Large free sets in universal algebras

We prove that for each universal algebra $(A,\mathcal A)$ of cardinality $|A|\ge 2$ and an infinite set $X$ of cardinality $|X|\ge|\mathcal A|$, the $X$-th power $(A^X,\mathcal A^X)$ of the algebra $(A,\mathcal A)$ contains a free subset $\mathcal F\subset A^X$ of cardinality $|\mathcal F|=2^{|X|}$. This generalizes the classical Fichtenholtz-Kantorovitch-Hausdorff result on the existence of an independent family $\mathcal I\subset\mathcal P(X)$ of cardinality $|\mathcal I|=|\mathcal P(X)|$ in the Boolean algebra $\mathcal P(X)$ of subsets of an infinite set $X$.

Proof. Assuming that a strongly C-independent set B is not C-independent, we can find a point b ∈ B such that b ∈ C(B \ {b}). Then b ∈ C({b}) ∩ C(B \ {b}) = C({b} ∩ (B \ {b})) = C(∅), which is not possible as B ⊂ X \ C(∅). Proposition 1.1 can be reversed for hull operators of matroid type. We say that a hull operator C : P(X) → P(X) • is a closure operator if C(C(A)) = C(A) for each subset A ⊂ X; • has finite supports if for each subset A ⊂ X and a point x ∈ C(A) there is a finite subset F ⊂ A with x ∈ C(F ); • has the MacLane-Steinitz exchange property if for any subset A ⊂ X and points x, y ∈ X \ C(A) the inclusion x ∈ C(A ∪ {y}) is equivalent to y ∈ C(A ∪ {x}); • is of matroid type if C is a closure operator with finite supports and the Maclane-Steinitz exchange property. It is well-known (and easy to see) that the operator of taking the linear hull in a linear space is of matroid type. The following proposition is well-known in the theory of independence (see [14], [15], and [6]). We present a short proof for the convenience of the reader. Proposition 1.2. Let C : P(X) → P(X) be a hull operator of matroid type on a set X.
(1) A subset B ⊂ X is C-independent if and only if it is strongly C-independent.
(2) Each C-independent subset lies in a maximal C-independent subset of X.
(3) Each maximal C-independent subset B ⊂ X has hull C(B) = X.
Proof. 1. Proposition 1.1 implies that each strongly C-independent subset of X is C-independent. Now assume that a set B is C-independent. Then for each b ∈ B we get b / ∈ C(B \ {b}) ⊃ C(∅) and hence B ∩ C(∅) = ∅. To show that B is strongly C-independent, it remains to check that C(B 1 ) ∩ C(B 2 ) ⊂ C(B 1 ∩ B 2 ) for any subsets B 1 , B 2 ⊂ B. This inclusion is trivial if B 1 ⊂ B 2 or B 2 ⊂ B 1 . So, we assume that B 1 ⊂ B 2 and B 2 ⊂ B 1 . Given any point c ∈ C(B 1 ) ∩ C(B 2 ), we should prove that c ∈ C(B 1 ∩ B 2 ). Assume conversely that c / ∈ C(B 1 ∩ B 2 ). Since the closure operator C has finite supports, there are minimal finite subsets F 1 ⊂ B 1 and F 2 ⊂ B 2 such that c ∈ C(F 1 ) ∩ C(F 2 ). It follows from F 1 ∩ F 2 ⊂ C(F 1 ∩ F 2 ) ⊂ C(B 1 ∩ B 2 ) ∋ h that c / ∈ F 1 ∩ F 2 and hence c / ∈ F 1 or c / ∈ F 2 . Without loss of generality, c / ∈ F 1 . It follows from C(F 1 ) ∩ C(F 2 ) ⊂ C(F 1 ∩ F 2 ), that F 1 ⊂ F 2 and F 2 ⊂ F 1 . Choose any point b ∈ F 1 \ F 2 and let A = which contradicts the C-independence of B.
