[proofplan] We prove by set induction, using Foundation, that every set is contained in some level $V_\alpha$ of the cumulative hierarchy. Given a set $x$, the inductive hypothesis gives ranks for all elements $y \in x$; Replacement collects these ranks into a set of ordinals. Taking the supremum of their successors produces one ordinal level containing every element of $x$, and the well-ordering of the ordinals then gives the least such level, which is $\operatorname{rank}(x)$. [/proofplan] [step:Use Foundation to reduce rank existence for $x$ to rank existence for its elements] Let $P(x)$ be the assertion that there exists an ordinal $\alpha$ such that $x \subseteq V_\alpha$. We prove $P(x)$ for every set $x$ by set induction, which is the induction principle obtained from the Axiom of Foundation. Fix a set $x$, and assume the induction hypothesis that $P(y)$ holds for every $y \in x$. Thus, for each $y \in x$, there exists an ordinal $\beta$ such that $y \subseteq V_\beta$. [guided] Define the property $P$ on sets by declaring that $P(x)$ means \begin{align*} \exists \alpha \in \operatorname{Ord}\text{ such that }x \subseteq V_\alpha. \end{align*} The Axiom of Foundation implies set induction: to prove $P(x)$ for all sets $x$, it is enough to prove that whenever $P(y)$ holds for every element $y \in x$, then $P(x)$ holds. So fix a set $x$ and assume the induction hypothesis: \begin{align*} \forall y \in x,\ \exists \beta \in \operatorname{Ord}\text{ such that }y \subseteq V_\beta. \end{align*} This is exactly where Foundation enters the proof: it lets us assume rank-boundedness for all elements of $x$ before proving rank-boundedness for $x$ itself. [/guided] [/step] [step:Define the least bounding level for each element of $x$] For each $y \in x$, define \begin{align*} \rho(y)=\min\{\beta \in \operatorname{Ord}: y \subseteq V_\beta\}. \end{align*} This is well-defined because the induction hypothesis makes the displayed class nonempty, and the ordinals are well-ordered. Thus $\rho(y)$ is an ordinal and $y \subseteq V_{\rho(y)}$ for every $y \in x$. [guided] For each element $y \in x$, define the class $A_y$ by \begin{align*} A_y:=\{\beta \in \operatorname{Ord}: y \subseteq V_\beta\}. \end{align*} The induction hypothesis says exactly that $A_y$ is nonempty. Since $A_y$ is a nonempty class of ordinals, the well-ordering of the ordinals gives a unique least member. Define \begin{align*} \rho(y):=\min A_y. \end{align*} Thus $\rho(y)$ is an ordinal, $y \subseteq V_{\rho(y)}$, and no smaller ordinal has this property. The uniqueness of this least ordinal matters because Replacement applies to functional assignments. We are not choosing arbitrary witnesses; we have defined a unique ordinal $\rho(y)$ for each $y \in x$. [/guided] [/step] [step:Use Replacement to collect the element ranks into a set of ordinals] By the Axiom Schema of Replacement applied to the functional formula defining $\rho(y)$, the collection \begin{align*} R_x:=\{\rho(y): y \in x\} \end{align*} is a set. Since each $\rho(y)$ is an ordinal, $R_x$ is a set of ordinals. [guided] The formula “$\gamma=\rho(y)$” defines a unique ordinal $\gamma$ for each $y \in x$. Therefore Replacement applies and produces the image set \begin{align*} R_x:=\{\rho(y): y \in x\}. \end{align*} This is the exact point where Replacement enters the proof. Without Replacement, we would know that every element $y \in x$ has some least bounding ordinal, but we would not be entitled to collect all those ordinals into a single set whose supremum can be taken. [/guided] [/step] [step:Take the supremum of successor ranks to bound all elements of $x$] Define \begin{align*} \alpha:=\sup\{\rho(y)+1: y \in x\}. \end{align*} Since $R_x$ is a set of ordinals, the set $\{\rho(y)+1:y\in x\}$ is also a set of ordinals, and its supremum $\alpha$ is an ordinal. For every $y \in x$, we have $\rho(y)+1 \leq \alpha$. Since $y \subseteq V_{\rho(y)}$, the