Prove that no ordinal is an element of itself

Question

I would like to show that the successor set of an ordinal: $s(\alpha) =\alpha \cup\{\alpha\}$ is an ordinal. To do so, I need to show that that $S(\alpha)$ is strictly well-ordered by $\in$.

I can't seem to show that $\alpha \notin \alpha$.

Answer 1

If $\alpha\in\alpha$, then $\alpha$ is not strictly well-ordered by $\in$, since $x=\alpha$ is an element of $\alpha$ such that $x\in x$.

(Alternatively, if you include the axiom of regularity in your axioms, then no set $x$ can satisfy $x\in x$, since then $\{x\}$ would violate the axiom of regularity.)