# Let $\alpha$ be an ordinal then $\alpha=\cup\alpha$ or $\alpha=s(\cup\alpha)$.

Let $\alpha$ be an ordinal then $\alpha=\cup\alpha$ or $\alpha=s(\cup\alpha)$.

Attempt:

Since $\alpha$ is transitive $\alpha=\cup\alpha$ (here we are done) or $\alpha\supsetneq\cup\alpha$.

If $\alpha\supsetneq\cup\alpha$, since $\cup\alpha$ is an ordinal we have $\cup\alpha\in\alpha$. Regularity axiom implies $s(\cup\alpha)=\alpha$ (here we are done) or $s(\cup\alpha)\in\alpha$.

But I can't see any contradiction assuming $(\cup\alpha\in\alpha) \land (s(\cup\alpha)\in\alpha)$.

-

To get a contradiction from $s(\bigcup\alpha)\in\alpha$, note that, by definition of $s$, we also have $\bigcup\alpha\in s(\bigcup\alpha)$. So $\bigcup\alpha$ is a member of a member of $\alpha$. By definition of $\bigcup$, we have $\bigcup\alpha\in\bigcup\alpha$, contradicting the axiom of regularity.

-
Thanks, I tried to find the contradiction even noticed that $\cup a \in s (\cup a) \in a$ and I did'nt see any contradiction. – Gastón Burrull Jun 20 '13 at 2:05

HINT: Show that if $\alpha=s(\beta)$, then $\bigcup\alpha=\beta$, and if there is no $\beta$ such that $\alpha=s(\beta)$, then $\bigcup\alpha=\alpha$.

-
If $\alpha=s(\beta)$ and $x\in\cup\alpha$ then exists $u$ such that $x\in u$ and $u\in\beta$ or $u=\beta$, that implies $x\in\beta$. Then $\cup\alpha\subset\beta$. I'll try the other inclusion now. – Gastón Burrull Jun 20 '13 at 2:00
@Gastón: Looks good so far. – Brian M. Scott Jun 20 '13 at 2:02
$x\in \beta\in s(\beta)=\alpha$ implies $x\in\cup\alpha$ then $\beta\subset \cup\alpha$. – Gastón Burrull Jun 20 '13 at 2:08
If $\alpha\neq s(\beta)$ for any ordinal $\beta$ in particular $\alpha\neq s(\cup\alpha)$ so $\alpha\in s(\cup \alpha)$ or $\cup\alpha\in s(\cup\alpha)\in\alpha$ but the second one implies $\cup\alpha\in\cup\alpha$ and the first one implies $\alpha\in\cup\alpha\subset\alpha$ (a contradiction) or $\alpha=\cup\alpha$. – Gastón Burrull Jun 20 '13 at 2:20
@Gastón: Looks good. And from this way of thinking about it you get to see exactly when $\bigcup\alpha=\alpha$: when $\alpha$ is a limit ordinal. – Brian M. Scott Jun 20 '13 at 2:25

Here is an alternative approach.

• First show that if $A$ is a set of ordinals then $\bigcup A=\sup A$.
• Show that if $\alpha$ is a limit ordinal or zero then $\bigcup\alpha=\alpha$.
• Show that if $\alpha$ is a successor ordinal then $\alpha$ (as a set of ordinals) has a maximum, which is the predecessor of $\alpha$, and $\sup\alpha=\max\alpha=\alpha-1$.
-