Cardinalities of Lattices of Topologies of Unars and Some Related Topics

In this paper we find cardinalities of lattices of topologies of un-countable unars and show that the lattice of topologies of a unar cannor be countably infinite. It is proved that under some finiteness conditions the lattice of topologies of a unar is finite. Furthermore, the relations between the lattice of topologies of an arbitrary unar and its congruence lattice are established. Let A = A, Ω be an arbitrary algebra. A topology on the set A, under which every operation from Ω is continuous is called a topology on the algebra A. It is known [5] (p. 69) that the topologies on an algebra A form a lattice under set inclusion. Let us call this lattice the lattice of topologies of the algebra A. Denote this lattice by (A). Let now A = A, f be a unar, i. e. an algebra with one unary operation f (see [6]). For any element a ∈ A and any positive integer n we put f 0 (a) = a and f n (a) = f (f n−1 (a)). Throughout the paper we shall denote by N the set of all positive integers and N 0 = N ∪ {0}. A unar generated by one element a is called monogenic and it is denoted by (a). A monogenic unar with the generator a and with defining relation f n (a) = f n+m (a), n ∈ N 0 , m ∈ N is denoted by C n m. The unar C 0 m is termed a cycle of length m. An element a of the unar A is cyclic if the subunar generated by this element is cyclic. The set of all cyclic elements of

Let A = A, Ω be an arbitrary algebra.A topology on the set A, under which every operation from Ω is continuous is called a topology on the algebra A. It is known [5] (p.69) that the topologies on an algebra A form a lattice under set inclusion.Let us call this lattice the lattice of topologies of the algebra A. Denote this lattice by (A).
Let now A = A, f be a unar, i. e. an algebra with one unary operation f (see [6]).For any element a ∈ A and any positive integer n we put f 0 (a) = a and f n (a) = f (f n−1 (a)).Throughout the paper we shall denote by N the set of all positive integers and N 0 = N ∪ {0}.
A unar generated by one element a is called monogenic and it is denoted by (a).A monogenic unar with the generator a and with defining relation f n (a) = f n+m (a), n ∈ N 0 , m ∈ N is denoted by C n m .The unar C 0 m is termed a cycle of length m.An element a of the unar A is cyclic if the subunar generated by this element is cyclic.The set of all cyclic elements of a unar A is denoted by Z(A).An element a of a unar The disjoint union of two unars B and C is denoted by B + C. Unars B and C are components of the unar B + C. A unar having no proper components is called connected.The set of all connected components of an arbitrary unar A is denoted by c(A).
Proposition 1.The lattice of all topologies on the set c(A) of connected components of an arbitrary unar A = A, f is isomorphic to some principal ideal of the lattice (A).

P roof. Define binary relation η on the set A by setting
for any elements x, y ∈ A. It is clear that η ∈ Con(A) and the factor unar A/η is a union of one-element cycles.Moreover the lattice (A/η) of topologies of the unar A/η coincides with the lattice of all topologies on the set A/η.By [2] (Theorem 3) the lattice of all topologies on the set c(A) is isomorphic to a principal ideal of (A) because |c(A)| = |A/η|.

Observe that
the lattice R(Y ) of all topologies on a nonvoid subset Y of an arbitrary set X can be embedded into the lattice R(X) of all topologies on the set X as a principal ideal.
In fact, fix a point y 0 ∈ Y and define a mapping ψ : R(Y ) → R(X) in the following way.Let σ ∈ R(Y ).Denote be ψ(σ) the family of subsets of the set X such that T ∈ ψ(σ) if and only if either T ∈ σ and Y ⊆ T and y 0 ∈ T .Then ψ is an isomorphism of R(Y ) onto the principal ideal of R(X) generated by the topology ψ(σ 1 ), where σ 1 is the discrete topology on X.
From Proposition 1, we can deduce Lemma 1.Let A = A, f be an arbitrary unar and K be a nonvoid subset of the set c(A) of connected components of the unar A. Then the lattice R(K) of all topologies on the set K is isomorphic to a principal ideal of the lattice (A).

