Axiom of regularity and ordinal ranks

Question

I am trying to prove that the following two statements are equivalent:

1. Axiom of regularity
2. $\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.

Answer 1

The proof you suggest is wrong.

The axiom of regularity doesn't imply that every set is well-ordered by inclusion. This is false in every possible aspect. The axiom of regularity says that $\in$ is well-founded. This means that every non-empty set $x$ has $z\in x$ such that $z\cap x=\varnothing$.

The proof should be by $\in$-induction, which we can preform since $\in$ is well-founded.

Suppose that for all $y\in x$, there is some ordinal $\alpha$ such that $y\in V_\alpha$. Let $\alpha_y$ be the least such ordinal, then $\{\alpha_y\mid y\in x\}$ is a set of ordinals, therefore there is some $\alpha$ larger than all these ordinals. It follows that $x\subseteq V_\alpha$ which means that $x\in \mathcal P(V_\alpha)=V_{\alpha+1}$ as wanted.

In the other direction, suppose that every $x$ is an element of some $V_\alpha$. Let $x$ be non-empty, and consider for every $y\in x$, $\alpha_y$ the least ordinal such that $y\in V_{\alpha_y}$. Now pick some $z$ such that $\alpha_z=\min\{\alpha_y\mid y\in x\}$. And I'll leave it to you to show that $z\cap x=\varnothing$.