(Alexander: In light of our exchange of comments, I have slightly revised the first sentence of my response to better reflect my intention.)

Perhaps you will find the following alternative formalization of the familiar informal characterization of a von Neumann Ordinal to be of interest.

A set $\alpha$ is a *von Neumann ordinal* if and only if there is a well-ordering $<_\alpha$ of $\alpha$ such that for each $y \in \alpha$, $y=\{x \in \alpha: x <_{\alpha}y \}$.

Proof. It is evident that an ordinal defined à la von Neumann is a von Neumann ordinal in the above sense. To establish the converse, let $x,y,z \in \alpha$. (i) If $x \in y$ and $y\in z$, then $x<_{\alpha} y$ and $y<_{\alpha} z$, which implies $x<_{\alpha} z$, which in turn implies $x\in z$. Moreover, (ii) since exactly one of $x=y$, $x<_{\alpha} y$ and $y<_{\alpha} x$ is the case, it follows that exactly one of $x=y$, $x\in y$ and $y\in x$ is the case, which shows $\alpha$ is totally order by $\in$. Now let $A$ be a subset of $\alpha$. Then $A$ has a $<_\alpha$-least member; and since $\in$ totally orders $\alpha$, $A$ has an $\in$-least member, which implies $\in$ well-orders $A$. To complete the proof note that every element of $\alpha$ is a subset of $\alpha$.

**Edit**: The above alternative definition of a von Neumann ordinal comes from p. 254 of my paper *All numbers great and small*, in **Real Numbers, Generalizations of the Reals, and Theories of Continua**, edited by P. Ehrlich, Kluwer Academic Publishers, Dordrecht, pp. 239-258.

`$\textit{strict}$`

again to italicise text. If you want it in (La)TeX, use`\emph{...}`

and on stackexchange sites and MO use`_underscores_`

$\endgroup$ – David Roberts Sep 18 at 0:04