Independent and free sets in universal algebras

0.19 MB·English

by  Taras Banakh

INDEPENDENT AND FREE SETS IN UNIVERSAL ALGEBRAS
TARASBANAKH,ARTURBARTOSZEWICZ,SZYMONGL A¸B
Abstract. We discuss notions of independent and free sets in universal algebras, and more generally, in spaces
endowedwithahulloperator. OurmainresultsaysthatforeachuniversalalgebraA=(A,A)ofcardinality|A|≥2
and an infinite set X of cardinality |X| ≥|A|, the X-power AX = (AX,AX) of the algebra A contains an A-free
subsetF ⊂AX ofcardinality|F|=2|X|. Thisgeneralizes aclassicalFichtenholtz-Katorovitch-Hausdorff resulton
2
theexistence oflargeindependent familiesofsubsetsofagiveninfiniteset.
1
0
2

p
Introduction
e
S
It is well-known that each maximal linearly independent subset of a linear space X generates X. A natural

8 notion of independence can be also defined for other algebraic structures, in particular, for universal algebras.
2 Unfortunately, even for relative simple universal algebras (like groups) the notion of independence does not work

] as good as for linear spaces. For example, any infinite linear space V over the two-element field F2 = {0,1} can
O be embedded into a non-commutative groupG of cardinality |G|=2|V| so that each maximallinearly independent
L subsetB ⊂V remainsmaximalindependentinGandhencedoesnotgeneratethewholegroupG. Thisfact(proved
inProposition1.3)showsthattheproblemofconstructinglargeindependentsetsingeneralalgebraicsettingisnot
.
h
trivial.
t
a A general notion of algebraic independence was actually introduced by Marczewski in [7] and studied in subse-
m
quent papers [8], [9]. A subset B of an abstract algebra (A,A) is called ([7]) independent provided each function
[ from B in A can be extended to a homomorphism defined on the algebra generated by B. This notion is a gen-

eralization of the linear and algebraic independence of numbers, linear independence of vectors, independence of
1
sets andelements ofBooleanalgebrasandmany others. Inour paper,Marczewski’sindependent sets willbe called
v
4 “free”. The notion of a “hull” (or “closure”) was introduced by Birkhoff [3] and also considered by Marczewski
4 (e.g. in [8]). Other concepts of independence considered in our paper can be also found in [8]. For the reader’s
4 convenience we recall some proofs and considerations in the modern notation.
6
OurmainresultisTheorem4.3sayingthatforeachuniversalalgebraA=(A,A)ofcardinality|A|≥2andeach
.
9 infinite set X of cardinality |X|≥|A|, the X-power AX =(AX,AX) of the algebra A contains an independent set
0 F ⊂ AX of cardinality |F| = 2|X|. This generalizes a classical Fichtenholtz-Katorovitch-Hausdorff result on the
2
existence of an independent family F ⊂ P(X) of cardinality |F| = 2|X| in the power-set P(X) of X. The other
1
: famous generalizationof Fichtenholtz-Katorovitch-Hausdorfftheorem is the Balcar-Franˇektheoremon the number
v
of independent elements in complete Boolean algebras [1].
i
X
r 1. Hull operators and (strongly) independent sets
a
By a hull operator on a set X we understand a function H :P(X)→P(X) defined on a family of all subsets of
X, which is monotone in the sense that for any subsets A⊆B ⊆X we get A⊆H(A)⊆H(B)⊆X.
Let H :P(X)→P(X) be a hull operator on a set X. A subset B ⊂X is called
• H-independent if b∈/ H(B\{b}) for each b∈B;
• strongly H-independent if B∩H(∅)=∅ and H(B )∩H(B )=H(B ∩B ) for any subsets B ,B ⊂B.
1 2 1 2 1 2
Proposition 1.1. Let X be a set and H be a hull operator on X. Each strongly H-independent set B ⊂ X is
H-independent.
Proof. Assuming that a strongly H-independent set B is not H-independent, we can find a point b ∈ B such
that b ∈ H(B \{b}). Then b ∈ H({b})∩H(B \{b}) = H({b}∩(B \{b})) = H(∅), which is not possible as
B ⊂X \H(∅). (cid:3)
Proposition1.1canbereversedforhulloperatorsofmatroidtype. WesaythatahulloperatorH :P(X)→P(X)
• is idempotent if H(H(A))=H(A) for each subset A⊂X;
1991 Mathematics Subject Classification. 05B35,08A99, 17A50.
Keywords and phrases. Hulloperator,universalalgebra,(strongly)independent set,freeset.
The first author has been partially financed by NCN grant DEC-2011/01/B/ST1/01439. The second and the third authors have
beensupportedbythePolishMinistryofScienceandHigherEducation GrantNoNN201414939(2010-2013).
1
2 TARASBANAKH,ARTURBARTOSZEWICZ,SZYMONGL A¸B
• has finite supports if for each subset A ⊂ X and a point x ∈ H(A) there is a finite subset F ⊂ A with
x∈H(F);
• has the MacLane-Steinitz exchange property if for any subset A ⊂ X and points x,y ∈ X \H(A) the
inclusion x∈H(A∪{y}) is equivalent to y ∈H(A∪{x});
• is of matroid type if H is idempotent, has finite supports and has the Maclane-Steinitz exchange property.
It is well-known(and easy to see) that the operatorof taking the linear hull in a linear space is of matroidtype.
Proposition 1.2. Let H :P(X)→P(X) be a hull operator of matroid type on a set X.
(1) A subset B ⊂X is H-independent if and only if it is strongly H-independent.
(2) Each H-independent subset lies in a maximal H-independent subset of X.
