Alternative proof that $|V_\alpha| \leq \beth_\alpha$: limit case Of course, the general result is that
$$\forall \alpha, |V_{\omega + \alpha}| = \beth_\alpha.$$
I wonder if it's possible to prove that
$$\forall \alpha, |V_\alpha| \leq \beth_\alpha$$
without using the addition of ordinals numbers.
The proof, of course, should go by induction as follows:


*

*$V_0 = 0$ and $\beth_0 = \aleph_0$, of course, $0 < \aleph_0$

*If $|V_\alpha| \leq \beth_\alpha$, then $|V_{\alpha + 1}| = |\mathcal{P}(V_\alpha)| = 2^{|V_\alpha|} \leq 2^{\beth_\alpha} = \beth_{\alpha + 1}$.

*Now, here comes the problem. Assume that $\lambda$ is limit, and that $\forall \alpha < \lambda, |V_\alpha| \leq \beth_\alpha.$ 
$V_\lambda = \bigcup_{\alpha < \lambda} V_\alpha$ and $\beth_\lambda = \bigcup_{\alpha < \lambda} \beth_\alpha$.
However, I don't see the way to prove it. There is probably a fundamental set theoretic piece I'm missing. Of course, it's not true in general that 
$|X| = \bigcup_{x \in X} |x|$.
Counterexample: $X = \{\mathbb{R},\mathbb{N}\}$. 
 A: There is a general theorem that states the following thing : Let $\kappa$ be an infinite cardinal. if $X_\alpha, \alpha<\lambda$ are sets  such that $|X_\alpha| < \kappa$ for all $\alpha$ and $\beta<\alpha \implies X_\beta \subset X_\alpha$, then $|\displaystyle\bigcup_{\alpha<\lambda} X_\alpha| \leq \kappa$.
Assuming we can prove that, we can apply this to $X_\alpha = V_\alpha$, $\kappa = \beth_\lambda$,
 ($|V_\alpha|\leq \beth_\alpha < \beth_\lambda$ since $\lambda$ is limit), and get the desired conclusion.
Now to prove this, we consider for $\xi <\lambda, Y_\xi = X_\xi \setminus \displaystyle\bigcup_{\alpha<\xi}X_\alpha$.
Well order each $Y_\xi$ and consider the well-ordered sum $\displaystyle\sum_{\xi<\lambda} Y_\xi$, isomorphic to the ordinal $\beta$, which is obviously equipotent with $X=\displaystyle\bigcup_{\alpha<\lambda}X_\alpha$.
Now consider $\beta_\xi$ the ordinal isomorphic to $\displaystyle\sum_{\alpha\leq\xi} Y_\alpha$ (which is equipotent to $X_\xi$). Clearly, $\beta = \sup_{\xi<\lambda} \beta_\xi$. Since each $\beta_\xi$ is equipotent to $X_\xi$ and is therefore smaller than $\kappa$, we have $\beta \leq \kappa$, and so $|X|\leq \kappa$, which is what we wanted.
