Algebraic General Topology. Vol 1: Paperback / E-book || Axiomatic Theory of Formulas: Paperback / E-book 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
 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 ).
I will denote (atoms
A
a) the set of atoms below element a of a lattice A.
In  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 )*
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  and atomistic by theorem
48 in , 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  the set of ﬁlters on a boolean lattice is complete.
So our formula is well-deﬁned.
Proof Using results from :
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 .
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 .
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
 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 