[proofplan] We fix $\alpha$ and prove the inclusion $V_\alpha \subseteq V_\beta$ by transfinite induction on $\beta \geq \alpha$. The base case $\beta = \alpha$ is immediate. The successor step $\beta \to \beta + 1$ uses the one-step identity $V_\beta \subseteq V_{\beta+1}$, which itself follows from the transitivity of $V_\beta$: every $x \in V_\beta$ satisfies $x \subseteq V_\beta$, hence $x \in \mathcal{P}(V_\beta) = V_{\beta+1}$. The limit step is direct from the definition $V_\lambda = \bigcup_{\gamma < \lambda} V_\gamma$. Chaining the inductive hypothesis $V_\alpha \subseteq V_\beta$ with $V_\beta \subseteq V_{\beta+1}$ (or with the limit union) closes the induction. [/proofplan] [step:Recall the recursive definition of the cumulative hierarchy and fix the induction variable] The cumulative hierarchy $(V_\gamma)_{\gamma \in \operatorname{Ord}}$ is defined by transfinite recursion on the ordinals by \begin{align*} V_0 &= \varnothing, \\ V_{\gamma + 1} &= \mathcal{P}(V_\gamma), \\ V_\lambda &= \bigcup_{\gamma < \lambda} V_\gamma \quad (\lambda \text{ limit}). \end{align*} Fix an arbitrary ordinal $\alpha$. We must prove \begin{align*} (\forall \beta \geq \alpha)\quad V_\alpha \subseteq V_\beta. \end{align*} We proceed by [transfinite induction](/theorems/1463) on $\beta$, starting at $\beta = \alpha$. [/step] [step:Establish the one-step inclusion $V_\beta \subseteq V_{\beta+1}$ via transitivity of $V_\beta$] We first prove a lemma that will drive the successor step. [claim:EachLevelIsTransitive] For every ordinal $\beta$, the set $V_\beta$ is transitive, i.e., \begin{align*} x \in V_\beta \implies x \subseteq V_\beta. \end{align*} [/claim] [proof] We argue by transfinite induction on $\beta$. **Base case ($\beta = 0$).** $V_0 = \varnothing$, which is vacuously transitive: the implication $x \in V_0 \implies x \subseteq V_0$ has no instances. **Successor step.** Suppose $V_\beta$ is transitive. Let $x \in V_{\beta + 1} = \mathcal{P}(V_\beta)$. Then by the definition of the power set, $x \subseteq V_\beta$. We must show $x \subseteq V_{\beta + 1}$. Let $y \in x$; then $y \in V_\beta$, and by the induction hypothesis $y \subseteq V_\beta$, so $y \in \mathcal{P}(V_\beta) = V_{\beta+1}$. Thus $x \subseteq V_{\beta + 1}$. **Limit step.** Suppose $\lambda$ is a limit ordinal and $V_\gamma$ is transitive for every $\gamma < \lambda$. Let $x \in V_\lambda = \bigcup_{\gamma < \lambda} V_\gamma$, so $x \in V_\gamma$ for some $\gamma < \lambda$. By the induction hypothesis $x \subseteq V_\gamma \subseteq V_\lambda$, so $x \subseteq V_\lambda$. This completes the induction, and $V_\beta$ is transitive for every ordinal $\beta$. [/proof] From the claim we deduce the one-step inclusion. Let $x \in V_\beta$. By transitivity (Claim \ref{EachLevelIsTransitive}), $x \subseteq V_\beta$, i.e., $x \in \mathcal{P}(V_\beta) = V_{\beta+1}$. Hence \begin{align*} V_\beta \subseteq V_{\beta+1} \end{align*} for every ordinal $\beta$. [guided] The successor step of the main theorem reduces to the following question: is $V_\beta \subseteq V_{\beta+1}$? Recall $V_{\beta+1} = \mathcal{P}(V_\beta)$, the power set of $V_\beta$. So we are asking: for $x \in V_\beta$, is $x$ a *subset* of $V_\beta$? The answer is "yes" precisely because of the **transitivity** of $V_\beta$ — the property that every element of $V_\beta$ is also a subset of $V_\beta$. Transitivity of each $V_\beta$ is a fundamental structural fact about the cumulative hierarchy and deserves its own lemma, which we now prove by transfinite induction on $\beta$. **Base ($\beta = 0$).** $V_0 = \varnothing$, and the implication "$x \in V_0 \implies x \subseteq V_0$" is vacuously true (there is no $x \in \varnothing$). **Successor.** Assume $V_\beta$ is transitive; we show $V_{\beta+1}$ is transitive. Take $x \in V_{\beta+1} = \mathcal{P}(V_\beta)$, which by definition of power set means $x \subseteq V_\beta$. We want $x \subseteq V_{\beta+1}$, so pick any $y \in x$. Then $y \in V_\beta$. By the inductive hypothesis (transitivity of $V_\beta$), $y \subseteq V_\beta$. Hence $y \in \mathcal{P}(V_\beta) = V_{\beta+1}$. Since this holds for every $y \in x$, we have $x \subseteq V_{\beta+1}$, as desired. **Limit.