(3) For each maximal H-independent subset B ⊂X its hull H(B)=X.
Proof. This proposition is known for finite sets X, see [10, §1.4]. For convenience of the reader we present a proof
for an arbitrary set X.
1. Proposition 1.1 implies that each strongly H-independent subset of X is H-independent. Now assume that
a set B is H-independent. Then for each b ∈ B we get b ∈/ H(B \{b}) ⊃ H(∅) and hence B ∩H(∅) = ∅. To
show that B is strongly H-independent, it remains to check that H(B )∩H(B ) ⊂ H(B ∩B ) for any subsets
1 2 1 2
B ,B ⊂ B. This inclusion is trivial if B ⊂ B or B ⊂ B . So, we assume that B 6⊂ B and B 6⊂ B . Given
1 2 1 2 2 1 1 2 2 1
any point h ∈ H(B )∩H(B ), we should prove that h ∈ H(B ∩B ). Assume conversely that h ∈/ H(B ∩B ).
1 2 1 2 1 2
Since the hull operator H has finite supports, there are minimal finite subsets F ⊂ B and F ⊂ B such that
1 1 2 2
h∈H(F )∩H(F ). It followsfrom F ∩F ⊂H(F ∩F )⊂H(B ∩B )6∋h that h∈/ F ∩F andhence h∈/ F or
1 2 1 2 1 2 1 2 1 2 1
h∈/ F . Withoutlossofgenerality,h∈/ F . ItfollowsfromH(F )∩H(F )6⊂H(F ∩F ),thatF 6⊂F andF 6⊂F .
2 1 1 2 1 2 1 2 2 1
Chooseanypointb∈F \F andletA=F \{b}. The H-independenceofthe setB implies b∈/ H(B\{b})⊃H(A)
1 2 1
andtheminimalityofthe setF impliesh∈/ H(A). Since h∈H(F )=H(A∪{b}),theMacLane-Steinitzexchange
1 1
property of H guarantees that
b∈H(A∪{h})⊂H(A∪H(F ))⊂H(H(A∪F ))=H(A∪F )⊂H(B\{b}),
2 2 2
which contradicts the H-independence of B.
2. The second statement can be easily proved using Zorn’s Lemma and the fact that the hull operator H has
finite supports.
3. Let B be a maximal H-independent subset in X. If H(B) 6= X, then we can choose a point x ∈ X \H(B).
By the maximality of B the set B = B ∪{x} is not H-independent. Consequently, there is a point b ∈ B such
x x
that b∈H(B \{b}). If b6=x, then consider the set A=B\{b} and observe that b∈H(B \{b})=H(A∪{x}).
x x
Then the MacLane-Steinitz exchange property implies x∈H(A∪{b})=H(B), which contradicts the choice of x.
So, b=x and x=b∈H(B \{b})=H(B), which contradicts the choice of x. (cid:3)
x
Unfortunately many hull operators that naturally appear in algebra are not of matroid type. The simplest
example is the grouphull operator. It assignsto eachsubset A of a groupG the subgroupH(A)⊂G generatedby
A. Observethattheset{2,3}isH-independentinthegroupofintegersZbutfailstobestronglyH-independentin
Z. Eachmaximalstrongly H-independent subset B in the groupof rational Q is a singletonand hence H(B)6=Q.
The following proposition yields an example of a (non-commutative) group G of cardinality continuum containing
a maximal H-independent subset which is countable and hence does not generate the whole group.
Proposition 1.3. Each infinite linear space V over the two-element field F = {0,1} embeds into a group G
2
of cardinality |G| = 2|V| such that for the operator of group hull H : P(G) → P(G) each maximal strongly H-
independent subset of V remains maximal H-independent in G.
Proof. Since the field F is finite, the linear space V has a Hamel basis of cardinality |V|. Since each permutation
2
of points of the Hamel basis induces a linear automorphism of the linear space V, the linear automorphism group
GL(V) of V has cardinality |GL(V)|=2|V|.
Let G = V ⋊GL(V) be the semidirect product of the groups (V,+) and GL(V). Elements of the group G are
ordered pairs (v,f) ∈ V ×GL(V) and for two pairs (v,f),(u,g) ∈ V ⋊GL(V) their product is defined by the
formula
(v,f)·(u,g)=(v+f(u),f ◦g).
The inverse element to a pair (v,f)∈G is the pair (−f−1(v),f−1).
Identify the group V with the normal subgroup V ×{id } of G. Here by id : V → V we denote the identity
V V
automorphism of V.
Let M ⊂ V be a maximal strongly H-independent subset of the additive group V. Since for each subset B of
V the group hull H(B) of B coincides with its linear hull, the operator of group (=linear) hull in V is of matroid
type. By Proposition 1.1, the maximal strongly H-independent subset M of V is maximal H-independent in V.
INDEPENDENT AND FREE SETS IN UNIVERSAL ALGEBRAS 3
We need to check that M remains maximal H-independent in the group G. Given any element g ∈ G\M we
need to show that the union M ∪{g} is not H-independent in G. If g ∈ V, then the set M ∪{g} ⊂ V is not
H-independent in G by the maximality of the H-independent set M in V.
So, we assume that g ∈/ V. In this case g = (u,f) ∈ V ×GL(V) for some non-identity automorphism f of V.
Observe that g−1 =(−f−1(v),f−1) and for each v ∈V we get
gvg−1 =(u,f)·(v,id )·(−f−1(u),f−1)=(f(v),id )=f(v) and
V V
g−1vg =(−f−1(u),f−1)·(v,id )·(u,f)=(f−1(v),id )=f−1(v).
