If $\alpha$ is an ordinal number, prove that $\bigcup(\alpha+1)=\alpha$ The statement corresponds to  Theorem 124 of the appendix of Kelley's book General Topology:

If $\alpha$ is an ordinal number, then $\bigcup(\alpha+1)=\alpha$.

My attempt:
$$\bigcup(\alpha+1)=\{\gamma:\exists\beta\in \alpha+1:\gamma\in\beta\}. $$
Because $\alpha+1$ is an ordinal number, the above equation can be rewritten as
$$\bigcup(\alpha+1)=\{\gamma:\gamma\in\alpha+1\}, $$
which is precisely $\alpha+1$.
Any help?
Thanks
 A: By definition, $\alpha + 1 = \alpha \cup \{\alpha\}$. Therefore, 
\begin{align}
\bigcup(\alpha+1) &= \{\gamma:\exists\beta\in \alpha+1:\gamma\in\beta\} = \{\gamma:\exists\beta\in \alpha \cup \{\alpha\} :\gamma\in\beta\} \\
&= \{\gamma :\gamma\in\alpha\} = \alpha
\end{align}
where the equality $\{\gamma:\exists\beta\in \alpha \cup \{\alpha\} :\gamma\in\beta\} = \{\gamma :\gamma\in\alpha\}$ holds because:
\begin{align}
 \gamma \in \beta \text{ for some } \beta \in \alpha \cup \{\alpha\} &\iff \text{either } \gamma \in \beta \text{ for some } \beta \in \alpha, \text{ or } \gamma \in \alpha \\
&\iff \gamma \in \alpha  \quad\text{ ($\Rightarrow$ holds by transitivity of }\alpha\text{).}
\end{align}
In particular, note that $\alpha \notin \bigcup (\alpha+1)$ because $\alpha \notin \alpha$. The proof attempt in the OP is not valid because its last step  would mean that $\alpha \in \alpha$.
A: Hint: You can "debug" your argument by setting $\alpha$ to be some small value, like $\alpha=1$. Run through all of your statements and inferences with this value. At some point you will find an incorrect statement, and it will point out where your argument is wrong.
