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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.
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