2. The second statement can be easily proved using Zorn's Lemma and the fact that the closure operator C has finite supports.
3. Let B be a maximal C-independent subset in X. If C(B) = X, then we can choose a point x ∈ X \ C(B). By the maximality of B the set B Then the MacLane-Steinitz exchange property implies x ∈ C(A ∪ {b}) = C(B), which contradicts the choice of Unfortunately many hull operators that naturally appear in algebra are not of matroid type. The simplest example is the group hull operator. It assigns to each subset A of a group G the subgroup C(A) ⊂ G generated by A. Observe that the set {2, 3} is C-independent in the group of integers Z but fails to be strongly Cindependent in Z. Each maximal strongly C-independent subset B in the group of rational Q is a singleton and hence C(B) = Q. The following proposition yields an example of a (non-commutative) group G of cardinality continuum containing a maximal C-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 2 = {0, 1} embeds into a group G of cardinality |G| = 2 |V | such that for the operator of group hull C : P(G) → P(G) each maximal C-independent subset of V remains maximal C-independent in G.
Proof. Since the field F 2 is finite, the linear space V has a Hamel basis of cardinality |V |. Since each permutation 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 ( Let M ⊂ V be a maximal C-independent subset of the additive group V . Since for each subset B of V the group hull C(B) of B coincides with its linear hull, the operator of group (=linear) hull in V is of matroid type. We need to check that M remains maximal C-independent in the group G. Given any element g ∈ G \ M we need to show that the union M ∪ {g} is not C-independent in G. If g ∈ V , then the set M ∪ {g} ⊂ V is not C-independent in G by the maximality of the C-independent set M in V .
So, we assume that g / ∈ V . In this case Being a maximal linearly independent subset of V , the set M is a Hamel basis in V . Since f = id V , there is a point a ∈ M with f (a) = a. Since M is a Hamel basis of the linear space V , each point v ∈ V can be written as the sum ΣB = b∈B b of a unique finite subset B ⊂ M . In particular, a = ΣA and f (a) = ΣF for some finite subsets 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.

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 : A Sα → A defined on a finite power A Sα of A, where S α is a finite subset of ω called the support of the operation α. An algebraic operation α : A Sα → A is constant if α(A Sα ) 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 * : A E → A F , σ * : f → f • σ, called a substitution operator. Then for any algebraic operation α : A Sα → A with support S α = F the composition β = α • σ * is a well-defined algebraic operation on A with support S β = E.
A family A of operations on a set A is called • unital if A contains the identity operation id A : • ∅-regular if for each constant operation α ∈ A there is an operation β ∈ A with empty support S β = ∅ such that β(A ∅ ) = α(A Sα ); • stable under substitutions (briefly, substitution-stable) if for any function σ : F → E between finite subsets of ω and any algebraic operation α ∈ A with supp(α) = F the algebraic operation α • σ * : • stable under compositions if for any finite subset S ⊂ ω and algebraic operations α ∈ A and α i ∈ A, i ∈ S α , with supports supp(α i ) = S for all i ∈ E the composition α • (α i ) i∈Sα : A S → A of the diagonal product (α i ) i∈Sα : A S → A Sα and α belongs to A; • a clone if A is unital, ∅-regular, and stable under substitutions and compositions. The clone of a universal algebra A = (A, A) is the universal algebraĀ = (A,Ā) endowed with the smallest cloneĀ that contains the operation family A. The cloneĀ is equal to the unionĀ = n∈ω A n of operation families A n , n ∈ ω, defined by induction.
The definition implies that this hull operation has finite supports. If the operation family A is ∅-regular, then for each constant algebraic operation α ∈ A, the singleton α(A nα ) lies in the A-hull A(∅) of the empty subset of A.
be a universal algebra whose operation family A is unital and stable under substitutions. 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 α → B such that a = α(x).
