Stages in the constructible universe I want to proof for (countable) $\alpha$ that $|L_\alpha|=|\alpha|$.
Since I can only define countably many things from it in one step, I thought I'd do induction over $\alpha$, but how do I find a bijection between $L_{\alpha+1}$ and $L_{\alpha}$.
Furthermore, $|L_{\omega_1}|=|\omega_1|$ would just follow directly from the definition of $\omega_1$?
 A: For the successor case, there's no need to explicitly whip up a bijection - although you can do so if you want. Remember that $L_{\alpha+1}$ is the set of definable-with-parameters subsets of $L_\alpha$. This means that the cardinality of $L_{\alpha+1}$ is at most the product of the number of finite tuples of parameters from $L_\alpha$ and the number of formulas in the language of set theory. The latter is always $\aleph_0$, and the former is always $\vert L_\alpha\vert$ (restricting attention to $\alpha\ge\omega$ for simplicity here). This gives $$\vert L_\alpha\vert\le \vert L_{\alpha+1}\vert\le\max\{\aleph_0, \vert L_\alpha\vert\}$$ (the first inequality is trivial and the second inequality follows from the definition of cardinal multiplication once infinite sets are involved), which is enough to show that $L_{\alpha+1}$ has the same cardinality as $L_\alpha$ whenever $\alpha$ is infinite.
Meanwhile, if $\lambda$ is a limit ordinal, then $$L_\lambda=\bigcup_{\alpha<\lambda}L_\alpha$$ by definition and so, picking a cofinal sequence $(\gamma_i)_{i<cf(\lambda)}$ in $\lambda$, we get $$\vert L_\lambda\vert\le\sum_{i<cf(\lambda)}\vert L_{\gamma_i}\vert.$$ If $\lambda$ is countable, then since the countable union of countable sets is countable we get $\vert L_\lambda\vert\le\aleph_0$ inductively; if $\lambda=\omega_1$, we get the upper bound $\vert L_{\omega_1}\vert\le\aleph_1$, which is optimal since $\omega_1\subset L_{\omega_1}$.
