Why are all von Neumann ordinals contained in each other? I'm a high school student and I'm giving a lecture in my high school math class on ordinal numbers, and I would like to prove that the von Neumann ordinals are well-ordered by set membership.
The definition of von Neumann ordinal that I'm using is as follows. An ordinal is a set $A$ such that the elements of $A$ are well-ordered by $\in$ and such that $\forall x (x\in A\implies x\subset A)$.
In order to prove that the von Neumann ordinals themselves are well-ordered, I will first show that they are totally ordered. To do this, I first show that the ordering is transitive, i.e. if $A\in B$ and $B\in C$ then $A\in C$. I then want to show that the ordering is trichotomous, i.e. for all $A$ and $B$, exactly one of the following is true, either $A\in B$ or $A=B$ or $B\in A$. I am having trouble showing this last part.
In other words, I would like to show, given only the definition of von Neumann ordinal written above, that any pair of ordinals are either equal, or one is a member of the other. Any help would be appreciated.
 A: Observe that for given ordinals $A$ and $B$ the following claims hold true:


*

*If $A$ is a proper subset of $B$, then $A \in B$.

*$A \cap B$ is an ordinal.


Now, given ordinals $A \neq B$ consider $A \cap B \subseteq A,B$. If $A \cap B = A$, then $A \subseteq B$, so that either $A = B$ or $A \in B$. Analog $A \cap B = B$ implies either $B = A$ or $B \in A$. If $A \cap B$ is a proper subset of both $A$ and $B$, then $A \cap B \in A \cap B$, which contradicts the axiom of regularity.
A: For any ordinals $A$ and $B$, below statements hold true:


*

*$A\subsetneq B\implies A\in B$. (I presented a proof for this at If $\alpha\ne\beta$ are ordinals and $\alpha\subset\beta$ show $\alpha\in\beta$)

*$A\cap B$ is an ordinal. (It's quite easy to prove this)


For $A=B$, the theorem is trivially true. For $A\neq B$, let $z=A\cap B$, hence $z$ is an ordinal. We will prove this case by contradiction.
Assume $A\not\subsetneq B$ and $A\not\subsetneq B$. Since $z\subsetneq A$ and $z$ is an ordinal, then $z\in A$. Similarly, $z\in B$. Thus $z\in A\cap B=z$, which contradicts the fact that $z$ is ordinal and hence well-ordered under $\in$. As a result, either $A\subsetneq B$ or $B\subsetneq A$.
