# (Proof Verification) Indecomposable (Limit) Ordinals

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

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 $$\omega^{\eta+1}=\sup\{\omega^\eta\cdot n:n\in\omega\}$$.