Can we show for ordinal numbers that $\alpha\in\beta$ implies $\alpha\subsetneqq\beta$ without axiom of regularity?

Question

Von Neumann definition of ordinals is that an ordinal is a transitive set which is well-ordered by $\in$. (To be more precise, the relation $\in$ gives a strict well-order on $\beta$. Although there are probably various other reformulations of this definition.)

If we take this as a definition, we can show that $\alpha\in\beta$ implies $\alpha\subsetneqq\beta$.

From transitivity of $\beta$ we get $\alpha\subseteq\beta$. At the same time we have $\alpha\ne\beta$, otherwise we would get $\beta\in\beta$, which contradicts the axiom of regularity.

Question: Is there a different argument which avoids using regularity? Or is a proof in ZF (or ZFC) without axiom of regularity impossible?

Answer 1

According to the link you posted the definition of Von Neumann doesn't state an ordinal is just a transitive set. It states:

"A set $S$ is an ordinal if and only if $S$ is strictly well-ordered with respect to set membership and every element of $S$ is also a subset of $S$"

So an ordinal has to be transitive and also has to be well ordered with respect to $\in$. So $\beta\notin\beta$ for all ordinals simply follows from the definition of strict ordering, you don't need regularity here.

Answer 2

Since $\in$ is a strict well-ordering, it is irreflexive by definition (recall that a strict partial order is irreflexive and transitive). If $\beta\in\beta$, then $\beta\in\beta$, and therefore $\in$ is not irreflexive.