Algebraic General Topology. Vol 1: Paperback / E-book || Axiomatic Theory of Formulas: Paperback / E-book

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Filters on posets and generalizations
Victor Porton
78640, Shay Agnon 32-29, Ashkelon, Israel
Abstract
They are studied in details properties of filters on lattices, filters on posets, and
certain generalizations thereof. Also it’s done some more genera l lattice theory
resear ch. There are posed several open problems. Detailed study of filters is
required for my ongoing research which will be published as Algebraic General
Topology” series.
Keywords: filters, ideals, lattice of filters, pseudodifference,
pseudocomplement
A.M.S. subject classification: 54A20, 06A06, 06B99
This is a pre print (incorporating errata) of an article [12] published in In-
ternational Journal of Pure and Applied Mathema tics (IJPAM).
Contents
1 Preface 3
2 Notation and basic results 3
2.1 Intersecting and joining elements . . . . . . . . . . . . . . . . . . 4
2.2 Atoms of a poset . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.3 Difference and complement . . . . . . . . . . . . . . . . . . . . . 6
2.4 Center of a la ttice . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.5 Galois connections . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.6 Co-Bro uwerian Lattices . . . . . . . . . . . . . . . . . . . . . . . 11
3 Straight maps and separation subsets 14
3.1 Straight maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2 Separation subsets and full sta rs . . . . . . . . . . . . . . . . . . 16
3.3 Atomically separable lattices . . . . . . . . . . . . . . . . . . . . 17
Email address: porton@narod.ru (Victor Porton)
URL: http://www.mathematics21.org (Victor Porton)
Preprint submitted to Elsevier January 14, 2014
4 Filtrators 19
4.1 Core part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.2 Filtrators with separable core . . . . . . . . . . . . . . . . . . . . 21
4.3 Intersecting and joining with an element o f the core . . . . . . . 22
5 Filters 22
5.1 Filters on posets . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5.2 Filters on meet-semilattice . . . . . . . . . . . . . . . . . . . . . . 23
5.3 Character iz ation of finitely mee t-closed filtrato rs . . . . . . . . . 24
6 Filter objects 24
6.1 Definition of filter objects . . . . . . . . . . . . . . . . . . . . . . 25
6.2 Order of filter objects . . . . . . . . . . . . . . . . . . . . . . . . 25
7 Lattice of filter objects 26
7.1 Minimal and maximal f.o. . . . . . . . . . . . . . . . . . . . . . . 26
7.2 Primary filtrator is filtered . . . . . . . . . . . . . . . . . . . . . . 26
7.3 Formulas for meets and joins of filter objects . . . . . . . . . . . 26
7.4 Distributivity of the la ttice of filter objects . . . . . . . . . . . . 28
7.5 Separability of core for primary filtrators . . . . . . . . . . . . . . 29
7.6 Filters over boolean lattices . . . . . . . . . . . . . . . . . . . . . 30
7.7 Distributivity for an element of boolean co re . . . . . . . . . . . 30
8 Generalized filter base 31
9 Stars 32
9.1 Free stars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
9.2 Stars of elements of filtra tors . . . . . . . . . . . . . . . . . . . . 33
9.3 Stars of filters on boolea n lattices . . . . . . . . . . . . . . . . . . 34
9.4 More about the lattice of filters . . . . . . . . . . . . . . . . . . . 36
10 Atomic filter objects 37
10.1 Prime filtrator elements . . . . . . . . . . . . . . . . . . . . . . . 38
11 Some criteria 39
12 Quasidifference and quasicomp lement 43
13 Complements and core parts 46
13.1 Core part and atomic elements . . . . . . . . . . . . . . . . . . . 48
14 Distributivity of core part over lattice operations 48
15 Fr´echet filter 50
16 Complementive fil ter objects and factoring by a filter 51
17 Number of filters on a s et 53
2
18 Partitioning filter objects 53
19 Open problems 54
19.1 Partitioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
19.2 Quasidifference . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
19.3 Non-formal problems . . . . . . . . . . . . . . . . . . . . . . . . . 56
Appendix A Some counter-examples 56
Appendix A.1 Weak and strong partition . . . . . . . . . . . . . . 57
Appendix B Logic of Generalizations 59
Appendix B.1 The formalistic . . . . . . . . . . . . . . . . . . . . 59
1. Preface
This article is intended to collect in one document the known properties of
filters on posets (and so me generalizations thereof, namely filtrators” defined
below).
It seems that until now were published no reference on the theory of filters.
This text is to fill the gap.
This text will also serve as the reference base for my further articles. This
text provides a definitive place to refer as to the collection of theorems about
filters.
Detailed study of filters is required for my ongoing research w hich will be
published as ”Algebraic General Topology” series.
In place of studying filters in this article are instea d researched what the au-
thor calls “filter objects”. Filter objects a re basically the lattice of filters orde red
reverse to set inclusion, with principal filters equated with the poset element
which generates them. (See be low for formal definition of “filter objects”.)
Although our primary interest are properties of filters on a set, in this work
are instead re searched the mor e gene ral theory of “filtra tors” (see below).
This article also contains some original research:
filtrators ;
straight maps and separation subsets;
other minor results, such as the theory of free stars.
2. Notation and basic results
We denote PS the set of all subsets of a set S.
hfi X
def
= {f x | x X} for any set X and function f.
3
2.1. Intersecting and joining element s
Let A be a poset.
Definition 1 I will call elements a and b of A intersecting and denote a 6≍ b
when exists not least element c such that c a c b.
Definition 2 a b
def
= ¬(a 6≍ b).
Obvious 1 If A is a meet-semilattice then a 6≍ b iff a b is non-least.
Obvious 2 a
0
6≍ b
0
a
1
a
0
b
1
b
0
a
1
6≍ b
1
.
Definition 3 I will call elements a and b of A joining and denote a b when
not exists not greatest element c such that c a c b.
Definition 4 a 6≡ b
def
= ¬(a b).
Obvious 3 Intersecting is the dual of non- joining.
Obvious 4 If A is a join-semilattice then a b iff a b is its greatest element.
Obvious 5 a
0
b
0
a
1
a
0
b
1
b
0
a
1
b
1
.
2.2. Atoms of a poset
Definition 5 An atom of the poset is an element which has no non-least subele-
ments.
Remark 1 This definition is valid even for posets without lea st element.
I will denote (atoms
A
a) or just (a toms a) the set of atoms contained in
element a of a poset A.
Definition 6 A poset A is called atomic when atoms a 6= for every non-least
element a A.
Definition 7 Atomistic poset is such poset that a =
S
atoms a for every
element a of this poset.
Proposition 1 Let A be a poset. If a is an atom of A and B A then a
B a 6≍ B.
Proof
a B a a a B, thus a 6≍ B because a is not least.
a 6≍ B implies existence of non-least element x such that x B and x a.
