3 Alexandroff closure operators
We call any map σ:𝒫(X)→𝒫(X) a set operator on X.
We denote by OP(X) the collection of all set operators on X, and for any x∈X, we will write σ(x) instead of σ({x}).
For any σ∈OP(X) and A∈𝒫(X), we will consider the following set systems on X:
𝒪σ,A:={σ(x):x∈A},𝒞σ,X:={Y∈𝒫(X):σ(Y)=Y},𝒯σ,X:={X∖Y:Y∈𝒞σ,X},
𝒰σ,X:={Y∈𝒫(X):y∉σ(Y∖{y})for ally∈Y},
𝒵σ,X:={Y∈𝒫(X):y,y′∈Yimplies thatσ(y)⊆σ(y′)orσ(y′)⊆σ(y)},
ℛσ,X(A):={B∈𝒫(A):σ(B)=σ(A),B′⫋Bimplies thatσ(B′)⫋σ(A)}.
In particular, if 𝒯 is a given topology on X, then the set system 𝒯cl𝒯,X coincides with the topology 𝒯, and we will write simply 𝒞𝒯,X instead of 𝒞cl𝒯,X, which is obviously the closed subset family of 𝒯.
Let us note that Y∈𝒵σ,X if and only if the set system is a chain (i.e. it is linearly ordered) with respect to the set-theoretical inclusion.
We recall that σ∈OP(X) is called a closure operator on X if, for all Y,Z∈𝒫(X), we have that Y⊆σ(Y) (σ is extensive), Y⊆Z implies that σ(Y)⊆σ(Z) (σ is increasing) and
σ(σ(Y))=σ(Y) (σ is idempotent).
We will denote by CLOP(X) the collection of all closure operators on X.
We also recall that if σ∈CLOP(X), then 𝒞σ,X coincides with the set system {σ(Y):Y∈𝒫(X)}, and 𝒞σ,X is a closure system.
As a consequence, (𝒞σ,X,⊆) is also a complete sub-lattice of the complete lattice (𝒫(X),⊆) such that, for any 𝒬⊆𝒞σ,X, we have ⋀𝒬=⋂𝒬 and ⋁𝒬=σ(⋃𝒬).
When σ∈CLOP(X), it is usual to call the elements Y∈𝒞σ,X the closed subsets of X, and σ(Y) the closure of Y for any Y∈𝒫(X).
Furthermore, when σ∈CLOP(X), the members of the set systems ℛσ,X(A) are usually called the σ-reducts of A, and their role becomes relevant in various branches of mathematics and theoretical computer science, as we said in the introductory section.
Let us note that if σ∈CLOP(X), then 𝒪σ,X⊆𝒞σ,X, and moreover,
σ(x)∥σ(x′)
if and only if
x∉σ(x′) and x′∉σ(x)
for any x,x′∈X.
Remark 3.1.
If σ∈CLOP(X) and Y∈𝒞σ,X, the condition σ(Y)=Y implies that the restriction σ|𝒫(Y) is a closure operator on Y.
However, when we will be able to consider such a restriction on the closed subset Y, we still use the symbol σ instead of σ|𝒫(Y).
For example, we will write simply 𝒞σ,Y instead of 𝒞σ|𝒫(Y),Y, and so also for all next similar cases.
We now give the fundamental notion of Alexandroff closure operator.
Definition 3.2.
We say that σ∈CLOP(X) is an Alexandroff closure operator on X if
σ(Y)=⋃{σ(y):y∈Y} for any Y∈𝒫(X).
We denote by CLOPa(X) the collection of all Alexandroff closure operators on X.
Let us recall that a given topology 𝒯 on X is an Alexandroff topology if and only if its closed subset family 𝒞𝒯,X is union closed, that is ℱ⊆𝒞𝒯,X implies that ⋃ℱ∈𝒞𝒯,X (see [21]).
Based on the above characterization of an Alexandroff topology, we obtain the following result, whose proof is a simply verify.
Proposition 3.3.
If σ∈CLOPa(X), then Tσ,X is an Alexandroff topology on X such that clTσ,X=σ.
Conversely, if T is an Alexandroff topology on X, then clT is an Alexandroff closure operator on X.
We will demonstrate in the following result the first basic result concerning Alexandroff closure operators relatively to the set systems introduced before.
Theorem 3.4.
Let σ∈CLOPa(X).
Then the following conditions hold:
if Y,Z∈𝒫(X),
σ(Z)=σ(Y) and y∈Y, then there exists z∈Z such that y∈σ((Y∖{y})∪{z});
let Y∈𝒫(X) and 𝒬=(σ(Y))𝒞σ,X↑; then ℐ(𝒬)={σ(Y∪{x}):x∈X∖σ(Y)};
𝒰σ,X is a finitary abstract simplicial complex and
(3.1)𝒰σ,X={Y∈𝒫(X):y∈Y𝑎𝑛𝑑T∈𝒪σ,Y∖{y}implies thaty∈T}=⋃{ℛσ,X(A):A∈𝒞σ,X};
for each A∈𝒞σ,X, it results that ℛσ,X(A)⊆max(𝒰σ,X∩𝒫(A)); moreover, when ℛσ,X(A)≠∅, it results that ℛσ,X(A) is equicardinal.
Proof.
(i): Since σ∈CLOPa(X) and Y⊆σ(Y)=σ(Z), there exists an element z∈Z such that y∈σ(z); therefore,
y∈σ(z)⊆σ((Y∖{y})∪{z})=⋃{σ(y′):y′∈(Y∖{y})∪{z}}.
(ii): Let us fix x∈X∖σ(Y) and consider the subset
B:={z∈X:σ(Y∪{x})=σ(Y∪{z})}.
Set A:=σ(Y∪{x})∖B.
Then
(3.2)σ(Y)⊆A⫋σ(Y∪{x}).
We claim that A∈𝒞σ,X.
Assume by contradiction the existence of z∈σ(A)∖A.
Clearly, we get z∈B.
Furthermore, it results that
(3.3)σ(A)=σ(Y∪{x}).
In fact, σ(A)⊆σ(Y∪{x}) by (3.2).
Conversely, again by (3.2) and by our choice of z, we get Y∪{z}⊆σ(A), whence σ(Y∪{z})=σ(Y∪{x})⊆σ(A).
This proves (3.3).
Now, in view of the above part (i), there must be some a∈A such that
x∈σ(((Y∪{x})∖{x})∪{a})=σ(Y∪{a}),
whence Y∪{x}⊆σ(Y∪{a}), and moreover, σ(Y∪{x})⊆σ(Y∪{a}).
The reverse inclusion clearly holds.
So we deduce that σ(Y∪{x})=σ(Y∪{a}), i.e. a∈B, which contradicts the definition of the subset A.
In this way, we showed that σ(A)=A, i.e. A∈𝒞σ,X.
At this point, it is immediate to verify that if A′∈𝒬 is such that A′⫋σ(Y∪{x}), then A′⊆A.
This shows that
{σ(Y∪{x}):x∈X∖σ(Y)}⊆ℐ(𝒬).
On the other hand, let Z∈ℐ(𝒬).
Thus Z∈𝒬 and ℐ𝒬(Z)≠∅.
In particular, |ℐ𝒬(Z)|=1.
Denote by Z′ the element of ℐ𝒬(Z).
Thus
(3.4)σ(Y)⊆Z′⫋Z.
Assume now by contradiction that Z≠σ(Y∪{x}) for any x∈X∖σ(Y).
By (3.4), there exists z∈Z∖σ(Y).
Therefore, the family
𝒞:={σ(Y∪{z}):z∈Z∖σ(Y)}
is non-empty and no element of 𝒞 agrees with Z.
As Z′∈ℐ𝒬(Z), it results that σ(Y∪{z})⊆Z′ for any subset σ(Y∪{z})∈𝒞.
So we get Z⊆Z′, which is clearly an absurd.
This proves that
ℐ(𝒬)⊆{σ(Y∪{x}):x∈X∖σ(Y)}.
(iii): By the definition of 𝒰σ,X, it results that Y∈𝒰σ,X if and only if y∈σ(Y∖{y}) for each y∈Y, which is equivalent to require that y∉σ(y′) for each y′∈Y∖{y}.
Clearly, such a condition is also equivalent to require that there is no T∈𝒪σ,Y∖{y} such that y∉T; therefore
𝒰σ,X={Y∈𝒫(X):y∈YandT∈𝒪σ,Y∖{y}implies thaty∈T}.
Let us now prove that 𝒰σ,X=⋃{ℛσ,X(A):A∈𝒞σ,X}.
To this aim, let Y∈𝒰σ,X, and consider any of its proper subsets Y′.
Then Y′⊆Y∖{y} for some y∈Y.
Thus σ(Y′)⊆σ(Y∖{y})≠σ(Y).
This suffices to show that Y∈REDσ(σ(Y)).
Conversely, let Y∈ℛσ,X(A) for some A∈𝒞σ,X, and fix y∈Y.
Set Y′:=Y∖{y}.
As σ(Y′)≠σ(A), it must necessarily be y∉σ(Y′)=σ(Y∖{y}), i.e. Y∈𝒰σ,X.
This concludes the proof of (3.1).
Now, let us prove that 𝒰σ,X is a finitary abstract simplicial complex.
Clearly, ∅∈𝒰σ,X.
Let Y∈𝒰σ,X, and consider Y′⫋Y.
Assume by contradiction that Y′∉𝒰σ,X.
Hence there exists y′∈Y′ such that y′∈σ(Y′∖{y′}).
But since σ is a monotone set operator, we get y′∈σ(Y∖{y′}), contradicting the fact that Y∈𝒰σ,X.
This proves that 𝒰σ,X is an abstract simplicial complex.
To prove finitarity, let Y∈𝒫(X) be such that F∈𝒰σ,X for each F⊆fY, and assume by contradiction that Y∉𝒰σ,X.
Hence there exists y∈Y such that
y∈σ(Y∖{y})=⋃{σ(y′):y′∈Y∖{y}}.
In particular, there must be some z∈Y∖{y} such that y∈σ(z).
Take now the subset Y′:={y,z}⊆Y.
In view of the above argument, we have that y∈σ(z)=σ(Y′∖{y}), contradicting our assumption on the finite subsets of Y.
This entails Y∈𝒰σ,X.
(iv): Let us firstly show that ℛσ,X(A)⊆Max(𝒰σ,X∩𝒫(A)).
For this, let B∈ℛσ,X(A).
In view of the previous part (iii), it results that B∈𝒰σ,X.
Assume that B⫋B′ for some B′∈𝒰σ,X∩𝒫(A).
In particular, we must necessarily have B′∈ℛσ,X(A).
Then we easily get σ(B′)=σ(B), contradicting the fact that B′∈ℛσ,X(A).
Thus B∈ℛσ,X(A).
We now claim that ℛσ,X(A) is equicardinal.
Assume firstly X to be finite.
Fix A∈𝒫(X), and consider two subsets U={u1,…,un},V={v1,…,vm}∈ℛσ,X(A) such that m<n.
Consider now the element v1∈V.
In view of the above part (i), there exists an element of U, say u1, such that v1∈σ((V∖{v1})∪{u1}).
Set
V(1):=V and V(2):=(V∖{v1})∪{u1}={u1,v2,…,vm}.
It is immediate to verify that σ(V(2))=σ(V(1))=σ(V)=σ(U).
Therefore, we may use again the previous part (i) to the subsets V(2) and U and to the element v2 in order to find u2∈U such that v2∈σ((V(2)∖{v2})∪{u2}).
Set
V(3):=(V(2)∖{v2})∪{v2}={u1,u2,v3,…,vm}.
Hence σ(V(3))=σ(U), and we may iterate the above procedure until we reach a subset
V(m):={u1,…,um}⫋U
such that σ(V(m))=σ(U).
Thus we conclude that U∉ℛσ,X(A), contradicting our choice of U.
This proves the claim when X is finite.
Let now X be non-finite, U,V∈ℛσ,X(A) and U also non-finite.
First of all, we claim that also the subset V must be non-finite.
To this regard,
since V⊆σ(V)=σ(U)=⋃{σ(u):u∈U}, for each v∈V, there exists uv∈U such that v∈σ(uv).
Let now Y:={uv:v∈V}.
Clearly, it results V⊆σ(Y) and Y⊆U, whence σ(Y)=σ(V)=σ(U)=σ(A).
As U∈ℛσ,X(A), we deduce that Y=U.
Now, if V were finite, even the subset Y (and so U) would be finite, which is in contrast with our assumption on U, so that also V is a non-finite subset.
Moreover, since Y=U, the map
v∈V↦uv∈U
is surjective, whence |U|≤|V|.
Reasoning as before after exchanging the role of U and V, we also get the relation |V|≤|U|.
∎
3.1 Alexandroff topologies induced by monoid actions
Let S be a monoid with identity 1S.
We recall that an ordered pair (X,θ) is called a (left) S-set if θ:S×X→X is a map, denoted by θ(s,x):=sx for any (s,x)∈S×X such that
(ab)x=a(bx) for all a,b∈S and all x∈X.
In such a case, θ is called a (left) S-action on X.
We will denote by LSETS the collection of all (left) S-sets (X,θ) of S.
When it is not necessary to assign a specific symbol θ to the action of S on X, we simply say that X is a (left) S-set, instead of saying (X,θ) is a (left) S-set, and we will write X∈LSETS.
Let us recall here that LSETS is the object class of a category 𝐋𝐬𝐞𝐭𝐒 whose morphisms between any two S-sets (X,θ) and (Y,ρ) are the maps f:X→Y such that f(θ(s,x))=ρ(s,f(x)) for all (s,x)∈S×X.
If X∈LSETS and x∈X, let Sx:={sx:s∈S}.
Now, for any Y∈𝒫(X), we set
φS(Y):=⋃{Sy:y∈Y}
so that we obtain a set operator φS on X such that φS(x)=Sx for any x∈X.
The following result may be easily shown.
Proposition 3.5.
We have that φS∈CLOPa(X).
Remark 3.6.
In what follows, when a monoid S and X∈LSETS are given, any set system on X whose definition depends by the set operator φS will be written using the symbols S and X.
For example, we will write 𝒪S,X instead of 𝒪φS,X, 𝒞S,X instead of 𝒞φS,X, 𝒯S,X instead of 𝒯φS,X, and so on for any further construction induced by φS.
Definition 3.7.
If X∈LSETS, we call 𝒯S,X the S-topology on X.
Hence, by Proposition 3.5, it follows that the S-topology of any X∈LSETS is an Alexandroff topology.
