How to know if an intersection contains another intersection?

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If I and J are two sets formed by intersections, how to know if I contains J, knowing that:

  • I = i1 i2 ... in
  • J = j1 j2 ... jn
  • nor I nor J are empty, therefore nox or jx the saint

I know the answer is ternary, it can be True, False or Undefined... As a matter of fact, I already know part of the answer, but since I am a single limited mind, Maybe there’s still some case I’m not able to visualize. Maybe someone with a more solid mathematical foundation can put an end to this.

Edition 1: Available operations

Sets are not enumerable. It is possible to use operators between sets. In my interface I have defined Contains(other), Intersects(other) and IsEmpty(), each operation can return a third value in addition to Yes and Not, that is Undefined... but for the tests carried out between the sets ix and jx we can assume a boolean response, because the undefined value would simply be propagated. There are also operations that return sets Intersection(other) and Union(other).

In short:

  • it is not possible to list
  • there are test operations each returning Yes, Not and Undefined:
    • Contains(other)
    • Intersects(other)
    • IsEmpty()
  • there are operations that return other sets
    • Intersection(other)
    • Union(other)
  • What is known about the sets ix and jx, And what is testable about these sets? For example, if all sets were finite and enumerable, it would simply be a matter of seeing element by element... On the other hand, if this is not possible, but you can establish predicates in relation to the elements (e.g..: i2 contains j4, yes or no? ), what are these predicates, and to what elements do they apply? I can for example test whether ix intersects with jx, but at the same time I can’t test whether ix ∩ iy intersects with jz? What are the applicable restrictions?

  • @mgibsonbr I updated the question with additional information on possible operations. About testing whetherix ∩ iy intersects with jz, is possible... will depend on the instance returned by the operation ix ∩ iy.

1 answer

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Here’s what I believe is the answer, or at least part of it:

I contains J for sure

One of the ways to make sure I contains J is below (maybe there are more conditions, I’m not sure)

  • all in contain at least one of the jn

    Note that the first intersection contains the second intersection in all cases listed below.

    1. example with 3 sets forming I and 3 forming J

      exemplo com 3 conjuntos em I e 3 em J => resultado

    2. example with 3 sets forming I and 2 forming J

      inserir a descrição da imagem aqui => inserir a descrição da imagem aqui

    3. example with 2 sets forming I and 3 forming J

      inserir a descrição da imagem aqui => inserir a descrição da imagem aqui

  • it’s not enough that all the jn are contained in one of then. See a counter-example:

    contra-exemplo => resultado

    How is it possible to see all the jn are contained in any in, meanwhile I does not contain J

  • the negative does not guarantee that I does not contain J. See a counter-example:

    contra-exemplo da negativa => resultado do contra-exemplo

    As it is possible to see, not all in contains some jn, meanwhile Icontains J even so. In the case presented above

I does not contain J for sure

I only know one way to ensure that I does not contain J:

  • any in no intersection with any jn

    If any ix is disjoint from any jk, as I is a subset of that ix and J is a subset of that jk, then I and J are also disjoint.

    disjuntos

    In this example the sets outlined in red are disjoint, and therefore the yellow intersection and the blue intersection are also disjoint.

  • The negative does not guarantee that I contains J:

    inserir a descrição da imagem aqui

    Although there is an intersection between all i and all j, I does not contain J... I intercept J, but does not. The yellow intersection does not contain the blue intersection.

Undefined

The cases that cannot be certain are indefinite. It is possible that there are more cases where it is possible to be sure, both positive and negative. But I have no way to prove this mathematically.

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