Because a is an atom, we have x = a. So a B.
4
Theorem 1 If A is a distributive lattice then
atoms(a b) = atoms a atoms b
for every a, b A.
Proof For any atomic element c
c atoms(a b)
c (a b) is not least
(c a) (c b) is not least
c a is not least or c b is not least
c atoms a c ato ms b.
Theorem 2 a toms
T
S =
T
hatomsi S whenever
T
S is defined for every S
PA where A is a poset.
Proof For any atom c
c atoms
\
S
c
\
S
a S : c a
a S : c atoms a
c
\
hatomsi S.
Corollary 1 atoms(a b) = atoms a atoms b for arbitrary meet-semilattice.
Theorem 3 A complete boolean lattice is atomic iff it is atomistic.
Proof
Obvious.
Let A be an atomic boolean lattice. Let a A. Suppose b =
S
atoms a a.
If x atoms(a \ b) then x a \ b and so x a and hence x b. But we
have x = x b (a \ b) b = 0 what contradicts to our s upposition.
5
2.3. Difference and complement
Definition 8 Let A be a distributive lattice with least element 0. The differ-
ence (denoted a \ b) of elements a and b is such c A that b c = 0 and
a b = b c. I will call b substractive from a when a \ b exists.
Theorem 4 If A is a distributive lattice with least element 0, there exists no
more than one difference of elements a, b A.
Proof Let c and d are both differences a \ b. Then b c = b d = 0 and
a b = b c = b d. So
c = c (b c) = c (b d) = (c b) (c d) = 0 (c d) = c d.
Analogously, d = d c. Consequently c = c d = d c = d.
Definition 9 I will call b complementive to a when t here exist s c A such
that b c = 0 and b c = a.
Proposition 2 b is complementive to a iff b is substractive from a and b a.
Proof
Obvious.
We deduce b a from b c = a. Thus a b = a = b c.
Proposition 3 If b is complementive to a then (a \ b) b = a .
Proof Because b a by the previous proposition.
Definition 10 Let A be a bounded distributive lattice. The complement (de-
noted ¯a) of element a A is such b A that a b = 0 and a b = 1.
Proposition 4 If A is a bounded distributive lattice then ¯a = 1 \ a.
Proof b = ¯a b a = 0 b a = 1 b a = 0 1 a = a b b = 1 \ a.
Corollary 2 If A is a bounded distributive lattice then exists no more than one
complement of an element a A.
Definition 11 An element of bounded distributive latt ice is called comple-
mented when its complement exists.
Definition 12 A distributive lattice is a complemented lattice iff every its
element is complemented.
6
Proposition 5 For a distributive latt ice (a \ b) \ c = a \ (b c) if a \ b and
(a \ b) \ c are defined.
Proof ((a \ b) \ c) c = 0; ((a \ b) \ c) c = (a \ b) c; (a \ b) b = 0;
(a \ b) b = a b.
We need to prove ((a\b) \c)(bc) = 0 and ((a\ b)\c)(bc) = a(bc).
In fact,
((a \ b) \ c) (b c) =
(((a \ b) \ c) b) (((a \ b) \ c) c) =
(((a \ b) \ c) b) 0 =
((a \ b) \ c) b
(a \ b) b = 0,
so ((a \ b) \ c) (b c) = 0;
((a \ b) \ c) (b c) =
(((a \ b) \ c) c) b =
(a \ b) c b =
((a \ b) b) c =
a b c.
2.4. Center of a lattice
Definition 13 The center Z(A) of a bounded distributive lattice A is the set
of its complemented elements.
Remark 2 For definition of center of non-distributive lattices see [3].
Remark 3 In [9] the word center and the notation Z(A) is used in a different
sense.
Definition 14 A complete lattice A is join infinite distributive when x
S
S =
S
hx∩i S; complete lattice is meet infinite distributive when x
T
S =
T
hx∪i S for all x A and S PA.
Definition 15 Infinitely distributive complete lattice is a complete lattice
which is both join infinite distributive and meet infinite distributive.
Definition 16 A sublattice K of a complete lattice L is a closed sublattice of
L if K contains the meet and the join of any its nonempty subset.
Theorem 5 Center of a infinitely distributive lattice is its closed sublattice.
7
Proof See [6].
Remark 4 See [7] for a more strong result.
Theorem 6 The center of a bounded distributive lattice constitutes its sublat-
tice.
Proof Let A be a bounded distributive lattice and Z(A) is its center. Let
a, b Z(A). Conseq ue ntly ¯a,
¯
b Z(A). Then ¯a
¯
b is the complement of a b
because
(a b) a
¯
b) = (a b ¯a) (a b
¯
b) = 0 0 = 0 and
(a b) a
¯
b) = (a ¯a
¯
b) (b ¯a
¯
b) = 1 1 = 1.
So a b is complemented, analogously a b is complemented.
Theorem 7 The center of a bounded distributive lattice constitutes a boolean
lattice.
Proof Because it is a distributive complemented lattice.
2.5. Galois connections
See [1] and [5] for more detailed treatment of Galois connections.
Definition 17 Let A and B be two posets. A Galois connection between A
and B is a pair of functions f = (f
; f
) with f
: A B and f
: B A such
that:
x A, y B : (f
x
B
y x
A
f
y).
f
is called upper adjoint of f
and f
is called lower adjoint of f
.
Theorem 8 A pair (f
; f
) of functions f
: A B and f
: B A is a
Galois connection iff both of the following:
1. f
and f
are monotone.
2. x
A
f
f
x and f
f
y
B
y for every x A and y B.
Proof
2. x
A
f
f
x since f
x
B
f
x; f
f
y
B
y since f
y
A
f
y.
1. Let a, b A and a
A
b. Then a
A
b
A
f
f
b. So by definition
f
a f
b that is f
is monotone. Analogously f
is monotone.
f
x
B
y f
f
x
A
f
y x
A
f
y. The other direction is analogous.
8
Theorem 9
1. f
f
f
= f
.
2. f
f
f
= f
.
Proof
1. Let x A. We have x
A
f
f
x; consequently f
x
B
f
f
f
x. On the
other hand, f
f
f
x
B
f
x. So f
f
f
x = f
x.
2. Analogous ly.
Proposition 6 f
f
and f
f
are idempotent.
Proof f
f
is idempotent bec ause f
f
f
f
y = f
f
y. f
f
is similar.
Theorem 10 Each of two adjoints is uniquely determined by the other.
Proof Let p and q be both upper adjoints of f. We have for all x A and
y B:
x p(y) f(x) y x q(y).
For x = p(y) we obtain p(y) q(y) and for x = q(y) we obtain q(y) p(y). So
p(y) = q(y).
Theorem 11 Let f be a function from a poset A t o a poset B.
