For any two ordinals $x$ and $y$, either $x\in y$, or $x=y$, or $y\in x$

Question

Definition: $A$ is ordinal if and only if $A$ is transitive and well-ordered under $\in$.

Theorem: For any two ordinals $x$ and $y$, either $x\in y$, or $x=y$, or $y\in x$.

A corollary that I proved and is available to use: If $A$ is an ordinal, then any member of $A$ is an ordinal too.

I have tried hard, but to no avail. Please give me some hints on this theorem!

Answer 1

For given ordinals $x$ and $y$ the following claims hold true:

1. $x\subsetneq y\implies x\in y$. (I presented a proof for this at If $\alpha\ne\beta$ are ordinals and $\alpha\subset\beta$ show $\alpha\in\beta$)
2. $x\cap y$ is an ordinal. (It's quite easy to prove this)

For $x=y$, the theorem is trivially true. For $x\neq y$, let $z=x\cap y$, hence $z$ is an ordinal. We will prove this case by contradiction.

Assume $x\not\subsetneq y$ and $y\not\subsetneq x$. Since $z\subsetneq x$ and $z$ is an ordinal, then $z\in x$. Similarly, $z\in y$. Thus $z\in x\cap y=z$, which contradicts the fact that $z$ is ordinal and hence well-ordered under $\in$. As a result, either $x\subsetneq y$ or $y\subsetneq x$.