In what follows, whenever we have a given monoid S and a corresponding S-set X, we always implicitly assume that X is an Alexandroff topological space with respect to its S-topology.
In such a way, we can transfer on S any given topological property, formally:
Definition 3.8.
Let Ptop be a given topological property.
We will say then that the monoid Sinduces the topological propertyPtop on LSETS, in symbols S is 𝔗-Ptop, if any X∈LSETS satisfies the property Ptop.
As said in the introductory section, from any Alexandroff topological space X, we may construct a monoid and an action on X whose associated closure operator φS agrees with that of the starting topological closure operator.
Theorem 3.9.
Let X be an Alexandroff topological space and σ its induced topological closure operator.
Then there exists a monoid S which acts on X for which σ=φS.
Proof.
Let us consider the set S of all maps f:X→X such that f(x)∈σ(x) for all x∈X.
Then S is a monoid with respect to the usual map composition.
In fact, obviously, idX∈S.
Moreover, let f,g∈S and x∈X.
Then f(x)∈σ(x), and therefore
(g∘f)(x)=g(f(x))∈σ(f(x))⊆σ(σ(x))=σ(x)
since g∈S and σ is an increasing and idempotent set operator.
We now consider the usual (left) action (f,x)∈S×X↦f(x)∈X.
Then, with respect to such an action,
σ(z)=Sz for any z∈X.
In fact, let z∈X be arbitrary and fixed.
Then the inclusion Sz={f(z):f∈S}⊆σ(z) is an immediate consequence of the definition of S.
Conversely, let y∈σ(z), and we take the map fy:X→X defined by
fy(x):={yifx=z,xotherwise.
Then fy∈S and fy(z)=y so that y∈Sz.
Thus σ(z)⊆Sz, and hence σ(z)=Sz.
Now, for any Y∈𝒫(X), since σ∈CLOPa(X), we have that
σ(Y)=⋃{σ(y):y∈Y}=⋃{Sy:y∈Y}:=φS(Y),
and the thesis follows.
∎
We close this section introducing a specific action of the additive monoid S=ℕ (so that, in such a case, 1S agrees with 0), which we will be used in further parts of the present paper.
Let a∈Seq0(ℤ).
If (an)n∈ℕ⊆2ℤ+1, then X will contain all the possible finite truncations of a; otherwise, in the case where am is the first even integer of a, X will contain all the k-truncations of a with k≤m-1, and moreover, the truncation a|m if am≠0 and am+1=0 and also the truncation a|m+1 if am+1≠0.
Formally, we have
Seq0§(ℤ)=Seq§(ℤ)∩Seq0(ℤ),
Seq0§,*(ℤ)={a∈Seq0§(ℤ):n≥1andan=0implies thatan+1=an+2=⋯=0},
Seq0§,*(ℤ|odd):={a∈Seq0§,*(ℤ):n≥2andan≠0implies that{a0,a1,…,an-2}⊆2ℤ+1}.
Now, for any element a=(a0,a1,…,am,0,…)∈Seq0§,*(ℤ|odd), we have
ζk≥(a)=∑i=kmai fork=0,…,m,and ζm+1≥(a)=0;
therefore we may partition ℤ+ into disjoint integer intervals taking
]0,ζm≥(a)],]ζm≥(a),ζm-1≥(a)],…,]ζ2≥(a),ζ1≥(a)],]ζ1≥(a),+∞].
Notice that when a=(a0,0,0,…), the above partition of ℤ+ consists of the single integer interval ℤ+.
At this point, let us consider the map θ:(n,a)∈ℕ×Seq0§,*(ℤ|odd)↦na∈Seq0§,*(ℤ|odd) which, if a=(a0,a1,…,am,0,…)∈Seq0§,*(ℤ|odd), is defined by
(3.5)0a:=a,
1a:=(a0,a1,…,am-1,0,…),
na:={(a0,a1,…,ak-1,ζk≥(a)-n,0,…)if there exists 1≤k≤msuch thatn∈]ζk+1≥(a),ζk≥(a)],(ζ0≥(a)-n,0,…)ifn∈]ζ1≥(a),+∞],
if n≥2.
In the following result, we shall prove that θ as defined in (3.5) is an action.
Proposition 3.10.
Let X=Seq0§,*(Z|𝑜𝑑𝑑) and a=(a0,…,am,0,…)∈X.
Then
if n≥1 and (n-1)a=(c0,c1,…,cq,0,…) with q≥1, then
na={(c0,c1,…,cq-1,cq-1,0,…)𝑖𝑓cq≥1,(c0-1,0,…)𝑖𝑓c1=⋯=cq=0;
let b=(b0,…,bm′,0,…)∈X with m≤m′;
then a∈ℕb if and only if ai=bi for each i=0,…,m-1 and am≤bm.
Proof.
(i): It suffices to show that (n1+n2)a=n1(n2a).
To this regard, let us firstly observe that
(n1+n2)a={(a0,…,al-1,ζl≥(a)-n1-n2,0,…)if there exists 1≤l≤msuch thatn1+n2∈]ζl+1≥(a),ζl≥(a)],(ζ0≥(a)-n1-n2,0,…)ifn1+n2∈]ζ1≥(a),+∞]
On the other hand, let b:=n2a.
We distinguish two cases.
If n2∈]ζj+1≥(a),ζj≥(a)], through the element b, we obtain a new partition of ℕ given as follows:
]0,ζj≥(b)],]ζj≥(b),ζj-1≥(b)],…,]ζ2≥(b),ζ1≥(b)],]ζ1≥(b),+∞],
where ζr≥(b):=ζr≥(a)-n2 for any r=1,…,j.
Notice moreover that ζ0≥(b)=ζ0≥(a)-n2.
Now, we may have either n1∈]ζk+1≥(a),ζk≥(a)] for some 1≤k≤m, or n1∈]ζ1≥(a),+∞].
When n1∈]ζ1≥(a),+∞], we get
n1(n2a)=(ζ0≥(b)-n1,0,…)=(ζ0≥(a)-n1-n2,0,…)=(n1+n2)a.
Moreover, if n1∈]ζk+1≥(a),ζk≥(a)] for some 1≤k≤m, we have either
n1∈]ζk′+1≥(b),ζk′≥(b)]=]ζk′+1≥(a)-n2,ζk′≥(a)-n2],
or n1∈]ζ1≥(b),+∞].
In the first situation, we necessarily get k′=l and hence
n1(n2a)=(a0,…,al-1,ζl≥(b)-n1,0,…)=(a0,…,al-1,ζl≥(a)-n1-n2,0,…)=(n1+n2)a,
while in the second, we get n1(n2a)=(ζ0≥(b)-n1,0,…)=(ζ0≥(a)-n1-n2,0,…)=(n1+n2)a.
If n2∈]ζ1≥(a),+∞[, we have n2a=(ζ0≥(a),0,…).
Thus, applying (3.5), we get
n1(n2a)=(ζ0≥(a)-n1-x2,0,…)=(n1+n2)a.
(ii): Let us firstly assume that n∈]ζk+1≥(a),ζk≥(a)] for some k∈{1,…,m}.
Two cases may occur.
If n<ζk≥(a), we have n+1∈]ζk+1≥(a),ζk≥(a)]; then
na=(a0,a1,…,ak-1,ck,0,…),
ck=ζk≥(a)-n,
and we get
(n+1)a=(a0,a1,…,ak-1,ζk≥(a)-(n+1),0,…)=(a0,a1,…,ak-1,ck-1,0,…).
If n=ζk≥(a), we have n+1∈]ζk+1≥(a),ζk≥(a)]; hence na=(a0,a1,…,ak-1,0,0,…), and subsequently we get
(n+1)a=(a0,a1,…,ak-2,ζk-1≥(a)-(n+1),0,…)=(a0,a1,…,ak-2,ζk-1≥(a)-ζk≥(a)-1,0,…)=(a0,a1,…,ak-2,ak-1-1,0,…).
On the other hand, if n∈]ζ1≥(a),+∞[, then we have na=(c0,0,…), where c0:=ζ0≥(a)-n.
Moreover, as n+1∈]ζ1≥(a),+∞[, we conclude that
(n+1)a=(ζ0≥(a)-(n+1),0,…)=(ζ0≥(a)-n-1,0,…)=(c0-1,0,…).
(iii): Let us assume that a∈ℕb so that there exists n∈ℕ such that a=nb.
This implies that
n=ζm+1≥(b)+(bm-am),ai=bifor eachi=0,…,m-1andam≤bm.
Conversely, let a,b∈X be such that ai=bi for each i=0,…,m-1 and am≤bm; just take
n:=ζm+1≥(b)+(bm-am)
in order to get the relation a=nb.
∎
4 Attractive closed subsets and first related results
In this section, we introduce the notion of attractive closed subset and next investigate some basic properties of this kind of subsets and of the corresponding set system, paving a particular attention to its maximal members which will have a relevant role in the whole analysis of the Alexandroff closure operators.
From now on, in all our general results, we will assume that X is a non-empty Alexandroff topological space and that σ∈CLOPa(X) is its associated topological closure operator.
On the other hand, when we treat specific cases of Alexandroff spaces induced by monoid actions, we will specify what is our space X and the corresponding monoid which acts on X.
Definition 4.1.
Let Y∈𝒞σ,X.
We say that Y is attractive if for any y1,y2∈Y there exists y∈Y such that y1,y2∈σ(y) and denote by 𝒜σ,X the set system of all attractive subsets of X.
Let us note that σ(x)∈𝒜σ,X for any x∈X; therefore 𝒪σ,X⊆𝒜σ,X⊆𝒞σ,X, and hence
also 𝒜σ,X is a covering of X since 𝒪σ,X has such a property.
Remark 4.2.
In what follows, we will also use the notations 𝒪σ,Z and 𝒜σ,Z, when Z is a closed subset of X, because in such a case we can consider the restriction of σ on 𝒫(Z) again an Alexandroff closure operator on the topological subspace Z.
Moreover, in such a case, it is clear that 𝒜σ,Z⊆𝒜σ,X.
In this work, a fundamental role is played by the families of maximal attractive subsets which are also coverings of X.
In the next definition, we formalize such a notion.
Definition 4.3.
Let ℱ be a set system on X.
We say that ℱ is a maximal attractive covering of X if ℱ⊆Max(𝒜σ,X) and ℱ is a covering of X.
We denote by MACσ(X) the collection of all maximal attractive coverings of X.
Now, we give some specific classes of Alexandroff topological spaces which we shall use in the remaining part of the paper.
Definition 4.4.
We say that the Alexandroff space X (or equivalently its topology) is
essentially attractive if 𝒪σ,X=𝒜σ,X,
essentially Noetherian if the set system 𝒪σ,X is Noetherian,
essentially Artinian if the set system 𝒪σ,X is Artinian,
MAC-union independent if there exists some ℱ∈MACσ(X) which is union independent,
MA-union dependent if the set system Max(𝒜σ,X) is union dependent,
MA-union undetermined if X is not MAC-union independent and X is not MA-union dependent.
Through monoids, we may provide various classes of Alexandroff spaces which are essentially Noetherian or essentially Artinian.
Let us note that if S is a group, then either Sx=Sx′ or they are disjoint; therefore we can never have a proper inclusion of the type Sx⫋Sx′.
Hence any group is trivially both essentially Artinian and essentially Noetherian.
Now, we see an example of a monoid which is essentially Noetherian but not essentially Artinian.
Let us consider the monoid ℕ0, whose ground set is ℕ and its operation is ss′:=max{s,s′}, so that 1ℕ0=0.
Then we have the following result.
Proposition 4.5.
Let S=N0.
Then the following conditions hold:
S is 𝔗-essentially Noetherian;
S is not 𝔗-essentially Artinian.
Proof.
(i): We assume by contradiction the existence of W∈LSETS for which the set system 𝒪S,W is not Noetherian.
Then there exists a set system ℱ={Sxn:n∈ℕ}⊆𝒪S,W which is strictly increasing, i.e.
Sx0⫋Sx1⫋Sx2⫋⋯.
Let now {sn:n≥1}⊆S be such that x0=s1x1, x1=s2x2, x2=s3x3,….
Then it must be s1>s2.
In fact, if it were s1≤s2, then s1s2=s2, and therefore x0=s1x1=s1(s2x2)=(s1s2)x2=s2x2=x1, which is in contrast with the strict inclusion Sx0⫋Sx1.
Analogously, with the same procedure, we can see that s2>s3.
Then, proceeding in such a way, we obtain an infinite strictly decreasing sequence s1>s2>s3>⋯, and this in contrast with the condition s1∈ℕ.
(ii): Let us consider the set W=ℕ and the S-action
(s,n)∈S×W↦sn:=max{s,n}∈W.
Then W∈LSETS, and it results that S0=ℕ⫌S1=ℕ∖{0}⫌S2=ℕ∖{0,1}⫌⋯.
Hence W is not 𝔗-essentially Noetherian.
∎
On the other hand, we now provide an example of a monoid which is essentially Artinian but is not essentially Noetherian.
Let us consider the totally ordered set (ℕ∞:=ℕ∪{∞},<), where ∞ is a formal symbol and the usual total order < on ℕ is extended on ℕ∞ by setting n<∞ for any n∈ℕ.
On ℕ∞, we consider the binary operation ss′:=min{s,s′} for any s∈ℕ∞.
Then ℕ∞ becomes a commutative monoid for which 1ℕ∞=∞.
Proposition 4.6.
Let S=N∞.
Then the following conditions hold:
S is 𝔗-essentially Artinian;
S is not 𝔗-essentially Noetherian.
Proof.
(i): We assume by contradiction that there exists W∈LSETS for which the set system 𝒪S,W is not Artinian.
Then there exists a set system ℱ={Sxn:n∈ℕ}⊆𝒪S,W which is strictly decreasing, i.e.
Sx0⫌Sx1⫌Sx2⫌⋯.
Let now {sn:n≥0}⊆S be such that x1=s0x0, x2=s1x1, x3=s2x2,….
Since Sx0⫌Sx1, it follows that s0≠∞; therefore s0∈ℕ.
We note now that it must be s1<s0.
In fact, if it were s1≥s0, then s1s0=s0, and therefore x2=s1x1=s1(s0x0)=(s1s0)x0=s0x0=x1, which is in contrast with the strict inclusion Sx1⫌Sx2.
Analogously, with the same procedure, we can see that s2<s1.
Then, proceeding in such a way, we obtain an infinite strictly decreasing sequence s0>s1>s2>⋯,
in contrast with the condition s0∈ℕ.
(ii): Consider the set W=ℕ and the S-action
(s,n)∈S×W↦sn:=min{s,n}∈W.