Proof. Since A is unital, A(B) = β∈A α(B S β ) and we can find an operation β ∈ A and a function z : S β → B such that a = β(z). Let x : S → z(S β ) ⊂ B be any bijective map defined on a finite subset S ⊂ ω. Consider the function σ = x −1 • z : S β → S, which induces the substitution operator σ * : A S → A S β . Since A is stable under substitutions, the operation α = β • σ * : A S → A belongs to A and has support S α = S. Moreover, 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: be a universal algebra andĀ = (A,Ā) be its clone. For each subset B ⊂ A the subalgebra generated by B coincides with theĀ-hullĀ(B) of B.
Proof. Let B denote the subalgebra of A generated by B. The inclusion B ⊂Ā(B) will follow as soon as we check that theĀ-hullĀ(B) of B is a subalgebra of the universal algebra A, that isĀ(B) contains all constants and it is closed under all operations from A. We need to check that A(Ā(B)) ⊂Ā(B). Take any element y ∈ A(Ā(B)) and find an operation α ∈ A and a function x : S α →Ā(B) such that y = α(x). For every i ∈ S α the point x(i) belongs to theĀ-hullĀ(B) of B and hence can be written as x(i) = α i (z i ) for some algebraic operation α i ∈Ā and some function z i : S αi → B. Choose a finite subset S ⊂ ω of cardinality |S| = i∈Sα |S αi | and for every i ∈ S α choose an injective function σ i : SinceĀ is a clone, it is closed under substitutions. Consequently, for every i ∈ S α the operation Since the function familyĀ is closed under compositions, the operation β = α • (β i ) i∈Sα : A S → A belongs toĀ. Since To prove thatĀ(B) ⊂ B , we use the decompositionĀ = n∈ω A n ofĀ into the countable union of the operation families A n , n ∈ ω, defined at the beginning of Section 2 right after the definition of clone. Since A(B) = n∈ω A n (B), it suffices to check that A n (B) ⊂ B for every n ∈ ω. This will be done by induction on n ∈ ω. Since Assume that for some n ∈ ω we have proved that A n (B) ⊂ B . The inclusion A n+1 (B) ⊂ B will follow as soon as we check that β(x) ∈ B for each operation β ∈ A n+1 and a function x : If β ∈ A n+1 \ A n , then by the definition of the operation family A n+1 , the following three cases are possible: If B = ∅, then we can take any function y : Let A = (A, A) be a universal algebra, X be a non-empty set, and A X be the set of all functions from X to A. For every x ∈ X denote by δ x : A X → A, δ x : f → f (x), the projection onto x-th coordinate.
Observe that each algebraic operation α : A Sα → A induces an algebraic operation α X : (A X ) Sα → A X on the set A X of all functions from X to A. The operation α X assigns to each function f : For a universal algebra A = (A, A) its X-th power is the pair A X = (A X , A X ) consisting of the X-th power of A and the operation family A X = {α X } α∈A . Now let us consider an important example of a universal algebra 2 = (2, B), called the Boolean clone. It consists of the doubleton 2 = {0, 1} and the family B of all possible algebraic operations on 2. In the next section, we shall see that the powers 2 X of the Boolean clone play an important role in studying independent and free sets in powers of arbitrary universal algebras.

Independent and free sets in universal algebras
In this section we shall be interested in three independence notions in universal algebras.
for any algebraic operation α ∈ A and any function x ∈ B Sα ⊂ A Sα . The notion of a (strongly) A-independent set is induced by the operator of A-hull. On the other hand, the notion of an A-free set is specific for universal algebras and has no hull counterpart.
In fact, A-free sets in universal algebras were introduced by E.Marczewski [9], [10], [11]. Because of that such set are sometimes called M -independent, see [6] and references therein.
