Why is every limit ordinal a multiple of $\omega$?

Question

An ordinal $\alpha$ is a limit ordinal if and only if $\alpha = \omega\cdot\beta$ for some $\beta$. I have checked through transfinite induction that $\omega\cdot\beta$ is always a limit ordinal, but I can't find a proof for the other implication.

Answer 1

Suppose that $\alpha$ is the least limit ordinal which is not $\omega\cdot\beta$.

Show that the set of limit ordinals below $\alpha$ is unbounded, and show that it has to be a set of the form $\{\omega\cdot\gamma\mid\gamma<\delta\}$ for some limit ordinal $\delta$, now what do you know about the recursive definition of ordinal multiplication when $\delta$ is a limit ordinal?

Answer 2

HINT: Suppose $\lambda$ is a limit ordinal. Let $S$ be the set of limit points in $\lambda$ - that is, $S=\{x\in \lambda: \forall y<x, \exists z(y<z<x)\}$ (note that this will include $0$).

Now, $S$ is a subset of an ordinal, so there is an order-preserving bijection between $S$ and some ordinal $\beta$. Can you show that $\omega\cdot \beta=\lambda$? (In particular, think about simple cases, e.g. $\lambda=\omega\cdot\omega$ - do you see why the statement is true for these?)

Answer 3

This result is an easy corollary of the ordinal division theorem, https://proofwiki.org/wiki/Division_Theorem_for_Ordinals.

By this theorem, if $\alpha$ is an ordinal then there are unique ordinals $\gamma$ and $\eta$ such that $\alpha=(\omega\cdot\gamma)+\eta$ and $\eta\in\omega$. If $\eta=\phi $ then we are done. Assume $\eta\neq\phi $. Then as $\eta\in\omega, \eta=\delta^+$ for some $\delta\in\omega$. And then $\alpha=(\omega\cdot\gamma)+\delta^+=((\omega\cdot\gamma)+\delta)^+$ , a contradiction as this would mean $\alpha$ is not a limit ordinal. Hence $\eta=\phi $ and we have the result.