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We define the von Neumann Hierarchy ($V_\alpha$)$_{\alpha \in Ord}$ by recursion:

$a)$ $V_0=\emptyset$

$b)$ $V_\alpha$$_+$$_1$$=$$\mathcal P(V_\alpha)$

$c)$ $V_\gamma=$$\bigcup\limits_{\alpha<\gamma} V_{\alpha}$.

I was wondering how to prove that in that case, the next holds:

$\forall x$$\subset\bigcup\limits_{\alpha \in Ord} V_{\alpha}$ $\exists \beta$ $x \subset V_\beta$.

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  • $\begingroup$ Assume the contrary and use the rank function to define a map from a power set of a `counterexample' to the class of ordinals. You will have a cofinal subclass of the class of ordinals, a contradiction. $\endgroup$ – Hanul Jeon Feb 18 '17 at 10:40
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Replacement, replacement, replacement.

If $x$ is a subset of $V$, then by definition, for every $y\in x$ there is some $\alpha$ such that $y\in V_\alpha$. Now consider the least such $\alpha$. This defines a function, apply Replacement and the fact that every set of ordinals is bounded from above.

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