The definitions imply that the notions of A-independent and A-free sets are "monotone" which respect to 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 : A 1 → A, β : A 0 → A with α(A 1 ) = β(A 0 ). Observe that each subset of A is strongly A-independent while each subset B ⊂ A of cardinality |B| ≥ 2 fails to be strongly A ′ -independent for the subfamily A ′ = {α}. Indeed, take two non-empty disjoint subsets B 1 , B 2 ⊂ B and observe that Now we shall characterize A-free sets in universal algebras. Our first characterization follows immediately from the definition and was noticed by Marczewski in [10]. a universal algebra (A, A) is A-free if and only if for any algebraic operations α, β ∈ A and functions x ∈ B Sα , y ∈ B S β 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 characterization can be improved as follows. 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 α = S β = S for some finite set S ⊂ ω. We need to show that α(x) = β(x) for each injective function x ∈ B S .
Since α = β, there is a function y ∈ B S such that α(y) = β(y). Using the injectivity of x, choose a function f : To prove the "if" part, assume that the set B is not A-free. Applying Proposition 3.3, find algebraic operations α, β ∈ A and functions x ∈ B Sα , y ∈ B S β such that α(x) = β(y) and α(f • x) = β(f • y) for some function f : B → A. Fix any bijective function z : S → x(S α ) ∪ y(S β ) ⊂ B defined on a finite subset S ⊂ ω. Consider the functions σ α = z −1 • x : S α → S and σ β : z −1 • y : S β → S, which induce the substitution operators σ * α : A S → A Sα and σ * β : A S → A S β . Since the operation family A is stable under substitutions, the algebraic operationsα = α • σ * α : A S → A andβ = β • σ * β : A S → A belong to the family A. Observe that 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.
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 1 , The universal algebra (A, A) is unital and hence contains the identity algebraic operation α : 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 ∅ = B S β . Then for every i ∈ {1, 2}, the choice off i guarantees Next, we prove that A(B 1 ) ∩ A(B 2 ) = A(B 1 ∩ B 2 ) for any subsets B 1 , B 2 ⊂ B. This equality is trivial if B 1 ⊂ B 2 or B 2 ⊂ B 1 . So we assume that both complements B 1 \ B 2 and B 2 \ B 1 are not empty. Assume that A(B 1 )∩A(B 2 ) = A(B 1 ∩B 2 ) and find a point a ∈ A(B 1 )∩A(B 2 )\A(B 1 ∩B 2 ). For the point a ∈ A(B 1 )∩A(B 2 ), there are algebraic operations α, β ∈ A and functions x ∈ B Sα 1 , y ∈ B S β 2 such that α(x) = a = β(y). Using a similar 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(A Sα ) = γ(A ∅ ). Then a = α(x) = γ(∅) ∈ A(∅) ⊂ A(B 1 ∩ B 2 ), which contradicts the choice of a.
Next, we prove that the intersection x(S α ) ∩ y(S β ) ⊂ B 1 ∩ B 2 is not empty. Assuming the converse and using the fact that the operation α is not constant, find a function x ′ ∈ A Sα such that α(x ′ ) = β(y). Since the subsets x(S α ) and y(S β ) of B are disjoint, we can find a function f : which is a contradiction.
Thus the intersection x(S α ) ∩ y(S β ) ⊂ B 1 ∩ B 2 is not empty and we can choose a function f : Since the set B is A-free, the function f can be extended to a functionf : A(B) → A such thatf (γ(z)) = f • γ(z) for each γ ∈ A and z ∈ B Sγ . In particular, This contradiction completes the proof of the strong A-independence of the set B.
Proposition 3.5 shows that for a subset of a unital ∅-regular substitution-stable universal algebra (A, A) we have the following implications: The first implication cannot be reversed as shown by the following simple example.
Example 3.6. There is a clone (A,Ā) containing an infinite stronglyĀ-independent subset B ⊂ X such that eachĀ-free subset of A is empty.
