Find all pairs of elements such that they are not covered by any given sets Given a universe $U$ and a family sets $F$ over $U$, I wanna find all pairs $(x, y)$ where $x, y \in U$, such that $x$ and $y$ are not in any set of $F$ at the same time, i.e., $\not\exists S \in F, x\in S \wedge y\in S$.
Also, I try to avoid compute $U \times U$, because this is an abstract math problem from other scenarios. If I use $U \times U$, that costs too much.
What's formal name of this problem?
Is it possible to find all such pairs systematically using Cartesian product of two sets with set algebra?
For example, for the two sets ($|F|=2$) case, let's say $F=\{A, B\}$, we can find all such pairs by
$(A \setminus B) \times (B \setminus A)$ and $(U \setminus (A \cup B)) \times (A \cup B)$
For $|F|=3$, using Venn diagram can give the following possible solution, which basically to find every pair of pieces which are not covered by any single set.
Venn diagram
$(A \setminus (B \cup C)) \times (B \setminus (A \cup C))$
$\cup$
$(A \setminus (B \cup C)) \times ((B \cap C) \setminus A)$
$\cup$
$(B \setminus (A \cup C)) \times (C \setminus (A \cup B))$
$\cup$
$(B \setminus (A \cup C)) \times ((A \cap C) \setminus B)$
$\cup$
$(C \setminus (A \cup B)) \times (A \setminus (B \cup C))$
$\cup$
$(C \setminus (A \cup B)) \times ((A \cap B) \setminus C)$
$\cup$
$(U \setminus (A \cup B \cup C)) \times (A \cup B \cup C)$
However, there is another better solution with much less set operations:
$A \setminus (B \cup C) \times (B \setminus A)$
$\cup$
$C \setminus (A \cup B) \times (A \setminus C)$
$\cup$
$B \setminus (A \cup C) \times (C \setminus B)$
$\cup$
$(U \setminus (A \cup B \cup C)) \times (A \cup B \cup C)$
For more general $|F| = k$ cases, that's complicated.

$$(U \times U) \setminus (\bigcup_{f \in F} f \times f)$$
EDIT: the idea is to consider all pairs of points and then for each set in $F$, subtract out the pairs of points such that both belong in it.
