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

Question

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:)

Answer 1

I would recommend showing the implications 1⇒3⇒2⇒1, and 1⇒4⇒1.

The following are (fairly complete) outlines of those you haven't completed.

(3⇒2) Given $\beta < \alpha$, note that as $\operatorname{type}(\beta) = \beta < \alpha$, by (3) it follows that $\operatorname{type}(\alpha \setminus \beta) = \alpha$. Then $\alpha = \operatorname{type} ( \beta + ( \alpha \setminus \beta ) ) = \operatorname{type} (\beta) + \operatorname{type}(\alpha \setminus \beta) = \beta + \alpha$.

(2⇒1) Recall that ordinal addition is strictly monotone in the right summand, and so using (2) it follows that given $\beta , \gamma < \alpha$ we have $\beta + \gamma < \beta + \alpha = \alpha$.

(1⇒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 $\omega^\delta \cdot \gamma + \zeta = \alpha$. By applying (1) it follows that $\zeta = 0$ and $\gamma = 1$.

(4⇒1) By transfinite induction on $\delta > 0$ we can show that $\beta + \gamma < \omega^\delta$ whenever $\beta , \gamma < \omega^\delta$.

• The base case $\delta = 1$ just says that the sum of two finite ordinals is a finite ordinal.

• If $\delta = \zeta + 1$, suppose that $\beta , \gamma < \omega^\delta = \omega^{\zeta+1} = \omega^\zeta \cdot \omega = \sup_{n < \omega} \omega^\zeta \cdot n$. Then there must be an $n < \omega$ such that $\beta , \gamma < \omega^\zeta \cdot n$, and we then have $$ \beta + \gamma < \omega^\zeta \cdot n + \omega^\zeta \cdot n = \omega^\zeta \cdot (n+n) < \omega^\zeta \cdot \omega = \omega^\delta. $$

• If $\delta > 1$ is a limit ordinal, and $\beta , \gamma < \omega^\delta = \sup_{\zeta < \delta} \omega^\zeta$, then there must be a $\zeta < \delta$ such that $\beta , \gamma < \omega^\zeta$, and by the induction hypothesis it follows that $\alpha + \beta < \omega^\zeta \leq \omega^\delta$.

Answer 2

(@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 $\beta+\gamma<\omega^{\xi'}$.