I think there is no question similar to this one (at least the editor desn't find any).
My purpose is to find $\bigcap \mathcal O$, where $\mathcal O$ is the class of all ordinals:
$$\mathcal O = \{\alpha : \alpha\mbox{ is an ordinal} \}. $$
The definition of $\bigcap X$ is:
$$\bigcap X=\{z:\forall x\; (x\in X \Rightarrow z\in x)\}. $$
Thus
$$ \bigcap \mathcal O = \{z:\forall \alpha \; (\alpha\in\mathcal O \Rightarrow z\in\alpha)\} =\{z:\forall \alpha \; (\alpha\in\mathcal O \Rightarrow z\subset\alpha)\}. $$
It is easy to see that $\{\emptyset\}\subseteq \bigcap \mathcal O$. So probably $\{\emptyset\}=\bigcap \mathcal O$ but I don't know how to prove the another inclusion.
Moreover, I would expect that $\bigcap \mathcal O$ was $0$, which is the first ordinal. It seems me quite strange that $\bigcap \mathcal O=1$.
Note that $\emptyset$ is an ordinal. So the intersection $\bigcap \mathcal{O}$ is an intersection of a family which contains the empty set, so is empty.