Proof. See [1].
See [1] for definition and properties of full stars and core sta rs. There it is also described a
bijection betwee n free stars on a boolean lattice and fi lters on that lattice (theorems 43, 44, and
45 in [1]).
3 Orde r of free stars
The set of free stars is ordered by the pa rtial order defined by the formula S ⊑ T ⇔ S ⊆ T .
Obvious 5. The minimal free sta r is ∅.
Proposition 6 . The maximal free star is A \ {0} (or A if 0 doesn’t exist).
Proof. Let’s denote our star S (that is S = A \ {0} or S = A depe ndently on existence of 0).
It is enough to prove that S is a free star.
That the least element is not in S is obvious. That S is an upper set is obvious.
The only thing remained to prove is
∀Z ∈ A: (Z ⊒ X ∧ Z ⊒ Y ⇒ Z ∈ S) ⇒ X ∈ S ∨ Y ∈ S.
If X or Y are non-least, it is obvious. Then only case r e mained to consider is X = Y = 0. In this
case our formula takes the form:
∀Z ∈ A: Z ∈ S ⇒ X ∈ S ∨ Y ∈ S what is true because ∀Z ∈ A: Z ∈ S is false (0
Z).
Proposition 7 . If S is a set of free stars on a join-semilattice, then
S
S is a free star.
Proof. Let S is a set of f ree stars on a join-semilatt ic e .
Obviously
S
S does not co ntain the least element (if it exists) and
S
S is an upper set.
It is remained to prove that X ⊔ Y ∈
S
S ⇒ X ∈
S
S ∨ Y ∈
S
S.
Let X ⊔ Y ∈
S
S. Then there e xists T ∈ S such that X ⊔ Y ∈ T . Hence X ∈ T ∨ Y ∈ T and thus
X ∈
S
S ∨ Y ∈
S
S.
Conjecture 8. This proposition cannot be strengthened for arbitrary posets instead of join-
semilattices.
Next two corollaries from the above proposition:
Corollary 9.
F
S =
S
S if S is a set of free stars on a join-semilattice.
Corollary 10. The set of free stars on a join-semilattice is a complete lattice.
4 Informal open problems
For which kinds of posets the corresponding poset of free stars is :
1. a distributive lattice?
2. a co-brouwe rian lattice?
3. an atomic poset?
4. an atomistic poset?
5. a separable poset? (see [1] for a definition)
6. an atomically separable poset? (see [1] for a definition)
Answers to these questions for the poset of all subsets of a fixed set are obv iously all true, due the
mentioned above bijective corres pondence with the set of filte rs on that poset and results in [1].
Can these results be significantly strengthened?
Bib liography
[1] Victor Porton. Filters on posets and generalizations. International Journal of Pure and Applied Mathe-
matics, 74(1):55–119, 2012.
2 Section