V V
Being a maximal linearly independent subset of V, the set M is a Hamel basis in V. Since f 6= id , there is a
V
pointa∈M withf(a)6=a. SinceM isaHamelbasisofthe linearspaceV, eachpointv ∈V canbe writtenasthe
sum ΣB = b of a unique finite subset B ⊂M. In particular, a=ΣA and f(a)=ΣF for some finite subsets
Pb∈B
A,F ⊂M. Since F 6=A, there is a point v ∈(A\F)∪(F \A).
If v ∈A\F, then consider the set A =A\{v}⊂M \{v} and observe that
v
v =a−ΣA =f−1(f(a))−ΣA =g−1·(ΣF,id )·g·(ΣA ,id )−1∈H({g}∪M \{v}),
v v A v A
which implies that M ∪{g} is not H-independent in G.
If v ∈F \A, then consider the set F =F \{v}⊂M \{v} and observe that
v
v =ΣF −ΣF =f(a)−ΣF =g·(ΣA,id )·g−1·(ΣF ,id )−1∈H({g}∪M \{v}),
v v V v V
which implies that M ∪{g} is not H-independent in G. (cid:3)
So, in general, hull operators generated by algebraic structures need not be of matroid type, which makes the
problem of constructing large (strongly) independent sets non-trivial. We shall be interested in hull operators
induced by the structure of universal algebra (which includes as partial cases the structures of group, linear space,
linear algebra, etc.) Such hull operators will be defined and studied in the next section.
2. Universal algebras
A universal algebra is a pair A =(A,A) consisting of a set A and a family A of algebraic operations on A. An
algebraic operation on a set A is any function α:ASα →A defined on a finite power ASα of A, where Sα is a finite
subset of ω called the support of the operation α. An algebraic operation α : ASα → A is constant if α(ASα) is a
singleton. In particular, each algebraic operation α:A∅ →A with empty support is constant.
Observe that any function σ : F → E between finite subsets of ω induces a dual function σ∗ : AE → AF,
σ∗ :f 7→f ◦σ, called a substitution operator. Then for any algebraic operation α:ASα →A with support Sα =F
the composition β =α◦σ∗ is a well-defined algebraic operation on A with support S =E.
β
A family A of operations on a set A is called
• unital if A contains the identity operation id :A1 →A, id :(a )7→a ;
A A 0 0
• ∅-regulariffor eachconstantoperationα∈Athere is anoperationβ ∈Awith empty supportS =∅such
β
that β(A∅)=α(ASα);
• stable under substitutions (briefly, substitution-stable) if for any function σ :F →E between finite subsets
ofω andanyalgebraicoperationα∈Awith supp(α)=F the algebraicoperationα◦σ∗ :AE →Abelongs
to A;
• stable under compositions ifforanyfinite subsetS ⊂ω andalgebraicoperationsα∈Aandα ∈A, i∈S ,
i α
with supports supp(α ) = S for all i ∈ E the composition α◦(α ) : AS → A of the diagonal product
i i i∈Sα
(αi)i∈Sα :AS →ASα with α belongs to A;
• a clone if A is unital, ∅-regular,and stable under substitutions and compositions.
The clone of a universalalgebraA=(A,A) is the universalalgebraA¯ =(A,A¯) endowedwith the smallestclone A¯
that contains the operation family A. The clone A¯ is equal to the union A¯= A of operation families A ,
Sn∈ω n n
n∈ω, defined by induction. Let A ={id } and
0 A
An+1 =An∪{α∈AA∅ :∃β ∈An α(A∅)=β(ASβ)}∪
∪{α◦σ∗ :α∈A , σ :S →F is a function into a finite subset F ⊂ω}∪
n α
∪{α◦(αi)i∈Sα :α∈A, (αi)i∈Sα ∈(An)Sα, ∀i,j ∈Sα Sαi =Sαj}
for n∈ω. This implies that the clone A¯of A has cardinality |A¯|≤max{|A|,ℵ }.
0
Each universal algebra A = (A,A) possesses the canonical hull operator A(·) : P(A) → P(A) assigning to each
subset B ⊂A its A-hull
A(B)=B∪ [ α(BSα).
α∈A
The definition implies that this hull operation has finite supports.
4 TARASBANAKH,ARTURBARTOSZEWICZ,SZYMONGL A¸B
If the operation family A is ∅-regular, then for each constant algebraic operation α ∈ A, the singleton α(Anα)
lies in the A-hull A(∅) of the empty subset of A.
Proposition 2.1. Let A=(A,A) be a universal algebra whose operation family A is unital and stable under substi-
tutions. Then for each subset B ⊂A and a point a∈A(B) there is an algebraic operation α∈A and an injective
function x:S →B such that a=α(x).
α
Proof. Since A is unital, A(B) = Sβ∈Aα(BSβ) and we can find an operation β ∈ A and a function z : Sβ → B
such that a = β(z). Let x : S → z(S ) ⊂ B be any bijective map defined on a finite subset S ⊂ ω. Consider
β
the function σ = x−1 ◦z : Sβ → S, which induces the substitution operator σ∗ : AS → ASβ. Since A is stable
under substitutions, the operation α = β ◦ σ∗ : AS → A belongs to A and has support S = S. Moreover,
α
α(x)=β◦σ∗(x)=β(x◦σ)=β(z)=a. (cid:3)
A subset B ⊂ A is called a subalgebra of a universal algebra A = (A,A) if A(B) ⊂ B, i.e., B coincides with its
A-hull A(B). Since the intersection of subalgebras is a subalgebra, for each subset B ⊂ A there is the smallest
subalgebra of A that contains B. This subalgebra is called the subalgebra generated by B and admits the following
simple description:
Proposition 2.2. Let A = (A,A) be a universal algebra and A¯ = (A,A¯) be its clone. For each subset B ⊂ A the
subalgebra generated by B coincides with the A¯-hull A¯(B) of B.