definition of the cumulative hierarchy gives $y \in V_{\rho(y)+1}$. By monotonicity of the hierarchy, $V_{\rho(y)+1} \subseteq V_\alpha$, hence $y \in V_\alpha$. Therefore $x \subseteq V_\alpha$. [guided] Now we need one ordinal level that contains every element of $x$. For each $y \in x$, we know \begin{align*} y \subseteq V_{\rho(y)}. \end{align*} By the successor-stage definition of the cumulative hierarchy, \begin{align*} V_{\rho(y)+1}=\mathcal{P}(V_{\rho(y)}), \end{align*} so $y \subseteq V_{\rho(y)}$ implies \begin{align*} y \in V_{\rho(y)+1}. \end{align*} Because Replacement gave us the set of ordinals $R_x=\{\rho(y):y\in x\}$, we may form the set of successor ordinals \begin{align*} S_x:=\{\rho(y)+1:y\in x\}. \end{align*} Define \begin{align*} \alpha:=\sup S_x. \end{align*} Then $\alpha$ is an ordinal and $\rho(y)+1\leq \alpha$ for every $y\in x$. The cumulative hierarchy is monotone in the ordinal index, so \begin{align*} V_{\rho(y)+1}\subseteq V_\alpha. \end{align*} Hence each $y\in x$ belongs to $V_\alpha$. Therefore \begin{align*} x\subseteq V_\alpha. \end{align*} This proves the induction step. [/guided] [/step] [step:Extract the least bounding ordinal and define the rank] By set induction, for every set $x$ there exists an ordinal $\alpha_0$ such that $x \subseteq V_{\alpha_0}$. Define the class $B_x$ by \begin{align*} B_x:=\{\alpha \in \operatorname{Ord}: x \subseteq V_\alpha\}. \end{align*} Then $B_x$ is nonempty because $\alpha_0 \in B_x$. To justify taking the least element of this class, use Separation on the set $\alpha_0+1$ to form \begin{align*} C_x:=\{\alpha \in \alpha_0+1: x \subseteq V_\alpha\}. \end{align*} The set $C_x$ is a nonempty set of ordinals, so it has a least element $\delta$. Since $C_x \subseteq B_x$ and every element of $B_x$ either lies in $\alpha_0+1$ or is at least $\alpha_0+1$, the ordinal $\delta$ is also the least element of $B_x$. Define \begin{align*} \operatorname{rank}(x):=\delta=\min B_x. \end{align*} Thus $\operatorname{rank}(x)$ is a well-defined ordinal and satisfies \begin{align*} x \subseteq V_{\operatorname{rank}(x)}. \end{align*} Uniqueness follows from uniqueness of the least element in a well-ordered set of ordinals. Therefore every set has a rank. [guided] By the induction step already proved, for every set $x$ there exists an ordinal $\alpha_0$ such that \begin{align*} x \subseteq V_{\alpha_0}. \end{align*} Define the class $B_x$ of all ordinal bounds for $x$ by \begin{align*} B_x:=\{\alpha \in \operatorname{Ord}: x \subseteq V_\alpha\}. \end{align*} This class is nonempty because $\alpha_0 \in B_x$. We now justify the class-minimum step using only set-level well-ordering. Since $\alpha_0$ is an ordinal, $\alpha_0+1$ is a set. By Separation applied inside the set $\alpha_0+1$, the collection \begin{align*} C_x:=\{\alpha \in \alpha_0+1: x \subseteq V_\alpha\} \end{align*} is a set. It is nonempty because $\alpha_0 \in C_x$. Since $C_x$ is a nonempty set of ordinals, the well-ordering of ordinals gives a unique least element; call it $\delta$. The ordinal $\delta$ is not merely least in $C_x$; it is least in the whole class $B_x$. Indeed, let $\beta \in B_x$. If $\beta \in \alpha_0+1$, then $\beta \in C_x$, so $\delta \leq \beta$ by the definition of $\delta$. If $\beta \notin \alpha_0+1$, then $\alpha_0 < \beta$, while $\delta \leq \alpha_0$ because $\delta \in C_x \subseteq \alpha_0+1$; hence again $\delta \leq \beta$. Therefore $\delta$ is the least member of $B_x$. Define \begin{align*} \operatorname{rank}(x):=\delta=\min B_x. \end{align*} Then $\operatorname{rank}(x)$ is an ordinal and, because it belongs to $B_x$, it satisfies \begin{align*} x \subseteq V_{\operatorname{rank}(x)}. \end{align*} The uniqueness of $\operatorname{rank}(x)$ follows from uniqueness of the least element of the set $C_x$, equivalently from antisymmetry of the ordinal order. This completes the proof that every set has a rank. [/guided] [/step]