How can I tell if some set is a von Neumann stage $V_\alpha$ for some ordinal $\alpha$?

Question

For some set $X$, is there some statement that's equivalent to "there exists an ordinal $\alpha$ such that $X$ is the von Neumann stage $V_\alpha$ in the von Neumann hierarchy of sets", but talks about what $X$ needs to look like instead of referring to the transfinite recursion definition of $V_\alpha$?

Trying to come up with one I only got this one:

A set $X$ is a von Neuman stage iff there is a subset $A \subseteq X$ such that

1. $X$ is closed under subsets (that is, for every $x \in X$ and every subset $y \subseteq x$, $y \in X$).
2. For every $a \in A$, $a$ is closed under subsets.
3. For every $x \in X$, there is some $a \in A$ such that $x \subseteq a$.
4. For every $a \in A$ and every $x \in a$, there is some $b \in a \cap A$ such that $x \subseteq b$.

(The intuition behind this is that $X$ will be some von Neumann stage and $A$ the set of all von Neumann stages before it.)

But

• I'm not actually sure it's correct (and the proof looks like it'd be long),
• the property is more verbose than expected, and
• this property is "less indirect" than one with the unbounded quantifier "there exists an ordinal $\alpha$ such that ..." it still has a quantifier bounded over $P(X)$, but I'm curious if there's an equivalent property that only uses quantifiers bounded over $X$ or any descendant of $X$. (Intuitively I want to know whether, to find out whether $X$ is a von Neumann stage, you only need to check relationships between $X$ and its elements and elements of elements and so on, or whether you need to check something for every subset of $X$.)

Are there any simpler or less indirect equivalent properties?

Answer 1

It is easy to remove the "indirectness" from successor stages, i.e.

A set $Y$ is a successor stage if and only if there exists $X∈Y$ such that $P(X)=Y$ and $X$ satisfies the properties you wrote.

Inside of $Y$, the existence of the set $A$ for $X$ is first order property.

You can remove the "indirectness" from limit stages by noting that $Y$ is a limit stage if and only if $Y$ is union of (all) previous stages:

A set $Y$ is a limit stage if and only if for all $X∈Y$, $X$ satisfies your properties implies that for all $a∈X$, $a∈Y$ and for all $b∈Y$ there exists $X∈Y$ such that $X$ satisfies your properties and $b∈X$.

The above still doesn't remove the indirectness because we need to show that "$X$ satisfies your condition" can be expressed using FOL in $Y$. So to complete that we add the condition "$Y$ is closed under powerset":

A set $Y$ is a limit stage if and only if $Y$ is a closed under powerset for all $X∈Y$, $X$ satisfies your properties implies that for all $a∈X$, $a∈Y$ and for all $b∈Y$ there exists $X∈Y$ such that $X$ satisfies your properties and $b∈X$.

With this the statement "$X$ satisfies your properties" only uses sets from $Y$.

Hence, because a set $X$ is a Von Neumann stage iff it is either successor stage or a limit case we completely classified Von Neumann stages only uses quantifier over the closure of $X$