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

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.

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?
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?)
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.