Characterization of limit ordinals

An ordinal $\alpha$ is limit ordinal if and only if there exists ordinal $\beta$ such that $\alpha=\omega\beta$.

Proof. $\Leftarrow)$ If $\beta=1$ then $\alpha=\omega$ so it is limit. Assume $\omega\gamma$ is limit for all $\gamma<\beta$. We will prove that $\omega\beta$ is limit.
First assume $\beta$ is limit. Then $\omega\beta=\sup\{\omega\gamma \ \big| \ \gamma<\beta\}$. Let $\theta<\omega\beta$, then $\theta<\omega\gamma'$ for some $\gamma'<\beta$. But $\omega\gamma'$ is limit thus $\theta+1<\omega\gamma'$ so $\theta+1<\omega\beta$ hence $\omega\beta$ is limit.
Now let $\beta=\beta'+1$ for some $\beta'$. Thus $\omega\beta=\omega\beta'+\omega$. If $\theta<\omega\beta=\omega\beta'+\omega=\sup\{\omega\beta'+n\ \big| \ n<\omega\}$ then $\theta<\omega\beta'+n_0$ for some $n_0$. And now $\theta+1<\omega\beta'+n_0+1<\sup\{\omega\beta'+n\ \big| \ n<\omega\}$ hence $\omega\beta$ is limit.
$\Rightarrow)$ Let $\beta=\sup\{\gamma\ \big| \ \omega\gamma\le\alpha\}$. We have that $\omega\beta\le \alpha$. Assume $\omega\beta<\alpha$. Since $\alpha$ is limit we find that $\omega\beta+1<\alpha$ and inductively $\omega\beta+n<\alpha$ for all $n<\omega$. Hence $\omega\beta+\omega\le \alpha$. This means that $\omega(\beta+1)\le \alpha$ which is a contradiction with the definition of $\beta$.

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