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.


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 $z\cap x=\varnothing$.

  • $\begingroup$ Sorry for being so obtuse, but thanks for setting me straight. I believe I understand that $\alpha_z$, the smallest such $\alpha_y$, cannot possibly contain any of the other $y$'s in $x$, since they were all formed at stages higher than $V_{\alpha_z}$. But how does one avoid $z \cap x \neq z$ without presuming regularity? I must be missing something conceptually. $\endgroup$ – user102851 Feb 6 '14 at 1:54
  • 1
    $\begingroup$ @user102851: Suppose that $y\in x\cap z$, since $z\in V_{\alpha_z}$ and that is the least such ordinal it has to be the case that $\alpha_z=\beta+1$ (since at limit stages we don't add set, we instead aggregate those sets we have so far), so $z\subseteq V_\beta$. It follows that $y\in V_\beta$ so $\alpha_y<\alpha_z$. Contradiction. $\endgroup$ – Asaf Karagila Feb 6 '14 at 9:15

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 $x\in X$, $\mathrm{rank}(x)\in \mathrm{rank}(X)$.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy