This question is about von Neumann's informal definition of ordinals as "sets of all smaller ordinals" and was discussed [in this math.stackexchange question](https://math.stackexchange.com/questions/3189295/motivation-of-the-von-neumann-definition-of-ordinals/3357377#3357377).


When trying to formalize this definition, I came to this: let's consider an arbitrary class $\mathrm{No}$ with the property $\forall s\in \mathrm{No} \Rightarrow s=\{x\in \mathrm{No}: x\subset s\}$, where $\subset$ denotes a *strict* embedding. It's easy to prove that every $s\in \mathrm{No}$ is *transitive*, but I faced some problems proving that $\mathrm{No}$ is *well-ordered* by $\in$, though I feel it's true. I tried to use induction on $\operatorname{card}(\mathrm{No})$, but it seemed inconvenient even for the finite cardinals. Maybe someone knows a nice proof of this fact? 



Compact definition of ordinals
(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.