Then W∈LSETS, and it results that S0={0}⫋S1={0,1}⫋S2={0,1,2}⫋⋯.
Hence W is not 𝔗-essentially Noetherian.
∎
In the next result, we will firstly provide some basic properties of the set system 𝒜σ,X, and next, we shall show that, whenever X is essentially Noetherian, then the set systems 𝒪σ,X and 𝒜σ,X agree.
On the other hand, the role of the condition of being essentially Artinian will be highlighted starting from Proposition 5.3.
Proposition 4.7.
The following conditions hold:
the set system 𝒜σ,X is chain union closed, i.e. if ℬ⊆𝒜σ,X is a chain, then ⋃ℬ∈𝒜σ,X;
for any Y∈𝒜σ,X, there exists M∈Max(𝒜σ,X) such that Y⊆M;
the set system Max(𝒜σ,X) is a covering of X;
if X is essentially Noetherian, then X is essentially attractive;
if Y∈𝒜σ,X and x∈Y, then Y=⋃{σ(z):z∈Y𝑎𝑛𝑑x∈σ(z)};
if x∣𝒪σ,Xy, then there exists (M,M′)∈Max(𝒜σ,X)2,≠ such that x∈M∖M′ and y∈M′∖M.
Proof.
(i): Let ℬ⊆𝒜σ,X be a chain.
Since 𝒞σ,X is union closed, we have ⋃ℬ∈𝒞σ,X.
Let now z1,z2∈⋃ℬ.
Hence there exists Z∈ℬ containing both z1 and z2.
In particular, since Z is an attractive subset of X, there exists z∈Z such that z1,z2∈σ(z).
In particular, z∈⋃ℬ, and we conclude that ⋃ℬ∈𝒜σ,X.
(ii): Let Y∈𝒜σ,X, and consider the set system 𝒜σ,X(Y):={Z∈𝒜σ,X:Y⊆Z}.
Clearly, 𝒜σ,X(Y) is non-empty because Y∈𝒜σ,X.
Moreover, if 𝒞 is a chain in (𝒜σ,X(Y),⊆), then ⋃𝒞∈𝒜σ,X(Y) by the previous part (i).
Therefore, by Zorn’s lemma, the poset (𝒜σ,X(Y),⊆) contains a maximal element M.
Clearly, M∈Max(𝒜σ,X), and hence the thesis follows.
(iii): We claim that, for each element x∈X, there exists some M∈Max(𝒜σ,X) such that x∈M.
To this regard, if x∈X, then x∈σ(x) and σ(x)∈𝒜σ,X.
So the result follows as a direct consequence of the previous part (ii).
(iv): Let us assume that 𝒪σ,X is Noetherian.
We must show that 𝒜σ,X⊆𝒪σ,X.
Let us suppose by contradiction that there exists Y∈𝒜σ,X such that Y∉𝒪σ,X.
Let y1∈Y.
By the hypothesis on Y, it results that σ(y1)⫋Y.
Let therefore y1′∈Y∖σ(y1).
Since Y∈𝒜σ,X, there exists y2∈Y such that {y1,y1′}⊆σ(y2).
Again, since σ(y2)⫋Y, we may continue the procedure.
In such a way, we can construct an ascending chain of the type
σ(y1)⫋σ(y2)⫋⋯⫋σ(yn)⫋σ(yn+1)⫋⋯⫋Y⊆X,
and this is in contrast with the hypothesis on 𝒪σ,X.
Hence the thesis follows.
(v): Let x∈Y be a fixed element, and take some y∈Y.
Since Y is a attractive subset, there exists z∈Y such that x,y∈σ(z).
Thus Y⊆⋃{σ(z):z∈Y,x∈σ(z)}.
The reverse inclusion is obvious.
(vi): Let x,y∈X be such that x∣𝒪σ,Xy.
As x∈σ(x) and σ(x)∈𝒜σ,X, in view of the above part (ii), there exists M∈Max(𝒜σ,X) such that x∈M.
Similarly, there exists M′∈Max(𝒜σ,X) such that y∈M′.
Now, notice that y∉M; otherwise we would find z∈M such that {x,y}⊆σ(z), contradicting the fact that x∣𝒪σ,Xy.
Similarly, x∉M′, and the thesis holds.
∎
Remark 4.8.
By part (iii) of Proposition 4.7 and since the set system Max(𝒜σ,X) is a covering of X, we deduce that X is MAC-union independent when it is union independent.
We determine now the family of all the attractive subsets and its maximal members relatively to the monoid S=ℕ and to its action (see Proposition 3.10) on the set X=Seq0§,*(ℤ|odd) defined in (3.5).
To this regard, we firstly set
Λ1:={a∈Seq0(ℤ):{an}n≥0⊆2ℤ+1},
Λ2:={a∈X:for allm≥0,am≠0andam+1=0implies thatam=1},
and Λ:=Λ1∪Λ2.
In the next result, we shall show that any member of 𝒜ℕ,X belongs to 𝒵ℕ,X.
Proposition 4.9.
Let X=Seq0§,*(Z|𝑜𝑑𝑑).
Then AN,X⊆ZN,X.
Proof.
Let T∈𝒜ℕ,X.
Then T∈𝒞ℕ,X, i.e. T=⋃{ℕt:t∈T}.
There is nothing to prove if T∈𝒪ℕ,X.
Let now a=(a0,…,al,0,…),b=(b0,…,bm,0,…)∈T.
We claim that a∈ℕb or b∈ℕa.
Without loss of generality, assume that l≤m.
As T∈𝒜ℕ,X, there must exist c=(c0,…,ck,0,…)∈T such that a,b∈ℕc.
So l≤m≤k, and in view of part (iii) of Proposition 3.10, we conclude that ai=bi=ci for each i=0,…,l-1.
At this point, if l=m, then we clearly have either a∈ℕb or b∈ℕa depending on whether either al≤bl or bl≤al.
On the other hand, if l<m, we clearly have b∉ℕa in view of part (iii) of Proposition 3.10.
Assume by contradiction that a∉ℕb.
Thus al>bl.
Nevertheless, we have cl≥al>bl and cl=bl again by part (iii) of Proposition 3.10.
This is an absurd.
So we proved that a∈ℕb or b∈ℕa.
In other terms, for any a,b∈T, it results that either ℕa⊆ℕb or ℕb⊆ℕa.
∎
Let us determine the attractive subsets with respect to the action of ℕ on Seq0§,*(ℤ|odd).
Proposition 4.10.
Let X=Seq0§,*(Z|𝑜𝑑𝑑).
Then T∈AN,X∖ON,X
if and only if T assumes one of the following forms:
⋃{ℕa|m:m∈ℕ}, where a=(a0,…,am,am+1,…)∈Λ1;
ℕa|m-1∪{(a0,…,am-1,k,0,…):k∈ℤ+}, where a=(a0,…,am-1,1,0,…)∈Λ2.
In particular, it results that
(4.1)Max(𝒜ℕ,X)=𝒜ℕ,X∖𝒪ℕ,X.
Proof.
Let us prove that if T assumes one among the forms given in the statement, then T∈𝒜ℕ,X∖𝒪ℕ,X.
Let a=(a0,…,am,am+1,…)∈Λ1.
We want to prove that Ta:=⋃{ℕa|m:m∈ℕ}∈𝒜ℕ,X∖𝒪ℕ,X.
First of all, we have that Ta∈X.
In fact, if z∈Ta, then there exist m,n∈ℕ such that z=n(a0,…,am,0,…).
At this point, just notice that a|m∈X, whence z=na|m∈X.
We firstly claim that φℕ(Ta)=Ta.
To this regard, it suffices to show that φℕ(Ta)⊆Ta.
Let x∈φℕ(Ta).
Thus there exists y∈Ta such that x∈ℕy.
In other terms, there exists n1∈ℕ such that x=ny.
Furthermore, the fact that y∈Ta implies the existence of two integers n2,m∈ℕ such that y=n2a|m.
Thus
x=n1y=n1(n2a|m)=(n1+n2)a|m∈ℕa|m⊆Ta
because Ta∈𝒫(X).
Thus Ta∈𝒞ℕ,X for each a∈Λ1.
Now, we claim that Ta∈𝒜ℕ,X∖𝒪ℕ,X.
To this regard, we must show that, for each y1,y2∈Ta, there exists y∈Ta such that y1,y2∈ℕy.
Let y1=n1a|m1 and y2=n2a|m2 for some n1,m1,n2,m2∈ℕ.
Take m:=Max(m1,m2)=m1 (the other case is similar), and set y:=a|m.
Then one may easily check that y1=n1y and y2=(n2+ζm2+1≥(y))y, whence y1,y2∈ℕy.
Thus Ta∈𝒜ℕ,X∖𝒪ℕ,X.
Let now a=(a0,…,am-1,1,0,…)∈Λ2.
We want to prove that
Wa:=ℕa|m-1∪{(a0,…,am-1,k,0,…):k∈ℤ+}∈𝒜ℕ,X∖𝒪ℕ,X.
For this, take w∈Wa.
Two cases may occur: if w=na|m-1 for some n∈ℕ, then w∈X in view of the fact that a∈X and na∈X for each n∈ℕ; while if w=(a0,…,am-1,k,0,…) for some k∈ℤ+, we clearly have w∈X.
We claim that Wa∈𝒞ℕ,X.
Let x∈φℕ(Wa).
Then there exist n1∈ℕ and y∈Wa such that x=n1y.
Now, to say that y∈Wa means that either there exists n2∈ℕ such that y=n2a|m-1 or y=(a0,…,am-1,k,…) for some k∈ℤ+.
Notice that, in the first case, we get
x=n1(n2a|m-1)=(n1+n2)a|m-1.
However, in both the above situations, we easily get x∈Wa.
This proves that φℕ(Wa)=Wa, i.e. Wa∈𝒞ℕ,X.
At this point, we claim that Wa∈𝒜ℕ,X∖𝒪ℕ,X.
Three cases may occur:
if y1=n1a|m-1 and y2=n2a|m-1 for some n1,n2∈ℕ, just take y:=(a0,…,am-1,0,…)∈Wa;
if y1=n1a|m-1 and y2=(a0,…,am-1,k,0,…) for some n1∈ℕ and k∈ℤ+, just take y:=y2 since y1=(n1+k)y=(n1+k)y2;
if y1=(a0,…,am-1,k1,0,…) and y2=(a0,…,am-1,k2,0,…), take y=(a0,…,am-1,k,0,…), where k:=max{k1,k2}.
Therefore, Wa∈𝒜ℕ,X∖𝒪ℕ,X.
In this way, we showed that if T assumes one of the two forms given in the statement, then it belongs to 𝒜ℕ,X∖𝒪ℕ,X.
Conversely, let T∈𝒜ℕ,X∖𝒪ℕ,X.
Hence T=⋃{ℕt:t∈T} since T∈𝒞ℕ,X.
Moreover, in view of Proposition 4.9, it results that T∈𝒵ℕ,X, i.e. the set system {ℕt:t∈T} is a chain.
Thus T may assume one among the wanted forms:
ℕa|m-1∪{(a0,…,am-1,k,0,…):k∈ℤ+},
where a=(a0,…,am-1,1,0,…)∈Λ2;
⋃{ℕa|m:m∈ℤ+},
where a=(a0,a1,…)∈Λ1.
At this point, equation (4.1) follows easily.
∎
5 Some characterizations for union dependence
In this section, we investigate the property of an arbitrary Alexandroff topological space of being or not MA-union dependent.
For instance, given a maximal attractive covering for X, the fact that X is MA-union dependent space X ensures that the aforementioned covering is weakly union dependent.
Furthermore, the condition of being MA-union dependent may be characterized in various ways, while non-MA-union dependence is in general not equivalent to MAC-union independence.
Nevertheless, the last case holds from a categorial point of view since we shall prove that a monoid S is not MA-union dependent if and only if it is MAC-union independent, i.e. any left S-set is not MA-union dependent if and only if any left S-set admits a maximal attractive covering which is union independent.
The following basic result shall be used several times in the present paper.
Proposition 5.1.
Let Y,Z∈Cσ,X be such that X=Y∪Z.
Then
C∈𝒜σ,X implies that C⊆Y or C⊆Z;
Y=σ(X∖Z) implies that there exists ℱ⊆Max(𝒜σ,X) such that Y=⋃ℱ.
Proof.
(i): Let us suppose that C⊈Y.
Then there exists an element b∈C such that b∈Z and b∉Y.
Since C is attractive, for any c∈C, there exists an element bc∈M such that b,c∈σ(bc).
We note that bc∉Y for any c∈C.
In fact, if it were bc∈Y, then b∈σ(bc)⊆Y since Y∈𝒞σ,X.
Therefore, bc∈Z for all c∈C.
Hence, since also Z∈𝒞σ,X, it follows that c∈σ(bc)⊆Z for all c∈C.
This shows that C⊆Z.
(ii): Let a∈Y.
In view of our assumption, there exists xa∈X∖Z such that a∈σ(xa).
In view of part (iii) of Proposition 4.7, there exists Ma∈Max(𝒜σ,X) such that xa∈Ma.
Then, since Ma∈𝒞σ,X, we have that a∈σ(xa)⊆Ma.
Moreover, since xa∈X∖Z, we also have that Ma⊈Z.
Then, by the previous part (i), we have that
(5.1)a∈Ma⊆Y.
At this point, if we take ℱ:={Ma:a∈Y}, by (5.1), we have that ℱ⊆Max(𝒜σ,X) and Y=⋃ℱ.
∎
Let us now provide another technical result where some basic properties of the set system Max(𝒜σ,X)ud shall be studied.
Lemma 5.2.
The following conditions hold:
if M∈𝒜σ,X∖𝒪σ,X and x∈M, there exists ux∈M∖{x} such that σ(x)⫋σ(ux);
if M∈Max(𝒜σ,X)ud and x∈M, then there exist zx∈X∖M and yx∈M such that σ(yx)∥σ(zx) and x∈σ(zx).
Proof.
(i): Suppose the thesis to be false.
Hence there exists x∈M such that, for any z∈M, it results that
(5.2)σ(x)=σ(z) or σ(x)⊈σ(z).
Let now y∈M be arbitrary.
Since M∈𝒜σ,X and x,y∈M, there must be some zx,y∈M such that
(5.3){x,y}⊆σ(zx,y).
In view of (5.2) and (5.3), we get σ(x)=σ(zx,y), and therefore y∈σ(x).
By the arbitrariness of y, we deduce that M⊆σ(x), and thus M=σ(x), contradicting our assumption on M.
(ii): Assume the claim to be false.
Hence there exists some M∈Max(𝒜σ,X)ud∩𝒪σ,X.