1. Both:
1. If f is monotone and g(b) = max {x A | fx b} is defined for every
b B then g is the upper adjoint of f .
2. If g : B A is the upper adjoint of f then g(b) = max {x A | fx b}
for every b B.
2. Both:
1. If f is monotone and g(b) = min {x A | f x b} is defined for every
b B then g is the lower adjoint of f.
2. If g : B A is the lower adjoint of f then g(b) = min {x A | fx b}
for every b B.
Proof We will prove only the first as the second is its dual.
9
1. Let g(b) = max {x A | fx b} for every b B. Then
x gy x max { x A | fx y} fx y
(because f is monotone) and
x gy x max {x A | fx y} fx y.
So fx y x gy that is f is the lower adjoint o f g.
2. We have
g(b) = max {x A | fx b}
fgb b x A : (fx b x gb)
what is true by properties of adjoints.
Theorem 12 Let f be a function from a poset A t o a poset B.
1. If f is an upper adjoint, f preserves all existing infima in A.
2. If A is a complete lattice and f preserves all infima, then f is an upper
adjoint of a function B A.
3. If f is a lower adjoint, f preserves all existing suprema in A.
4. If A is a complete lattice and f preserves all suprema, then f is a lower
adjoint of a function B A.
Proof We will prove only first two items be c ause the rest items are similar.
1. Let S PA and
T
S exists . f
T
S is a lower bound for h f i S because f is
order-preserving. If a is a lower bound for hfi S then x S : a fx that is
x S : x ga where g is the lower adjoint of f . Thus
T
S ga and hence
f
T
S a. So f
T
S is the greatest lower bound fo r hfi S.
2. Let A be a complete lattice and f preser ves all infima. Let g(a) =
T
{x A | fx a}.
Since f prese rves infima, we have
f(g(a)) =
\
{f(x) | x A, f(x) a} a.
g(f(b)) =
T
{x A | f x fb} b.
Obviously f is monotone and thus g is also mo notone.
So f is the upper adjoint of g.
Corollary 3 Let f be a function from a complete lattice A to a poset B. Then:
1. f is an upper adjoint of a function B A iff f preserves all infima in A.
2. f is a lower adjoint of a function B A iff f preserves all suprema in A.
10
2.6. Co-Brouwerian Lattices
Definition 18 Let A be a poset. Let a A. Pseudocomplement of a is
max {c A | c a} .
If z is pseudocomplement of a we will denote z = a
.
Definition 19 Let A be a poset. Let a A. Dual pseudocomplement of a
is
min {c A | c a} .
If z is dual pseudocomplement of a we will denote z = a
+
.
Definition 20 Let A be a join-semilattice. Let a, b A. Pseudodifference of
a and b is
min {z A | a b z} .
If z is a pseudodifference of a and b we will denote z = a \
b.
Remark 5 I do not require that a
is undefined if there are no pseudocom-
plement of a and likewise for dual pseudocomplement and pseudodifference. In
fact below I will define quasicomplement, dual quasicomplement, and quasidif-
ference which will generalize pseudo-* counterparts. I will denote a
the more
general case of quasicomplement than of pseudocomplement, and likewise for
other notation.
Obvious 6 Dual pseudocomplement is the dual of pseudocomplement.
Definition 21 Co-brouwerian lattice is a lattice for which is defined pseu-
dodifference of any two its elements.
Proposition 7 Every non-empty co-brouwerian lattice A has least element.
Proof Let a be an arbitrary lattice element. Then a\
a = min {z A | a a z} =
min A. So min A exists.
Definition 22 Co-Heyting lattice is co-brouwerian latt ice with greatest ele-
ment.
Theorem 13 For a co-brouwerian lattice a is an upper adjoint of \
a
for every a A.
Proof g(b) = min {x A | a x b} = b \
a exists for every b A and thus
is the lower adjoint of a .
Corollary 4 a, x, y A : (x\
a y x a y) for a co-brouwerian lattice.
11
Definition 23 Let a, b A where A is a complete lattice. Quasidifference a\
b
is defin ed by the formula
a \
b =
\
{z A | a b z} .
Remark 6 The more detailed theory of quasidifference (as well as quasicom-
plement and dual quasicomplement) will be considered below.
Lemma 1 (a \
b) b = a b for elements a, b of a meet infinite distributive
complete lattice.
Proof
(a \
b) b =
\
{z A | a b z} b =
\
{z b | z A, a b z} =
\
{t A | t b, a t} =
a b.
Theorem 14 The following are equivalent for a complete latt ice A:
1. A is meet infinite distributive.
2. A is a co-brouwerian lattice.
3. A is a co-Heyting lattice.
4. a has lower adjoint for every a A.
Proof
(2)(3) Obvious (taking in acc ount completeness of A).
(4)(1) Let \
a be the lower adjoint of a. Let S PA. For every y S
we have y (a y) \
a by properties o f Galois connections; consequently
y (
T
ha∪i S) \
a;
T
S (
T
ha∪i S) \
a. So
a
\
S ((
\
ha∪i S) \
a) a
\
ha∪i S.
But a
T
S
T
ha∪i S is obvious.
(1)(2) Let a\
b =
T
{z A | a b z}. To prove that A is a co- brouwerian
lattice is enough to prove that a b (a \
b). But it follows from the
lemma.
12
(2)(4) a \
b = min {z A | a b z}. So a is an upper adjoint of
\
a.
(1)(4) Because a preserves all meets.
Corollary 5 Co-brouwerian lattices are distributive.
The following theorem is essentially borrowed from [8 ]:
Theorem 15 A lattice A with least element 0 is co-brouwerian with pseudodif-
ference \
iff \
is a binary operation on A sat isfying the following identities:
1. a \
a = 0;
2. a (b \
a) = a b;
3. b (b \
a) = b;
4. (b c) \
a = (b \
a) (c \
a).
Proof
We have
c b \
a c a a (b \
a) = a b b;
c a b c = c (c \
a) (a \
a) (c \
a) = (a c) \
a b \
a.
So c b \
a c a b that is a is an upper adjoint of \
a. By
a theorem above our lattice is co-brouwerian. By an other theorem above
\
is a ps eudodifference.
1. Obvious.
2.
a (b \
a) =
a
\
{z A | b a z} =
\
{a z | z A, b a z} =
a b.
3. b(b\
a) = b
T
{z A | b a z} =
T
{b z | z A, b a z} =
b.
13
4. Obviously (bc)\
a b\
a and (bc)\
a c\
a, thus (bc )\
a
(b \
a) (c \
a). We have
(b \
a) (c \
a) a =
((b \
a) a) ((c \
a) a) =
(b a) (c a) =
a b c
b c.
From this by the definition of adjoints: (b \
a)(c\
a) (b c)\
a.