Proof. Let hBi denote the subalgebra of A generated by B. The inclusion hBi ⊂ A¯(B) will follow as soon as we
checkthatthe A¯-hull A¯(B) ofB isa subalgebraofthe universalalgebraA, thatisA¯(B)containsallconstantsand
it is closed all operations from A. We need to check that A(A¯(B)) ⊂ A¯(B). Take any element y ∈ A(A¯(B)) and
findanoperationα∈A anda functionx:S →A¯(B) suchthaty =α(x). Foreveryi∈S the pointx(i) belongs
α α
to the A¯-hull A¯(B) of B and hence can be written as x(i)=α (z ) for some algebraic operation α ∈A¯ and some
i i i
function z : S → B. Choose a finite subset S ⊂ ω of cardinality |S| = |S | and for every i ∈ S choose
i αi Pi∈Sα αi α
an injective function σ :S →S so that σ (S )∩σ (S )=∅ for i6=j. Each function σ induces the surjective
i αi i αi j αj i
substitution operatorsσi∗ :AS →ASαi, σi∗ :f 7→f◦σi. Considera unique function z :S →B suchthat z◦σi =zi
for all i∈S .
α
Since A¯ is a clone, it is closed under substitutions. Consequently, for every i ∈ S the operation β = α ◦σ∗ :
α i i i
AS → A belongs to A¯. It follows that β (z)= α ◦σ∗(z)=α(z◦σ ) =α (z )= x(i). Since the function family A¯
i i i i i i
is closed under compositions, the operation β =α◦(β ) :AS →A belongs to A¯. Since
i i∈Sα
β(z)=α (β (z)) =α (x(i)) =α(x)=y,
(cid:0) i i∈Sα(cid:1) (cid:0) i∈Sα(cid:1)
the point y = β(z) belongs to A¯(B). So, A(A¯(B)) ⊂ A¯(B) and A¯(B) is a subalgebra of A, which implies
hBi⊂A¯(B).
To prove that A¯(B) ⊂ hBi, we use the decomposition A¯ = A of A¯ into the countable union of the
Sn∈ω n
operation families A , n ∈ ω, defined at the beginning of Section 2 right after the definition of clone. Since
n
A¯(B) = A (B), it suffices to check that A (B) ⊂ hBi for every n ∈ ω. This will be done by induction on
Sn∈ω n n
n∈ω.
Since A = {id }, A (B) = B ⊂ hBi. Assume that for some n ∈ ω we have proved that A (B) ⊂ hBi. The
0 A 0 n
inclusion A (B) ⊂ hBi will follow as soon as we check that β(x) ∈ hBi for each operation β ∈ A and a
n+1 n+1
function x:S →B. If β ∈A , then β(f)∈A (B)⊂hBi.
β n n
If β ∈A \A , then by the definition of the operation family A , the following three cases are possible:
n+1 n n+1
1) Sβ =∅ and there is a constant operation α∈An such that β(A∅)=α(ASα). If B =∅, then x∈BSα implies
S =∅=S and hence β(x)=α(x)∈A (B)⊂hBi.
α β n
If B 6=∅, then we can take any function y :S →B and conclude that β(x)=α(y)∈A (B)⊂hBi.
α n
2)β =α◦σ∗forsomeoperationα∈A andsomefunctionσ :S →S . Considerthefunctiony =x◦σ :S →B
n β α β
and observe that β(x)=α◦σ∗(x)=α(x◦σ)∈A (B)⊂hBi.
n
3) β =α◦(αi)i∈Sα for some operations α∈A and (αi)i∈Sα ∈ASnα with Sαi =Sβ for all i∈Sα. Since x∈BSβ,
the inductive assumption A (B) ⊂ hBi guarantees that for every i ∈ S the point y(i) = α (x) ∈ A (B) belongs
n α i n
to hBi. Consider the function y : S →hBi, y : i7→ y(i)= α (x). Then β(x) =α (α (x)) = α((y(i)) ) =
α i (cid:0) i i∈Sα(cid:1) i∈Sα
α(y)∈A(hBi)⊂hBi. The last inclusion A(hBi)⊂hBi follows from the fact that hBi is a subalgebra of A. (cid:3)
Let A=(A,A) be a universal algebra, X be a non-empty set, and AX be the set of all functions from X to A.
For every x∈X denote by δ :AX →A, δ :f 7→f(x), the projection onto x-th coordinate.
x x
Observe that each algebraic operation α : ASα → A induces an algebraic operation αX : (AX)Sα → AX on
the set AX of all functions from X to A. The operation αX assigns to each function f : S → AX the function
α
αX(f):X →A defined by αX(f):x7→α(δ ◦f)for x∈X. Writing the function f is coordinatesas f =(f ) ,
x i i∈Sα
we get that αX(f)=αX (f ) is the function assigning to each x∈X the point α (f (x)) .
(cid:0) i i∈Sα(cid:1) (cid:0) i i∈Sα(cid:1)
INDEPENDENT AND FREE SETS IN UNIVERSAL ALGEBRAS 5
For a universal algebra A = (A,A) its X-th power is the pair AX = (AX,AX) consisting of the X-th power of
A and the operation family AX ={αX} .
α∈A
Nowletus consideranimportantexampleofa universalalgebra2=(2,B),calledthe Boolean clone. Itconsists
of the doubleton 2={0,1}and the family B of all possible algebraicoperations on 2. In the next section, we shall
seethatthepowers2X oftheBooleancloneplayanimportantroleinstudyingindependentandfreesetsinpowers
of arbitrary universal algebras.
