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?)