Theorem 16 (
S
S) \
a =
S
{x \
a | x S} for a A and S PA where A
is a complete co-brouwerian lattice.
Proof Because lower adjoint preserves all suprema.
Theorem 17 (a \
b) \
c = a \
(b c) for elements a, b, c of a complete
co-brou werian lattice.
Proof a \
b =
T
{z A | a b z}.
(a \
b) \
c =
T
{z A | a \
b c z}.
a \
(b c) =
T
{z A | a b c z}.
It’s left to prove a \
b c z a b c z.
Let a \
b c z. Then a b b c z by the lemma and consequently
a b c z.
Let a b c z. Then a \
b (b c z) \
b c z by a theorem above.
3. Straight maps and separation subsets
3.1. Straight maps
Definition 24 Let f be a monotone map from a meet-semilattice A to some
poset B. I call f a straight m ap when
a, b A : (f a fb fa = f (a b)).
Proposition 8 The following statements are equivalent for a monotone map
f:
1. f is a straight map.
2. a, b A : (f a fb fa f(a b)).
3. a, b A : (f a fb fa 6 f(a b)).
14
4. a, b A : (f a f(a b) f a * f b).
Proof
(1)(2)(3) Due fa f(a b).
(3)(4) Obvious.
Remark 7 The definition of str aight map can be generalized for any poset A
by the formula
a, b A : (fa fb c A : (c a c b f a = fc)).
This generalization is not yet researched however.
Proposition 9 Let f be a monotone map from a meet-semilattice A to some
poset B. If
a, b A : (f (a b) = fa fb)
then f is a straight map.
Proof Let fa fb. Then f (a b) = f a fb = fa.
Proposition 10 Let f be a monotone map from a meet-semilattice A to some
poset B. If
a, b A : (f a fb a b)
then f is a straight map.
Proof fa fb a b a = a b f a = f(a b).
Theorem 18 If f is a straight monotone map from a meet-semilattice A then
the following statements are equivalent:
1. f is an injection.
2. a, b A : (f a fb a b).
3. a, b A : (a b f a fb).
4. a, b A : (a b f a 6= fb).
5. a, b A : (a b f a + fb).
6. a, b A : (f a fb a 6 b).
Proof
15
(1)(3) Let a, b A. Let fa = fb a = b. Let a b. f a 6= fb because
a 6= b. f a fb because a b. So f a fb.
(2)(1) Let a, b A. Let fa fb a b. Let fa = fb. Then a b b a
and consequently a = b.
(3)(2) Let a, b A : (a b fa f b). Let a * b. Then a a b. So
fa f(a b). If f a f b then fa f(a b) what is a contradiction.
(3)(5)(4) Obvious.
(4)(3) Because a b a b fa fb.
(5)(6) Obvious.
3.2. Separation subset s and full stars
Definition 25
Y
a = {x Y | x 6≍ a} for an element a of a poset A and
Y PA.
Definition 26 Full star of a is ⋆a =
A
a.
Proposition 11 If A is a meet-semilattice, then is a straight monotone map.
Proof Monotonicity is obvious. Let ⋆a * (a b). Then it exists x ⋆a such
that x / (a b). So x a / ⋆b but x a ⋆a and conseq ue ntly a * b.
Definition 27 A separation subset of a poset A is such its subset Y that
a, b A : (
Y
a =
Y
b a = b).
Definition 28 I call separable such poset that is an injection.
Obvious 7 A poset is separable iff it has separation subset.
Definition 29 A poset A has disjunction property of Wallman iff for any
a, b A either b a or there exists a non-least element c b such that a c.
Theorem 19 For a meet-semilattice with least element the following statements
are equivalent:
1. A is separable.
2. a, b A : (⋆a ⋆b a b).
3. a, b A : (a b a ⋆b).
4. a, b A : (a b a 6= ⋆b).
16
5. a, b A : (a b a + ⋆b).
6. a, b A : (⋆a ⋆b a 6 b).
7. A conforms to Wallman’s disjunction property.
8. a, b A : (a b c A \ {0} : (c a c b)).
Proof
(1)(2)(3)(4)(5)(6) By the above theorem.
(8)(4) Let the property (8) holds. Let a b. Then it exists ele ment c b
such that c 6= 0 and c a = 0. But c b 6= 0. So ⋆a 6= ⋆b.
(2)(7) Let the property (2) holds. Let a * b. Then a * ⋆b that is exists
c ⋆a such that c / ⋆b, in other words c a 6= 0 and c b = 0. Let
d = c a. Then d a and d 6= 0 and d b = 0. So disjunction property
of Wallman holds.
(7)(8) Obvious.
(8)(7) Let b * a. Then ab b that is a
b where a
= ab. Consequently
c A \ {0} : (c a
c b). We have c a = c b a = c a
. So c b
and c a = 0. Thus Wallman’s disjunction property holds.
3.3. Atomically separable lattices
Proposition 12 atoms is a straight monotone map (for any meet-semilattice).
Proof Monotonicity is obvious. The rest follows from the formula
atoms(a b) = atoms a atoms b
(the corolla ry 1).
Definition 30 I will call atomically separable such a poset that atoms is
an injection.
Proposition 13 a, b A : (a b atoms a atoms b) iff A is atomically
separable for a poset A.
Proof
Obvious.
17
Let a 6= b for example a * b. Then a b a; atoms a atoms(a b) =
atoms a atoms b and thus atoms a 6= atoms b .
Let atoms a 6= atoms b for example atoms a * atoms b . Then atoms(a
b) = atoms a atoms b atoms a and thus a b a a nd so a * b
consequently a 6= b.
Proposition 14 Any atomistic poset is atomically separable.
Proof We need to prove that atoms a = atoms b a = b. But it is obvious
because
a =
[
atoms a and b =
[
atoms b.
Theorem 20 If a lattice with least element is atomic and separable then it is
atomistic.
Proof Suppose the contrary that is a
S
atoms a. Then, because our lattice
is separable, exists c A such that c a 6= 0 and c
S
atoms a = 0. There
exist atom d c such that d c a. d
S
atoms a c
S
atoms a = 0. But
d atoms a. Contradiction.
Theorem 21 Any atomistic lattice is atomically separable.
Proof L et A be an atomistic lattice. Let a, b A, a b. Then
S
atoms a
S
atoms b and consequently atoms a atoms b.
Theorem 22 Let A be an atomic meet-semilattice with least element. Then the
following statements are equivalent:
1. A is separable.
2. A is atomically separable.
3. A conforms to Wallman’s disjunction property.
4. a, b A : (a b c A \ {0} : (c a c b)).
Proof
(1)(3)(4) Proved above.
(2)(4) Let our semilattice be atomically separable. Let a b. Then atoms a
atoms b and so exists c atoms b such that c / atoms a. c 6= 0 and c b;
c * a, from which (taking in account that c is an atom) c b and ca = 0.