3. Independent and free sets in universal algebras
In this section we introduce and study three independence notions in universal algebras.
A subset B ⊂A of a universal algebra A=(A,A) is called
• A-independent if b∈/ A(B\{b}) for all b∈B;
• strongly A-independent if B∩A(∅)=∅ and A(B )∩A(B )=A(B ∩B ) for any subsets B ,B ⊂B;
1 2 1 2 1 2
• A-free if for any function f : B → A there is a function f¯: A(B) → A such that f¯◦α(x) = α(f ◦x) for
any algebraic operation α∈A and any function x∈BSα ⊂ASα.
Thenotionofa(strongly)A-independentsetisinducedbytheoperatorofA-hull. Ontheotherhand,thenotion
of an A-free set is specific for universal algebras and has no hull counterpart.
The definitions imply that the notions of A-independent and A-free sets are “monotone” which respect to A:
Proposition 3.1. If a subset B ⊂ A of a universal algebra (A,A) is A-independent (A-free), then it is A′-
independent (A′-free) for each operation family A′ ⊂A.
On the other hand, the strong A-independence is not monotone with respect to A.
Example 3.2. Take a set A of cardinality |A|≥2 and consider an operation family A={α,β} consisting of two
constant operations α : A1 → A, β : A0 → A with α(A1) = β(A0). Observe that each subset of A is strongly
A-independentwhileeachsubsetB ⊂Aofcardinality|B|≥2failstobestronglyA′-independentforthesubfamily
A′ ={α}. Indeed,taketwonon-emptydisjointsubsetsB ,B ⊂B andobservethatA′(B )∩A′(B )={α(A1)}=6
1 2 1 2
∅=A′(∅).
Now we shall characterize A-free sets in universal algebras. On first characterization follows immediately from
the definition.
Proposition 3.3. A subset B ⊂A of a universal algebra (A,A) is A-free if and only if for any algebraic operations
α,β ∈A and functions x∈BSα, y ∈BSβ the equality α(x)=β(y) implies that α(f◦x)=β(f◦y) for any function
f :B →A.
For unital substitution-stable universal algebras this characterizationcan be improved as follows.
Proposition 3.4. A subset B ⊂ A of a unital substitution-stable universal algebra (A,A) is A-free if and only if
for any distinct algebraic operations α,β ∈ A with S = S the inequality α(x) 6= β(x) holds for each injective
α β
function x∈BSα =BSβ.
Proof. To prove the “only if” part, assume that the set B is A-free. Fix two distinct algebraic operations α,β ∈A
such that S = S = S for some finite set S ⊂ ω. We need to show that α(x) 6= β(x) for each injective function
α β
x∈BS.
Since α 6= β, there is a function y ∈ BS such that α(y) 6= β(y). Using the injectivity of x, choose a function
f : B → A such that f|x(S) = y ◦x−1|x(S). Since B is A-free, there is a function f¯ : A(B) → A such that
f¯(γ(z))=γ(f◦z)foranyγ ∈Aandz ∈ASγ. Inparticular,f¯(α(x))=α(f◦x)=α(y)6=β(y)=β(f◦x)=f¯(β(x)),
which implies α(x)6=β(x).
To prove the “if” part, assume that the set B is not A-free. Applying Proposition3.3, find algebraicoperations
α,β ∈ A and functions x ∈ BSα, y ∈ BSβ such that α(x) = β(y) and α(f ◦x) 6= β(f ◦y) for some function
f : B → A. Fix any bijective function z : S → x(S )∪y(S ) ⊂ B defined on a finite subset S ⊂ ω. Consider
α β
the functions σ = z−1 ◦ x : S → S and σ : z−1 ◦ y : S → S, which induce the substitution operators
α α β β
σ∗ : AS → ASα and σ∗ : AS → ASβ. Since the operation family A is stable under substitutions, the algebraic
α β
operationsα˜ =α◦σ∗ :AS →Aandβ˜=β◦σ∗ :AS →Abelongtothe family A. Observethatα˜(z)=α◦σ∗(z)=
α β α
α(z◦σ )=α(x)=β(y)=β(z◦σ )=β˜(z). On the other hand,
α β
α˜(f ◦z)=α◦σ∗(f ◦z)=α(f ◦z◦σ )=α(f ◦x)6=β(f ◦y)=β(f ◦z◦σ )=β˜(f ◦z)
α α β
implies that α˜ 6=β˜. (cid:3)
By Proposition 1.1, each strongly A-independent subset B ⊂A of a universal algebra (A,A) is A-independent.
Proposition 3.5. Let (A,A) be a unital ∅-regular substitution-stable universal algebra of cardinality |A|≥2. Each
A-free subset B ⊂A is strongly A-independent.
6 TARASBANAKH,ARTURBARTOSZEWICZ,SZYMONGL A¸B
Proof. First we prove that B∩A(∅)=∅. Assume conversely that B∩A(∅) contains some point b.
Since |A| ≥ 2, we can find two functions f ,f : B → A such that f (b) 6= f (b). Since B is A-free, for every
1 2 1 2
i∈{1,2}, there is a function f¯i :A(B)→A such that f¯i◦α(x)=α(fi◦x) for any α∈A and x∈BSα.
Theuniversalalgebra(A,A)isunitalandhencecontainstheidentityalgebraicoperationα:A1 →A,α:(a)7→a.