** Assume $\lambda$ is a limit ordinal and $V_\gamma$ is transitive for every $\gamma < \lambda$. Let $x \in V_\lambda = \bigcup_{\gamma < \lambda} V_\gamma$. By definition of the union, there is some $\gamma < \lambda$ with $x \in V_\gamma$. By the inductive hypothesis, $x \subseteq V_\gamma$. But $V_\gamma \subseteq V_\lambda$ (every element of $V_\gamma$ lies in the union), so $x \subseteq V_\lambda$. The induction is complete, and we conclude every $V_\beta$ is transitive. As a corollary — the actual content we need — pick any $x \in V_\beta$. Transitivity gives $x \subseteq V_\beta$, hence $x \in \mathcal{P}(V_\beta) = V_{\beta+1}$. Thus \begin{align*} V_\beta \subseteq V_{\beta+1}. \end{align*} This is where the successor rule $V_{\beta+1} = \mathcal{P}(V_\beta)$ "pays off": taking the power set at each successor stage automatically enlarges the hierarchy, because the transitivity of the previous stage converts elements to subsets. [/guided] [/step] [step:Run the transfinite induction on $\beta \geq \alpha$] Having established the one-step inclusion, we prove $V_\alpha \subseteq V_\beta$ for all ordinals $\beta \geq \alpha$ by transfinite induction on $\beta$. **Base case ($\beta = \alpha$).** The inclusion $V_\alpha \subseteq V_\alpha$ is immediate. **Successor step.** Assume $V_\alpha \subseteq V_\beta$ for some $\beta \geq \alpha$. By the one-step inclusion from Step 2, $V_\beta \subseteq V_{\beta + 1}$. Composing the two inclusions gives \begin{align*} V_\alpha \subseteq V_\beta \subseteq V_{\beta + 1}, \end{align*} so $V_\alpha \subseteq V_{\beta+1}$. **Limit step.** Let $\lambda$ be a limit ordinal with $\lambda > \alpha$, and assume $V_\alpha \subseteq V_\gamma$ for every ordinal $\gamma$ with $\alpha \leq \gamma < \lambda$. By the recursive definition, \begin{align*} V_\lambda = \bigcup_{\gamma < \lambda} V_\gamma. \end{align*} Pick any $\gamma_0$ with $\alpha \leq \gamma_0 < \lambda$ (such $\gamma_0$ exists: e.g., $\gamma_0 = \alpha$, since $\alpha < \lambda$ by hypothesis). Then $V_\alpha \subseteq V_{\gamma_0}$ by the inductive hypothesis, and $V_{\gamma_0} \subseteq V_\lambda$ because $V_{\gamma_0}$ is one of the sets in the union defining $V_\lambda$. Chaining: \begin{align*} V_\alpha \subseteq V_{\gamma_0} \subseteq V_\lambda. \end{align*} This completes the transfinite induction. For every ordinal $\beta \geq \alpha$, we have $V_\alpha \subseteq V_\beta$, proving the theorem. [guided] We now run the main induction, having the one-step inclusion $V_\beta \subseteq V_{\beta+1}$ in hand from Step 2. Fix $\alpha$ once and for all. We prove by transfinite induction on $\beta$ that \begin{align*} \beta \geq \alpha \implies V_\alpha \subseteq V_\beta. \end{align*} **Base ($\beta = \alpha$).** Reflexivity of inclusion gives $V_\alpha \subseteq V_\alpha$. **Successor.** Suppose the claim holds at $\beta$, i.e., $V_\alpha \subseteq V_\beta$. For the next stage $\beta + 1$: by Step 2, $V_\beta \subseteq V_{\beta+1}$. Chaining, \begin{align*} V_\alpha \subseteq V_\beta \subseteq V_{\beta+1}, \end{align*} so the claim also holds at $\beta + 1$. **Limit.** Suppose $\lambda > \alpha$ is a limit ordinal and the claim holds at every ordinal $\gamma$ with $\alpha \leq \gamma < \lambda$. We must show $V_\alpha \subseteq V_\lambda$. Where does $V_\lambda$ come from? By the recursive definition, \begin{align*} V_\lambda = \bigcup_{\gamma < \lambda} V_\gamma. \end{align*} So it suffices to exhibit some $\gamma < \lambda$ for which $V_\alpha \subseteq V_\gamma$, because then $V_\alpha \subseteq V_\gamma \subseteq V_\lambda$. The simplest choice is $\gamma = \alpha$ itself: we have $\alpha < \lambda$ (because $\lambda > \alpha$), and $V_\alpha \subseteq V_\alpha$ by reflexivity (no induction hypothesis needed for this case). Hence $V_\alpha \subseteq V_\alpha \subseteq V_\lambda$. Why did we need the hypothesis $\lambda > \alpha$? Without it, the case $\lambda = \alpha$ could be a limit, but then "$\beta \geq \alpha$" and "$\beta = \alpha$" coincide, and this is handled by the base case — we only need to run the successor/limit steps for $\beta > \alpha$. The induction is complete. For every $\beta \geq \alpha$, $V_\alpha \subseteq V_\beta$, which is the theorem. [/guided] [/step]