Question: Given a set $X$ and subsets $S_1, S_2 \subseteq X$, what is $S_1 \cap S_2$ called when $S_1 \cup S_2 = X$?
More generally, given a set $X$ and subsets $\{S_{i}\}_{i \in \mathcal{I}}$, with $S_{i_1} \cup S_{i_2} = X$ whenever $i_1 \not= i_2$, what is $\bigcap_{i \in \mathcal{I}} S_i$ called?
If there is no general terminology specific to power sets, then what about for general Boolean or bounded lattices?
Note: To be accepted, answers should ideally include at least one reference using the term. (Multiple references, or a standard textbook, would be fantastic however.) I've thought of possible ad hoc names myself (e.g. "codisjoint intersection" or "(pairwise) exhaustive intersection"), but I am asking here about whether standard terminology exists.
Context: Note that $S_1 \cup S_2 = X$ if and only if $S_1^c \cap S_2^c = \emptyset$ and that $(S_1 \cap S_2)^c = S_1^c \cup S_2^c$. So basically I am asking for the name of an intersection when the complement of the intersection is a disjoint union (of the complements).
From an order-theoretic perspective, $^c$ takes the original poset-lattice of subset inclusion of $X$ to its dual. So this notion is (order-theoretically) dual to the notion of disjoint union. Using typical notation for a bounded lattice, a disjoint union corresponds to the join $a_1 \lor a_2$ of two elements whose meet is the least element, $a_1 \land a_2 = 0$, whereas I am asking for the name of the dual notion, the meet $b_1 \land b_2$ of two elements whose join is the greatest element, $b_1 \lor b_2 = 1$.
I feel as if I've seen this notion used implicitly in the literature before, but it's pretty simple, so I don't recall it being given a name in the sources I've seen.
Possibly related questions:
Why is it obvious that if the intersection of all sets is empty then its complements cover the whole space?
Technically this about jointly exhaustive sets rather than pairwise exhaustive sets, plus it's not asking about terminology.
Name for collection of sets whose intersection is empty but where sets are not necessarily pairwise disjoint
Again this answer is talking about the jointly exhaustive case, rather than pairwise exhaustive, plus it's not asking about terminology either.
