This is problem 5 in Kunen's Chapter 1 exercises.

Suppose $\alpha$ is a limit ordinal. I want to show the following equivalences:

(1) $\forall \beta, \gamma < \alpha$, $\beta + \gamma < \alpha$

(2) $\forall \beta < \alpha$, $\beta + \alpha = \alpha$

(3) $\forall A \subseteq \alpha$, $ot(A) = \alpha$ or $ot(\alpha \setminus A) = \alpha$ ($ot$ means order type).

(4) $\exists \delta \alpha = \omega^{\delta}$

(1) $\implies$ (2) is fairly easy: just pass to the limit on $\gamma$. (2) $\implies$ (1) is also pretty easy: if $\beta + \gamma = \alpha$ then $\beta + (\gamma + 1) > \alpha$ and since $\gamma + 1 < \alpha$ as $\alpha$ is a limit, this contradicts (2).

(3) $\implies$ (2) is fairly easy. Just take $A = \beta$.

I can finish the equivalences by showing (4) $\iff$ (1) + (2) and (4) $\implies$ (3)

(4) $\implies$ (3): An induction argument. Suppose $\alpha = \omega^{\delta + 1}$. Then we may express the situation as $X_1 \cup X_2 = \alpha$. Then by induction we have that for every $n \in \omega$, $ot(X_i \cap [\omega^{\delta} \cdot n, \omega^{\delta} \cdot (n+1))) = \omega^{\delta}$ for at least one of $i = \{1, 2\}$. Then one of the $i$ has infinitely many $n \in \omega$ such that $ot(X_i \cap [\omega^{\delta} \cdot n, \omega^{\delta} \cdot (n+1))) = \omega^{\delta}$ and then for this $i$ we have that $ot(X_i) \geq \omega^{\delta}\cdot n$ for every $n \in \omega$, and thus $ot(X_i) = \omega^{\delta + 1}$.

If $\delta$ is a limit, then apply the inductive hypothesis on all $\zeta < \delta$. Then let $A_i = \{\zeta < \delta : ot(X_i \cap \omega^{\zeta}) = \omega^{\zeta} \}$ for $i = \{1, 2\}$. As $A_1 \cup A_2= \delta$, we have for some $i$ that $sup(A_i) = \delta$. Then again we see that $ot(X_i) \geq \omega^{\zeta}$ for all $\zeta < \delta$ and thus $ot(X_i) = \omega^{\delta}$.

(4) $\implies (1) + (2)$: I suppose I don't have to do this since I have that $(4) \implies (3) \implies (2)$ so I'll omit it to save space.

$(1) + (2) \implies (4)$: Let $A = \{\delta \in OR : \omega^{\delta} \leq \alpha \}$. Then take $\eta = sup(A)$. We have that $\omega^{\eta} \leq \alpha$ as well. Suppose for contradiction that $\omega^{\eta} < \alpha$. Then $\omega^{\eta} + \alpha = \alpha$ by assumption. Then it is easy to see by induction that $\omega^{\eta}\cdot n + \alpha = \alpha$ for every $n \in \omega$. Here's a step I'm not sure of: this implies that $\omega^{\eta}\cdot \omega + \alpha = \alpha$. I think such limits can be passed only on the right side of the addition by definition. So it would have been fine if it was $\alpha + \omega^{\eta} \cdot n$ but I'm not sure about $\omega^{\eta} \cdot n + \alpha$. In other words, is $sup_{n \in \omega}(\omega^{\eta} \cdot n + \alpha) = \omega^{\eta} \cdot \omega + \alpha$? Assuming this is true however, this gives us $\omega^{\eta}\cdot \omega + \alpha = \omega^{\eta + 1} + \alpha = \alpha$ which means that $\eta \in A$ contradicting $\eta = sup(A)$.


1 Answer 1


Your concern about that step in the last part is not misplaced: $n+\omega=\omega$ for all $n\in\omega$, but $\omega+\omega\ne\omega$. (There is also a comparatively minor difficulty with $A$: to employ Comprehension, you really need to limit $\delta$ to a set of ordinals, which basically means coming up with an ordinal $\eta$ such that $\omega^\eta>\alpha$.)

Suppose that $\alpha$ is not a power of $\omega$. Ordinal exponentiation is continuous in the exponent, so if $\alpha$ is not a power of $\omega$, then it is not the limit of powers of $\omega$. Show that there is a largest $\omega^\eta<\alpha$, so that $\omega^\eta<\alpha<\omega^{\eta+1}$. Let $\beta$ be the order type of $\alpha\setminus\omega^\eta$, so that $\alpha=\omega^\eta+\beta$, and conclude that $\beta=\alpha$. Use your induction argument to show that $\omega^\eta\cdot n<\alpha$ for each $n\in\omega$. Then get a contradiction by observing that $\omega^{\eta+1}=\sup\{\omega^\eta\cdot n:n\in\omega\}$.


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