Then M=σ(x) for some x∈X.
Now, since M∈Max(𝒜σ,X)ud, we have that M⊆⋃(Max(𝒜σ,X)∖{M}); therefore there exists some Z∈Max(𝒜σ,X), with Z≠M, such that y∈Z.
Then M=σ(y)⫋Z, which is in contrast with the maximality of M in 𝒜σ,X.
(iii): Let x∈M be fixed.
Assume firstly by contradiction that,
(5.4)for allz∈X∖M,x∉σ(z).
Since M∈Max(𝒜σ,X)ud, we have that M⊆⋃(Max(𝒜σ,X)∖{M}); therefore there exists M′∈Max(𝒜σ,X) such that M≠M′ and x∈M′.
Now, in view of the conditions on M′, we clearly infer that M′⊈M.
Let therefore ux∈M′∖M.
Since {x,ux}⊆M′, there exists zx∈M′ such that {x,ux}⊆σ(zx).
If zx∈M, we would have that ux∈σ(zx)⊆M, contradicting the choice of ux.
Therefore, zx∈X∖M and x∈σ(zx), in contrast with (5.4).
Let now zx∈X∖M such that x∈σ(zx).
We claim the existence of yx∈M such that yx∉σ(zx).
In fact, if this were false, we would have that y∈σ(zx) for each y∈M, whence M⊆σ(zx).
However, since zx∈X∖M, then M⫋σ(zx)∈𝒜σ,X, contradicting the maximality of M in 𝒜σ,X.
Finally, notice that the condition zx∉σ(yx) holds since zx∈X∖M, while yx∈M.
∎
The condition for an Alexandroff topological space of being essentially Artinian is very useful in order to find for each element x of a subset of M∈Max(𝒜σ,X)ud two elements, in M and in its complement respectively, whose closures both contain x.
Proposition 5.3.
Let X be essentially Artinian, M∈Max(Aσ,X)ud and x∈M.
Then there exist zx′∈X∖M and yx′∈M such that σ(yx′)∥σ(zx′) and x∈σ(yx′)∩σ(zx′).
Proof.
Let x∈M be fixed.
Take yx and zx as in part (iii) of Lemma 5.2, and choose zx′:=zx.
Now, if x∈σ(yx), we take yx′:=yx to obtain the thesis.
Therefore, let us assume x∉σ(yx).
Since M∈𝒜σ,X, there exists u1∈M such that {x,yx}⊆σ(u1).
Now, if u1∉σ(zx′), the thesis holds taking yx′:=u1.
We may then assume u1∈σ(zx′).
Since u1∈M and zx′∉M, it follows that zx′∉σ(u1)⊆M.
Thus
σ(zx′)⫌σ(u1).
By part (ii) of Lemma 5.2, we have that M∉𝒪σ,X.
So apply part (i) of Lemma 5.2 to M and to the element u1 in order to get an element u2∈M such that σ(u1)⫋σ(u2).
In particular, x∈σ(u2).
If u2∉σ(zx′), we take yx′:=u2 in order to obtain the thesis.
Therefore, we may assume u2∈σ(zx′).
Moreover, since u2∈M and zx′∉M, we have that zx′∉σ(u2)⊆M.
Thus we get
σ(zx′)⫌σ(u2)⫌σ(u1).
Now, since X is essentially Artinian, the previous procedure cannot be infinite, otherwise we would obtain an infinite proper descending chain in 𝒪σ,X.
So there exist n∈ℤ+ and un∈M such that un∉σ(zx′) and x∈σ(un).
Set yx′:=un in order to get the thesis.
∎
We consider now the following subset (possibly empty) of X, which we will be relevant in order to describe some various general results of the present paper:
Uσ,X:={x∈X:there is exactly oneY∈Max(𝒜σ,X)such thatx∈Y}.
As Max(𝒜σ,X) is a covering of X, notice that
(5.5)X∖Uσ,X={x∈X:there exists(Y,Z)∈Max(𝒜σ,X)2,≠such thatx∈Y∩Z}.
In the next result, in reference to the monoid ℕ and to its action on Seq0§,*(ℤ|odd) defined in (3.5), we will characterize all the sequences of Seq0§,*(ℤ|odd) belonging to Uℕ,Seq0§,*(ℤ|odd).
To this regard, for any a=(a0,…,am,0,…)∈Seq0§,*(ℤ|odd), we set
(5.6)γ(a)={(a0,…,am-1,am+1,1,0,…)ifm=0,orm>0and{am-1,am}⊆2ℤ+1,(a0,…,am,1,0,…)ifam-1∈2ℤ+1andam∈2ℤ,(a0,…,am-1,1,0,…)ifam-1∈2ℤ.
Furthermore, in the next result, in order to simply notations within the proof, we shall use the map Ψ:Λ2→𝒫(Seq0(ℤ)) defined as follows:
(5.7)Ψ(a)=ℕa|m-1∪Γm-1(a)
for any a=(a0,…,am-1,1,0,…)∈Λ2.
Then we obtain the following characterization for Uℕ,X.
Proposition 5.4.
Let X=Seq0§,*(Z|𝑜𝑑𝑑) and a=(a0,…,am,0,…)∈X.
Then
a∈Uℕ,X if and only if am-1∈2ℤ.
Proof.
(i): immediate by both (5.7) and (5.6).
(ii): Let b∈Λ2.
In view of (5.6), it easily follows that γ(b)=(b0,…,bm-1,1,0,…)=b.
Conversely, let b∈Fix(Λ2).
Then γ(b)=b.
As b=(b0,…,bm-1,bm,0,…), in view of (5.6), we must necessarily have bm-1∈2ℤ and bm=1, i.e. b∈Λ2.
(iii): In view of the previous part (i), we have that γ(a)∈Λ2.
Now, by both (5.6) and (5.7), we easily get a∈Ψ(γ(a)).
(iv): Let a=(a0,…,am-1,1,0,…)∈X with am-1∈2ℤ.
In view of the previous part (iii), it results that a∈Ψ(γ(a)), where γ(a)=(a0,…,am-1,1,0,…).
Assume now that a∈Ψ(b) for some b∈Λ such that b≠γ(a).
It cannot be b∈Λ1; otherwise the condition a=nb|m′ for some n,m′∈ℕ would imply that b contains some even term, contradicting our choice of b.
So it must follow b∈Λ2, i.e. b=(b0,…,bl-1,1,0,…).
Now, if a∈{(b0,…,bl-1,k,…):k∈ℤ+}, we must necessarily have that l=m and, in particular, that b=γ(a), contradicting our choice of b.
Therefore, there must necessarily be an integer n such that a=nb|l-1.
In view of part (iii) of Proposition 3.10 and of the fact that am-1∈2ℤ, the latter equality holds if and only if l-1=m and bi=ai for any i=0,…,m-1 and am<bm.
Nevertheless, since bm=1, we conclude that am is a positive integer strictly less than 1, absurd.
This proves that there exists only a member of Max(𝒜ℕ,X), namely Ψ(γ(a)), containing a, whence a∈Uℕ,X.
Conversely, assume that a∈Uℕ,X.
Suppose by contradiction that am-1∈2ℤ+1.
In view of the previous part (iii), we get a∈Ψ(γ(a)), where
γ(a)={(a0,…,am-1,am+1,1,0,…)ifm=0oram∈2ℤ+1,(a0,…,am,1,0,…)ifam∈2ℤ.
Now, if γ(a)=(a0,…,am-1,am+1,1,0,…), by part (iii) of Proposition 3.10, it may be easily shown that a∈⋃{ℕb|s:s∈ℕ} for each b∈Λ1 such that b=(a0,…,am-1,bm,…) and bm>am; otherwise, if γ(a)=(a0,…,am,1,0,…), just notice that a∈Ψ(b) for each b=(b0,b1,…)∈Λ1, where ai=bi for each i=0,…,m-1 and bm=am+k with k∈2ℤ+1.
In both the previous situations, we get a∉𝕌ℕ,X, contradicting our assumption.
∎
At this point, we study the main properties of the maximal attractive coverings of an Alexandroff topological space.
Proposition 5.5.
Let F∈MACσ(X).
Then the following conditions hold:
if ℱ is union independent, for any Y∈ℱ we have that
Y⊈⋃(Max(𝒜σ,X)∖{Y});
if ℱ is union independent and 𝒢∈MACσ(X), then ℱ⊆𝒢;
if ℱ is union independent and 𝒢∈MACσ(X) is also union independent, then ℱ=𝒢;
ℱ is union independent if and only if Y∩Uσ,X≠∅ for any Y∈ℱ;
if X is MA-union dependent, then ℱ is weakly union dependent.
Proof.
(i): We assume that ℱ is union independent.
Let B:=⋃(ℱ∖{Y}).
Then Y,B∈𝒞σ,X in view of the fact that ℱ⊆Max(𝒜σ,X), and it results that
(5.8)X=Y∪B.
We assume by contradiction that Y⊆⋃(Max(𝒜σ,X)∖{Y}).
Let then K∈Max(𝒜σ,X)∖{Y}.
Since K and Y are both maximal members of 𝒜σ,X and K≠Y, it follows that
K⊈Y.
Therefore, by (5.8) and by part (i) of Proposition 5.1, we deduce that K⊆B.
Hence, by the arbitrariness of K in Max(𝒜σ,X)∖{Y}, we have that
Y⊆⋃(Max(𝒜σ,X)∖{Y})⊆B=⋃(ℱ∖{Y}),
which is in contrast with the hypothesis that ℱ is union independent.
(ii): Let ℱ be union independent and 𝒢∈MACσ(X).
Let Y∈ℱ, and let us assume by contradiction that Y∉𝒢 so that 𝒢⊆Max(𝒜σ,X)∖{Y}.
Then, since 𝒢 is a covering of X, it follows that
Y⊆X=⋃𝒢⊆⋃(Max(𝒜σ,X)∖{Y}),
which is in contrast with the thesis of the previous part (i).
(iii): It is an immediate consequence of the previous part (ii).
(iv): Let ℱ be union independent and Y∈ℱ.
By the above part (i), we have that Y⊈⋃(Max(𝒜σ,X)∖{Y}).
Therefore, there exists an element x∈Y such that x∉F for all F∈Max(𝒜σ,X) such that F≠Y.
Thus Y is the only maximal element of 𝒜σ,X containing x, and hence we get x∈Y∩Uσ,X.
Hence Y∩Uσ,X≠∅.
Conversely, we assume that Y∩Uσ,X≠∅ for any Y∈ℱ and, by contradiction, that ℱ is not union independent.
Therefore, there exists F∈ℱ such that
(5.9)F⊆⋃(ℱ∖{F}).
On the other hand, by our hypothesis, we also have F∩Uσ,X≠∅ so that there exists an element
(5.10)x∈F∩Uσ,X.
Then, by (5.9) and (5.10), we obtain a subset K∈ℱ, K≠F such that x∈F∩K.
Hence we found two distinct maximal elements of the set system 𝒜σ,X which contains the element x∈Uσ,X, and this in contrast with the definition of the subset Uσ,X.
This shows that ℱ is union independent.
(v): It is an immediate consequence of the previous part (i).
∎
Remark 5.6.
By part (iii) of Proposition 5.5, we see that, whenever X is MAC-union independent, there exists a unique independent set system ℱ∈MACσ(X).
We will denote it by the symbol 𝔖σ,X.
Definition 5.7.
If X is MAC-union independent, we call the above set system 𝔖σ,X the standard MAC-union independent system of X.
We now exhibit the relation between essential attractiveness with that of being MAC-union independent.
Proposition 5.8.
If X is essentially attractive, then X is also MAC-union independent and
𝔖σ,X=Max(𝒜σ,X)=Max(𝒪σ,X).
Proof.
Since Max(𝒜σ,X) is a covering of X, it is sufficient to show that Max(𝒜σ,X) is an independent set system of X.
Let therefore Y∈Max(𝒜σ,X).
In view of our assumption, we have that Y=σ(y) for some y∈X.
Let us assume by contradiction that Y⊆⋃Max(𝒜σ,X)∖{Y}.
Then there exists some Z∈Max(𝒜σ,X), Z≠Y, such that y∈Z.
Now, since 𝒪σ,X=𝒜σ,X, there exists some z∈X such that Z=σ(z).
This implies that σ(y)⫋σ(z), which is in contrast with the maximality of σ(y) in 𝒪σ,X.
Hence the thesis follows.
∎
In the next result, we will characterize the condition for an Alexandroff topological space X of being MA-union dependent showing, for each element x of X, the existence of two elements not belonging to the closure of a same point of X and whose closures both contain X.
In particular, such a condition may be also expressed in terms of the subset Uσ,X.
Theorem 5.9.
The following conditions are equivalent:
for any x∈X, there exist (Y,Z)∈Max(𝒜σ,X)2,≠ such that x∈Y∩Z;
for any x∈X, there exist y,z∈X such that y∣𝒪σ,Xz and x∈σ(y)∩σ(z);
Proof.
(i) ⇒ (ii):
Let x∈X.
Since Max(𝒜σ,X) is a covering of X, there exists Y∈Max(𝒜σ,X) such that x∈Y.
Now, since Max(𝒜σ,X) is dependent, we have that
x∈Y⊆⋃(Max(𝒜σ,X)∖{Y}).
Therefore, there exists some Z∈Max(𝒜σ,X), with Z≠Y, such that x∈Z.
(ii) ⇒ (i):
We must show that Max(𝒜σ,X) is dependent.
Let therefore Y∈Max(𝒜σ,X).
To this regard, let us note that, for every x∈Y, there exists a maximal element Zx∈Max(𝒜σ,X), with Zx≠Y, such that x∈Zx; otherwise there would be some element of y∈Y for which Y is the only member of Max(𝒜σ,X) containing y, contradicting the assumption in (ii).
Therefore, it results that
Y⊆⋃{Zx:x∈Y}⊆⋃(Max(𝒜σ,X)∖{Y}).
Hence Max(𝒜σ,X) is dependent.
(iii) ⇒ (ii):
Let x∈X.
In view of our assumption, there exist y,z∈X such that x∈σ(y)∩σ(z) and {y,z}⊈T for all T∈𝒪σ,X.
Let Y,Z∈Max(𝒜σ,X) be such that σ(y)⊆Y and σ(z)⊆Z.
Then x∈Y∩Z.
Moreover, Y≠Z.
In fact, if by contradiction, it were Y=Z, then {y,z}⊆σ(y)∪σ(z)⊆Y, and since Y is attractive, there exists w∈Y such that {y,z}⊆σ(w)∈𝒪σ,X, which is in contradiction with our hypothesis.
