Proof. Take
f =
(X ; Y) | X ∈ F(A), Y ∈ F(B),
\
X and
\
Y are infinite
and
g = f ∪ {(X ; Y) | X ∈ F(A), Y ∈ F(B), X ⊒ a, Y ⊒ b}
where a and b are nontrivial ultrafilters on A and B correspo ndingly, c is the funcoid defined by
the relation
[c]
∗
= δ = {(X; Y ) | X ∈ PA, Y ∈ PB , X and Y are infinite}.
First prove that f is a pseudofuncoid. The formulas ¬(I f 0) and ¬(0 f I) are obvious. We have
I ⊔ J f K ⇔
T
(I ⊔ J ) and
T
Y are infinite⇔
T
I ∪
T
J and
T
Y are infinite⇔(
T
I or
T
J is
infinite) ∧
T
Y is infinite⇔(
T
I and
T
Y are infinite)∨ (
T
J and
T
Y are infinite)⇔ I f K ∨ J f K.
Similarly K f I ⊔ J ⇔ K f I ∨ K f J . So f is a pseudofuncoid.
Let now prove that g is a pseudofuncoid. The formulas ¬(I g 0) and ¬(0 g I) are obvious. Let
I ⊔ J g K. Then either I ⊔ J f K and then I ⊔ J g K or I ⊔ J ⊒ a and then I ⊒ a ∨ J ⊒ a thus
having I g K ∨ J g K. So I ⊔ J g K ⇒ I g K ∨ J g K. The reverse implication is obvious. We have
I ⊔ J g K ⇔ I g K ∨ J g K and similarly K g I ⊔ J ⇔ K g I ∨ K g J . So g is a pseudofuncoid.
Obviously f
g (a g b but not a f b).
It remains to prove f ∩ (P × P) = g ∩ (P × P) = [c]∩(P × P). Really, f ∩ (P × P) = [c]∩(P × P)
is obvious. If (↑
A
X; ↑
B
Y ) ∈ g ∩ (P × P) then either (↑
A
X; ↑
B
Y ) ∈ f ∩ (P × P) or X ∈ up a, Y ∈ up b,
so X and Y are infinite and thus (↑
A
X; ↑
B
Y ) ∈ f ∩ (P × P). So g ∩ (P × P) = f ∩ (P × P).
Remark 8. The above c ounter-example shows that pseudofuncoids (and more generally, any
staroids on filters) are “second class” objects, they are not full-fledged because they don’t bijectively
correspond to funcoids and the ele gant funcoids theory does not apply to them.
From the above it follows that staroids on filters do not correspond (by restriction) to staroids
on principal filters (or staroids on se ts).
4 Complete staroids and multifuncoids
4.1 Complete free stars
Definition 9. Let A be a poset. Complete free sta rs on A are s uch S ∈ P A that the least element
(if it e xists) is not in S and for every T ∈ PA
∀Z ∈ A: (∀X ∈ T : Z ⊒ X ⇒ Z ∈ S) ⇔ T ∩ S
∅.
Obvious 10. Every complete free star is a free star.
Proposition 11. S ∈ PA where A is a poset is a complete free star iff all the following:
1. The least element (if it exists) is not in S.
2. ∀Z ∈ A: (∀X ∈ T : Z ⊒ X ⇒ Z ∈ S) ⇒ T ∩ S
∅.
3. S is an upper set.
Proof.
⇒. (1) and (2) are obvious. S is an upper s et because S is a free star.
⇐. We need to prove that
∀Z ∈ A: (∀X ∈ T : Z ⊒ X ⇒ Z ∈ S) ⇐ T ∩ S
∅.
Let X
′
∈ T ∩ S. Then ∀X ∈ T : Z ⊒ X ⇒ Z ⊒ X
′
⇒ Z ∈ S because S is an upper set.
Proposition 12. Let S be a complete lattice. S ∈ P A is a complete fre e star iff all the following:
1. The least element (if it exists) is not in S.
2 Section 4