[proofplan] We prove the stronger statement $(\forall x)(\exists \alpha)(x \in V_{\alpha+1})$, equivalently $(\forall x)(\exists \alpha)(x \subseteq V_\alpha)$, by **$\in$-induction** — the principle of induction along the well-founded membership relation, available under the Axiom of Foundation. The induction hypothesis supplies, for every element $y \in x$, a rank $\operatorname{rank}(y)$ with $y \in V_{\operatorname{rank}(y) + 1}$. Collecting these ranks, we set $\alpha = \sup\{(\operatorname{rank}(y))^+ : y \in x\}$, which is an ordinal by the Replacement Axiom (the collection of ranks is a set-indexed family of ordinals). The monotonicity of the cumulative hierarchy then packs every $y \in x$ into $V_\alpha$, giving $x \subseteq V_\alpha$ and therefore $x \in V_{\alpha+1}$. [/proofplan] [step:Reduce to proving $(\forall x)(\exists \alpha)(x \subseteq V_\alpha)$ by $\in$-induction] It suffices to prove that for every set $x$, \begin{align*} \text{there exists an ordinal } \alpha \text{ with } x \subseteq V_\alpha. \end{align*} Indeed, $x \subseteq V_\alpha$ is equivalent to $x \in \mathcal{P}(V_\alpha) = V_{\alpha+1}$, so the existence of $\alpha$ with $x \subseteq V_\alpha$ implies the existence of an ordinal (namely $\alpha + 1$) with $x \in V_{\alpha+1}$, as required by the statement of the theorem. We prove $(\forall x)(\exists \alpha)(x \subseteq V_\alpha)$ by $\in$-induction. The principle of [$\in$-induction](/theorems/1494), which follows from the [Axiom of Foundation](/page/Axiom%20of%20Foundation), asserts: to prove $(\forall x)\, \varphi(x)$, it suffices to show that for every set $x$, \begin{align*} \bigl[(\forall y \in x)\, \varphi(y)\bigr] \implies \varphi(x). \end{align*} We apply this with $\varphi(x) := (\exists \alpha \text{ ordinal})(x \subseteq V_\alpha)$. [guided] The theorem asserts that every set $x$ eventually appears in the cumulative hierarchy. The natural induction principle is one that reflects the "bottom-up" construction of the hierarchy: if every element of $x$ appears by some stage, then so does $x$, one stage later. This is exactly the structure of **$\in$-induction**. $\in$-induction is the principle that, to prove a property $\varphi$ holds of every set, it suffices to show: whenever $\varphi$ holds of every element $y \in x$, then $\varphi$ holds of $x$ itself. In symbols, \begin{align*} \bigl[(\forall x)\bigl((\forall y \in x)\, \varphi(y) \implies \varphi(x)\bigr)\bigr] \implies (\forall x)\, \varphi(x). \end{align*} This principle is a consequence of the [Axiom of Foundation](/page/Axiom%20of%20Foundation), which guarantees that the membership relation $\in$ is well-founded (every non-empty set has an $\in$-minimal element). Without Foundation, $\in$-induction can fail. We apply $\in$-induction with the property \begin{align*} \varphi(x) := (\exists \alpha \text{ ordinal})(x \subseteq V_\alpha). \end{align*} Why this formulation, rather than directly $x \in V_\alpha$? Because the inductive step is naturally about subsets: knowing that each element $y \in x$ lies in some $V_{\alpha_y}$ tells us $x \subseteq \bigcup_y V_{\alpha_y}$, which has the shape "$x \subseteq$ some cumulative level," not "$x \in$ some level." The equivalence $x \subseteq V_\alpha \iff x \in V_{\alpha+1}$ converts freely between the two. [/guided] [/step] [step:Form the ordinal $\alpha = \sup\{(\operatorname{rank}(y))^+ : y \in x\}$ using Replacement] Fix a set $x$. Assume the $\in$-inductive hypothesis: for every $y \in x$, there exists an ordinal $\beta$ with $y \subseteq V_\beta$, equivalently $y \in V_{\beta+1}$. For each $y \in x$, the set of ordinals $\beta$ with $y \subseteq V_\beta$ is non-empty (by the induction hypothesis) and is a class of ordinals, hence has a least element (every non-empty class of ordinals has a minimum). Define \begin{align*} \operatorname{rank}: x &\to \operatorname{Ord} \\ y &\mapsto \operatorname{rank}(y) := \min\{\beta \in \operatorname{Ord} : y \subseteq V_\beta\}. \end{align*} Note that $\operatorname{rank}$ is well-defined by the induction hypothesis. By the [Axiom of Replacement](/page/Axiom%20of%20Replacement), the image $\{\operatorname{rank}(y) : y \in x\}$ is a set. Let \begin{align*} \alpha := \sup\bigl\{(\operatorname{rank}(y))^+ : y \in x\bigr\}, \end{align*} where $(\operatorname{rank}(y))^+ := \operatorname{rank}(y) + 1$ is the ordinal successor. Since $\{(\operatorname{rank}(y))^+ : y \in x\}$ is a set of ordinals (obtained from the previous set by applying the successor operation, again a set by Replacement), its supremum is an ordinal, namely \begin{align*} \alpha = \bigcup\bigl\{(\operatorname{rank}(y))^+ : y \in x\bigr\}. \end{align*} [guided] The induction hypothesis tells us: for each individual $y \in x$, some ordinal $\beta$ exists with $y \subseteq V_\beta$. But this is a pointwise statement — we need a **single** ordinal $\alpha$ that works for the whole set $x$. The strategy is to take the least ordinal that works for each $y$ and then combine them. **Step 1: define ranks.