Given the definition of an ordinal to be well-ordered by $\in$ and transitive, I am interested with proving the following:
I know and understand the following which easily proves it, but uses the Axiom of Choice:
Is there a proof that uses ZF axioms and not the Axiom of Choice?
[sorry for posting pictures, but I rather typeset in a LaTeX editor with macros + \newcommands]
Suppose $X$ is totally ordered by $\in$. We must prove that $X$ is well ordered by $\in$.
Therefore assume that $A\subseteq X$ is non-empty. We seek to prove that $A$ has an $\in$-least element.
Simply apply the Axiom of Regularity to $A$! This gives $a\in A$ such that $a\cap A=\varnothing$. However, then $a$ must be a $\in$-minimal element of $A$, because what that means is exactly that there is no $b\in A$ with $b\in a$, or in other words that $a\cap A=\varnothing$. Because $X$ is totally ordered by $\in$, a minimal element is necessarily a least element, as required.