Is L[A,B] = L[C] for some set C? Suppose that instead of talking about $L[A]$, where we add one predicate to the constructible hierarchy, we add two, so we now have $L[A,B]$.


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*Is there always a set $C$ such that $L[A,B] = L[C]$?

*If there is such a set $C$, can we characterize its relation to $A$ and $B$ somehow?

*Assuming a positive answer to Q1, is this true for infinitely many predicates? And does the characterization from Q2 generalize?

 A: Unless I'm missing something, $(A\times\{1\})\sqcup (B\times\{0\})$ does the job. In general, if you have sets $A_1, ..., A_n$ just pick distinct constructible sets $a_1, ..., a_n$ and look at $\bigsqcup (A_i\times\{a_i\})$.
It's important to note that the choice of tuple $a_1,..., a_n$ doesn't matter since $n$ is finite. When we have infinitely many objects, this becomes problematic since there need not be any natural way to index our predicates. Indeed, you will not get a model of ZFC in general this way! For example, $L(\mathbb{R})$ can be thought of as the infinite extension $L[r: r\in\mathbb{R}]$.

EDIT: As Andres points out below, what I've written above may be very misleading, so let me elaborate:
Suppose I have a set $A$, which I'm really thinking of as a set of things I want to adjoin to $L$ individually - in the $L[-]$ sense, not the $L(-)$ sense. I'm going to define a transitive class $L[a: a\in A]$ as follows:


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*$L[a: a\in A]$ is $\bigcup_{\alpha\in Ord} M_\alpha$.

*$M_0=\emptyset$.

*$M_\lambda=\bigcup_{\alpha<\lambda}M_\alpha$ for $\lambda$ limit.

*$M_{\alpha+1}$ is the set of all subsets of $M_\alpha$ which are definable in some structure of the form $(M_\alpha; \in, p_1, ..., p_n)$, where each $p_i$ is a predicate corresponding to some $a_i\in A$. 
Note that I'm only using finitely many parameters from $A$ when I create a new set, and I'm never (explicitly) throwing in $A$ itself.
Alright, so what can we say about $L[a: a\in A]$?
I think the key observation is that in general we will not have $A\in L[a: a\in A]$. E.g. if $A\subset L$ then $L[a: a\in A]=L$. So this is a vastly weaker construction than it may appear at first. On the other hand, sometimes we will have $A\in L[a: a\in A]$, and $A=\mathbb{R}$ (or better yet, $A=2^\omega$) is one of those times, the point being that $A$ itself is definable in an appropriate way. 
The proof that $L[X]$ satisfies choice breaks down of course in this context, and indeed these are choiceless models in general: this is clear once we notice $L(\mathbb{R})=L[a: a\in \mathbb{R}]$. On the other hand, $L[a: a\in A]$ is always a model of ZF, by the usual bounding cheats:


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*Given $x\in L[a: a\in A]$, let $x\in M_\alpha$. For each $\beta>\alpha$ the set $\mathcal{P}(x)[\beta]=\{y\in M_\beta: y\subseteq x\}$ is in $M_{\beta+1}$. The sequence $(\mathcal{P}(x)[\beta])_{\beta\in Ord}$ is increasing, hence eventually stabilizes since $x$ has fewer-than-$Ord$-many subsets, and this eventual value is $\mathcal{P}(x)^{L[a: a\in A]}$.

*Replacement is a bit more complex, since whether or not $\varphi(x, y)$ holds can alternate as we move up the hierarchy. However, by a straightforward induction argument, fixing $\varphi, x, y$ the set of $\alpha$ such that $M_\alpha\models\varphi(x, y)\iff L[a: a\in A]\models\varphi(x, y)$ contains a (class) cub. Given $z\in L[a: a\in A]$ and $\varphi$ (with parameters suppressed) satisfying the hypotheses of replacement, let $\alpha$ be large enough that $z\in M_\alpha$ and for every $x\in z$ there is some $y\in M_\alpha$ with $L[a: a\in A]\models \varphi(x, y)$; then since the intersection of set-many class clubs is a class club, there is some $\beta>\alpha$ such that $\varphi(x, y)^{M_\beta}\iff\varphi(x, y)^{L[a: a\in A]}$ for $x\in z, y\in M_\alpha$. The desired solution to replacement is then in $M_\beta$.
So nothing too awful happens. Now, in general this sort of construction can only be really interesting if the elements of $A$ are in general not even subsets of $L$, since otherwise it's subsumed by the $L(-)$ construction, but it does seem mildly neat and it certainly is useful in this case in terms of building intuition.