** For each $y \in x$, the collection of ordinals $\{\beta : y \subseteq V_\beta\}$ is non-empty by the induction hypothesis. Every non-empty class of ordinals has a minimum (ordinals are well-ordered by $\in$), so we can define \begin{align*} \operatorname{rank}(y) := \min\{\beta \in \operatorname{Ord} : y \subseteq V_\beta\}. \end{align*} This gives a function $\operatorname{rank}: x \to \operatorname{Ord}$. **Step 2: collect ranks into a set.** Here is where the [Axiom of Replacement](/page/Axiom%20of%20Replacement) enters. Replacement asserts that if $x$ is a set and $\operatorname{rank}$ is a (definable) class function, then the image $\{\operatorname{rank}(y) : y \in x\}$ is also a set. Without Replacement, we would only know this "image" is a class, and we could not take its supremum. With Replacement, the image is a legitimate set of ordinals. **Step 3: form the supremum.** The supremum of a set $S$ of ordinals is $\sup S = \bigcup S$ — the union of a set of ordinals is an ordinal, and it is the least upper bound. We define \begin{align*} \alpha := \sup\bigl\{(\operatorname{rank}(y))^+ : y \in x\bigr\}. \end{align*} Why do we take $(\operatorname{rank}(y))^+ = \operatorname{rank}(y) + 1$ rather than $\operatorname{rank}(y)$ itself? We need strict inequality $\operatorname{rank}(y) < \alpha$ for the final step. Using $(\operatorname{rank}(y))^+$ ensures $\operatorname{rank}(y) < (\operatorname{rank}(y))^+ \leq \alpha$. [/guided] [/step] [step:Show that every $y \in x$ satisfies $y \in V_\alpha$, using monotonicity] Let $y \in x$. By the definition of $\operatorname{rank}(y)$, we have $y \subseteq V_{\operatorname{rank}(y)}$, equivalently \begin{align*} y \in V_{\operatorname{rank}(y) + 1} = V_{(\operatorname{rank}(y))^+}. \end{align*} By definition of $\alpha$ as a supremum, $(\operatorname{rank}(y))^+ \leq \alpha$. We now apply the monotonicity of the cumulative hierarchy: by the [theorem that the $V_\gamma$ are increasing](/theorems/1499), since $(\operatorname{rank}(y))^+ \leq \alpha$, \begin{align*} V_{(\operatorname{rank}(y))^+} \subseteq V_\alpha. \end{align*} Combining, \begin{align*} y \in V_{(\operatorname{rank}(y))^+} \subseteq V_\alpha, \end{align*} so $y \in V_\alpha$. Since $y \in x$ was arbitrary, $x \subseteq V_\alpha$. [guided] We now complete the $\in$-inductive step. We need to show $x \subseteq V_\alpha$, so pick any $y \in x$ and verify $y \in V_\alpha$. By construction of $\operatorname{rank}(y)$, \begin{align*} y \subseteq V_{\operatorname{rank}(y)}, \end{align*} which (by definition of the successor stage $V_{\operatorname{rank}(y)+1} = \mathcal{P}(V_{\operatorname{rank}(y)})$) is equivalent to \begin{align*} y \in V_{(\operatorname{rank}(y))^+}. \end{align*} We want to promote this to $y \in V_\alpha$. The tool we need is **monotonicity of the cumulative hierarchy**: if $\gamma \leq \delta$ then $V_\gamma \subseteq V_\delta$. This is the theorem that [the $V_\gamma$ are increasing](/theorems/1499). Apply this with $\gamma = (\operatorname{rank}(y))^+$ and $\delta = \alpha$. The hypothesis $\gamma \leq \delta$ holds because, by construction, $\alpha$ is the supremum (and hence an upper bound) of the set $\{(\operatorname{rank}(y))^+ : y \in x\}$. Hence \begin{align*} V_{(\operatorname{rank}(y))^+} \subseteq V_\alpha, \end{align*} and in particular $y \in V_\alpha$. Since $y \in x$ was arbitrary, $x \subseteq V_\alpha$ — which is the inductive conclusion at $x$. [/guided] [/step] [step:Conclude by closing the $\in$-induction] We have shown: assuming $(\forall y \in x)(\exists \beta)(y \in V_{\beta+1})$, there exists an ordinal $\alpha$ with $x \subseteq V_\alpha$, i.e., $x \in V_{\alpha+1}$. By $\in$-induction, $(\forall x)(\exists \alpha)(x \in V_{\alpha+1})$. Since $\{V_\gamma : \gamma \in \operatorname{Ord}\}$ is a class of sets, we have shown that every set belongs to some $V_\gamma$, i.e., \begin{align*} V := \bigcup_{\gamma \in \operatorname{Ord}} V_\gamma \end{align*} contains every set. Combined with the obvious converse — every element of some $V_\gamma$ is a set — we obtain the set-theoretic universe identity \begin{align*} V = \bigcup_{\gamma \in \operatorname{Ord}} V_\gamma. \end{align*} In particular, the rank function \begin{align*} \operatorname{rank}: V &\to \operatorname{Ord} \\ x &\mapsto \min\{\alpha \in \operatorname{Ord} : x \subseteq V_\alpha\} \end{align*} is well-defined on every set, giving a stratification of $V$ by ordinals. [/step]