I am trying to prove that the following two statements are equivalent:
- Axiom of regularity
- $\forall x \exists \alpha (\alpha $ is an ordinal and $ x \in V_\alpha)$
I believe I understand how to prove $(1) \implies (2)$:
By regularity, $x$ is well-ordered by inclusion, and since every well-ordered set is isomorphic to a unique ordinal, $\exists \beta$ such that $(x, \in) \cong (\beta, \in)$. Now, $\alpha = \beta + 1$ is an ordinal. It is clear then that $x \in V_\alpha$, since $\beta < \alpha$.
I'm a complete amateur, so let me know if this reasoning is just totally incorrect.
However, I'm not sure how to get anything in the other direction and was hoping that I could get some feedback here. Thanks.