I want to prove that if $X\subset \mathcal O$ is a non-empty subclass of the class of ordinals numbers, then
$$ \bigcap X \subset X .$$
The author reasons as follows:
1.- Since $\mathcal O$ is transitive, $X$ is an ordinal.
2.- Thus, every element of $X$ is also a proper subset of $X$.
3.- And also every element $z$ of every element $x\in X$ will also belong to $X$ and hence
$$ \bigcap X \subset X . $$
I understand the proof except for step 1. I know that $\mathcal O$ is an ordinal and thus transitive. So if $\alpha\in \mathcal O$ then $\alpha\subset \mathcal O$. But the converse may not be true. I'm thinking for example in $X=\mathcal O\setminus \{\emptyset\}$. Then $\emptyset\in\{\emptyset\}\in X$, but $\emptyset\notin X$. So $\{\emptyset\}\in X$ but $\{\emptyset\}\not\subset X$.
What do you think? Is the author right? Or is my counterexample valid? And how can I prove that $\bigcap X\subset X$?
Thanks
Edit The theorem is all completely wrong. The statement is that if $X\subset \mathcal O$ is a non-empty subclass of the class of ordinals numbers, then
$$ \bigcap X \in X .$$
The author proves (wrongly) that $\bigcap X\subset X$, which I don't know if it is right.
My aplogies.
Easier counterexample of (1): obviously, X =$\{0,2\}\subset\mathcal O$ but isn't an ordinal because $2 = \{0,1\}\in X$ but $2 = \{0,1\}\not\subset X$.