Partitioning of a lattice element: a conjecture

Let \mathfrak{A} is a complete lattice. Let a\in\mathfrak{A}.

I will call weak partitioning of a a set S\in\mathscr{P}\mathfrak{A}\setminus\{0\} such that

\bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall x\in S: x\cap^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}(S\setminus\{x\}) = 0.

I will call strong partitioning of a a set S\in\mathscr{P}\mathfrak{A}\setminus\{0\} such that

\bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall A,B\in\mathscr{P}S: (A\cap B=0\Rightarrow \bigcup{}^{\mathfrak{A}}A\cap{}^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}B = 0).

Question: Do exist complete lattices for which weak partitioning and strong partitioning are not the same?

If this conjecture (that it is the same for arbitrary complete lattices) is indeed false for arbitrary complete lattices, we must find the cases when it is true. (I strongly suspect that it is true for distributive complete lattices.)

3 comments

  1. Pingback: Proposal: Partitioning a lattice element « Victor Porton's Math Blog
  2. Pingback: Complete lattice generated by a partitioning of a lattice element « Victor Porton's Math Blog
  3. Pingback: Complete lattice generated by a partitioning – finite meets « Victor Porton's Math Blog

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