Properties of the constructible closure of a set Given any set $A$ define


*

*$L_0(A)=\{A\}\cup \operatorname{tr cl}(A)$

*$L_{\alpha+1}(A)=\mathcal{D}^+(L_\alpha(A))$

*$L_\gamma (A)=\bigcup_{\alpha<\gamma}L_\alpha(A)$ if $\gamma$ is a limit ordinal.


where $\mathcal{D}^+(L_\alpha(A))$ denotes the subsets of $L_{\alpha}(A)$ defined with parameters in $L_{\alpha}(A)$ . 
I'm having struggles with the proofs of the following two facts:


*

*$L_\alpha\subset L_\alpha(A)$ for every $\alpha$

*$|L_\alpha(A)|=\max\{|\operatorname{tr cl}(A)|,|\alpha|\}$


Could somebody help me?
Thanks in advance.
 A: Claim: $L_\alpha \subseteq L_\alpha(A)$ for all $\alpha \in \operatorname{Ord}$.
Proof by induction on $\alpha$. For $\alpha = 0$ and limit ordinals, this is immediate. So let $\alpha = \beta +1$. We have $L_\alpha(A) \cap \operatorname{Ord} \ge L_\alpha \cap \operatorname{Ord} = \alpha$ and thus $\beta \in L_\alpha(A)$. This also yields $L_\beta \in L_\alpha(A)$. Now $L_\alpha$ can be charaterized as the closure of $L_\beta \cup \{L_\beta\}$ under Gödel Operations restricted to subsets of $L_\beta$ (see for example chapter 13 in Jech's 3rd millennium Set Theory). Since $L_\alpha(A)$ is closed under Gödel Operations restricted to subsets of $L_\alpha(A)$, it follows that $L_\alpha \subseteq L_\alpha(A)$. q.e.d.
Now consider $A = \{\emptyset\}$. Then $L_0(A) = \{A\} \cup \operatorname{trcl}(A) = \{ \{\emptyset\}, \emptyset\}$. Thus your 2nd claim fails in this case. However
Claim: $\mid L_\alpha(A) \mid \le \max \{ \mid \operatorname{trcl}(\{A\}) \mid , \mid \alpha \mid, \aleph_0 \}$.
Proof by induction on $\alpha$. Again if $\alpha = 0$ or $\alpha$ is a limit, this is immediate. Let $\alpha = \beta +1$. Then $\mid L_\alpha(A) \mid \le \mid L_\beta(A) \mid \ \cdot \aleph_0 \le \max \{ \mid \operatorname{trcl}(\{A\}) \mid , \mid \beta \mid, \aleph_0 \} = \max \{ \mid \operatorname{trcl}(\{A\}) \mid , \mid \alpha \mid, \aleph_0 \}$. q.e.d.