(i) ⇒ (iii):
We will prove the contrapositive proposition.
Let us assume therefore that there exists an element x∈X such that
(5.11)ify,z∈Xandx∈σ(y)∩σ(z)⟹there existsTy,z∈𝒪σ,Xsuch that{y,z}⊆Ty,z.
We must prove that the set system Max(𝒜σ,X) is not dependent.
To this regard, we consider the subset
Ux:=⋃{T∈𝒪σ,X:x∈T}.
Clearly, x∈Ux; therefore, if we are able to show that Ux∈Max(𝒜σ,X) and
(5.12)x∉⋃(Max(𝒜σ,X)∖{Ux}),
this implies that Max(𝒜σ,X) is not dependent, and the thesis follows.
It is obvious that Ux∈𝒞σ,X.
Let now u1,u2∈Ux.
Then there exist y1,y2∈X such that u1∈σ(y1), u2∈σ(y2) and x∈σ(y1)∩σ(y2).
Therefore, by (5.11), there exists an element w∈X such that {y1,y2}⊆σ(w).
This implies that x∈σ(w).
Hence w∈Ux, and also
{u1,u2}⊆σ(y1)∪σ(y2)⊆σ(w).
This shows that Ux∈𝒜σ,X.
Let now K∈𝒜σ,X be such that Ux⊆K, and u∈K arbitrary.
Then, since K is attractive and {u,x}⊆K, there exists an element t∈K such that {u,x}⊆σ(t).
This implies that u∈σ(t)∈𝒪σ,X and x∈σ(t), therefore u∈Ux, so that Ux=K.
Hence Ux∈Max(𝒜σ,X).
Finally, let N∈Max(𝒜σ,X) be such that N≠Ux.
Let us assume by contradiction that x∈N.
Let z∈N be arbitrary.
Then, since N is attractive, there exists an element y∈N such that {x,z}⊆σ(y).
This implies that z∈Ux, and therefore N⊆Ux.
Then, since N is maximal in 𝒜σ,X, we have that N=Ux, which is a contradiction.
This proves (5.12), and hence the thesis.
(ii) ⇒ (iv):
obvious.
(iv) ⇒ (ii):
Let x∈X.
Since Max(𝒜σ,X) is a covering of X and Uσ,X=∅, condition (ii) is verified.
∎
We conclude this subsection showing that if a maximal attractive subset Y is contained in the union of some other maximal attractive subsets, then Y agrees with one among the members of such a union.
Furthermore, we shall prove that any finite subcollection of maximal attractive subset is union independent.
Proposition 5.10.
The following conditions hold:
if Y,Y1,…,Yn∈Max(𝒜σ,X) and Y⊆Y1∪⋯∪Yn, then Y=Yi for some i∈{1,…,n};
if ℱ⊆Max(𝒜σ,X) and |ℱ|<∞, then ℱ is union independent.
Proof.
(i): Let Z:=Y1∪…∪Yn.
If n=1, in view of the maximality of both Y and Y1, we conclude that Y=Y1.
Moreover, if n=2, we get Z=Y1⊆Y2, and since Y1,Y2∈𝒞σ,Z, in view of part (i) of Proposition 5.1, we get Y⊆Y1 or Y⊆Y2, whence Y=Y1 or Y=Y2 because of the maximality of Y, Y1 and Y2.
Assume that the thesis holds for any integer less than or equal to n-1, and assume that Z=Y1∪…∪Yn and Y⊆Z.
Set A:=Y1 and B:=Y2∪⋯∪Yn.
Then Z∈𝒞σ,X because 𝒞σ,X is union closed.
Moreover, we get A,B∈𝒞σ,Z, A∪B=Z and Y∈𝒜σ,Z is such that Y⊆A∪B.
Therefore, by part (i) of Proposition 5.1, we have that Y⊆A or Y⊆B.
In the first case, by the maximality of Y and Y1, we have Y=Y1.
Otherwise, Y⊆Y2∪⋯∪Yn, and in view of the inductive hypothesis, we get Y=Yi for some i∈{2,…,n}.
(ii): Let us assume by contradiction that ℱ is not union independent.
Then there exists Y∈ℱ such that Y⊆Y1∪⋯∪Yn, where {Y1,…,Yn}=ℱ∖{Y}, which is in contrast with the previous part (i).
∎
5.1 Some MA-union dependent spaces induced by monoid actions
We conclude this section by showing an application of Theorem 5.9 when S is a monoid and X∈LSETS.
To this aim, we first associate with any strictly increasing set system (if it exists) of closed subsets of X a new S-set as follows.
Assume that ℱ={Yn:n≥0}⊆𝒞S,X is strictly increasing, and take Y=⋃ℱ.
Then the set Y×{0,1}ℤ+ becomes an S-set by means of the S-action given by s(y,ϵ):=(sy,ϵ) for any s∈S and (y,ϵ)∈Y×{0,1}ℤ+.
Moreover, relatively to the equivalence relation ∼ℱ defined in (2.1) we have that
(y,ϵ)∼ℱ(y′,ϵ′)⟹s(y,ϵ)∼ℱs(y′,ϵ′);
therefore also the quotient set (Y×{0,1}ℤ+)/∼ℱ becomes an S-set by setting
s[y,ϵ]ℱ:=[sy,ϵ]ℱ.
In the next result, we show that, whenever the set system 𝒪S,X is not Noetherian, we can always associate with any strictly increasing set subsystem of 𝒪S,X an MA-union dependent space.
Proposition 5.11.
Let F={Sxn:n≥0}⊆OS,X be strictly increasing.
Then the S-set (⋃F×{0,1}Z+)/∼F is an MA-union dependent topological space.
Proof.
Let Y=⋃ℱ and Z:=(Y×{0,1}ℤ+)/∼ℱ.
In order to prove that Z is MA-union, by part (iii) of Theorem 5.9, it suffices to prove that, for any arbitrary choice of an element z∈Z, there exist two elements z′,z′′∈Z such that
(5.13)z∈φS(z′)∩φS(z′′) and z′∣𝒪S,Zz′′.
Let therefore z=[y,ϵ]ℱ for some pair (y,ϵ)∈Y×{0,1}ℤ+.
Two cases may occur.
(a) Assume firstly that y∈Sxn∖Sxn-1 for some n≥1.
Let us take the elements
(5.14)y′=y′′=xn+1∈Sxn+1∖Sxn
and the maps ϵ′,ϵ′′∈{0,1}ℤ+ defined as follows:
ϵ′(i)={ϵ(i)if 1≤i≤n,1ifi=n+1,0otherwise, ϵ′′(i)={ϵ(i)if 1≤i≤n,0otherwise.
Let now z′:=[y′,ϵ′]ℱ and z′′:=[y′′,ϵ′′]ℱ.
Clearly, z′≠z′′.
Moreover, since Sxn⫋Sxn+1=Sy′=Sy′′, there exist s′,s′′∈S such that y=s′y′=s′′y′′∈Sxn∖Sxn-1.
So, in view of the definition of the relation ∼ℱ, we get
(y,ϵ)∼ℱ(s′y′,ϵ′) and (y,ϵ)∼ℱ(s′′y′′,ϵ′′),
whence z∈φS(z′)∩φS(z′′).
Finally, assume by contradiction the existence of T∈𝒪S,Z such that {z′,z′′}⊆T.
Let T=Sz*, where z*=[y*,ϵ*]ℱ∈Z for some (y*,ϵ*)∈Y×{0,1}ℤ+.
Then the condition {z′,z′′}⊆Sz* implies the existence of r′,r′′∈S such that y′=r′y*, y′′=r′′y* and
(5.15)(y′,ϵ′)∼ℱ(r′y*,ϵ*),(y′′,ϵ′′)∼ℱ(r′′y*,ϵ*).
In view of both (5.14) and (5.15) and of the relation ∼ℱ, we get ϵ′(i)=ϵ*(i)=ϵ′′(i) for any i=1,…,n+1, which contradicts the fact that ϵ′(n+1)≠ϵ′′(n+1).
Hence (5.13) holds.
(b) Assume that y∈Sx0.
In this assumption, the proof goes back to that provided in the previous case.
In fact, it suffices to repeat the above argument taking y′=y′′=x1∈Sx1∖Sx0 and two maps ϵ′,ϵ′′∈{0,1}ℤ+ such that ϵ′(1)≠ϵ′′(1).
∎
In the following result, we see that, although the condition of non-MA-union dependence is not equivalent to the condition of MAC-union independence relatively to a single space X∈LSETS, to a global level this happens.
Theorem 5.12.
The following conditions for a monoid S are equivalent:
S is not 𝔗-MA-union dependent;
S is 𝔗-essentially Noetherian;
S is 𝔗-essentially attractive;
S is 𝔗-MAC-union independent.
Proof.
(i) ⇒ (ii)
follows by Proposition 5.11.
(ii) ⇒ (iii)
is a direct consequence of part (iv) of Proposition 4.7.
(iii) ⇒ (iv)
follows by Proposition 5.8.
(iv) ⇒ (i)
is a direct consequence of part (v) of Proposition 5.5.
∎
6 2MA-decompositions of X
In this section, we study bi-partitions of X by means of two subspaces, one of which is MAC-union independent and the other is MA-union dependent.
In Theorem 6.4 we will see that such a type of decomposition is uniquely determined and that its two components are effectively proper subspaces when X in MA-union undetermined.
To an interpretative level, this means that the condition of MA-union indetermination corresponds effectively to an intermediate and balanced situation between MAC-union independence and MA-union dependence on X.
Definition 6.1.
We say that an ordered pair (Y1,Y2)∈𝒞σ,X2 is a 2MA-decomposition of X, which is denoted by X:=(Y1|Y2), if the following conditions hold:
Y1 is MAC-union independent;
Y2 is MA-union dependent;
Y1=σ(X∖Y2) and Y2=σ(X∖Y1).
We say that a 2MA-decomposition (Y1,Y2) of X is
proper if ∅⫋Y1⫋X and ∅⫋Y2⫋X,
unique if, for any further 2MA-decomposition (Y1′,Y2′) of X, we have that Y1=Y1′ and Y2=Y2′.
Remark 6.2.
Let us note that if a pair (Y1,Y2)∈𝒞σ,X2 satisfies above the identities given in (d3), then
(∅⫋Y1⫋Xor∅⫋Y2⫋X) implies that (∅⫋Y1⫋Xand∅⫋Y2⫋X),
Y1 and Y2 are regularly closed.
We set now
(6.1)X~:=σ(Uσ,X),X^:=σ(X∖X~),
(6.2)ℋσ,X:={Y∈Max(𝒜σ,X):Y∩Uσ,X≠∅},
𝒬σ,X:={K∈Max(𝒜σ,X):K∩(X∖X~)≠∅}.
Let us note that X~,X^∈𝒞σ,X; moreover, since Max(𝒜σ,X) is a covering of X, by the definition of ℋσ,X, we have that
(6.3)Uσ,X=∅⇔ℋσ,X=∅.
In the next result, we provide some useful basic properties of the subsets and of the set systems we introduced above.
We will use such properties in Theorem 6.4.
Lemma 6.3.
The following conditions hold:
if x∈Uσ,X,
y∈X and x∈σ(y), then y∈Uσ,X;
if K∈ℋσ,X, then K=σ(K∩Uσ,X);
ℋσ,X is union independent and ℋσ,X∈MACσ(X~);
X~ is MAC-union independent, and its standard union independent system is ℋσ,X;
X is MAC-union independent if and only if X=X~;
if X is MAC-union independent, then X is not MA-union dependent;
if K∈𝒬σ,X, then K=σ(K∩(X∖X~));
X^ is MA-union dependent.
Proof.
(i): Let x∈Uσ,X and y∈X be such that x∈σ(y).
Let us assume by contradiction that y∉Uσ,X.
Then, by (5.5), there exists (T,Z)∈Max(𝒜σ,X)2,≠ such that y∈T∩Z.
Since T,Z∈𝒞σ,X, this implies that x∈σ(y)⊆T∩Z, which is in contrast with the choice of x in Uσ,X.
(ii): If K∈ℋσ,X, then K∩Uσ,X≠∅.
Let x∈K∩Uσ,X.
Since K is attractive, for any w∈K, there exists an element yw∈K such that {x,w}⊆σ(yw).
Now, since x∈Uσ,X, by the previous part (i), we have that yw∈Uσ,X.
Then the conditions yw∈K∩Uσ,X and w∈σ(yw) imply that w∈σ(K∩Uσ,X), and hence K⊆σ(K∩Uσ,X).
Conversely, since K is a fixed point of σ, we also have
σ(K∩Uσ,X)⊆σ(K)=K, and the thesis follows.
(iii): By the previous part (ii), we see that any member K∈ℋσ,X has the form K=σ(K∩Uσ,X), which is a subset of σ(Uσ,X)=X~.
Therefore, ℋσ,X⊆𝒫(X~).
This implies that any K∈ℋσ,X is a member of 𝒜σ,X~ which is maximal in 𝒜σ,X⊇𝒜σ,X~; therefore we get K∈Max(𝒜σ,X~).
(iv): We assume first that Uσ,X=∅.
Then also ℋσ,X is empty by (6.3); therefore, in such a case, the thesis follows because σ(∅)=∅.
We can assume therefore Uσ,X≠∅.
By the previous part (ii), it is clear that ⋃ℋσ,X⊆X~.
Conversely, in view of the definition of Uσ,X and ℋσ,X, we have that Uσ,X⊆⋃ℋσ,X.
Moreover, since ℋσ,X⊆𝒞σ,X and 𝒞σ,X is union closed, we have ⋃ℋσ,X∈𝒞σ,X (i.e. ⋃ℋσ,X is a fixed point of σ).
Hence X~=σ(Uσ,X)⊆σ(⋃ℋσ,X)=⋃ℋσ,X,
and the thesis follows.
(v): Let K∈ℋσ,X.
Then K∩Uσ,X≠∅.
Let x∈K∩Uσ,X≠∅.
Then K is the only member of Max(𝒜σ,X) containing x; therefore x∉⋃(ℋσ,X∖{K}), and hence K⊈⋃(ℋσ,X∖{K}).
This shows that the set system ℋσ,X is independent.
Moreover, by the previous parts (iii) and (iv), it is clear that ℋσ,X∈MACσ(X).
(vi): It is a direct consequence of Remark 5.6 and of the previous part (v).
(vii): If X=X~, by the previous part (vi), it follows that X is MAC-union independent.
Conversely, let X be MAC-union independent and ℱ=𝔖σ,X its standard independent system.
Then, by part (v) of Proposition 5.5, we have
(6.4)Y∩Uσ,X≠∅ for allY∈ℱ.
