Comparing ordinal sizes I was trying to understand the proof to the following theorem, specifically (iii). (The same theorem is listed here: Prove that ordinals are comparable, but the proofs seem to be different.)

Suppose $\alpha, \beta, \gamma$ are ordinals. Then, (iii) exactly one of $\alpha<\beta, \alpha=\beta, \beta<\alpha$ holds, and (iv) if $X$ is a non-empty set of ordinals, then $X$ has a least element $\delta$ and $\delta=\bigcap X$.

Proof:
(iii) Certainly we already know that no more than one of these holds. Consider $\alpha \cap \beta .$ This is also an ordinal (see (iv), or do as exercise). If, for example $\alpha \cap \beta=\alpha$ then $\alpha \subseteq \beta$ and so $\alpha \leq \beta .$ So for a contradiction, suppose $\alpha \cap \beta$ is a proper subset of both $\alpha$ and $\beta .$ Thus (by a previous lemma) $\alpha \cap \beta \in \alpha$ and $\alpha \cap \beta \in \beta .$ So $\alpha \cap \beta \in \alpha \cap \beta:$ a contradiction.
I'm having a lot of difficulty understanding this proof.

*

*Why does $\alpha \cap \beta \in \alpha$ and $\alpha \cap \beta \in \beta$ imply that $\alpha \cap \beta \in \alpha \cap \beta$ ? Surely, $\alpha \cap \beta \subset \alpha$, $\alpha \cap \beta \subset \beta$ implies that $\alpha \cap \beta \subseteq \alpha \cap \beta$, and not necessarily $\alpha \cap \beta \subset \alpha \cap \beta$? (In this course, $\subset$ means a strict subset)


*I don't comprehend what the argument is. I get that if $\alpha \cap \beta=\alpha$, then $\alpha \subseteq \beta$ and so $\alpha \leq \beta .$ But what does this contribute to proving the theorem?
 A: *

*For any $x$, $A$ and $B$, if $x \in A$ and $x \in B$, then $x \in A \cap B$. The statement you are querying is an instance of that (with $x = \alpha \cap \beta$, $A = \alpha$ and $B = \beta$).


*I think words "for example" in the text of the proof are a bit misleading. The form of the argument is that one of the following must hold:
$$
\begin{array}{cl}
\mbox{(a)} & \alpha \cap \beta = \alpha\\
\mbox{(b)} & \alpha \cap \beta = \beta \\
\mbox{(c)} & \alpha \cap \beta \subsetneq \alpha \mbox{ and } \alpha \cap \beta \subsetneq \beta
\end{array}
$$
The "for example" statement is saying that to deal with (a) and (b) , it is sufficient, by symmetry, just to consider (a).
A: *

*By definition of intersection, any element $x$ satisfies $x\in\alpha\cap\beta\iff x\in\alpha\land x\in\beta$. This is applied to $x=\alpha\cap\beta$.


*We have proved that $\alpha\cap\beta$ can't be a proper subset of both $\alpha$ and $\beta$, thus either $\alpha\cap\beta=\alpha$, that is, $\alpha\le\beta$ or $\alpha\cap\beta=\beta$, that is, $\beta\le\alpha$ must hold.
Stating otherwise, if $\alpha\ne\beta$ then either $\alpha<\beta$ or $\beta<\alpha$.
