Show that if $\delta$ is a ordinal number then $\bigcup{\delta} \neq \delta \iff \exists \gamma \in \mbox{Ord} : \gamma + 1 = \delta$. An ordinal number $A$ is a pair $(A,<)$ with a well-ordering $<$ such that $\forall a \in A$ we have $a = A_a = \lbrace x \in A | x < a \rbrace$.
Now I want to show that for an ordinal number $\delta$ it holds that:
$\bigcup{\delta} \neq \delta \iff \exists \gamma \in \mbox{Ord} : \gamma + 1 = \delta$.
I was able to prove the direction  $\Leftarrow$. Now for $\Rightarrow$ I tried using an argument by contradiction by assuming that $\forall \gamma \in Ord : \gamma+ 1 \neq \delta$ so either $\gamma + 1 = \gamma \cup \lbrace \gamma \rbrace \leq \gamma$ or $\gamma + 1 \geq \gamma$. But this didn't get me further.
 A: Suppose $\bigcup \delta \neq \delta$. Let $\gamma \in \delta \setminus \bigcup \delta$. We claim that $\delta = \gamma + 1$. It's clear that $\gamma \cup \{\gamma\} = \gamma + 1 \subseteq \delta$, because $\gamma \in \delta$ and $\delta$ is transitive. Next, let $x \in \delta$ be arbitrary. Then $x < \gamma$ or $x = \gamma$ or $x > \gamma$. We proceed in cases.
Case 1 If $x < \gamma$, then $x \in \gamma$, so $x \in \gamma \cup \{\gamma\} = \gamma + 1$.
Case 2 If $x = \gamma$, then $x \in \gamma \cup \{\gamma\} = \gamma + 1$.
Case 3 If $x > \gamma$, then $\gamma \in x$. But $x \in \delta$, so this implies $\gamma \in \bigcup \delta$, which is a contradiction. Thus, this case is impossible.
In total, we can conclude $\delta \subseteq \gamma + 1$. QED
A: If $\delta$ is not a successor, then it is a limit ordinal, in which case $\bigcup \delta =\delta:$
We always have $\bigcup \delta\subseteq \delta,$ for if $x$ belongs to the union then there is a $y\in \delta$ such that $x\in y$ and now transitivity implies that $x\in \delta.$
On the other hand, suppose $x\in \delta.$ Since $\delta$ is a limit ordinal, $x\cup\{x\}\neq \delta.$  If $y\in x,$ then $y\in \delta$ by transitivity so $ x\cup\{x\}\subseteq \delta,$ so also $ x\cup\{x\}\in\delta.$ And since $x\in x\cup\{x\},$ it follows by definition of the union that $x\in \bigcup \delta.$