For this operation we get f¯(b)=f¯ ◦α(b)=α(f (b))=f (b) and hence f¯(b)6=f¯(b).
i i i i 1 2
On the other hand, the inclusion b ∈ A(∅) yields a 0-ary operation β ∈ A such that b = β(∅) where ∅ : ∅ → B
is the unique element of the power B∅ = BSβ. Then for every i ∈ {1,2}, the choice of f¯i guarantees that
f¯(b)=f¯ ◦β(∅)=β(f ◦∅)=β(∅)=b and hence f¯(b)=b=f¯(b), which contradicts the inequality f¯(b)6=f¯(b)
i i i 1 2 1 2
proved earlier. Hence B∩A(∅)=∅.
Next, we prove that A(B ) ∩A(B ) = A(B ∩ B ) for any subsets B ,B ⊂ B. This equality is trivial if
1 2 1 2 1 2
B ⊂ B or B ⊂ B . So we assume that both complements B \B and B \B are not empty. Assume that
1 2 2 1 1 2 2 1
A(B )∩A(B )6=A(B ∩B )andfind a point a∈A(B )∩A(B )\A(B ∩B ). For the pointa∈A(B )∩A(B ),
1 2 1 2 1 2 1 2 1 2
therearealgebraicoperationsα,β ∈Aandfunctionsx∈BSα,y ∈BSβ suchthatα(x)=a=β(y). Usingasimilar
1 2
reasoning as in the proof of Proposition 3.4 we may assume that x and y are injective
Let us show that the operation α is not constant. In the opposite case, the ∅-regularity of A, yields a 0-ary
operation γ ∈ A such that α(ASα) = γ(A∅). Then a = α(x) = γ(∅) ∈ A(∅) ⊂ A(B1∩B2), which contradicts the
choice of a.
Next,weprovethattheintersectionx(S )∩y(S )⊂B ∩B isnotempty. Assumingtheconverseandusingthe
α β 1 2
fact that the operationα is not constant,find a function x′ ∈ASα suchthat α(x′)6=β(y). Since the subsets x(Sα)
and y(S ) of B are disjoint, we can find a function f :B →A such that f|x(S )=x′◦x−1 and f|y(S )=y◦y−1.
β α β
Then
f¯(a)=f¯(α(x)) =α(f ◦x)=α(x′)6=β(y)=β(f ◦y)=f¯(β(y))=f¯(a),
which is a contradiction.
Thus the intersection x(S )∩y(S )⊂B ∩B is not empty and we can choose a function f :B →A such that
α β 1 2
f|y(S ) = id and f(x(S )\y(S )) ⊂ B ∩B . Such a choice of f guarantees that f ◦x ∈ B ∩B and then
β y(Sβ) α β 1 2 1 2
α(f ◦x)∈A(B ∩B )6∋a implies that a6=α(f ◦x). Since the set B is A-free, the function f can be extended to
1 2
a function f¯:A(B)→A such that f¯(γ(z))=f ◦γ(z) for each γ ∈A and z ∈BSγ. In particular,
f¯(a)=f¯(β(y))=β(f ◦y)=β(y)=a6=α(f ◦x)=f¯(α(x))=f¯(a).
This contradiction completes the proof of the strong A-independence of the set B. (cid:3)
Proposition3.5showsthatfor asubsetofa unital ∅-regularsubstitution-stable universalalgebra(A,A) wehave
the following implications:
A-free ⇒ strongly A-independent ⇒ A-independent.
The first implication cannot be reversedas shown by the following simple example.
Example 3.6. Thereis aclone (A,A¯)containinganinfinite stronglyA¯-independentsubsetB ⊂X suchthateach
A¯-free subset of A is empty.
Proof. Consider the linear algebra c consisting of all functions x:ω →R with finite support supp(x)={n∈ω :
00
x(n)6=0}⊂ω. ThisalgebraisendowedwiththeoperationfamilyA={+,·}∪{α :t∈R}consistingoftwobinary
t
operations (of addition and multiplication) and continuum many unary operations α : ~x 7→ t·~x of multiplication
t
by a real number t. The clone A¯of A contains the subfamily A′ consisting of all polynomials p(x)= n λ xi+1
Pi=0 i
of one variable, equal to zero at the zero function.
Claim 3.7. The set B = {x ∈ c : x(ω) ⊂ {0,1}, |supp(x)| = 1} of characteristic functions of singletons is
00
strongly A¯-independent.
Proof. It is easy to check that for each subset B′ ⊂B its A¯-hull A¯(B′) is equal to the linear subspace Rsupp(B′) =
{x∈ c :supp(x) ⊂ supp(B′)} where supp(B′) = supp(b). It follows that B∩A¯(∅)= B∩{0}= ∅ and for
00 Sb∈B′
any two subsets B ,B ⊂B we get
1 2
A¯(B )∩A¯(B )=Rsupp(B1)∩Rsupp(B2) =Rsupp(B1)∩supp(B2) =Rsupp(B1∩B2) =A¯(B ∩B ),
1 2 1 2
which means that the subset B is strongly independent. (cid:3)
Claim 3.8. Any non-empty subset B ⊂A is not A′-free and hence is not A¯-free.
Proof. Fix any function b ∈ B. This function has finite support F = supp(b). Then b and all its finite powers bn,
n > 0, belong to the finite-dimensional linear subspace RF = {x ∈ c : supp(x) ⊂ F} of c . Consequently, the
00 00
set {bn+1 : 0 ≤ n ≤ |F|} is linearly dependent, which allows us to find a non-zero vector (λ ,...,λ ) ∈ R|F|+1
0 |F|
such that |F| λ bi+1 = 0. This means that p(b) = 0 for the non-zero polynomial p(x) = |F| λ xi+1. Since the
Pi=0 i Pi=0 i
INDEPENDENT AND FREE SETS IN UNIVERSAL ALGEBRAS 7
polynomial p∈A′ is non-zero, there is a vector y ∈RF such that p(y)6=0. Let f :B →c be any function such
00
that f(b)=y.