Proof. Consider the linear algebra c 00 consisting of all functions x : ω → R with finite support supp(x) = {n ∈ ω : x(n) = 0} ⊂ ω. This algebra is endowed with the operation family A = {+, ·} ∪ {α t : t ∈ R} consisting of two binary operations (of addition and multiplication) and continuum many unary operations α t : x → t · x of multiplication by a real number t. The cloneĀ of A contains the subfamily A ′ consisting of all polynomials p(x) = n i=0 λ i x i+1 of one variable, equal to zero at the zero function.
which means that the subset B is strongly independent.
Claim 3.8. Any non-empty subset B ⊂ A is not A ′ -free and hence is notĀ-free.
Proof. Fix any function b ∈ B. This function has finite support F = supp(b). Then b and all its finite powers b n , n > 0, belong to the finite-dimensional linear subspace R F = {x ∈ c 00 : supp(x) ⊂ F } of c 00 . Consequently, the set {b n+1 : 0 ≤ n ≤ |F |} is linearly dependent, which allows us to find a non-zero vector (λ 0 , . . . , λ |F | ) ∈ R |F |+1 such that |F | i=0 λ i b i+1 = 0. This means that p(b) = 0 for the non-zero polynomial p(x) = |F | i=0 λ i x i+1 . Since the polynomial p ∈ A ′ is non-zero, there is a vector y ∈ R F such that p(y) = 0. Let f : B → c 00 be any function such that f (b) = y.
Assuming that the set B is A ′ -free, we could find a functionf : p(f (b)). This contradiction completes the proof.
As we saw in Proposition 1.2, the C-independence is equivalent to the strong C-independence for closure operators of matroid type. In contrast, the notions of an A-independent set and a A-free set are not be equivalent (even in presence of the MacLane-Steinitz exchange property). The following simple example is taken from [7, p.297]. Example 3.9. Consider the 2-element set A = {0, 1} and the set A = {·} containing the binary operation · : A × A → A of multiplication. It follows that each subset of A is an A-subalgebra, which implies that the set A is A-independent. By a trivial reason the A-hull operator on X has the MacLane-Steinitz exchange property and hence is a closure operator of matroid type. However, the set A is not A-free, as the unique non-identity bijection f : A → A is not a homomorphism of A.
According to [7], a universal algebra A = (A, A) is called an independence algebra if the operator ofĀhull has the MacLane-Steinitz exchange property and eachĀ-independent subset of A isĀ-free. Independent algebras were thoroughly studied in [7], [1] and much earlier in [12], [13] (as v * -algebras).

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 2 X = (2 X , B X ) 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 B X -free subsets of 2 X coincide with independent sets, well studied in Set Theory [8, §17].
Let us recall [8, p.83] that a family F of subsets of a set X is called independent if for any finite disjoint sets is not empty.
Identifying each subset F ⊂ X with its characteristic function χ F : X → 2 = {0, 1}, we can reformulate the notion of an independent family of sets in the language of an independent function family. Namely, a function family F ⊂ 2 X is independent if for any pairwise distinct functions f 0 , . . . , f n−1 ∈ F their diagonal product (f i ) i<n : X → 2 n is surjective.
Proposition 4.1. For any set X a function family F ⊂ 2 X is independent if and only it is B X -free in the clone 2 X .
To prove the "only if" part, assume that F is independent but not B X -free. By Proposition 3.4, there are two distinct operations α, β ∈ B such that S α = S = S β for some finite set S ⊂ ω and α X (ξ) = β X (ξ) for some injective function ξ : S → F . The function ξ can be written in the form (ξ i ) i∈S where ξ i = ξ(i) ∈ F . The independence of F guarantees that the diagonal product (ξ i ) i∈S : X → 2 S , δ = (ξ i ) i∈S : x → (ξ i (x)) i∈S of these functions is surjective.
By a classical Fichtenholtz-Katorovitch-Hausdorff Theorem 17.20 [8], the power-set P(X) of each infinite set X contains an independent family F ⊂ P(X) of cardinality |F | = |2 X | = |P(X)|. Reformulating this result with help of Proposition 4.1, we get the following result: For each infinite set X the power-clone 2 X = (2 X , B X ) contains a B X -free subset B ⊂ 2 X of cardinality |B| = |2 X |.
