# Von Neumann Universe V, rank and every set contained in V

I sought a consistent definition for the rank of set and I found the definition of Von Neumann universe to define the rank of a set as follows:

The cumulative hierarchy is defined by transfinite recursion over all ordinals by

• $V_0=\emptyset$,
• $V_{\alpha+1}=\mathcal P(V_\alpha)$, and
• $V_\alpha=\bigcup_{\beta<\alpha}V_\beta$

Then, the rank of a set $S$ is the smallest $\alpha$ such that $S \subseteq \alpha$. But I don't understand how does any mathematical set (for instance a real set) could be include in a $V_\alpha$ whereas $V_\alpha$ contains a construction of empty set like that

    V0 = { },
V1 = {{ }},
V2 = {{ }, {{ }}},
V3 = {{}, {{ }}, {{ }, {{ }}}},
etc.


I saw that Von Neumann set the natural set with that method, but in the ZFC theory is there a result that proves it exists a bijection between any set and that ranked construction of Von Neumann please?

I have never studied neither the set theory nor the ZFC theory before but I would like to understand that point.

First of all, your $V_3$ is wrong - it should have four elements. $V_3=\{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset,\{\emptyset\}\}\}$. (I've abbreviated "$\{\}$" as "$\emptyset$" for readability.)
A good exercise: what's the cardinality of $V_n$?
You ask where in $V$ do common mathematical objects, like $\{1\}$, lie. Well, the view of set theory is that mathematical objects are sets built up from the empty set. E.g. $0$ is $\{\}$, $1$ is $\{\{\}\}$, etc. - these are the finite ordinals. A rational number is an equivalence class of ordered triples of finite ordinals (sign, numerator, denominator), and a real number is a set of rational numbers. (This is merely one way to do the coding; there are other ways.) The role of set theory as a foundation of mathematics comes from the observation that all of mathematics can be developed within $V$. It's difficult to make this precise, since what is "all of mathematics?", but there is a lot of writing on this subject.
• @Wmog I'm not sure I understand your question, but maybe you're asking whether we can prove in ZFC that every set is in $V$? The answer is yes to this question, and the crucial axiom is Foundation: if $\alpha_0$ were a set not in $V$, it would have to have an element $\alpha_1$ not in $V$; and proceeding this way we can construct a sequence of sets $\alpha_0\ni \alpha_1\ni \alpha_2\ni ...$, contradicting Foundation. Does this answer your question? If this is not your question, can you elaborate on it? – Noah Schweber Mar 8 '17 at 19:17