What I want to prove is Theorem 121 of the appendix of Kelley's book General topology:
If $X\subset\mathcal O$ (here $\mathcal O$ denotes the class of all ordinals) is non-empty, then $\bigcap X\in X$. Moreover, it is the least element of $X$.
Since $X\subset\mathcal O$, the elements of $\bigcap X$ are ordinals and thus $\bigcap X$ is a transitive set. By a previous result, I know that for two ordinals $\alpha$ and $\beta$, $\alpha\in\beta$ or $\beta\in\alpha$ or $\alpha=\beta$, and only one is true. So I think this proves $\bigcap X$ is trichotomic. Hence, $\bigcap X$ is an ordinal.
But I need to proof that $\bigcap X\in X$, and I don't know how. I thought consider $\bigcap X\in\mathcal O \setminus X$ and try to get a contradiction. But I didn't kno how.
I also thought: if $\alpha\in\bigcap X$, then $\alpha\in \kappa$, for some $\kappa \in X$. So $\alpha\subset\bigcap X$ and $\alpha\subset\kappa$ and also:
$$ \alpha\subset \kappa\cap\left(\bigcap X \right) .$$
But any of this aideas convence myself $\bigcap X\in X$.
Any ideas for that?
The second part would be easier:
If there is $\kappa\in X$ such that $\kappa\in\bigcap X$, then
$$\kappa\in\alpha \qquad \forall \alpha \in X. $$
And $X$ hasn't got any element disjoint with itself.
Is that proof right?
Thanks in advance.
Remark. We are considering only proper inclusions. If $X=\mathcal O$, then
$$\bigcap X=\emptyset , $$
and the theorem follows eaisly.
By the same reason, we should consider $X\subset \mathcal O$ such that $\emptyset\notin X$.