Assuming that the set B is A′-free, we could find a function f¯: A′(B) →c such that f¯(p(b))=p(f ◦b). But
00
f¯(p(b))=06=p(y)=p(f(b)). This contradiction completes the proof. (cid:3)
(cid:3)
4. Free sets in powers of universal algebras
In this section we shall construct free sets of large cardinality in powers of universal algebras. We start with
studying free sets in a power 2X =(2X,BX) of the Boolean clone 2=(2,B)consisting of the doubleton 2={0,1}
and the family B of all possible algebraic operations on 2.
It turns out that BX-free subsets of 2X coincide with independent sets, well studied in Set Theory [6, §17].
Let us recall [6, p.83] that a family F of subsets of a set X is called independent if for any finite disjoint sets
F ,F ⊂F the intersection
1 2
\ F ∩ \ (X \F)
F∈F1 F∈F2
is not empty.
Identifying each subset F ⊂ X with its characteristic function χ : X → 2 = {0,1}, we can reformulate the
F
notion of an independent family of sets in the language of an independent function family. Namely, a function
family F ⊂ 2X is independent if for any pairwise distinct functions f ,...,f ∈ F their diagonal product
0 n−1
(f ) :X →2n is surjective.
i i<n
Proposition 4.1. For any set X a function family F ⊂ 2X is independent if and only it is BX-free in the clone
2X.
Proof. To prove the “if” part, assume that F is BX-free but not independent. Then there are pairwise distinct
functions ξ ,...,ξ ∈F ⊂2X whose diagonal product δ =∆ ξ :X →2n, δ :x7→(ξ (x)) is not surjective
0 n−1 i<n i i i<n
and hence the set F = 2n \δ(X) is not empty. Let α,β : 2n → 2 be two algebraic operations on 2 defined by
α−1(1) = F and β−1(1) = ∅. They induce the algebraic operations αX,βX : (2X)n → 2X on 2X. It follows that
for the function ξ :n→F, ξ :i7→ξ , we get αX(ξ)=βX(ξ).
i
On the other hand, Since αX 6= βX, there is a function ξ˜: n → 2X such that αX(ξ˜) 6= βX(ξ˜). Now choose any
function f : F → 2X such that f|ξ(n) = ξ˜◦ξ−1. Since F is BX-free, there is a function f¯: BX(F) → 2X such
that f¯(γX(ξ)) =γX(f ◦ξ) for any algebraic operation γ ∈B with S =n. In particular, f¯(αX(ξ))=αX(f ◦ξ)=
γ
αX(ξ˜)6=βX(ξ˜)=βX(f◦ξ)=f¯(βX(ξ)), whichcontradictsαX(ξ)=βX(ξ). Thiscontradictioncompletesthe proof
of the “if” part.
To prove the “only if” part, assume that F is independent but not BX-free. By Proposition 3.4, there are two
distinctoperationsα,β ∈B suchthatS =S =S forsomefinitesetS ⊂ω andαX(ξ)=βX(ξ)forsomeinjective
α β
functionξ :S →F. The functionξ canbe writtenintheform(ξ ) whereξ =ξ(i)∈F. The independence ofF
i i∈S i
guaranteesthatthe diagonalproduct(ξ ) :X →2S, δ =(ξ ) :x7→(ξ (x)) ofthese functions is surjective.
i i∈S i i∈S i i∈S
Then α6=β implies α◦δ 6=β◦δ, which yields a point x∈X such that α◦δ(x)6=β◦δ(x). Now observe that
αX(ξ)(x)=α((ξ (x)) )=α◦δ(x)6=β◦δ(x)=β((ξ (x)) )=βX(ξ)(x)
i i∈S i i∈S
which contradicts αX(ξ)=βX(ξ). (cid:3)
By a classical Fichtenholtz-Katorovitch-HausdorffTheorem 17.20 [6], the power-set P(X) of each infinite set X
contains an independent family F ⊂P(X) of cardinality |F|=|2X|=|P(X)|. Reformulating this result with help
of Proposition4.1, we get the following result:
Corollary 4.2. For each infinite set X the power-clone 2X = (2X,BX) contains a BX-free subset B ⊂ 2X of
cardinality |B|=|2X|.
In fact, Corollary 4.2 is a partial case of the following theorem, which is the main result of this paper.
Theorem 4.3. Assume that a universal algebra A = (A,A) has cardinality |A| ≥ 2. For each infinite set X of
cardinality |X| ≥ |A| the function algebra AX = (AX,AX) contains an AX-free subset F ⊂ AX of cardinality
|F|=2|X|.
Proof. Let A¯ be the clone of the operation family A. It has cardinality |A¯| ≤ max{|A|,ℵ } ≤ |X|. Since each
0
A¯X-free subset of AX is AX-free, we lose no generality assuming that A is a clone. In particular, A is unital,
∅-regular and substitution-stable.
For each finite subset S ⊂ω, consider the family
T ={(α,β,s)∈A×A×XS :S =S =S , α6=β}
S α β
8 TARASBANAKH,ARTURBARTOSZEWICZ,SZYMONGL A¸B
andobservethatithascardinality|T |≤|A×A×XS|≤|X|. Thenthe unionT = T whereS runsover
S SS∈[ω]<ω S
all finite subsets of ω also has cardinality |T|≤ |X| and hence admits an enumeration T = {(α ,β ,s ) : x ∈ X}
x x x
by points of the set X. By the definition of the family T, for every x ∈ X the algebraic operations α and β
x x
are distinct. Consequently, we can choose a function p : S → A defined on the set S = S = S such that
x x x αx βx
α (p )6=β (p ).
x x x x
Using Fichtenholtz-Katorovitch-HausdorffTheorem 17.20 [6], fix an independent subfamily U ⊂P(X) of cardi-
nality |U|=2|X|.
