Variation on partial Set Cover with penalties I'm looking at the following problem, which I'm hoping is perhaps a known variant of the Set Cover problem:

Input: We're given a universe $U$ and two disjoint subsets $A$ and $B$ such that $A \cup B=U$. We're also given a collection $C$ of $k$ sets $S_1,S_2,\ldots,S_k$ where $S_i \subseteq U$ for all $i=1\ldots k$. The sets $S_1,S_2,\ldots,S_k$ may or may not be disjoint, and the union of all $S_i$ may or may not equal the universe.

Solution form: A collection $C' \subseteq C$.

Objective function: Maximize $\left|\left(\bigcup\limits_{c \in C'} c\right) \cap A\right| - \left|\left(\bigcup\limits_{c \in C'} c\right) \cap B\right|$

In English, the objective function rewards solutions that cover a large number of elements in $A$ and penalizes those that cover a large number of elements in $B$. The best solution would be one that covers $A$ fully and does not cover $B$ at all. The worst solution covers $B$ fully and does not cover $A$ at all. The cardinality of $C'$ is not important.

Example: Universe $U=\left\{1,2,3,4,5\right\}$. Susbsets $A=\left\{1,3,4,5\right\}$ and $B=\left\{2\right\}$. Sets $S_1=\left\{1,2,3,5\right\}$, $S_2=\left\{1,4\right\}$ and $S_3=\left\{2,5\right\}$. Here the best solution is to take $S_1$ and $S_2$, which gives a measure value of $3$.

It's relatively trivial to adapt the classical greedy heuristic to get an approximate solution. My questions are 1) is this problem studied in any literature, and 2) do you think it would be possible to prove an approximation ratio for the greedy heuristic on it?

I had a look at A compendium of NP optimization problems and questions like this one, but haven't been able to find anything similar yet.
This answers question (2):

The greedy heuristic for Set Cover / Maximum Coverage always picks the set which contains the maximal number of uncovered elements.

Assuming your modification for the heuristic is picking the set which greedily increases the solution profit, you might end up in a lousy approximation ratio.

Consider the following example:
$$A = \{a_1,a_2,\ldots a_{k}\}, B=\{b_1,b_2,\ldots b_k\}$$
$$C = \{C_1\} \cup C_2$$
$$C_1= A\cup B \setminus\{b_1\}$$
$$ C_2 = \{\{a_{i},b_1\}|i\in[k]\}$$

The greedy heuristic will output $C'=\{C_1\}$, so the profit is $1$, while the optimal solution is $C_2$ which gives profit $k-1$.