Answer by Maxime Ramzi for Prove that $|V_\alpha|=|\operatorname{P}(\alpha)|$...
Suppose $\kappa$ is a cardinal and $\beth_\kappa >\kappa$. Then by definition of $\beth_\kappa$ (since $\kappa$ is a limit ordinal), $\beth_\gamma >\kappa$ for some $\gamma < \kappa$. This...
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$\kappa$ is a cardinal, $V_\alpha$ belongs to the Von Neumann hierarchy $\begin{cases} V_0=\emptyset \\ V_{\alpha+1}=P(V_\alpha) \\ V_\lambda=\underset{\gamma<\lambda}{\bigcup}V_\gamma \end{cases}$...
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