On topological basic algebras

Abstract

In this paper, the notions of (semi) topological basic algebra and (semi) topological implication basic algebra are introduced, along with evaluating their properties. Then, different operations are defined based on basic algebras and the relationship between semicontinuity and continuity of operations is considered. In addition, the separation axioms on (semi) topological basic algebras are investigated by considering some conditions implying that a (semi) topological basic algebra becomes a T_i- space, for i ∈ {0, 1, 2}. In the sequel, some relations between (weak) ideals and (weak) filters of basic algebras are obtained and (left) topological (implication) basic algebra is constructed by using the concepts of (weak) filters, which is a zero dimensional, normal, disconnected, locally compact and completely regular (left) topological space. Further, the notion of quotient basic algebras are presented along with evaluating the interaction of topological basic algebras and topological quotient basic algebras. Finally, it is proved that there is an implication basic algebra ${IB}^{*}$ with cardinality n + 1 and filter F^* = F ∪ {z^*}, which $z^{*} \notin IB$ for any implication basic algebra $IB$ of cardinality n and filter F. Accordingly, it is proved that there is at least one nontrivial regular and normal topological implication basic algebra of cardinality n.

Keywords

Basic algebra topological basic algebra continuous separation axioms topological quotient basic algebras 54A05 54A10 03G12 03G25

1 Introduction

Two algebraic structures including MV-algebras and orthomodular lattices are employed based on the logic of quantum mechanics. Chajda [10] introduced the basic algebras to generalize orthomodular lattices similarly in which MV-algebras generalize Boolean algebras. A basic algebra (B ; ⊕ , ∼ , 0, 1), is a bounded lattice whose sections are antitone involutions. MV-algebras coincide with associative basic algebras and orthomodular lattices can be identified with basic algebras satisfying y ⊕ (x ∧ y). In addition, lattice effect algebras, which are algebraic structures for evaluating the logical foundations of quantum mechanics can be regarded as a subclass of basic algebras. Historically, Abbot [1] studied implication reduct for Boolean algebra named implication algebra. Further, chajda [11] introduced implication basic algebras, which are common generalization of orthoimplication algebras and implication reducts of MV-algebras.

Most important objects of mathematics represent a combination of algebraic and topological structure. It is well-known that the topological structures of basic algebras can describe some properties of quantum measures, which can help to characterize some algebraic operations of basic algebras.

During the recent decade, some mathematicians studied algebraic logical structures with their topologies and investigated their properties. Borzooei et al. [2–4 , 18] defined (semi) topological BL-algebras and proved that other operations of BL-algebra A, ∨ and ∧, are continuous if (A, {⊙ , →} , U) is a topological BL-algebra. They found some conditions under which a semitopological BL-algebra is a T₀, T₁ or T₂-space. In addition, they studied metrizability on (semi) topological BL-algebras, along with their relationship between separation axioms and (semi) topological quotient BL-algebras. Further, Haveshki et al. [13] introduced the topology induced by uniformity on BL-algebras. Hoo [14] proposed topological notions on MV-algebras in order to define topological MV-algebras so that their operations ⊕, ⊖ , ∗ , … are continuous. Then, he considered linear topologies on an MV-algebra A, by family of ideals. furthermore, he defined I-adic topology and completion of A with respect to this topology, where I indicates an ideal of A. Recently, Najafi et al. [17] defined (para, quasi) topological MV-algebras and evaluated their related properties. The results indicated being a topological MV-algebra is regarded as the equivalent conditions for $(A, T)$ , where A shows an MV-algebra and $T$ is considered as a topology on A. Additionally, they obtained topologies by which an MV-algebra can be a (para, semi) topological. Luo et al. [16], showed that every weakly algebraic ideal of an effect algebra E induces a uniform topology (weakly algebraic ideal topology) where E is a first-countable, zero-dimensional, disconnected, locally compact and completely regular topological space, and the operation ⊕ of effect algebras is continuous with respect to these topologies.

The presented study aimed to introduce topologies leading to the operations of a basic algebra or implication basic algebra to be semicontinuous or continuous, along with evaluating their properties. To this aim, (semi) topological basic algebras and (semi) topological implication basic algebras are defined and some examples are presented. First, the various operations are defined based on basic algebras. Then, the relationship between semicontinuity and continuity operations is studied and we investigate that under which condition a (semi) topological basic algebra is a T_i- space for i ∈ {0, 1, 2} are found.

In general, the current research of topological logic algebras is mainly based on filters or ideals, which increased the motivation for conducting the present work.

