Given an ordinal $\alpha$, I need to show that its successor, i.e., $\alpha\cup\{\alpha\}$ is also an ordinal.
I just barely acquired the concept of ordinal and I am having some trouble.
The following is my work: Suppose $x\in\alpha\cup\{\alpha\}$, then we have either $x\in\alpha$ or $x=\alpha$. If $x\in\alpha$, since $\alpha$ is transitive, $x\subseteq\alpha$. If $x=\alpha$, then $x=\alpha\subseteq\alpha$. This implies that $\alpha\subseteq\alpha\cup\{\alpha\}$, hence, $x\subseteq\alpha\cup\{\alpha\}$ and that $\alpha\cup\{\alpha\}$ is transitive.
I now need to show that $\alpha\cup\{\alpha\}$ is strictly well-ordered, but I'm struggling and no idea how to go about this.
Hint: consider a nonempty subset $S\subset\alpha\cup\{\alpha\}$. And check what happens in the three cases:
(1) $\alpha\not\in S$.
(2) $\alpha\in S$, $S\setminus\{\alpha\}\ne\emptyset$.
(3) $S = \{\alpha\}$.