1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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, $\beta \in S$.

$\endgroup$
  • $\begingroup$ Much appreciated! $\endgroup$ – JustDroppedIn Jun 6 '18 at 9:23
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.