Ordinal question

Question

Does anybody know whether the following is true: Given a limit ordinal $\lambda$, $\lambda = \bigcup_{\alpha < \lambda} \alpha$. And if not so, is $\bigcup_{\alpha < \lambda} \alpha$ an ordinal?

Answer 1

Yes it's true. On the one hand suppose $\beta \in \lambda$. By definition of the ordinal numbers, $\beta$ is an ordinal and $\beta<\lambda$. Since $\lambda$ is limit, there exists $\alpha$ such that $\beta < \alpha < \lambda$. By definition of tbe ordinal ordering, $\beta \in \alpha$. So $\beta \in \bigcup_{\alpha < \lambda} \alpha$. Therefore $\lambda \subset \bigcup_{\alpha < \lambda} \alpha$. On the other hand, the converse inclusion is quite similar. Hence both are equal.