This implies that
(6.5)ℱ⊆ℋσ,X.
In fact, if by contradiction, (6.5) does not hold, there exists some K∈ℱ such that K∉ℋσ,X, so K∩Uσ,X=∅, which is in contrast with (6.4).
Then, since ℱ is a covering of X, by (6.5) and by the previous part (iv), we obtain that
X=⋃ℱ⊆⋃ℋσ,X=X~,
and hence X=X~.
(viii): If X is MAC-union independent, by the previous part (vii), we deduce that Uσ,X≠∅.
Hence Max(𝒜σ,X) is not dependent by Theorem 5.9.
(ix): Firstly, if X~=∅, by (6.1), we have X^=σ(X)=X so that σ(X∖X^)=σ(∅)=∅=X~.
If X~=X, again by (6.1), we have X^=∅, and therefore σ(X∖X^)=σ(X)=X=X~.
We can assume therefore ∅⫋X~⫋X.
Let y∈σ(X∖X^).
Then there exists an element x∈X∖X^=X∖σ(X∖X~)⊆X~ such that y∈σ(x).
By the definition of X~, there also exists an element u∈Uσ,X such that x∈σ(u).
Then we have that y∈σ(x)⊆σ(u), and hence y∈σ(Uσ,X):=X~.
This shows that σ(X∖X^)⊆X~.
Conversely, let us assume by contradiction that Uσ,X⊈X∖X^.
Then there exists an element z∈Uσ,X such that z∈X^:=σ(X∖X~).
Therefore, there exists an element w∈X∖X~ such that z∈σ(w).
By the previous part (i), this implies that w∈Uσ,X⊆σ(Uσ,X):=X~, which provides a contradiction.
Thus we have that Uσ,X⊆X∖X^, and therefore also X~=σ(Uσ,X)⊆σ(X∖X^).
Hence X~=σ(X∖X^).
(x): Let K∈𝒬σ,X.
Since K∩(X∖X~)⊆K and K∈𝒞σ,X, we have that
σ(K∩(X∖X~))⊆σ(K)=K.
Conversely, since K∩(X∖X~)≠∅, we fix an element x∈K∩(X∖X~).
Let now w∈K be an arbitrary element.
Since K∈𝒜σ,X, there exists an element yw∈K such that
(6.6){x,w}⊆σ(yw).
Now, if (by contradiction) yw∈X~, since X~∈𝒞σ,X, we have that x∈σ(yw)⊆X~, which is in contrast with the choice of x.
Therefore,
(6.7)yw∈K∩(X∖X~).
Then, by (6.6) and (6.7), we have that w∈σ(yw)⊆σ(K∩(X∖X~)).
Hence also K⊆σ(K∩(X∖X~)), and this proves part (x).
(xi): Let K∈𝒬σ,X.
By the previous part (x), we have that K=σ(K∩(X∖X~))⊆σ(X∖X~):=X^.
Hence ⋃𝒬σ,X⊆X^.
Conversely, let z∈X^.
Then there exists t∈X∖X~ such that z∈σ(t).
Since Max(𝒜σ,X) is a covering of X, there exists M∈Max(𝒜σ,X) such that t∈M so that z∈σ(t)⊆M.
Moreover, M∩(X∖X~)≠∅ because t∈M∩(X∖X~); therefore M∈𝒬σ,X.
So also the reverse inclusion X^⊆⋃𝒬σ,X is verified, and hence (xi) holds.
(xii): Let us first note that
(6.8)Uσ,X∩(X∖X~)=∅.
In fact, Uσ,X∩(X∖X~)⊆σ(Uσ,X)∩(X∖X~)=X~∩(X∖X~)=∅.
Therefore, by (5.5) and (6.8), it follows that,
(6.9)for allx∈X∖X~,there exists(Yx,Wx)∈Max(𝒜σ,X)2,≠such thatx∈Yx∩Zx.
By (6.9), we also deduce that Yx∩(X∖X~)≠∅ and Zx∩(X∖X~)≠∅; therefore, in view of the definition of 𝒬σ,X and of the above part (xi), we have that Yx,Zx∈𝒬σ,X⊆𝒫(X^).
This implies that Yx,Zx∈Max(𝒜σ,X^).
Hence (6.9) can be reformulated as follows:
(6.10)for allx∈X∖X~,there exists(Yx,Wx)∈Max(𝒜σ,X^)2,≠such thatx∈Yx∩Zx.
Let now w∈X^ be arbitrary.
By the definition of X^, there exists an element tw∈X∖X~ such that w∈σ(tw).
Then, by (6.10), we obtain w∈σ(tw)⊆Ytw∩Ztw, where Ytw and Ztw are two distinct elements of Max(𝒜σ,X^).
Therefore, by Theorem 5.9, we deduce that Max(𝒜σ,X^) is dependent.
∎
At this point, we shall prove that X admits a unique 2MA-decomposition, and moreover, when X is MA-union undetermined, we also show that such a decomposition is proper.
Theorem 6.4.
In reference to the previous notations, we have that
the 2MA-decomposition (X~,X^) of X is unique,
if X is MA-union undetermined, then the 2MA-decomposition (X~,X^) of X is proper.
Proof.
(i): It is a direct consequence of Definition 6.1, of (6.1) and of the parts (vi), (ix), (xii) of Lemma 6.3.
(ii): Let X=(Y1|Y2) be another 2MA-decomposition of X.
We must prove that
(6.11)X~:=σ(Uσ,X)=Y1.
We first show that X~⊆Y1.
To this regard, let x∈Y2, and assume by contradiction that x∈Uσ,X so that there exists a unique element K∈Max(𝒜σ,X) such that x∈K.
On the other hand, since Max(𝒜σ,Y2) is dependent, by Theorem 5.9, there exists (M,M′)∈Max(𝒜σ,Y2)2,≠ such that x∈M∩M′.
Since {M,M′}⊆𝒜σ,Y2⊆𝒜σ,X, by part (iii) of Proposition 4.7, there exist {N,N′}⊆Max(𝒜σ,X) such that M⊆N and M′⊆N′.
Then, since x∈M∩M′⊆N∩N′, by the uniqueness of K, we deduce that N=N′=K.
Now, since X=Y1∪Y2 and Y2=σ(X∖Y1), in view of Proposition 5.1 (ii), there exists ℱ⊆Max(𝒜σ,X) such that Y2=⋃ℱ.
Then, since x∈Y2, there exists some member F∈ℱ such that x∈F, and again by the uniqueness of K, it must be necessarily F=K, and this implies that K⊆⋃ℱ=Y2.
Therefore, we have that
K∈𝒜σ,Y2,M⊆K,M′⊆K,
and hence K=M=M′ by the maximality of both M and M′ in 𝒜σ,Y2.
This provides a contrast with the above condition M≠M′, and it proves that Y2∩Uσ,X≠∅.
Then, since X=Y1∪Y2, we have that Uσ,X⊆Y1∈𝒞σ,X.
Therefore, we obtain X~:=σ(Uσ,X)⊆σ(Y1)=Y1.
Conversely, let us consider the subset
Uσ,Y1={y∈Y1:there is exactly oneY∈Max(𝒜σ,Y1)such thaty∈Y}.
Since, by hypothesis, Y1 is MAC-union independent, by part (vii) of Lemma 6.3, we have that
(6.12)Y1=Y~1:=σ(Uσ,Y1).
Let now w∈Uσ,Y1 be arbitrary.
Since Uσ,Y1⊆Y1=σ(X∖Y2), there exists w′∈X∖Y2⊆Y1 such that
(6.13)w∈σ(w′).
Then, by part (i) of Lemma 6.3, we deduce that
(6.14)w′∈Uσ,Y1.
Moreover, since Y1=σ(X∖Y2), by part (ii) of Proposition 5.1, there exists 𝒢⊆Max(𝒜σ,X) such that
(6.15)Y1=⋃𝒢.
Then, since Uσ,Y1⊆Y1, by (6.14) and (6.15), there exists N∈𝒢 such that w′∈N⊆Y1; therefore
(6.16)w′∈N∈Max(𝒜σ,Y1)∩Max(𝒜σ,X).
Let now M be any member of Max(𝒜σ,X) such that w′∈M.
In view of part (i) of Proposition 5.1, we have that M⊆Y1 or M⊆Y2.
However, since w′∈M and w′∉Y2, it must be necessarily M⊆Y1.
This implies that
(6.17)w′∈M∈Max(𝒜σ,Y1)∩Max(𝒜σ,X).
Then, by (6.14), (6.16) and (6.17), we obtain that M=N.
This shows that N is the only member of Max(𝒜σ,X) which contains w′; hence
(6.18)w′∈Uσ,X.
Therefore, by (6.13) and (6.18), we deduce that w∈σ(w′)⊆σ(Uσ,X):=X~, and so
(6.19)Uσ,Y1⊆X~
by the arbitrariness of w.
Then, by (6.12) and (6.19), we obtain the reverse inclusion Y1=σ(Uσ,Y1)⊆σ(X~)=X~.
Hence (6.11) holds, and consequently we also have that
(6.20)Y2=σ(X∖Y1)=σ(X∖X~):=X^.
By (6.11) and (6.20), we obtain then the uniqueness of the 2MA-decomposition X=(X~|X^).
(iii): Since Max(𝒜σ,X) is not dependent, by Theorem 5.9, we have that Uσ,X≠∅, and therefore also X~:=σ(Uσ,X)≠∅.
By part (vi) of Lemma 6.3, the set X~ is MAC-union independent, but since X is not MAC-union independent by part (vii) of Lemma 6.3, it follows that X~⫋X.
Therefore, ∅⫋X~⫋X.
Then, by part (ii) of Remark 6.2, we also have ∅⫋X^⫋X.
This proves that (X~|X^) is a proper 2MA-decomposition of X if X is MA-union undetermined.
∎
7 MA-descending union dependence chains of X
In this section, we define and study a recursive construction of specific closed subsets of X, which is related to the notion of MA-union dependence of the Alexandroff topology on X.
Indeed, as we have seen in Theorem 5.9, one among the equivalent conditions so that the space X is MA-union dependent is the following: for each x∈X, there exist two distinct maximal attractive subsets whose intersection contains x itself.
More formally, we have that X is MA-union dependent if and only if
X={x∈X:there exists(M,M′)∈Max(𝒜σ,D0)2,≠such thatx∈M∩M′}.
Then we will start from such a result in order to define a descending chain of closed subsets of X that provides a sort of level of MA-union dependence of X in a way we will explain in what follows.
We first set D0:=X and
D1:=X∖Uσ,X={x∈X:there exists(M,M′)∈Max(𝒜σ,D0)2,≠such thatx∈M∩M′}.
Then it results that D1∈𝒞σ,X.
In fact, let x∈D1 and (M,M′)∈Max(𝒜σ,D0)2,≠ such that x∈M∩M′.
Since {M,M′}⊆𝒜σ,X⊆𝒞σ,X and 𝒞σ,X is intersection closed, we have that M∩M′∈𝒞σ,X.
Therefore, since σ is monotone, it follows that
(7.1)σ(x)⊆σ(M∩M′)=M∩M′.
Then, by definition of D1 and (7.1), we deduce that σ(x)⊆D1, and hence σ(D1)=⋃{σ(x):x∈D1}⊆D1 due to the arbitrariness of x in D1.
Hence D1∈𝒞σ,X.
Let now n≥1, and we assume by inductive hypothesis that
Dn:={x∈Dn-1:there exists(M,M′)∈Max(𝒜σ,Dn-1)2,≠such thatx∈M∩M′}∈𝒞σ,X.
Then, with same procedure used in the initial case, it results that
Dn+1:={x∈Dn:there exists(M,M′)∈Max(𝒜σ,Dn)2,≠such thatx∈M∩M′}∈𝒞σ,X.
Let now ω be the first ordinal limit, and we set
Dω:=⋂{Dn:n≥0}
so that Dω is closed, and we can repeat the above construction by substituting D0 with Dω.
Then, by transfinite induction, we can extend the previous definitions of Dn and Dω by setting
(7.2)Dα:={{x∈Dα-1:there exists(M,M′)∈Max(𝒜σ,Dα-1)2,≠such thatx∈M∩M′}ifα∈𝔑,⋂{Dβ:β<α}ifα∈𝔏,
for any α∈𝔒*.
Then we obtain a decreasing sequence of closed subsets
X=D0⊇D1⊇D2⊇⋯⊇Dα⊇Dα+1⊇⋯
which we call MA-descending union dependence chain of X, and we denote it by 𝒟σ,X.
The next result justifies the previous name for the chain 𝒟σ,X.
Proposition 7.1.
For any α∈O the following conditions are equivalent:
Dα is MA-union dependent or Dα=∅;
Proof.
(i) ⇒ (ii):
Let us assume that Dα≠∅.
By hypothesis, we have that
(7.3)Dα=Dα+1:={x∈Dα:there exists(M,M′)∈Max(𝒜σ,Dα)2,≠such thatx∈M∩M′}.
Therefore, as a direct consequence of part (ii) in Theorem 5.9 and (7.3), we deduce that Dα is non-empty and MA-union dependent.
(ii) ⇒ (iii):
If Dα=∅, then α stabilizes 𝒟σ,X.
Let therefore Dα be non-empty and MA-union dependent.
Then, again by part (ii) of Theorem 5.9, the identity in (7.3) holds.
Thus Dα+1 is non-empty and MA-union dependent.
Therefore, with the same reasoning, we deduce that Dα+1=Dα+2, and so on for any α+k, with k≥0 integer.
Moreover, for the next limit ordinal β>α+k, for all k≥0, we have again Dβ=⋂{Dα+k:k≥0}=Dα; therefore Dβ is non-empty and MA-union dependent.
Proceeding in such a way, we can continue the transfinite induction in order to have the thesis.
(iii) ⇒ (i):
obvious.
∎
The next theorem is the main result of the present section, and in it, we relate the 2MA-decomposition (X~|X^) to the chain 𝒟σ,X.
Theorem 7.2.
The following conditions hold:
X~=σ(X∖Dα) for any α∈𝔒*;
Dα+1=⋃{σ(y)∩σ(z):y,z∈Dα𝑎𝑛𝑑y∣𝒪σ,Dαz};
Proof.
(i): Let us assume first that X∖X~=∅.
In such a case, we have that X^:=σ(X∖X~)=σ(∅)=∅, and the thesis follows.
We can suppose therefore that X∖X~≠∅.
We will prove first that
(7.4)M⊆Dα for allM∈𝒬σ,Xand allα∈𝔒.
