If $\alpha$ is an ordinal number, prove that $\bigcup(\alpha+1)=\alpha$

Question

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

Answer 1

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 $\alpha \in \alpha$.

Answer 2

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.