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 A X = (A X , A X ) contains an A X -free subset F ⊂ A X of cardinality |F | = 2 |X| .
Proof. LetĀ be the clone of the operation family A. It has cardinality |Ā| ≤ max{|A|, ℵ 0 } ≤ |X|. Since each A X -free subset of A X is A X -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 and observe that it has cardinality |T S | ≤ |A × A × X S | ≤ |X|. Then the union T = S∈[ω] <ω T S where S runs over all finite subsets of ω also has cardinality |T | ≤ |X| and hence admits an enumeration T = {(α x , β x , s x ) : x ∈ X} by points of the set X. By the definition of the family T , for every x ∈ X the algebraic operations α x and β x are distinct. Consequently, we can choose a function p x : Using Fichtenholtz-Katorovitch-Hausdorff Theorem 17.20 [8], fix an independent subfamily U ⊂ P(X) of cardinality |U| = 2 |X| .
For each set U ∈ U define a function f U : X → A assigning to a point x ∈ X the point p x (i) where i is a unique point of the set s −1 x (U ) ⊂ U x if this set is a singleton, and an arbitrary point of A otherwise. We claim that the set F = {f U } U∈U ⊂ A X has cardinality |F | = 2 |X| and is A X -free. Proof. Given two distinct sets U, V ∈ U, we should prove that f U = f V . By the unitality, the operation family A contains the identity operation id : A 1 → A, id : (a) → a. Consider the embeddings σ 0 : 1 → 2 σ 0 : 0 → 0 and σ 1 : 1 → 2, σ 1 : 0 → 1.
The substitution-stability of A implies that the algebraic operations α = id • σ * 0 : A 2 → A, α : (a, b) → a, and β = id • σ * 1 : A 2 → A, β : (a, b) → b, belong to the operation family A. It follows from |A| ≥ 2 that α = β. 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 2 ⊂ T and hence is equal to (α x , β x , s x ) for some x ∈ X. Then p x ∈ A 2 is a function such that p x (0) = α x (p x ) = β x (p x ) = p x (1). The definition of the functions f U and f V guarantees that f U (x) = p x (0) = p x (1) = f V (x) and hence f U = f V .
Claim 4.5. The set F is A-free.
Proof. Assuming that F is not A-free and applying Proposition 3.4, find a finite subset S ⊂ ω, an injective function ξ : S → F and two distinct algebraic operations α, β ∈ A such that S α = S = S β and α X (ξ) = β X (ξ).
For every i ∈ S find a set U i ∈ U such that ξ(i) = f Ui . The independence of the family U guarantees the existence of a function s : S → X such that s(i) ∈ U i ∩ j∈S\{i} (X \ U j ). It follows that s −1 (U i ) = {i} for each i ∈ S.
The triple (α, β, s) belongs to the family T S ⊂ T and hence is equal to (α x , β x , s x ) for some point x ∈ X. For every i ∈ S the definition of the function f Ui guarantees that f Ui (x) = p x (i). Let δ x : A X → A, δ x : g → g(x), denote the x-th coordinate projection. We claim that δ x • ξ = p x . Indeed, for each i ∈ S we get δ x • ξ(i) = δ x (f Ui ) = f Ui (x) = p x (i).
Remark 4.6. For some concrete universal algebras, in particular, for the linear algebra F X of all functions from an infinite set X to a field F ∈ {R, C}, a free subalgebra A ⊂ F X with 2 |X| generators (and some additional properties) has been constructed in [4].
Problem 4.7. Let A = (A, A) be a universal algebra with |A| ≥ 2, X be an infinite set of cardinality |X| ≥ |A|, and F ⊂ A X be a maximal A X -free subset. Is |F | ≥ |2 X |?