So our semilattice conforms to the formula (4).
18
(4)(2) Let formula (4) holds. Then for any elements a b exists c 6= 0 such
that c b and ca = 0. B ecause A is atomic there exists atom d c . d
atoms b and d / atoms a. So atoms a 6= atoms b and a toms a atoms b.
Consequently atoms a atoms b.
4. Filtrators
Definition 31 I will call a filtrator a pair (A; Z) of a poset A and its subset
Z A. I call A the base of a filtrator and Z the core of a filtrator.
Definition 32 I will call a lattice filtrator a pair (A; Z) of a lattice A and its
subset Z A.
Definition 33 I will call a complete lattice filtrator a pair (A; Z) of a com-
plete lattice A and its subset Z A.
Definition 34 I will call a central filtrator a filtrator (A; Z(A)) where Z(A)
is the center of a bounded lattice A.
Remark 8 One use of filtra tors is the theory of filters where the base lattice
(or the lattice o f principal filters) is essentially considered as the core of the
lattice of filters. See below for a more exact formulation. Our primary interest
is the properties o f filters on sets (that is the filtrator of filters on a set), but
instead we will research more general theor y of filtrators.
Remark 9 An other important example of filtrators is filtrator of funcoids
whose base is the set of funcoids [11] and whose core is the set of binary relations
(or discrete funcoids).
Definition 35 I will call el ement of a filtrator an element of its base.
Definition 36 up a = {c Z | c a} where a A.
Definition 37 down a = {c Z | c a} where a A.
Obvious 8 up and down are dual.
The main purpose of this text is knowing properties of the core of a filtrator
to infer pro perties of the base of the filtrator, specifically properties of up a for
every element a.
Definition 38 I call a filtrator with join-closed core such filtrator (A; Z) that
S
Z
S =
S
A
S whenever
S
Z
S exists for S PZ.
19
Definition 39 I call a filtrator with meet-closed core such filtrator (A; Z) t hat
T
Z
S =
T
A
S whenever
T
Z
S exists for S PZ.
Definition 40 I call a ltrator with finitely join-closed core such filtrator
(A; Z) that a
Z
b = a
A
b whenever a
Z
b exists for a, b Z.
Definition 41 I call a filtrator with fini tely meet-closed core such filtrator
(A; Z) that a
Z
b = a
A
b whenever a
Z
b exists for a, b Z.
Definition 42 Filtered filtrator is a ltrator (A; Z) such that a A : a =
T
A
up a.
Definition 43 Prefiltered filtrator is a filtrator (A; Z) such that up is in-
jective.
Definition 44 Semifiltered filtrator is a filtrator (A; Z) such that
a, b A : (up a up b a b).
Obvious 9 Every filtered filtrator is semifiltered.
Every semifiltered filtrator is prefiltered.
Obvious 10 up is a straight map from A to the dual of the poset PZ if (A; Z)
is a semifiltered filtrator.
Theorem 23 Each semifiltered filtrator is a filtrator with join-closed core.
Proof Let (A; Z) be a semifiltered filtrator. Let S PZ and
S
Z
S is defined.
We need to prove
S
A
S =
S
Z
S. That
S
Z
S is an upper bound for S is obvious.
Let a A be a n upper bound for S. Enough to prove tha t
S
Z
S a. Really,
c up a c a x S : c x c
[
Z
S c up
[
Z
S;
so up a up
S
Z
S and thus a
S
Z
S because it is semifiltered.
4.1. Core part
Definition 45 The core part of an element a A is Cor a =
T
Z
up a.
Definition 46 The dual core part of an element a A is Cor
a =
S
Z
down a.
Obvious 11 Cor
is du al of Cor.
Theorem 24 Cor a a whenever Cor a exists for any element a of a filtered
filtrator.
20
Proof Cor a =
T
Z
up a
T
A
up a = a.
Corollary 6 Cor a down a whenever Cor a exists for any element a of a
filtered filtrator.
Theorem 25 Cor
a a whenever Cor
a exists for any element a of a filtrator
with join-closed core.
Proof Cor
a =
S
Z
down a =
S
A
down a a.
Corollary 7 Cor
a down a whenever Cor
a exists for any element a of a
filtrator with join-closed core.
Proposition 15 Cor
a Cor a whenever both Cor a and Cor
a exist for any
element a of a filt rator with join-closed core.
Proof Cor a =
T
Z
up a Cor
a because A up a : Co r
a A.
Theorem 26 Cor
a = Cor a whenever both Cor a and Cor
a exist for any ele-
ment a of a filtered filtrator.
Proof It is with join-closed core because it is semifiltered. So Cor
a Cor a.
Cor a down a. So Cor a
S
Z
down a = Cor
a.
Obvious 12 Cor
a = max down a for an element a of a filtrator with join-
closed core.
4.2. Filtrators with separable core
Definition 47 Let A be a filtrator. A is a filtrator w ith separable core when
x, y A : (x
A
y X up x : X
A
y).
Proposition 16 Let A be a filtrator. A is a filtrator with separable core iff
x, y A : (x
A
y X up x, Y up y : X
A
Y ).
Proof
Apply the definition twice.
Obvious.
Definition 48 Let A be a filtrator. A is a filtrator with co-separable core
when
x, y A : (x
A
y X down x : X
A
y).
21
Obvious 13 Co-separability is the dual of s eparability.
Proposition 17 Let A be a filtrator. A is a filtrator with co-separable core iff
x, y A : (x
A
y X down x, Y down y : X
A
Y ).
Proof By duality.
4.3. Intersecting and joining with an element of the core
Definition 49 I call down-aligned filtrator such a ltrator (A; Z) that A and
Z have common least element. (Let’s denote it 0.)
Definition 50 I call up-aligned filtrator such a filt rator (A; Z) that A and Z
have common greatest element. (Let’s denote it 1.)
Theorem 27 For a filtrator (A; Z) where Z is a boolean lattice, for every B Z,
A A:
1. B
A
A
B A if it is down-aligned, with finitely meet-closed and sepa-
rable core;
2. B
A
A
B A if it is u p-aligned, with finitely join-closed and co-
separable core.
Proof We will prove only the first as the second is dual.
B
A
A
A up A : B
A
A
A up A : B
A
A = 0
A up A : B
Z
A = 0
A up A :
B A
B up A
B A.
5. Filters
5.1. Filters on posets
Let A be a poset (pa rtially orde red set) with the partial order . I will call
it the base poset.
Definition 51 Filter base is a nonempty subset F of A such that
X, Y F Z F : (Z X Z Y ).
22
Obvious 14 A nonempty chain is a lter base.
Definition 52 Upper set is a subset F of A such that
X F, Y A : (Y X Y F ).