Since D0:=X, it is clear that (7.4) holds when α=0.
Let then δ∈𝔒* be a fixed ordinal, and we assume (by inductive hypothesis) that (7.4) holds for any ordinal τ<δ.
Let K∈𝒬σ,X be arbitrary and fixed.
We distinguish two cases.
If δ is a limit ordinal, then
K⊆⋂{Dτ:τ<δ}:=Dδ.
We can assume therefore that δ=τ+1 for some ordinal τ.
Then, for any w∈K∩(X∖X~), there exists Kw∈Max(𝒜σ,X) such that K≠Kw and w∈K∩Kw (otherwise K is the only member of Max(𝒜σ,X) containing w, so that w∈Uσ,X⊆X~, and this in contrast with the choice of w).
Now, since w∈Kw∩(X∖X~), in view of the definition of 𝒬σ,X, it follows that Kw∈𝒬σ,X.
By inductive hypothesis applied on τ, K and Kw, we have that K⊆Dτ and Kw⊆Dτ.
Therefore, since (K,Kw)∈Max(𝒜σ,X)2,≠, it follows that
(7.5)(K,Kw)∈Max(𝒜σ,Dτ)2,≠ and w∈K∩Kw.
Then, in view of the definition of Dτ+1 and (7.5), we deduce that w∈Dτ+1=Dδ, and due to the arbitrariness of w in K∩(X∖X~), we obtain that
(7.6)K∩(X∖X~)⊆Dδ.
At this point, since Dδ∈𝒞σ,X and K∈𝒬σ,X, by (7.6) and by part (x) of Lemma 6.3, we have that
K=σ(K∩(X∖X~))⊆σ(Dδ)=Dδ,
and this proves by induction that K⊆Dα for any ordinal α.
Hence (7.4) holds since K is arbitrary in 𝒬σ,X.
Now, by (7.4) and by part (xi) of Lemma 6.3, we deduce that
X^=⋃𝒬σ,X⊆Dα for any ordinal α.
(ii): Let α>0 be any ordinal.
Let us note that D1=X∖Uσ,X; therefore Uσ,X=X∖D1⊆X∖Dα, and consequently X~=σ(Uσ,X)⊆σ(X∖Dα).
Conversely, since X^⊆Dα, we have X∖Dα⊆X∖X^⊆σ(X∖X^)=X~, and hence σ(X∖Dα)⊆σ(X~)=X~.
(iii): Assume firstly the existence of y,z∈Dα such that y∣𝒪σ,Dαz and x∈σ(y)∩σ(z).
As σ(y),σ(z)∈𝒜σ,Dα, in view of part (ii) of Proposition 4.7, we can find two members M and M′ belonging to Max(𝒜σ,Dα) such that σ(y)⊆M and σ(z)⊆M′.
Now, z∉M; otherwise there exists w∈M such that {y,z}⊆σ(w), contradicting the fact that y∣𝒪σ,Dαz.
Thus M≠M′, and we get x∈Dα+1 by (7.2).
On the other hand, assume that, for any y,z∈Dα such that x∈σ(y)∩σ(z), there exists Ty,z∈𝒪σ,Dα such that {y,z}⊆Ty,z.
Consider the subset
Ux:=⋃{T∈𝒪σ,X:x∈T}.
Clearly, x∈Ux and Ux∈𝒞σ,Dα.
Let now u1,u2∈Ux.
Then there exist y1,y2∈Dα such that u1∈σ(y1), u2∈σ(y2) and x∈σ(y1)∩σ(y2).
Thus, by our assumption, we find w∈Dα such that {y1,y2}⊆σ(w), whence x∈σ(w), and therefore w∈Ux.
Furthermore, we get {u1,u2}⊆σ(y1)∪σ(y2)⊆σ(w), whence Ux∈𝒜σ,Dα.
Let now K∈𝒜σ,Dα be such that Ux⊆K, and take some u∈K.
Then, since K is attractive and {u,x}⊆K, there must be t∈K such that {u,x}⊆σ(t).
This implies that u∈σ(t)∈𝒪σ,X and x∈σ(t); therefore u∈Ux, so that Ux=K.
Hence Ux∈Max(𝒜σ,Dα).
Finally, let N∈Max(𝒜σ,Dα) be such that N≠Ux and containing x.
Let moreover z∈N be arbitrary.
Then, since N is attractive, there exists an element y∈N such that {x,z}⊆σ(y).
This implies that z∈Ux, and therefore N⊆Ux.
Therefore, in view of the maximality of N, we get N=Ux.
Thus x∉Dα+1.
(iv): If x∈Dα+1, in view of the previous part (iii), there must be y,z∈Dα such that x∈σ(y)∩σ(z) and y∣𝒪σ,Xz.
Nevertheless, as Dα∈𝒜σ,X, the condition y∣𝒪σ,Xz cannot hold.
Thus Dα+1=∅.
∎
In reference to the monoid ℕ and to its action on Seq0§,*(ℤ|odd) defined in (3.5), we will now analyze the behavior of the MA-descending union dependence chain for the set Seq0§,*(ℤ|odd) when ℕ acts on it.
Proposition 7.3.
Relatively to the N-set X=Seq0§,*(Z|𝑜𝑑𝑑), the following conditions hold:
X is MA-union independent;
D1 is MA-union dependent;
𝒟ℕ,X stabilizes for α=1, i.e. X⫌D1=D2=⋯, or equivalently, μ(𝒟ℕ,X)=1.
Proof.
(i): We must show that there exists an independent set system ℱ∈MACS(X).
To this regard, consider the set system ℱ:={Ψ(γ(b)):b∈X}.
By part (iii) of Proposition 5.4, it readily follows that
X=⋃{Ψ(γ(b)):b∈X};
therefore ℱ∈MACS(X).
Fix now Ψ(γ(a)) for some a∈X.
As γ(a)∈Ψ(γ(a)), if it were Ψ(γ(a))⊆⋃{Ψ(γ(b)):b∈X,b≠a}, we would have γ(a)∈Ψ(γ(b)) for some b∈X such that b≠a, contradicting part (ii) of Proposition 5.4.
This ensures that ℱ is independent.
(ii): By the above part (i) and by part (vii) of Lemma 6.3, we have that X=X~, whence
X^=σ(X∖X~)=σ(∅)=∅.
(iii): Let us firstly notice that the set
D1:={x∈X:there exists(M,M′)∈Max(𝒜σ,D0)2,≠such thatx∈M∩M′}
must assume the form
D1={a=(a0,…,am-1,am,0,…)∈X:am≠0,am-1∈2ℤ+1}
in view of part (iii) of Proposition 5.4.
Now, let a=(a0,…,am-1,am,0,…)∈D1.
We clearly have a∈Ψ(b), Ψ(b′)⊆D1 for b=(a0,…,an,1,0,…) and b′=(a0,…,2an+1,1,0,…).
This means that
Ψ(b),Ψ(b′)∈Max(𝒜ℕ,D1),
so these sets satisfy the condition given in part (ii) of Theorem 5.9, whence we conclude that D1 is MA-union dependent.
(iv): The thesis easily follows in view of the above part (iii) and of Proposition 7.1.
∎
At this point, for each integer q≥0, we construct a specific space X∈LSETℕ∞ for which μ(𝒟ℕ∞,X)=ω+q.
To this regard, let us set
Seq§,*(𝕂)={a∈Seq§(𝕂):an=0implies thatan+1=an+2=⋯=0},
Seq§,*(ℕ|0):={a∈Seq§,*(ℕ):ai=0for alli≥a0+1},
Seq§,*(ℕ|l):={a∈Seq§,*(ℕ):a0=landai=0for alli≥al+l+1}
for any l≥1, and
Seq§,*(ℕ|[0,q]):=Seq§,*(ℕ|0)∪Seq§,*(ℕ|1)∪…∪Seq§,*(ℕ|q)
for any q≥0 so that, in particular, Seq§,*(ℕ|[0,0])=Seq§,*(ℕ|0).
Now, for any q≥0, we define a map
ν:(s,a)∈ℕ∞×Seq§,*(ℕ|[0,q])↦sa∈Seq§,*(ℕ|[0,q])
as follows: if a=(a0,a1,…,an-1,an,0,0,…)∈Seq§,*(ℕ|[0,q]) and s∈ℕ∞, we set
(7.7)sa:={aifs∈[ζn≤(a),∞],(a0,a1,…,ar-1,ar,s-ζr≤(a),0,0,…)if 0≤r≤n-1ands∈[ζr≤(a),ζr+1≤(a)[,(s,0,0,…)ifs∈[0,a0[.
Remark 7.4.
Let a=(a0,…,an,0,…)∈Seq§,*(ℕ|[0,q]) and s∈ℕ∞.
Assume that
sa=(a0′,…,am′,0,…).
Then it may be easily verified that
ζm≤(sa)=min{s,ζn≤(a)}.
In the following result, we shall show that the map ν defined as in (7.7) is a monoid action.
Proposition 7.5.
Let X=Seq§,*(N|[0,q]), S=N∞ and b=(b0,…,bm,0,…)∈X.
Then
Sb={(b0,…,bn-1,an,0,…)∈X:0≤n≤m, 0≤an≤bn}.
Proof.
(i): Clearly, in view of (7.7), it results that ∞⋅a=a for each a∈X.
Let now s,s′∈S and
a=(a0,…,am,0,…)∈X.
We claim that
(7.8)(s⋅s′)a=s(s′a).
Without loss of generality, we may assume that s⋅s′=min{s,s′}=s.
Three cases may occur.
If s∈[ζm≤(a),∞], then s′∈[ζm≤(a),∞], and thus (s⋅s′)a=sa=a=s′a=s(s′a).
If s∈[ζr≤(a),ζr+1≤(a)[, then (s⋅s′)a=(a0,a1,…,ar-1,ar,s-ζr≤(a),0,…).
Then either s′∈[ζn≤(a),∞], and whence s(s′a)=sa=(a0,…,ar-1,ar,s-ζr≤(a),0,…)=(s⋅s′)a,
or there exists r≤r′≤n-1 such that s′∈[ζr′≤(a),ζr′+1≤(a)[, and hence we have s′a=(a0,a1,…,ar′-1,ar′,s′-ζr′≤(a),0,…).
Moreover, since s(s′a)=(a0,a1,…,ar-1,ar,s-ζr≤(s′a),0,…) and ζr≤(s′a)=ζr≤(a), again (7.8) holds.
If s∈[0,a0[, then (s⋅s′)a=sa=(s,0,…) in view of (7.7).
Thus, any way one chooses s′≥s, the zeroth component of s′a agrees either with a0 or with s′, whence we get s(s′a)=(s,0,…) again by (7.7).
Therefore, also in this third case, (7.8) holds.
(ii): Let us assume that a∈Sb, i.e. there exists s∈S such that a=sb.
We claim that n≤m, ai=bi for each i=0,…,n-1 and an≤bn.
We distinguish three cases:
if s∈[ζm≤(b),∞], then a=b, and in particular, the claim holds;
if there exists r=0,…,m-1 such that s∈[ζr≤(b),ζr+1≤(b)[, the equality a=sb forces r=n-1 and s=ζn-1≤(b)+an, whence in view of (7.7), we must have n≤m, ai=bi for each i=0,…,n-1 and an≤bn;
finally, the case s∈[0,b0[ forces a to be equal to the sequence a=(s,0,…).
Conversely, notice firstly that, in view of (7.7), we get (s,0,…)∈Sb for each s∈[0,b0[.
Let now a=(a0,…,an,0,…) be such that 1≤n≤m, ai=bi for each i=0,…,n-1 and an≤bn.
Let
s:=ζn≤(a)=ζn-1≤(a)+an=ζn-1≤(b)+an∈[ζn-1≤(b),ζn≤(b)[.
Hence, in view of (7.7), we get
sb=(b0,…,bn-1,s-ζn-1≤(b),0,…)=(b0,…,bn-1,an,0,…)=(a0,…,an-1,an,0,…)=a.∎
In the next result, we shall show that any member of 𝒜S,X belongs to 𝒵S,X, and moreover, we shall exhibit the specific form of the attractive subsets.
Lemma 7.6.
Let X=Seq§,*(N|[0,q]) and S=N∞.
Then
T∈𝒜S,X∖𝒪S,X if and only if T assumes one among the following forms:
T=Sa|k∪Γk(a) for some a∈Seq§,*(ℕ|0) and 0≤k≤a0-1,
T=Sa|l+r∪Γl+r(a) for some a∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0),
1≤l≤q and 0≤r≤al-1.
Proof.
(i): Let T∈𝒜S,X.
Then T∈𝒞S,X, i.e. T=⋃{Sb:b∈T}.
There is nothing to prove if T∈𝒪S,X.
Let now a=(a0,…,al,0,…), a′=(a0′,…,am′,0,…)∈T.
We claim that a∈Sa′ or a′∈Sa.
Without loss of generality, assume that l≤m.
Since T∈𝒜S,X, there must exist c=(c0,…,cp,0,…)∈T such that a,a′∈Sc.
By part (ii) of Proposition 7.5, it follows that l≤m≤p and ai=ai′=ci for each i=0,…,l-1.
On the one hand, if l=m, then either am≤am′, whence a∈Sa′, or am′≤am, whence a′∈Sa.
On the other hand, if l<m, we clearly have a′∉Sa again by part (ii) of Proposition 7.5.
Assume by contradiction that a∉Sa′.
Therefore, in order to satisfy the condition a∉Sa′, it must result that al>al′.
However, as al′=cl and al≤cl, we must also have al≤al′, which is an absurd.
Thus either a∈Sa′ or a′∈Sa.
In other terms, for any a,a′∈T, it results that either Sa⊆Sa′ or Sa′⊆Sa.
(ii): Suppose firstly that T assumes one among the three forms given in the statement.
Clearly, T∉𝒪S,X.
We claim that T∈𝒞S,X.
To this regard, it is straightforward to verify that T={(j,0,…):j∈ℤ+} belongs to 𝒜S,X.
Now, on the one hand, assume that T=Sa|k∪Γk(a) for some a∈Seq§,*(ℕ|0) and 0≤k≤a0-1.
Let x∈φS(T).
Hence there exist s∈S and y∈T such that x=sy.
Furthermore, the condition y∈T implies that either there exists s′∈S such that y=s′a|k or y=(a0,…,ak,j,0,…) for some j∈ℤ+.
In particular, in the first case, we get x=(s⋅s′)a|k=min{s,s′}a|k.
Thus we easily get x∈T.
Now, we shall demonstrate that T∈𝒜S,X∖𝒪S,X, i.e. for each y1,y2∈T, there exists y∈T such that y1,y2∈Sy.