For each set U ∈U define a function f :X →A assigning to a point x∈X the point p (i) where i is a unique
U x
point of the set s−1(U)⊂U if this set is a singleton, and an arbitrary point of A otherwise.
x x
We claim that the set F ={f } ⊂AX has cardinality |F|=2|X| and is AX-free.
U U∈U
Claim 4.4. The function f :U →F, f :U 7→f , is bijective and hence |F|=|U|=2|X|.
U
Proof. Given two distinct sets U,V ∈ U, we should prove that f 6=f . By the unitality, the operation family A
U V
contains the identity operation id:A1 →A, id:(a)7→a. Consider the embeddings
σ :1→2 σ :07→0 and σ :1→2, σ :07→1.
0 0 1 1
The substitution-stability of A implies that the algebraic operations α = id◦σ∗ : A2 → A, α : (a,b) 7→ a, and
0
β =id◦σ∗ :A2 →A, β :(a,b)7→b, belong to the operation family A. It follows from |A|≥2 that α6=β.
1
By the independence of U ∋ U,V, there is a function s : 2 →X such that s(0) ∈U \V and s(1)∈ V \U. The
triple (α,β,s) belongs to the family T ⊂ T and hence is equal to (α ,β ,s ) for some x∈ X. Then p ∈ A2 is a
2 x x x x
function such that p (0) = α (p ) 6= β (p ) = p (1). The definition of the functions f and f guarantees that
x x x x x x U V
f (x)=p (0)6=p (1)=f (x) and hence f 6=f . (cid:3)
U x x V U V
Claim 4.5. The set F is A-free.
Proof. AssumingthatF isnotA-freeandapplyingProposition3.4,findafinitesubsetS ⊂ω,aninjectivefunction
ξ :S →F and two distinct algebraic operations α,β ∈A such that S =S =S and αX(ξ)=βX(ξ).
α β
For every i ∈ S find a set U ∈ U such that ξ(i) = f . The independence of the family U guarantees the
i Ui
existence of a function s : S → X such that s(i) ∈ U ∩ (X \U ). It follows that s−1(U ) = {i} for each
i Tj∈S\{i} j i
i∈S.
Thetriple(α,β,s)belongstothefamilyT ⊂T andhenceisequalto(α ,β ,s )forsomepointx∈X. Forevery
S x x x
i∈Sthedefinitionofthefunctionf guaranteesthatf (x)=p (i). Letδ :AX →A,δ :g 7→g(x),denotethex-
Ui Ui x x x
thcoordinateprojection. Weclaimthatδ ◦ξ =p . Indeed,foreachi∈Swegetδ ◦ξ(i)=δ (f )=f (x)=p (i).
x x x x Ui Ui x
Applying the function δ to the equality αX(ξ) = βX(ξ), we get δ ◦αX(ξ) = δ ◦βX(ξ). On the other hand,
x x x
δ ◦αX(ξ)=α(δ ◦ξ)=α(p )6=β(p )=β(δ ◦ξ)=δ ◦βX(ξ), which is a desired contradiction. (cid:3)
x x x x x x
(cid:3)
Problem 4.6. Let A = (A,A) be a universal algebra with |A| ≥ 2, X be an infinite set of cardinality |X| ≥ |A|,
and F ⊂AX be a maximal AX-free subset. Is |F|≥|2X|?
Problem 4.7. Let A = (A,A) be a universal algebra such that the operator of A-hull has the MacLane-Steinitz
exchange property. Is each A-independent subset of A A-free?
References
[1] B.Balcar,F.Franˇek,Independent families incomplete Boolean algebras, Trans.Amer.Math.Soc.274:2(1982), 607–618.
[2] T.Banakh,I.Protasov,Partitions of groups and matroids into independent subsets,AlgebraDiscreteMath.10:1(2010), 1–7.
[3] G.Birkhoff,Lattice Theory,Amer.Math.Soc.NewYork,N.Y.,1948.
[4] S.Burris,H.P.Sankappanavar, A course inUniversal Algebra,Springer-Verlag,NewYork-Berlin,1981.
[5] R.Engelking,General Topology, HeldermannVerlag,Berlin,1989.
[6] W.Just,M.Weese,Discovering modern settheory. II. Set-theoretictools foreverymathematician,Amer.Math.Soc.,Providence,
RI,1997.
[7] E.Marczewski,Ageneralschemeofthenotionsofindependenceinmathematics,Bull.Acad.Polon.Sci.S´er.Sci.Math.Astr.Phys.
6(1958) 731–736.
[8] E.Marczewski,Independence and homomorphisms inabstract algebras, Fund.Math.50(1961/1962) 45–61.
[9] E.Marczewski,Independence in abstract algebras. Results and problems,Colloq.Math.14(1966)169–188.
[10] J.Oxley,Matroid Theory,OxfordUniv.Press,1992.
T.Banakh: Ivan FrankoUniversity of Lviv (Ukraine) andJanKochanowski Uniwersity in Kielce (Poland)
E-mail address: [email protected]
A.Bartoszewicz, S.G la¸b: Institute of Mathematics, Technical University of L o´d´z, Wo´lczan´ska 215, 93-005 L o´d´z,
Poland
E-mail address: [email protected], [email protected]