Definition 53 Filter is a subset of A which is both lter base and upper set . I
will denote the s et of filt ers f.
Proposition 18 If 1 is the maximal element of A then 1 F for any lter F .
Proof If 1 6∈ F then K A : K 6∈ F and so F is empty what is impossible.
Proposition 19 Let S be a filter base. If A
0
, . . . , A
n
S (n N), then
C S : (C A
0
... C A
n
).
Proof It can be easily proved by induction.
The dual of filters is called ideals. We do not use ideals in this work however.
5.2. Filters on meet-semilattice
Theorem 28 If A is a m eet-semilattice and F is a nonempty subset of A then
the following conditions are equivalent:
1. F is a lter.
2. X , Y F : X Y F and F is an u pper set.
3. X , Y A : (X, Y F X Y F).
Proof
(1)(2) Let F be a filter. Then F is a n upper set. If X , Y F then Z
X Z Y for s ome Z F . Because F is an upper set and Z X Y
then X Y F.
(2)(1) Let X, Y F : X Y F and F is an upper set. We need to prove
that F is a filter base. But it is obvious taking Z = X Y (we have also
taken in account that F 6= ).
(2)(3) Let X, Y F : X Y F and F is an upper set. Then
X, Y A : (X, Y F X Y F).
Let X Y F ; then X, Y F because F is an upper set.
23
(3)(2) Let
X, Y A : (X, Y F X Y F).
Then X, Y F : X Y F . Let X F a nd X Y A. Then
X Y = X F. Consequently X, Y F. So F is an upper set.
Proposition 20 Let A be a meet-semilattice. Let S be a lter base. If A
0
, . . . , A
n
S (n N), then
C S : C A
0
... A
n
.
Proof It can be easily proved by induction.
Proposition 21 If A is a meet-semilattice and S is a lter base, A A, then
hA∩i S is also a filter base.
Proof hA∩i S 6= because S 6= .
Let X, Y hA∩i S. Then X = A X
and Y = A Y
where X
, Y
S.
Exists Z
S such that Z
X
Y
. So X Y = AX
Y
AZ
hA∩i S.
5.3. Characterization of finitely meet-closed filtrators
Theorem 29 The following are equivalent for a filtrator (A; Z) whose core is a
meet-semilattice such that a A : up a 6= :
1. The filtrator is finitely meet-closed.
2. up a is a filter on Z for every a A.
Proof
(1)(2) Let X, Y up a. Then X
Z
Y = X
A
Y a. That up a is an upper
set is obvious. So taking in account that up a 6= , up a is a filter.
(2)(1) It is enough to prove that a A, B a A
Z
B for every A, B A.
Really:
a A, B A, B up a A
Z
B up a a A
Z
B.
6. Filter objects
I want to equate principal filters (see below) with the elements of the base
poset. Such thing can be done using the principle s described in the appendix
Appendix B. The formal definitions fo llow.
24
6.1. Definition of filter objects
Let A be a poset.
Definition 54 Let a
def
= {x A | x a} for every a A. Elements of the
set h↑i A are called principal filters.
Obvious 15 is an injection from A to f.
Let M be a bijection defined on f such that M = id
A
. (See the appendix
Appendix B for a proof that such a bijection exists.)
Definition 55 Let F = im M . I call elements of F as filter objects (f.o. for
short).
Remark 10 Below we will show that up A = M
1
A for each A F.
Obvious 16 = M
1
|
A
.
Obvious 17 M
1
is a bijection F f.
Proposition 22 A F.
Proof x A M x = x x im M x F.
6.2. Order of lter objects
Proposition 23 a b M
1
a M
1
b.
Proof a b ⇔↑ a b M
1
a M
1
b.
As a generalization of the last proposition we may define the order on F:
Definition 56 A B
def
= M
1
A M
1
B for all A, B A.
I will call the pair (F; A) the primary filtrator.
Theorem 30 For the primary ltrator (F; A) we have up A = M
1
A for each
A F.
Proof x up A x A M
1
x M
1
A ⇔↑ x M
1
A x M
1
A
for every x A.
So we have:
”up” is a bijection from F to f.
A B up A up B for each A, B F.
up a = a for every a A.
25
A filter object A is represented by the value of up A. We ar e not interested in
the internal structure of filter objects (which can be inferred from the appendix
Appendix B), but only in the value of up A. Thus the name “filter objects”
by analogy with an o bject in object oriented programming where an object is
completely char acterized by its methods, likewise a filter object A is completely
characterized by up A.
7. Lattice of filter objects
7.1. Minimal and maximal f.o.
Obvious 18 The filter object 0 = up
1
A (equal to the least element of the
poset A if this least exists) is the least element of the poset of filter objects.
Proposition 24 If there exists greatest element 1 of the poset A then it is also
the greatest element of the poset of filter objects.
Proof Take in account that filters a re nonempty.
Obvious 19 1. If the base poset has least element, then the primary filtrator
is down-aligned.
2. If the base poset has greatest element, then the primary filtrator is up-aligned.
7.2. Primary filtrator is filtered
Theorem 31 Every primary ltrator is filtered.
Proof We need to prove that A =
T
F
up A for every A F.
A is obviously a lower bound for up A.
Let B be a lower b ound for up A that is K up A : K B. Then
up A up B; A B. So A is the greates t lower b ound of up A.
7.3. Formulas for meets and joins of lter objects
Lemma 2 If f is an order embedding from a poset A to a complete lattice B
and S PA and exists such F A that f F =
S
B
hfi S, then
S
A
S exists and
f
S
A
S =
S
B
hfi S.
Proof f is an order isomorphism from A to B|
hfiA
. f F B|
hfiA
.
Consequently,
S
B
hfi S B|
hf iA
and
S
B|
hf iA
hfi S =
S
B
hfi S.
f
S
A
S =
S
B|
hfiA
hfi S because f is a n order isomorphism.
Combining, f
S
A
S =
S
B
hfi S.
Theorem 32 If A is a meet-semilattice with greatest element 1 then
S
F
S exists
and up
S
F
S =
T
PA
hupi S for every S PF.
26
Proof Taking in account the lemma it is enough to prove that exists F F
such that up F =
T
PA
hupi S, that is that R =
T
PA
hupi S is a filter.
R is nonempty because 1 R. Let A, B R; then ∀F S : A, B up F,
consequently ∀F S : A
A
B up F. Consequently A
A
B
T
PA
hupi S = R.
So R is a filter base. Let X R and X Y A; then ∀F S : X up F;
∀F S : Y up F; Y R. So R is an upper set.
Corollary 8 If A is a meet-semilattice with greatest element 1 then F is a
complete lattice.
Corollary 9 If A is a meet-semilattice with greatest element 1 then for any
A, B F
up(A
F
B) = up A up B.
Theorem 33 If A is a join-semilattice then F is a join-semilattice and for any
A, B F
up(A
F
B) = up A up B.
