How to prove that the von Neumann universe equals $V$?

Question

Do I understand correctly, that it is possible to prove in NBG set theory that the von Neumann universe, i.e. the union of $$ \begin{align} V_0 &= \varnothing, \\ V_{\alpha+1} &= P(V_{\alpha}), \quad\text{for all ordinals,}\quad\\ V_{\alpha} &= \bigcup_{\beta<\alpha} V_{\beta}, \quad\text{for limit ordinals} \end{align} $$ equals the class of all sets? How can this be proved?

This claim is stated and proved in the German language textbook Transfinite Zahlen, 2nd ed., by Heinz Bachmann, as item (h) on p. 27. However, Bachmann's proof is not in the context of NBG, but is instead in the context of showing this statement is true inside a Grothendieck universe in ZFC.

Answer 1

This follows from (and in fact, is equivalent to) the axiom of regularity. There are various equivalent ways to state the axiom of regularity (more on that below), but I will take the following statement: any nonempty class $X$ has an element $a$ such that $a\cap X=\emptyset$.

Now, consider the class $X$ of all sets that are not in $V_\alpha$ for any ordinal $\alpha$. If $X$ is nonempty, then by regularity there exists $a\in X$ such that $a\cap X=\emptyset$. That means that every element of $a$ is in $V_\alpha$ for some $\alpha$. For $b\in a$, let $f(b)$ be the least ordinal $\alpha$ such that $b\in V_\alpha$. By replacement and union, $\alpha=\bigcup\{f(b):b\in a\}$ is a set, and is an ordinal which is greater than or equal to $f(b)$ for all $b\in a$. Thus $b\in V_\alpha$ for all $b\in a$, which means $a\in P(V_\alpha)=V_{\alpha+1}$. But then $a\not\in X$, which is a contradiction. Thus $X$ must be empty.

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As I said above, there are other equivalent statements of the axiom of regularity and the proof may need modification depending on what statement you use. One version of the axiom of regularity is the very statement $V=\bigcup V_\alpha$, in which case there's nothing to prove. Another version is that any nonempty set $X$ has an element $a$ such that $a\cap X=\emptyset$. With that version, you need to use a trick because the class $X$ used in the proof above may not be a set; see Statements Equivalent to Axiom of Foundation for details.

Answer 2

Using the Axiom of Regularity, you can show that every set $x$ has a rank $\mathrm{rk}(x)\in\mathrm{Ord}$, where $\mathrm{rk}(x)$ is the smallest ordinal larger than $\mathrm{rk}(y)$ for every $y\in x$.

Suppose not, then either there is a sequence of elements in $x$ whose ranks are cofinal in $\mathrm{Ord}$, which makes $x$ a proper class, or $x$ must contain an element without rank (since if all $y\in x$ have a rank, we can take the union of their ranks to find an ordinal that is larger than or equal to all the ranks of the elements of $x$). Repeating this, we can construct a sequence $x\ni x_1\ni x_2\ni \dots$ of sets without rank, which contradicts the Axiom of Regularity.

For example, $\mathrm{rk}(\emptyset)=0$, $\mathrm{rk}(\{\emptyset\})=1$, and by transfinite induction you can show $\mathrm{rk}(\alpha)=\alpha$ for every ordinal $\alpha$.

It is not too difficult to see that then $x\in V_{\mathrm{rk}(x)+1}\subset V$, since $y\in x$ implies $y\in V_{\mathrm{rk}(x)}$ by induction, and thus $x\in\mathcal P(V_{\mathrm{rk}(x)})=V_{\mathrm{rk}(x)+1}$.

Without the Axiom of Regularity, there could be sets not in the Von Neumann universe, since there could exist infinite sequences $x\ni x_1\ni x_2\ni\dots$, but it is still possible to show that every well-founded set is in the Von Neumann universe (and vice versa).