Three cases may occur:
if y1=s1a|k and y2=s2a|k for some s1,s2∈S, just take y:=(a0,…,ak,0,…)∈T;
if y1=s1a|k and y2=(a0,…,ak,j,0,…) for some s1∈ℕ, j∈ℤ+, take y:=y2 since y1=s1y=s1y2;
if y1=(a0,…,ak,j1,0,…) and y2=(a0,…,ak,j2,0,…), just take y=(a0,…,ak,j,0,…), where j:=max{j1,j2}.
On the other hand, the proof for the case T=Sa|l+r∪Γl+r(a) for some a∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0), 1≤l≤q and 0≤r≤al-1 is similar to the previous argument.
This proves T∈𝒜S,X∖𝒪S,X when T assumes one among the forms given in the statement.
Conversely, let T∈𝒜S,X∖𝒪S,X.
Then, in view of the previous part (i), it results that T∈𝒵S,X, i.e. the set system {Sb:b∈T} is a chain, and furthermore, T=⋃{Sb:b∈T} since T∈𝒞S,X.
Let c=(c0,0,…)∈T.
Then c∈Sc′, where c′=(c0′,0,…) and c0′>c0.
As T∈𝒵S,X, we can always find a sequence c′′=(c0′′,0,…) such that Sc′⫋Sc′′ simply taking c0′′>c0′.
Hence T may assume the form {(j,0,…):j∈ℤ+}.
Let c=(c0,…,cp,0,…)∈T, where 1≤p≤c0.
In particular, notice that c∈Seq§,*(ℕ|0).
Clearly, we have that c∈Sc′, where c′=(c0,…,cp-1,cp′,0,…) and cp′>cp.
As T∈𝒵S,X, we can always find c′′=(c0,…,cp-1,cp′,0,…) such that Sc′⫋Sc′′ simply taking cp′′>cp′.
Set a:=c′, a0:=c0 and k:=p-1.
Hence we conclude that T=Sa|k∪Γk(a) for some a∈Seq§,*(ℕ|0) and 0≤k≤a0-1.
Let c=(c0,…,cl,cl+1,…,cl+r′,0,…)∈T.
In particular, notice that c∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0), where 1≤l≤q and 1≤r′≤l.
Clearly, c∈Sc′, where c′=(c0,…,cl,cl+1,…,cl+r′-1,cl+r′′,0,…) and cl+r′′>cl+r′.
As T∈𝒵S,X, we can always find c′′=(c0,…,cl,cl+1,…,cl+r′-1,cl+r′′′,0,…) such that Sc′⫋Sc′′ simply taking cl+r′′′>cl+r′′.
Set a:=c′, a0:=c0 and r:=r′-1.
Hence we conclude that T=Sa|l+r∪Γl+r(a) for some a∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0), 1≤l≤q and 0≤r≤al-1.
This concludes the proof.
∎
As an immediate consequence of Lemma 7.6, in the following result, we shall deduce that 𝒜S,X∖𝒪S,X agrees with the collection of all the maximal members of 𝒜S,X.
Theorem 7.7.
Let X=Seq§,*(N|[0,q]) and S=N∞.
Then
(7.9)Max(𝒜S,X)=𝒜S,X∖𝒪S,X.
Let us now characterize the membership of a sequence of X=Seq§,*(ℕ|[0,q]) in the subsets of the chain 𝒟ℕ∞,X.
Lemma 7.8.
Let X=Seq§,*(N|[0,q]), S=N∞, a=(a0,…,an,0,…)∈X, with an≠0.
Set l:=a0.
Then, for any i≥0, it results that a∈Di if and only if one among the following conditions holds:
Proof.
There is nothing to prove when taking D0=X, in view of the definition of Seq§,*(ℕ|[0,q]).
Let now a=(a0,…,an,0,…)∈X with an≠0, and assume that q<a0.
We shall demonstrate that a∈Di if and only if n≤(a0-i-1)+.
We will proceed by induction.
Suppose the claim to be true for Di.
Let us now assume that n≤(a0-i-1,0)+.
We claim that a∈Di+1.
We distinguish two cases:
we may have either a0≥i+1 or a0<i+1.
Let us investigate the first situation.
Take
b:=(a0,…,an-1,an+1,0,…) and c:=(a0,…,an,1,0,…).
Clearly, b∈X.
It also results that c∈X since the condition n≤a0-i-1 implies n+1≤a0-i≤a0.
Furthermore, we easily see that both b and c satisfy the conditions in (i).
So b,c∈Di by the inductive hypothesis.
By (7.7), we easily observe that a=(ζn≤(b)-1)b=ζn≤(a)b and a=(ζn+1≤-1)c=ζn≤(a)c, i.e. a∈Sb∩Sc.
Moreover, Sb∥Sc by part (ii) of Proposition 7.5.
Therefore, in view of part (iii) of Theorem 7.2, we conclude that a∈Di+1.
On the other hand, if a0<i+1, then n=0, and in particular, we get a=(a0,0,…).
Thus we have that a∈S(i+1,0,…) and (i+1,0,…)∈Di+1 by the previous argument, whence a∈Di+1.
Conversely, let a∈Di.
In view of the inductive hypothesis, it must be n≤max{a0-i,0}.
We will prove that n≤(a0-i-1)+.
To this aim, assume that n=a0-i>0.
As Di∈𝒞S,X, there exists b∈Di such that a∈Sb, where b=(a0,…,an-1,bn,0,…) and an≤bn in view of part (ii) of Proposition 7.5.
Let now c∈Di be such that a∈Sc.
If c∉Sb, then we get b∈Sc.
In fact, the condition a∈Sc implies that c=(a0,…,an-1,cn,…,cm,0,…) with an≤cn by part (ii) of Proposition 7.5, while the fact that c∉Sb ensures that cn>bn, whence b∈Sc by part (ii) of Proposition 7.5.
This means that there is only a subset M∈Max(𝒜S,X) containing a, and hence a∉Di+1.
Thus we proved that n≤max{a0-i-1,0}, as wanted.
Let now a=(a0,…,an,0,…)∈X, with an≠0 and such that a0=l≤q.
We shall demonstrate that a∈Di if and only if n≤l+(al-i)+.
We will proceed again by induction.
Suppose the claim to be true for Di.
Let us now assume that n≤l+(al-i-1)+.
We claim that a∈Di+1.
We distinguish two cases: we may have either al≤i+1 or al<i+1.
Let us investigate the first situation.
Take b:=(a0,…,an-1,an+1,0,…) and c:=(a0,…,an,1,0,…).
Clearly, b∈X.
It also results that c∈X since the condition n≤l+al-i-1 implies n+1≤l+al-i≤l+al.
Furthermore, we easily see that both b and c satisfy the conditions in (ii).
So b,c∈Di by the inductive hypothesis.
By (7.7), we easily observe that a=(ζn≤(b)-1)b=ζn≤(a)b and a=(ζn+1≤(c)-1)c=ζn≤(a)c, i.e. a∈Sb∩Sc.
Moreover, Sb∥Sc by part (ii) of Proposition 7.5.
Therefore, in view of part (iii) of Theorem 7.2, we conclude that a∈Di+1.
On the other hand, if al<i+1, we get n≤l.
Therefore, just observe that
a∈{S(a0,…,an-1,i+1,0,…)ifn=l,S(a0,…,an,…,al-1,i+1,0,…)ifn<l.
In both cases, applying the above argument to the elements
(a0,…,an-1,i+1,0,…) and (a0,…,an,…,al-1,i+1,0,…)
and using the fact that Di+1∈𝒞S,X, we easily conclude that a∈Di+1.
Vice versa, let a∈Di.
In view of the inductive hypothesis, it must be n≤l+(al-i)+.
We will prove that n≤l+(al-i-1,0)+.
To this aim, assume that n=l+al-i>l.
As Di∈𝒞S,X, there exists b∈Di such that a∈Sb, where b=(a0,…,an-1,bn,0,…) and an≤bn in view of part (ii) of Proposition 7.5.
Let now c∈Di be such that a∈Sc.
If c∉Sb, then we get b∈Sc.
In fact, by part (ii) of Proposition 7.5, the condition a∈Sc implies that c=(a0,…,an-1,cn,…,cm,0,…) with an≤cn, while the fact that c∉Sb ensures that cn>bn, whence b∈Sc again by part (ii) of Proposition 7.5.
This means that there is only a subset M∈Max(𝒜S,X) containing a, and hence a∉Di+1.
Thus we showed that n≤(a0-i-1)+.
This concludes the proof.
∎
In view of Lemma 7.8, we get the following characterization for US,X.
Proposition 7.9.
In the same hypotheses and notations of Lemma 7.8, we have a=(a0,…,an,0,…)∈US,X if and only if one among the following conditions holds:
Proof.
Let a=(a0,…,an,0,…), where an≠0, and set l:=a0.
In view of Lemma 7.8, it results that a∈US,X if and only if one among the following conditions holds:
In case (a), we clearly have a∈Seq§,*(ℕ|0), and in view of its definition, ai=0 for each i>l.
So n=l.
In case (b), we may have either
a∈Seq§,*(ℕ|0)∩Seq§,*(ℕ|l) or a∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0).
If a∈Seq§,*(ℕ|0)∩Seq§,*(ℕ|l), then al≠0 and n≤l.
On the other hand, if a∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0), it results that al≠0 and al+i=0 when i>al.
So n=l+al.
∎
In view of (7.9), of Proposition 7.9 and of (6.2), the following characterization for the subsets belonging to the set system ℋS,X easily holds.
Corollary 7.10.
Let X=Seq§,*(N|[0,q]) and S=N∞.
Then Y∈HS,X if and only if one among the following situations occurs:
Y=Sa|l-1∪Γl-1(a) for some a=(l,a1,…,al,0,…)∈Seq§,*(ℕ|0);
Y=Sa|l+al-1∪Γl+al-1(a) for some a=(l,a1,…,al,…,al+al,0,…)∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0).
At this point, in reference to the ℕ∞-set X=Seq§,*(ℕ|[0,q]), in the next result, we shall firstly exhibit the 2MA-decomposition of Seq§,*(ℕ|[0,q]) and next demonstrate that the chain 𝒟ℕ∞,X stabilizes for α=ω+q.
Theorem 7.11.
Let X=Seq§,*(N|[0,q]) and S=N∞.
Then
X is MAC-union independent;
Proof.
(i): Let a=(a0,…,an,0,…)∈X, where an≠0, and set l:=a0.
Assume firstly that a∈Seq§,*(ℕ|0).
Then n≤l.
Take b:=(a0,…,an,bn+1,…,bl,0,…) with bl≠0.
Hence a∈Sb|l-1∪Γl-1(b), i.e. a∈ℋS,X.
Assume now that a∈Seq§,*(ℕ|l)∖Seq§,*(ℕ|0).
Then l+1≤n≤l+al.
Take b:=(a0,…,an,0,…), with bl+al≠0.
Hence a∈Sa|l+bl-1∪Γl+bl-1(b), i.e. a∈ℋS,X.
(ii): It follows by the previous part (i) and by part (vii) of Lemma 6.3.
(iii): We firstly claim that
(7.10)Dω={(a0,0,…)∈X:a0∈ℕ}∪{(a0,…,an,0,…)∈X:0<n≤a0≤q}.
Let us recall that Dω=⋂{Di:i∈ℕ}.
In view of Lemma 7.8, the elements (a0,0,…) with a0>q belong to Di for each i∈ℕ.
So (a0,0,…)∈𝒟ω.
Moreover, again by Lemma 7.8, we deduce that (a0,1,0,…)∉Da0.
Thus a sequence a=(a0,…,an,0,…), where a0>q, belongs to Dω if and only if n=0.
Furthermore, let a=(a0,…,an,…)∈X be such that a0=s≤q.
Again by Lemma 7.8, we deduce that (a0,…,as,0,…)∈Di for each i∈ℕ and (a0,…,as,1,0,…)∉Das.
Therefore, we conclude that a=(a0,…,an,…)∈X such that a0≤q belongs to Dω if and only if n≤a0.
Thus (7.10) holds.
Let us compute Dω+1.
Notice that a=(a0,0,…), where a0>q, may belong only to S(a0′,0,…) with a0≤a0′, and hence to a unique member of Max(𝒜S,X).
Similarly, the sequence a=(q,a1,…,aq,0,…) belongs to S(q,a1,…,bq,0,…), where aq≤bq, and hence it belongs to a unique member of Max(𝒜S,X).
Let now a=(1,a1,0,…).
Then it results that a∈S(1,b1,0,…) for each a1≤b1, and hence it belongs to a unique member of Max(𝒜S,X).
On the other hand, let a=(a0,0,…) with a0≤q.
Hence a∈S(a0,a1,a2,0,…)∩S(a0,a1+1,0,…) and S(a0,a1,a2,0,…)∥(a0,a1+1,0,…).
Let now a=(l+1,a1,…,al,0,…) for some l=1,…,q-2.
Hence we get
a∈S(l+1,a1,…,al+1,0,…)∩S(l+1,a1,…,al,al+1,0,…),
where S(l+1,a1,…,al+1,0,…)∥S(l+1,a1,…,al,al+1,0,…).
Finally, if a=(q,a1,…,at,0,…), with t<q, we get
a∈S(q,a1,…,at+1,0,…)∩S(q,a1,…,at,1,0,…),
where S(q,a1,…,at+1,0,…)∥S(q,a1,…,at,1,0,…).
Thus we proved that
Dω+1={(a0,0,…)∈X:0≤a0≤q}∪{(1+l,a1,…,al,0,…)∈X:l=1,…,q-1}.
Now, extending inductively the above argument, we get
Dω+i={(a0,0,…)∈X:0≤a0≤q}∪{(i+l,a1,…,al,0,…)∈X:l=1,…,q-i}
for i=1,…,q-1.
In particular, it results that
Dω+q-1={(a0,0,…)∈X:0≤a0≤q}∪{(a0,a1,0,…)∈X:a0=q,a1≥1}.
Notice that Dω+q-1∈𝒜S,X.
In fact,
if a=(a0,0,…), b=(b0,0,…)∈Dω+q-1, then a,b∈Sc, where c=(max{a0,b0},0,…);
if a=(q,a1,0,…) and b=(q,b1,0,…), then a,b∈Sc, where c=(q,max{a1,b1},0,…);
if a=(a0,0,…) and b=(q,b1,0,…), then a,b∈Sb.
Thus we conclude that Dω+q=∅ in view of part (iv) of Theorem 7.2.
This proves that μ(𝒟S,X)=ω+q.
∎
and
Federico G. Infusino
