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Pseudodiﬀerence on atomistic co-brouwerian

lattices

Victor Porton

∗†

September 4, 2016

Abstract

I prove a conjecture about presenting pseudodiﬀerence of ﬁlters in

several equivalent forms from my earlier article, and more generally a

result for arbitrary atomistic co-brouwerian lattices.

A.M.S. subject classiﬁcation: 54A20, 06A06, 06B99

Keywords: ﬁlters, pseudodiﬀerence.

1 The problem

I will call the set of ﬁlter objects the set of ﬁlters ordered reverse to set theoretic

inclusion of ﬁlters, with principal ﬁlters equated to the corresponding sets. See

[1] for the formal deﬁnition of ﬁlter objects. I will denote (up a) the ﬁlter

corresponding to a ﬁlter object a. I will denote the set of ﬁlter objects (on U)

as F. So, a ⊆ b ⇔ up a ⊇ up b for a, b ∈ F.

F is actually a complete lattice (see [1]).

I will denote (atoms

A

a) the set of atoms below element a of a lattice A.

In [1] I’ve formulated the following open problem (problem 1):

Problem 1 Which of the following expressions are pairwise equal for all a, b ∈ F for

each lattice F of ﬁlters on a set U ? (If some are not equal, provide counter-examples.)

1.

T

F

z ∈ F | a ⊆ b ∪

F

z

(quasidiﬀerence of a and b);

2.

S

F

z ∈ F | z ⊆ a ∧ z ∩

F

b = ∅

(second quasidiﬀerence of a and b);

3.

S

F

(atoms

F

a \ atoms

F

b);

4.

S

F

a ∩

F

(U \ B) | B ∈ up b

.

∗

Email: porton@narod.ru

†

Web: http://www.mathematics21.org

1

2 A generalization and the proof

We will prove more general statements:

Theorem 1 For an atomistic co-brouwerian lattice A and a, b ∈ A the following

expressions are always equal:

1.

T

A

z ∈ A | a ⊆ b ∪

A

z

(quasidiﬀerence of a and b);

2.

S

A

z ∈ A | z ⊆ a ∧ z ∩

A

b = 0

(second quasidiﬀerence of a and b);

3.

S

A

(atoms

A

a \ atoms

A

b).

Proof Proof of (1)=(3):

a \

∗

b =

T

A

z ∈ A | a ⊆ b ∪

A

z

, so it’s enough to prove that

a \

∗

b =

A

[

(atoms

A

a \ atoms

A

b).

Really:

a \

∗

b =

A

[

atoms a

!

\

∗

b = (theorem 16 in [1])*

A

[

A \

∗

b | A ∈ atoms

A

a

=

A

[

A if A 6∈ atoms

A

b

0 if A ∈ atoms

A

b

| A ∈ atoms

A

a

=

A

[

A | A ∈ atoms

A

a, A 6∈ atoms

A

b

=

A

[

(atoms

A

a \ atoms

A

b).

* The requirement of theorem 16 that our lattice is complete is superﬂuous and

can be removed.

Proof of (2)=(3):

a \

∗

b is deﬁned because our lattice is co-brouwerian. Taking the above into

account, we have

a \

∗

b =

[

(atoms a \ atoms b) =

[

z ∈ atoms a | z ∩

A

b = 0

A

.

So

S

z ∈ atoms a | z ∩

A

b = 0

A

is deﬁned.

2

If z ⊆ a ∧ z ∩

F

b = 0

A

then z

0

=

S

x ∈ atoms z | x ∩

A

b = 0

A

is deﬁned.

z

0

is a lower bound for

z ∈ atoms a | z ∩

A

b = 0

A

.

Thus z

0

∈

z ∈ A | z ⊆ a ∧ z ∩

A

b = 0

A

and so

S

z ∈ atoms a | z ∩

A

b = 0

A

is an upper bound of

z ∈ A | z ⊆ a ∧ z ∩

A

b = 0

A

.

If y is above every z

0

∈

z ∈ A | z ⊆ a ∧ z ∩

A

b = 0

A

then y is above every

z ∈ atoms a such that z∩

A

b = 0

A

and thus y is above

S

z ∈ atoms a | z ∩

A

b = 0

A

.

Thus

S

z ∈ atoms a | z ∩

A

b = 0

A

is least upper bound of

z ∈ A | z ⊆ a ∧ z ∩

A

b = 0

A

,

that is

[

z ∈ A | z ⊆ a ∧ z ∩

A

b = 0

A

=

[

z ∈ atoms a | z ∩

A

b = 0

A

=

[

(atoms a\atoms b).

2

Note that F is co-brouwerian by corollary 11 in [1] and atomistic by theorem

48 in [1], so our theorem applies to the lattice F, and more generally to any

ﬁlters on a boolean lattice.

Proposition 1 For ﬁlters on boolean lattices the three above ways to express

quasidiﬀerence of a and b are also equal to

S

F

{a ∩

F

↑ B | B ∈ up b} (↑ X

denotes the principal ﬁlter induced by X).

Remark 1 By corollary 8 in [1] the set of ﬁlters on a boolean lattice is complete.

So our formula is well-deﬁned.

Proof Using results from [1]:

S

F

{z ∈ F | z ⊆ a ∧ z ∩ b = 0

F

} ⊆

S

F

{a ∩

F

↑ B | B ∈ up b} because

z ∈ {z ∈ F | z ⊆ a ∧ z ∩ b = 0

F

} ⇔ z ⊆ a ∧ z ∩ b = 0

F

⇔

z ⊆ a ∧ ∃B ∈ up b : z ∩ ↑ B = 0

F

⇔ z ⊆ a ∧ ∃B ∈ up b : z ⊆ ↑ B ⇔

∃B ∈ up b : (z ⊆ a ∧ z ⊆ ↑ B) ⇔ ∃B ∈ up b : z ⊆ a ∩

F

↑ B ⇒

z ⊆

F

[

{a ∩

F

↑ B | B ∈ up b}.

But obviously a ∩

F

↑ B ∈ {z ∈ F | z ⊆ a ∧ z ∩ b = 0

F

} and thus

a ∩

F

↑ B ⊆

F

[

{z ∈ F | z ⊆ a ∧ z ∩ b = 0

F

}

and so

S

F

{z ∈ F | z ⊆ a ∧ z ∩ b = 0

F

} ⊇

S

F

{a ∩

F

↑ B | B ∈ up b}. 2

The above proposition completes the proof of problem 1 in [1].

3

There is a little more general theorem in my unpublished book “Algebraic

General Topology. Volume 1” (available on the Web), currently at the end of

the section “Filtrators over Boolean Lattices”.

I present a part of this theorem here without a proof, as its (fairly technical,

not long however) proof is available in this my free e-book:

The below theorem uses terminology from [1].

Theorem 2 If (A; Z) is a complete co-brouwerian atomistic down-aligned lat-

tice ﬁltrator with binarily meet-closed and separable boolean core, then the three

expressions of pseudodiﬀerence of a and b in the above theorem are also equal

to

S

F

{a ∩

F

↑ B | B ∈ up b}.

References

[1] Victor Porton. Filters on posets and generalizations. International Jour-

nal of Pure and Applied Mathematics, 74(1):55–119, 2012. http://www.

mathematics21.org/binaries/filters.pdf.

4