Prove that $\forall\alpha\geq\omega$, $|L_\alpha|=|\alpha|$ without AC Without using Axiom of Choice, prove that
$$\forall\alpha\geq\omega,~~|L_\alpha|=|\alpha|,$$
in which $\alpha$ is an ordinals, $\omega$ is the set of natural numbers, $L_\alpha$ is the $\alpha$-th constructiable set, $|x|$ is the cardinals of $x$.
I am learning a set theory, and I have read a proof by using AC. The AC is used only when $\alpha$ is a limit ordinals during the transfinite inductive method for $\alpha$ . I wonder if there is a proof without AC. 　THANKS!
 A: First you can notice that for a limit ordinal $\delta$, $L_\delta\models V=L$ and in particular $L_\delta$ has a definable well-ordering of type $\delta$.
So the use of the axiom of choice can be removed and instead you can use the definable well-orderings for everything instead.
Then the proof is quite similar. First note that $L_\omega$ is countable; then proceed by induction. $L_{\alpha+1}$ has a bijection with $(L_\alpha)^{<\omega}\times\omega$ (by encoding formulas and parameters) and using the induction hypothesis $|\alpha|=|L_\alpha|=|L_{\alpha+1}|=|\alpha+1|$. When arriving to limit ordinals, use the fact that there is a definable well-ordering to conclude that $L_\delta$ is the union of uniformly well-ordered sets and conclude $|L_\delta|=|\delta|\cdot\sup\{|L_\alpha|\mid\alpha<\delta\}=|\delta|\cdot|\delta|=|\delta|$.
(All the cardinal arithmetic is true because it holds for ordinals without using choice.)
Note, by the way, that Gödel proved that the axiom of choice is at all consistent with the rest of the axioms of set theory by proving that it holds in $L$. So all the arguments about the construction of $L$ shouldn't, morally, appeal to the axiom of choice. And they don't.