Proof Taking in account the lemma it is enough to prove that R = up Aup B
is a filter.
R is nonempty because exist X up A and Y up B and R X
A
Y .
Let A, B R. Then A, B up A; so exists C up A such that C A C
B. Analogously exists D up B such that D A D B. Let E = C
A
D.
Then E up A and E up B; E R and E AE B. So R is a filter base.
That R is an upper set is obvious.
Theorem 34 If A is a distributive lattice then for S PF \ {∅}
up
\
F
S =
n
K
0
A
. . .
A
K
n
| K
i
[
hupi S where i = 0, . . . , n for n N
o
.
Proof Let’s denote the right part of the equality to be proven as R. First we
will prove that R is a filter. R is nonempty because S is nonempty.
Let A, B R. Then A = X
0
A
. . .
A
X
k
, B = Y
0
A
. . .
A
Y
l
where
X
i
, Y
j
S
hupi S. So
A
A
B = X
0
A
. . .
A
X
k
A
Y
0
A
. . .
A
Y
l
R.
Let R C A. Consequently (distributivity used)
C = C
A
A = (C
A
X
0
)
A
. . .
A
(C
A
X
k
).
X
i
up P for some P S; C
A
X
i
up P ; consequently C up P ; C R.
We have proved that R is a filter base and an uppe r set. So R is a filter.
Consequently the statement of our theorem is equivalent to
T
F
S = up
1
R.
Let A S. The n up A hupi S; up A
S
hupi S;
R
K
0
A
. . .
A
K
n
| K
i
up A where i = 0, . . . , n for n N
= up A.
27
Consequently A up
1
R.
Let now B F and ∀A S : A B. Then ∀A S : up B up A. up B
S
hupi S. From this up B T for any finite set T
S
hupi S. Consequently
up B
T
A
T . Thus up B R; B up
1
R.
Comparing we get
T
F
S = up
1
R.
Theorem 35 If A is a distributive lattice then for any F
0
, . . . , F
m
F (m N)
up(F
0
F
. . .
F
F
m
) =
K
0
A
. . .
A
K
m
| K
i
up F
i
, i = 0, . . . , m
.
Proof Let’s denote the right part of the equality to be proven as R. First we
will prove that R is a filter. Obviously R is nonempty.
Let A, B R. Then A = X
0
A
. . .
A
X
m
, B = Y
0
A
. . .
A
Y
m
where
X
i
, Y
i
up F
i
.
A
A
B = (X
0
A
Y
0
)
A
. . .
A
(X
m
A
Y
m
),
consequently A
A
B R.
Let R C A.
C = A
A
C = (X
0
A
C)
A
. . .
A
(X
m
A
C) R.
So R is a filter. C onsequently the statement of our theorem is equivalent to
F
0
F
. . .
F
F
m
= up
1
R.
Let P
i
up F
i
. Then P
i
R because P
i
= (P
i
A
P
0
)
A
. . .
A
(P
i
A
P
m
). So
up F
i
R; F
i
up
1
R.
Let now B F and i {0, . . . , m} : F
i
B. Then i {0, . . . , m} :
up F
i
up B.
L
i
up B for any L
i
up F
i
. L
0
A
. . .
A
L
m
up B. So up B R;
B up
1
R.
So F
0
F
. . .
F
F
m
= up
1
R.
Definition 57 I will call a lattice of filter objects on a set a set of lter
objects on the lattice of all subsets of a set. ( From the above it follows that it is
actually a complete lattice.)
7.4. Distributivity of the lattice of lter objects
Theorem 36 If A is a distributive lattice with greatest element, S PF and
A F then A
F
T
F
S =
T
F
A∪
F
S.
28
Proof Taking in account the previous subsectio n, we have:
up
A
F
\
F
S
=
up A up
\
F
S =
up A
n
K
0
A
. . .
A
K
n
| K
i
[
hupi S where i = 0, . . . , n for n N
o
=
n
K
0
A
. . .
A
K
n
| K
0
A
. . .
A
K
n
up A, K
i
[
hupi S where i = 0, . . . , n for n N
o
=
n
K
0
A
. . .
A
K
n
| K
i
up A, K
i
[
hupi S where i = 0, . . . , n for n N
o
=
n
K
0
A
. . .
A
K
n
| K
i
up A, K
i
[
{up X | X S} where i = 0, . . . , n for n N
o
=
n
K
0
A
. . .
A
K
n
| K
i
up A
[
{up X | X S} where i = 0, . . . , n for n N
o
=
n
K
0
A
. . .
A
K
n
| K
i
[
{up A up X | X S} where i = 0, . . . , n for n N
o
=
n
K
0
A
. . .
A
K
n
| K
i
[
up(A
F
X ) | X S
where i = 0, . . . , n for n N
o
=
n
K
0
A
. . .
A
K
n
| K
i
[
hupi
A
F
X | X S
where i = 0, . . . , n for n N
o
=
up
\
F
A
F
X | X S
.
Corollary 10 If A is a distributive lattice with greatest element, then F is also
a distributive lattice.
Corollary 11 If A is a distributive lattice with greatest element, then F is a
co-brou werian lattice.
7.5. Separability of core for primary filtrators
Theorem 37 A primary filtrator with least element, whose base is a distributive
lattice, is with separable core.
Proof Let A
F
B where A, B F.
up(A
F
B) =
A
A
B | A up A, B up B
.
So
A
F
B
0 up(A
F
B)
A up A, B up B : A
A
B = 0
A up A, B up B : A
F
B = 0
(used the theorem 23).
29
Theorem 38 Let (A; Z) be an up-aligned filtered filtrator whose core is a meet
infinite distributive complete lattice. Then this filtrator is with co-separable core.
Proof Our filtrator is with join-closed core.
Let a, b A. Cor a and Cor b exist since Z is a complete lattice.
Cor a down a and Cor b down b by the corollary 6 since our filtrator is
filtered. So we have
x down a, y down b : x
A
y = 1
Cor a
A
Cor b = 1 (by finite join-closedness of the core)
Cor a
Z
Cor b = 1
\
Z
up a
Z
\
Z
up b = 1 (by infinite distr ibutivity)
\
Z
x
Z
y | x up a, y up b
= 1
x up a, y up b : x
Z
y = 1 (by finite jo in-closedness of the core)
x up a, y up b : x
A
y = 1
a
A
b = 1.
7.6. Filters over boolean lattices
Theorem 39 If A is a boolean lattice then a \
F
B = a
F
B (where the comple-
ment is taken on A).
Proof F is distributive by the theorem 10. Our filtrator is with finitely meet-
closed core by the theorem 29 and with join-clo sed core by the theore m 23.
