Axiom of regularity and ordinal ranks 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.
 A: 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 $z\cap x=\varnothing$.
A: I best understand the equivalence from Enderton's Elements of Set Theory. The following is his proof that I typed up in my LaTeX editor (to use the macros + \newcommands). Even though this post is a year old, I want to share Enderton's nice proof. There are two properties of rank that's important in this proof:


*

*For each set $X$, if every element of $X$ has a rank (or is grounded), then $X$ is grounded.

*For each set $X$, if $X$ is grounded, then for each $x\in X$, $\mathrm{rank}(x)\in \mathrm{rank}(X)$.

