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.

In order to prove that any class $\mathrm{No}$ satisfying this definition is well-ordered by $\in$ it is crucial to use the axiom of foundation. Axiom of foundation tells us that $\in$ is a well-founded relation. Thus we only need to prove that any two elements of $\mathrm{No}$ are $\in$-comparable. This could be achieved by a proof by a contradiction, where we consider a $\in$-pointwise-minimal $\in$-incomparable pair $a,b\in\mathrm{No}$ to get to a contradiction. This is essentially the same as the arguments that use ranks of sets suggested in the comments (one needs axiom of foundation to prove that every set has a rank), but formulated in the terms of more elementary notions.

If one drops the axiom of foundation, then we could have models, where there are classes satisfying this definition, although not being well-ordered by $\in$-relation. For example, it is possible to have a model with two chains of sets $a_i=\{a_j\mid j>i\}$ and $b_i=\{b_j\mid j>i\}$ such that all $a_i$ and $b_i$ are pairwise distinct sets. Clearly, $\{a_i\mid i\in\omega\}\cup \{b_i\mid i\in\omega\}$ satisfies your definition, but is neither linearly ordered nor well-founded.

Okay, here is an expanded version of my comment. Let ${\rm No}$ be a class of sets with the property that every the elements of any $x \in {\rm No}$ are precisely the proper subsets of $x$ which belong to ${\rm No}$.

Theorem: ${\rm No}$ is either an ordinal or the class of all ordinals.

Proof: We will prove by induction on $\alpha$ that for any ordinal $\alpha$, if there is an element of ${\rm No}$ of rank $\alpha$ then there is exactly one such element, and it is $\alpha$ itself. As it follows immediately from the condition on ${\rm No}$ that any element of a set in ${\rm No}$ also belongs to ${\rm No}$, this means that ${\rm No}$ is a class of ordinals which contains every ordinal less than any ordinal it contains, which implies the theorem.

Let $\alpha$ be an ordinal and suppose the claim is true for all smaller ordinals. Let $x$ be an element of ${\rm No}$ whose rank is $\alpha$. By the definition of rank, it follows that the ranks of elements of $x$ are cofinal in $\alpha$. (This includes both the successor and limit cases.) But the induction hypothesis implies that every element of $x$ is an ordinal. As it is immediate from the condition on ${\rm No}$ that $x$ is transitive, this implies that $x$ is precisely the set of all ordinals less than $\alpha$, i.e., $x = \alpha$. ///