Prove that ordinals are comparable There is one important theorem about ordinals, which says if $\alpha$ and $\beta$ are ordinals, then either $\alpha\in\beta$, or $\alpha=\beta$, or $\beta\in\alpha$.
The preceding lemma states that either $\alpha\cap\beta=\alpha$ or $\alpha\cap\beta=\beta$, which is used in the proof of this theorem.
Proof: Suppose $\alpha\neq\beta$, then either $\alpha\subset\beta$ or $\beta\subset\alpha$, where $A\subset B$ means $A$ is a proper subset of $B$.
But how does $\alpha\subset\beta$ imply $\alpha\in\beta$?
 A: We show $\alpha \subset \beta \implies \alpha \in \beta$
Let $x$ be the $\in$ minimal element of $\beta \setminus \alpha$.
Let $\gamma$ be in $x$. If $\gamma$ is not in $\alpha$ then $\gamma$ is in $\beta \setminus \alpha$ by transitivity of $\in$. This is a contradiction to $x%$ being $\in$ minimal in $\beta \setminus \alpha$. This shows $x\subseteq \alpha$.
Let $\delta$ be in $\alpha$. Since there is no ordinal in $\beta \setminus \alpha$ that is in $x$, any ordinal in $x$ must be a member of $\alpha$, since by assumption, $\alpha \subset \beta$ and since $x \in \beta$, $x \subset \beta$. We prove $\delta \in x$.
Assume $\delta \not \in x$. Since $\delta$ is not equal to $x$, there is an $x_1 \in x\setminus \delta$. Now,  $\delta \not \in x_1$, otherwise, by transitivity of $\in$, $\delta \in x$. Also, $x_1$ cannot be equal to $\delta$, because then $\delta \in x$. So there is an element of $x_1$, say $x_2$ that is not in $\delta$. Continue this reasoning infinitely, you obtain a sequence of ordinals $x_k$, that has no $\in$ minimal element, a contradiction to the well-foundedness of $\in$. Thus $\delta \in x$.

Now $x=\alpha$, and $x\in \beta$ shows $\alpha \in \beta$.
A: For given ordinals $\alpha$ and $\beta$ the following claims hold true:


*

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

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


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