega$. $\beta$ and $\gamma$ must be bigger than $\omega^y$, otherwise $\beta+\gamma\le\omega^y$. So let $\beta=\omega^y \times i +j$, and $\gamma=\omega^y \times m + n$, where $i,j,m,n \in \omega$. Therefore, $\beta + \gamma\le \omega^y \times (i+1+m+1)\le \omega^\delta.$

Suppose that $\delta$ is a limit ordinal. Let $\omega^\delta=\sup\{\omega^\xi: \xi \in \delta\}$. Therefore for any $\beta, \gamma < \omega^\delta$, there exists a $\xi'$ such that $\beta, \gamma<\omega^{\xi'}$, and hence by the assuming, we have $\beta+\gamma<\omega^{\xi'}$.

-
The only problem I see is in the inductive step for $\delta$ a successor ordinal. It is not clear (to me, at least) how a case like $\beta \leq \omega^y < \gamma < \omega^y \cdot \omega$ could be immediately done away with. I would instead use the fact that $\beta , \gamma < \omega^y \cdot \omega = \sup_{n < \omega} \omega^y \cdot n$. This should enable you to find an upper bound on $\beta + \gamma$ (that is strictly less that $\omega^y \cdot \omega$). –  Arthur Fischer Aug 6 '12 at 9:04