Show that an ordinal is a limit ordinal if and only if it is $\omega\cdot\beta$ for some $\beta$.
Could someone please explain the existence of the greatest limit $\gamma_0 $ in Camilo Arosemena's comment.
I do not see how just because $\gamma \leq \alpha$ exists where there are no limits between $\gamma$ and $\alpha$, $\gamma_0 $ is supposed to exist.
On
He is showing by transfinite induction on $\alpha$ that if $0\ne \alpha=\cup \alpha$ then $\alpha=\omega \beta$ for some $\beta.$ The case $\alpha=\omega$ is obvious, so he is considering $\alpha >\omega.$ Now $\alpha > \omega$ implies that there exists $\gamma <\alpha$ with $0\ne \gamma =\cup \gamma.$ This splits into 2 cases: (i). There is a largest such $\gamma$ (e.g. when $\alpha=\omega \cdot 9),$ which he calls $\gamma_0,$ or.... (ii). There isn't (e.g. when $\alpha$ is a cardinal).
In the first case $\gamma_0$ is the largest limit ordinal less than $\alpha$ so there can't be any limit ordinal $\gamma$ with $\gamma_0<\gamma<\alpha.$
Consider the set $$ \Gamma = \{\beta < \alpha \mid \text{there is no limit ordinal $\delta$ with $\beta < \delta < \alpha)$}\}. $$ By the assumption underlying the case distinction, $\Gamma$ is non-empty. Its least element is a limit ordinal, for if it were a successor $\delta + 1$, then we would also trivially have $\delta \in \Gamma$. That least element is the $\gamma_0$ you are looking for.