# Proving that the class of all ordinals is well ordered under $\in$

I'm having some trouble proving the following statement:

Any set $S$ of ordinal numbers is well ordered" which is equivalent (i think) to the statement "The class $On$ of all ordinals is well ordered".

The ordering i'm talking about is obviously $\in$. Here's where I'm stuck:

I've already proven that given $\alpha, \beta$ ordinals it is either $\alpha\in\beta$ or $\alpha=\beta$ or $\beta\in\alpha$ therefore $On$ is totally ordered. Now let $S$ be a nonempty subset of $On$; we want to prove that $S$ has a minimum element; Let $\beta=\bigcap S$. It is obvious that if $\gamma\in S$, then $\beta\subset\gamma$ and since $\beta$ is a transitive set (intersection of transitive sets) we have that $\beta$ is an ordinal. What i can't prove is that $\beta \in S$. What am I not seeing here?

## 2 Answers

For every $\gamma \in S$, we know that $\beta \subseteq \gamma$, so $\beta \in \gamma$ or $\beta = \gamma$. If $\beta \in \gamma$ for every $\gamma \in S$, then $\beta \in \bigcap S = \beta$, contradiction. Therefore, $\beta = \gamma$ for some $\gamma \in S$. Therefore, $\beta \in S$.

• Much appreciated! – JustDroppedIn Jun 6 '18 at 9:23

For every $\gamma\in S$ we have $\beta\in\gamma\vee\beta=\gamma$.

So if $\beta\notin S$ then $\beta\in\gamma$ for every $\gamma\in S$ and consequently $\beta\in\cap S=\beta$.

A contradiction.