Elements a, b of an arbitrary unar
Lemma 2. Let A = A, f be an arbitrary unar and A 1 be an infinite set of pairwise non-cyclic incomparable elements of A. Then the lattice R(A 1 ) of all topologies on the set A 1 can be embedded into the lattice (A).We claim that ρ ∈ Con B. In fact let a / ∈ A 1 , and f (a) ∈ A 1 .Since a ∈ B, there exists an element c ∈ A 1 and an integer n ∈ N 0 such that a = f n (c).Hence, f (a) = f n+1 (c).It follows that n + 1 = 0 and n / ∈ N 0 , since f (a) ∈ A 1 and c ∈ A 1 .Every topology on the factor set B/ρ is a topology on the unar B/ρ because either f −1 (X) = ∅ or f −1 (X) = B/ρ holds for any subset X of the set B/ρ. Thus applying [2] (Theorems 2 and 3) and the equality |A 1 | = |B/ρ| we can conclude that the lattice R(A 1 ) of all topologies on the set A 1 can be embedded into the lattice (A).Lemma 3. Let A = A, f be an arbitrary infinite unar.Then either the lattice (F 1 ) or the lattice (C ∞ 1 ) can be embedded into the lattice (A) of all topologies of the unar A.

P roof.
If A contains a torsion-free element a, then (a) ∼ = F 1 , where (a) is the monogenic subunar of the unar A generated by the element a.By [2] (Theorem 3) the lattice (F 1 ) can be embedded into the lattice (A).
If the set c(A) is infinite or the inequality |f −1 ({a})| ≥ ℵ 0 holds for some a ∈ A, then the lattice R(X) of all topologies on a countable infinite set X is isomorphic to some sublattice of the lattice (A) by Lemmas 1 and 2. On the other hand, |F 1 | = ℵ 0 .Therefore, the lattice (F 1 ) of all topologies of F 1 can be embedded into the lattice R(X) and, hence, into the lattice (A).
Let A be periodic, |c(A)| < ℵ 0 , a set f −1 ({a}) finite for any element a ∈ A. Then there exists a subunar B = B, f of the unar A, which is isomorphic to Then ρ ∈ Con B and the factor unar B/ρ is isomorphic to C ∞ 1 .Consequently, the lattice (C ∞ 1 ) of all topologies of C ∞ 1 can be embedded into the lattice (A) by [2] (Theorems 2 and 3).
The least topology with respect to inclusion on the unary algebra A, containing a given family of subsets {A α ⊆ A| α ∈ I} will be called the topology on the algebra A generated by the set of elements {A α | α ∈ I}.This topology will be denoted by t({A α | α ∈ I}) and respectively by t(U ) if the family

of the set A Z(A).
P roof.Since t(X 1 ) = t(X 2 ), we conclude that X 2 ∈ t(X 1 ).Hence, the set X 2 is a union of finite intersections of some sets of the form f −i (X 1 ), where On the other hand, since X 1 ∈ t(X 2 ), the set X 1 is a union of finite intersections of some sets of the form f −j (X 2 ), where j ∈ N 0 .However, by (1), the condition f i+j (x) ∈ X 2 implies i+j = 0. Hence, i = 0 and so I = {0}.Consequently, X 1 ⊆ X 2 .Similarly, we can prove that X 2 ⊆ X 1 .Thus, X 1 = X 2 .
Let A be an arbitrary algebra and θ ∈ Con(A).θ-congruence classes form a base of some topology τ (θ) which we shall call the topology generated by the congruence θ.Proposition 2. There exists a set H of the cardinality 2 ℵ 0 of different Hausdorff topologies on the unar F 1 , such that for any topology σ ∈ H there exist topologies σ 1 , σ 2 ∈ H, for which σ 1 ≤ σ and σ ≤ σ 2 .P roof.Let x, y ∈ F 1 and k be an arbitrary fixed positive integer.Put It is not hard to see that ζ k ∈ Con F 1 .Let P (S) be the set of all subsets of the set S of all primes.We claim now that the mapping ϕ : P (S) → (F 1 ) given by It remains to construct an infinite increasing chain of elements of the lattice (A).Let  Let now the set A be finite.We claim that the mappping θ → τ (θ) from Con(A) into (A) is not surjective.Indeed, there exist such elements a, b ∈ A, that a / ∈ (b) because the unar A is not a cycle.Suppose that ρ is a congruence of A such that τ (ρ) = t({a}).Then [b] ρ ∈ t({a}).Hence, [b] ρ = A, because b / ∈ ∪ i∈N 0 f −i ({a}).Therefore, ρ is the universal relation and the topology t({a}) = τ (ρ) is anti-discrete.However, {a} ∈ t({a}).

P
roof.Denote by B = B, f the subunar of unar A generated by set A 1 .Define a binary relation ρ = {(a, b) ∈ B × B| a = b ∨ {a, b} ∩ A 1 = ∅} on the set B. Certainly the relation ρ is an equivalence.