When is the union of a set of ordinals a limit ordinal? Let $Y$ be a set of ordinals such that $\bigcup Y \not\in Y$. Then $\bigcup Y$ is a limit ordinal. Does this hold?
It's taken from Drake's Set Theory, An Introduction to Large Cardinals, page 114, I am having a hard time understanding his proof.
 A: If $Y$ is a set of ordinals, then $\bigcup Y$ is an ordinal: it is transitive (since $Y$ is a set of transitive set, therefore $\bigcup Y$ is a transitive set), and it is a set well-ordered by $\in$ (well, every element of an element of $Y$ is an ordinal, since $Y$ is a set of ordinals, and ordinals are themselves sets of ordinals; so $\bigcup Y$ is again a set of ordinals).
Moreover, each $\alpha\in Y$ satisfies that $\alpha\subseteq\bigcup Y$, by definition of union; and since for ordinals $\leq$ is the same as $\subseteq$ it follows that if $\eta$ is an upper bound for $Y$, then every $\alpha\in Y$ satisfies $\alpha\subseteq\eta$, and therefore $\bigcup Y\subseteq\eta$. So $\sup Y$ is exactly $\bigcup Y$.
Now, when is $\sup Y$ not in $Y$? Exactly when $Y$ does not have a last element, a maximum if you will. But this implies that $\bigcup Y$ is a limit ordinal. If $\bigcup Y=\alpha+1$, then maybe $\alpha+1=\max Y$, which is not the case because $\sup Y\notin Y$ by assumption; and therefore $\alpha$ has to be $\max Y$ which is a contradiction to the fact that $\sup Y=\bigcup Y=\alpha+1$.
(Note, by the way, that $\bigcup\{\omega\}=\omega$, so this is not an equivalent definition. A set of ordinals can have a last element which is a limit ordinal. But if the least ordinal not in $Y$ is a limit ordinal, then it is also the supremum of $Y$.)
A: Suppose $\bigcup Y$ is a successor ordinal, say the successor of some ordinal $\alpha$.  Then $\alpha\in \bigcup Y$, so $\alpha\in y$ for some $y\in Y$.  But $\alpha+1$ is the least ordinal containing $\alpha$, so $y\geq \alpha+1$.  Since $y\in Y$, $y\subseteq \bigcup Y=\alpha+1$, so $y\leq \alpha+1$.  Thus $y=\alpha+1=\bigcup Y$.  This means that $\bigcup Y$ is an element of $Y$.
