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

Question

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 $(\cup\alpha\in\alpha) \land (s(\cup\alpha)\in\alpha)$.

Answer 1

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.

Answer 2

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

Answer 3

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 $\sup\alpha=\max\alpha=\alpha-1$.