# Need a name for a common set operation

I have a series of algorithms which perform set Boolean operations converging to a certain type of partitioning of a given set of sets. To present these algorithms in a human readable form, it would be great if I could name a particular operation which is very common in my algorithms. Can someone tell me whether this operation has a name which I can use to simply my explanations?

It is the following: Given non-empty sets $$A$$ and $$B$$, which are subsets G, the set $$\{A\cap B, ~A\setminus B, ~B\setminus A\}\setminus$$ {$$\emptyset$$}. This set sometimes has 3 elements. But if $$A$$ and $$B$$ are equal, it has one element, and if $$A \subset B$$, it has two elements, because $$B\setminus A = \emptyset$$.

I'd like to give a name to this operation which takes two sets and returns a set of these 1, 2, or 3 elements. And I'd love to have a notation for it, as it occurs in several formulas and algorithms in my paper. I have considered "the simple closure", or the "trivial partition", or the "standard partition".

I welcome anyone's suggest, especially if there is already a standard name for the operation.

• I don't know of a name for this particular operation, nor do I see any reason to give this very specific operation a name. – 5xum Sep 27 '18 at 8:36
• I don't think there is a name for the operation, but you can call it "Jimmification of $A$ and $B$" if you'd like. Unfortunately, I don't think any attempt of naming this operation will be a hit. This operation is not commonly used as far as I am concerned. – Batominovski Sep 27 '18 at 9:19
• Batominovski, the reason that I'd like a notation, is simply because this term seems to appear often in equations with other things and thus obfuscates the "other" operations. Grouping it together into a single term, may make the equation easier to read. Of course there's always a risk in introducing more notation. If there were already a "standard" notation, I wouldn't have to apologize to the audience for introducing a new one. – Jim Newton Sep 27 '18 at 9:27

Yes it is the standard partition of $$A\bigcup B$$ into 3 disjoint sets but with the empty set omitted . Throwing out the empty set is new to me .
• Hi Stuart, you make a good point about $\emptyset$. I debated early on whether to always include the empty set, but it turns out easier for my application to omit it. The application is in type theory of dynamic programming languages. breaking type-based conditionals into non-intersecting branches of the code. I never break up these sets into the empty type because that would result in unreachable code... so the partitioning function always eliminates $\emptyset$. Your point is well taken however: the math might have been easier notation-wise if I didn't tack on this restriction. – Jim Newton Sep 27 '18 at 9:24