The intersection of a class of ordinals belongs to the class

Question

Let $C$ be a non-empty class of ordinals. I want to prove that $\bigcap C \in C.$ What I managed to prove so far :

1. That $\bigcap C$ is an ordinal.
2. If $\alpha \in C$, that $\bigcap (\alpha^+ \cap C) = \bigcap C.$
3. That $\bigcap (\alpha^+ \cap C)$ and $\alpha^+ \cap C$ are ordinals (and so I can compare them).

Actually, I try to prove the property for $\alpha^+ \cap C$ (which a set of ordinals) before proving it for $C$ itself but I am stuck here.

Answer 1

You need to assume $C$ is non-empty (otherwise $\bigcap C$ doesn't make sense).

Since the ordinals are well-ordered, every non-empty class of ordinals has a least element. Let $\alpha$ be the least element of $C$. I claim that $\alpha=\bigcap C$.

Since $\alpha\in C$, $\bigcap C\subseteq \alpha$. In the other direction, for all $\beta\in C$, $\alpha\leq \beta$, so $\alpha\subseteq \beta$. Thus $\alpha\subseteq \bigcap C$.

Answer 2

I'm learning Ordinals as well. How about this for a proof. I'm assuming you have proved $\bigcap C$ is an ordinal, as you have stated.

Suppose $c\in C$. Then $\bigcap C \subseteq c$. And as $c$ and $\bigcap C$ are ordinals, $\bigcap C\underline{\in}c$. Hence $\bigcap C$ is smaller than or equal to every element in $C$ (*).

Let $z=\bigcap C$. We know that $z^+$ is an ordinal. If $c\in C$ then by trichotomy, $z^+\underline{\in}\ c$ or $c \in z^+$. Suppose $z^+\underline{\in}\ c$ for all $c \in C$. Then $z\in c$ for all $c\in C$. Hence by definition, $z\in\bigcap C$, a contradiction. Hence there is some $c\in C$ for which $c\in z^+$. And so for this $c$, $c=z$ or $c\in z$. And if $c\in z$ this is a contradiction to (*). Hence $c=z$ and so $z\in C$.