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.
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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