Prove that $\alpha$ is a limit ordinal iff $\alpha = \sup(\alpha)$.
Here we’ve defined limit ordinal as any nonzero ordinal which is not a successor ordinal, and $\sup(\alpha)$ as the ordinal $\bigcup \alpha$.
I have that a nonzero ordinal $\alpha$ Is a limit ordinal iff $(\forall \lambda \in \alpha)(\lambda ^+ \in \alpha)$. I’m not sure if this is the wrong direction to take this in. Any help is appreciated.
Suppose $\alpha = sup(\alpha) = \bigcup \alpha$. If $\alpha = \beta + 1$, then $\beta \cup \{\beta\} = \bigcup (\beta \cup \{\beta\}) = \bigcup{\beta} \cup \beta$ and, as $\beta \in \alpha$, we have that $\beta \in \bigcup \beta$ or $\beta \in \beta$, a contradiction. So $\alpha$ can't be a successor ordinal.
Conversely, if $\alpha$ is a limit ordinal, we have that if $\beta \in \alpha$, then $\beta + 1 \in \alpha$ - otherwise we would have that $\alpha \leq \beta + 1$, from which follows contradiction. So, if $\beta \in \alpha$, $\beta \in \beta + 1 \in \alpha$, from which follows that $\beta \in sup(\alpha)$. We have then $\alpha \subseteq sup(\alpha)$, and it follows that $\alpha = sup(\alpha)$ (the other inclusion is trivial).