Let $\alpha$ be an ordinal.Then prove that $$\alpha \cup \{\alpha \} = \text {inf}\ \left \{\beta\ \mid\ \beta \gt \alpha \right \}.$$
I know that if $\alpha$ is an ordinal then so is $\alpha \cup \{\alpha \}.$ Now let $S = \{\beta\ \mid\ \beta \gt \alpha \}.$ So in order to prove the required result I need to first show that $\alpha \cup \{\alpha \}$ is a lower bound of $S.$ Now clearly for any $\beta \in S$ we have $\alpha \in \beta.$ So $\{\alpha \} \subseteq \beta.$ Also since $\beta$ is an ordinal so $\alpha \subsetneq \beta.$ Therefore $\alpha \cup \{\alpha \} \subseteq \beta.$ But this shows that either $\alpha \cup \{\alpha \} = \beta$ or $\alpha \cup \{\alpha \} \in \beta$ i.e. $\alpha \cup \{\alpha \} \lt \beta.$ So in any case we have $\alpha \cup \{\alpha \} \leq \beta.$ So $\alpha \cup \{\alpha \}$ is a lower bound of $S.$
Now we should prove that $\alpha \cup \{\alpha \}$ is the greatest lower bound of $S.$ So let $\gamma$ be any other lower bound of $S.$ Need only to show that $\gamma \leq \alpha \cup \{\alpha \}$ i.e. either $\gamma = \alpha \cup \{\alpha \}$ or $\gamma \in \alpha \cup \{\alpha \}.$ Since $\alpha \cup \{\alpha \}$ is an ordinal we need only to prove that $\gamma \subseteq \alpha \cup \{\alpha \}$ which is same as saying that $\gamma \not\gt \alpha$ i.e. $\alpha \not\in \gamma.$
How do I show that? Any help will be highly appreciated.
Thank you very much.