(a
F
B)
F
B = (a
F
B)
F
(B
F
B) = (a
F
B)
F
(B
A
B) = (a
F
B)
F
1 =
a
F
B.
(a
F
B)
F
B = a
F
(B
F
B) = a
F
(B
A
B) = a
F
0 = 0.
So a
F
B is the difference of a and B.
7.7. Distributivity for an element of boolean core
Lemma 3 Let F be the set of filter objects over a boolean lattice A.
Then A
F
is a lower adjoint of
A
F
for every A A.
Proof We will use the theorem 8.
That A
F
and A
F
are monotone is o bvious.
We need to prove (for every x, y F) that
x
A
F
(A
F
x) and A
F
(A
F
y) y.
Really,
A
F
(A
F
x) = (A
F
A)
F
(A
F
x) = (A
A
A)
F
(A
F
x) =
1
F
(A
F
x) = A
F
x x and A
F
(A
F
y) = (A
F
A)
F
(A
F
y) =
(A
A
A)
F
(A
F
y) = 0
F
(A
F
y) = A
F
y y.
30
Theorem 40 Let F be the set of lter objects over a boolean latt ice A.
A
F
S
F
S =
S
F
A
F
S for every A A and every set S PF.
Proof Direct consequence of the lemma.
8. Generalized filter base
Definition 58 Generalized filter base is a lter base on the set F.
Definition 59 If S is a generalized filter base and A =
T
F
S, then we will call
S a generalized base of filter object A.
Theorem 41 If A is a distributive lattice and S is a generalized base of lter
object F then for any element K of the base poset
K up F ∃L S : L K.
Proof
Because F =
T
F
S.
Let K up F. Then (taken in account distributivity of A and that S is
nonempty) exist X
1
, . . . , X
n
S
hupi S such that X
1
A
. . .
A
X
n
= K.
Consequently (by theorem 29) X
1
F
. . .
F
X
n
= K. Replacing every X
i
with such X
i
S that X
i
up X
i
(this is obviously possible to do), we
get a finite set T
0
S such that
T
F
T
0
K. From this exists C S such
that C
T
F
T
0
K.
Corollary 12 If A is a distributive latt ice with least element 0 and S is a
generalized base of filter object F then 0 S F = 0.
Proof Substitute 0 as K.
Theorem 42 Let A be a distributive lattice with least element 0 and S is a
nonempty set of filter objects on A such that F
0
F
. . .
F
F
n
6= 0 for every
F
0
, . . . , F
n
S. Then
T
F
S 6= 0.
Proof Consider the set
S
=
F
0
F
. . .
F
F
n
| F
0
, . . . , F
n
S
.
Obviously S
is nonempty a nd finitely meet-closed. So S
is a genera lized filter
base. Obviously 0 6∈ S
. So by properties of genera liz ed filter bases
T
F
S
6= 0.
But obviously
T
F
S =
T
F
S
. So
T
F
S 6= 0.
31
Corollary 13 Let A be a distributive lattice with least element 0 and let S PA
such that S 6= and A
0
A
. . .
A
A
n
6= 0 for every A
0
, . . . , A
n
S. Then
T
F
S 6= 0.
Proof Because A is finitely meet-closed (by the theore m 29).
9. Stars
9.1. Free stars
Definition 60 Let A be a poset. Free stars on A are su ch S PA that the
least element (if it exists) is not in S and for every X, Y A
Z A : (Z X Z Y Z S) X S Y S.
Proposition 25 S PA where A is a poset is a free star iff all of the following:
1. The least element (if it exists) is not in S.
2. Z A : (Z X Z Y Z S) X S Y S for every X, Y A.
3. S is an upper set.
Proof
(1) and (2) are obvious. Let prove that S is an upper set. Let X S and
X Y A. The n X S X S and thus Z A : (Z X Z X
Z S) that is Z A : (Z X Z S), and so Y S.
We need to prove that
Z A : (Z X Z Y Z S) X S Y S.
Let X S Y S. Then Z X Z Y Z S for every Z A
because S is an upper se t.
Proposition 26 Let A be a join-semilattice. S PA is a free star iff all of the
following:
1. The least element (if it exists) is not in S.
2. X Y S X S Y S for every X, Y A.
3. S is an upper set.
Proof
32
We need to prove only X Y S X S Y S. Let X Y S.
Because S is an upper se t, we have Z A : (Z X Y Z S)
and thus Z A : (Z X Z Y Z S) from which we conclude
X S Y S.
We need to pr ove Z A : (Z X Z Y Z S) X S Y S.
But this trivially follows from that S is an upper set.
Proposition 27 Let A be a join-semilattice. S PA is a free star iff the least
element (if it exists) is not in S and for every X , Y A
X Y S X S Y S.
Proof
We need to prove only that X Y S X S Y S what follows from
that S is an upper set.
We need to pr ove only that S is an upper set. Let X S and X Y A.
Then X S X S Y S X Y S Y S. So S is an upper
set.
9.2. Stars of elements of filtrators
Definition 61 Let (A; Z) be a ltrator. Core star of an element a of this
filtrator is
a =
x Z | x 6≍
A
a
.
Proposition 28 up a a for any non-least element a of a filtrator.
Proof For any element X Z
X up a a X a a X 6≍ a X a.
Theorem 43 Let (A; Z) be a distribut ive lattice filtrator with least element and
finitely join-closed core which is a join-semilattice. Then a is a free star for
each a A.
33
Proof For every A, B Z
A
Z
B a
A
A
B a
(A
A
B)
A
a 6= 0
(A
A
a)
A
(B
A
a) 6= 0
A
A
a 6= 0 B
A
a 6= 0
A a a.
That a doesn’t contain 0 is obvious.
Definition 62 I call a filtrator star-separable when its core is a separation
subset of its base.
9.3. Stars of filters on boolean lattices
In this section we will consider the set of filter objects F on a boolean lattice
A.
Theorem 44 If A is a boolean latt ice and A F then
1. A =
X | X A \ up A
;
2. up A =
X | X A \ A
.
Proof 1. For any K A (taking into account the theorems 29, 37, and 27)
K
X | X A \ up A
K A \ up A
K 6∈ up A
K + A
K 6≍
F
A
K A.
2. For any K A (taking into account the same theorems)
K
X | X A \ A
K A \ A
K 6∈ A
K
F
A
K A
K up A.
34
Corollary 14 If A is a boolean lattice, X up A X 6∈ A for every X A,
A F.
Corollary 15 If A is a boolean lattice, is an injection.
Theorem 45 If A is a boolean lattice, then for any set S PA exists filter
object A such that A = S iff S is a free star.
Proof
That 0 / S is obvious. For every A, B A
A
A
B S
(A
A
B)
F
A 6= 0
(A
F
B)
F
A 6= 0
(A
F
A)
F
(B
F
A) 6= 0
A
F
A 6= 0 B
F