It is known that 0 ≠ I ⊆ M is a congruence kernel of an MV-algebra M if and only if I is an ideal of MV-algebra i.e., if and only if it satisfies the conditions: (P1) a ⊕ b ∈ I for all a, b ∈ I, (P2) if a ∈ I and a ≥ b, then b ∈ I. Krňávek and Kühr [15] called such subsets pre-ideals. But for a basic algebra $B$ , these conditions are not enough for I to be a congruence kernel of $B$ . Chajda et al. [12] defined conditions (I1)-(I3) for $(0 \in) I \subseteq B$ . I is called an ideal, if I satisfied (I1), (I2) and (I3) and weak ideal, if it satisfied only (I1) and (I2). For commutative basic algebra $B$ , any weak ideal is an ideal. They defined weak congruence of a basic algebra that means an equivalence relation Θ on $B$ which satisfies the compatibility condition, (a, b) ∈ Θ if and only if (c ø a, c ø b) ∈ Θ (Θ is not a congruence). They proved that, if I is a (weak) ideal, Θ_I is a (weak) congruence which is defined by (x, y) ∈ Θ_I if and only if x ø y ∈ I and y ø x ∈ I and conversely, if Θ is a (weak) congruence, then (weak) congruence class [0] _Θ is an (weak) ideal. Chajda [8] introduced (weak)filters and (weak)congruences for basic algebras too, and found the same results.

In this paper, we find some relations between (weak)ideals and (weak)filters of basic algebras. Then we attempt to construct some topologies by employing (weak) filters and (weak) ideals that will turn it into a left topological basic algebra which it is a zero dimensional, normal, disconnected, locally compact and completely regular (left) topological space. Also, we present the notion of quotient basic algebras and study the interaction of topological basic algebras and topological quotient basic algebras. Finally, for any implication basic algebra $IB$ of cardinality n and filter F, we find an implication basic algebra ${IB}^{*}$ of cardinality n + 1 and filter F^* = F ∪ {z^*}, which $z^{*} \notin IB$ and we prove that there exists at least, one nontrivial regular and normal topological implication basic algebra of cardinality n.

From the algebraic point of view, basic algebras slightly generalize MV-algebras. In addition, MV-algebras coincide with associative basic algebra. Further, we observe that topological MV-algebras, which are MV-algebras with topologies such that all operations are continuous with respect to these topologies, generalize topological basic algebras in a certain sense. Let MV-algebra $M$ be as a basic algebra. Then, I is an ideal of $M$ if and only if I satisfies in (I1) which is called "pre-ideal". We can see that, pre-ideals, weak ideals and ideals of $M$ coincide each other and I is an ideal of $M$ as a basic algebra [12]. The same conditions are available for filters of MV-algebras [8]. More explicitly, if $(M, T)$ is a topological MV-algebra, then $(M, T)$ is a topological basic algebra, because, all of the operations defined like MV-algebras, x ⊙ y : = ∼ (∼ x ⊕ ∼ y), x ⊖ y : = x ⊙ ∼ y, x ∨ y : = (x ⊖ y) ⊕ y and x ∧ y : = ∼ (∼ x ∨ ∼ y).

2 Preliminaries

In this section, we recollect some definitions and results about topology and basic algebras which will be used in this paper. For more details, we refer to the [4 –12].

A basic algebra is an algebra $(B; \oplus, \sim, 0)$ of type (2, 1, 0) that satisfies the identities:

x ⊕ 0 = x,

∼ (∼ x) = x,

∼ (∼ x ⊕ y) ⊕ y = ∼ (∼ y ⊕ x) ⊕ x,

∼ (∼ (∼ (x ⊕ y) ⊕ y) ⊕ z) ⊕ (x ⊕ z) =1, where 1 = ∼0, for any $x, y, z \in B$ .

In any basic algebra $(B; \oplus, \sim, 0)$ , several operations are defined as follows, for any $x, y, z \in B$ ,

x → y : = ∼ x ⊕ y,

x ⊙ y : = ∼ (∼ x ⊕ ∼ y) = ∼ (x → ∼ y),

x ⊖ y : = ∼ (∼ x ⊕ y) = ∼ (x → y) = x ⊙ ∼ y,

x ø y : = ∼ (y ⊕ ∼ x) = ∼ (∼ y → ∼ x) = ∼ y ⊖ ∼ x,

x ⇝ y : = y ⊕ ∼ x = ∼ (x ø y) = ∼ y → ∼ x,

x ∨ y : = ∼ (∼ x ⊕ y) ⊕ y = (x ⊖ y) ⊕ y = (x → y) → y,

x ∧ y : = ∼ (∼ x ∨ ∼ y) = (x ⊕ ∼ y) ⊖ ∼ y.

It is evident that every basic algebra can be redefined in the above signatures. In particular, → allows to reduce the number of defining identities (B1)–(B4) to (BI1) (x → 0) →0 = x, (BI2) (x → y) → y = (y → x) → x, (BI3) (((x → y) → y) → z) → (x → z) =1, where 1 = 0 →0.

A basic algebra $B$ is called commutative if it satisfies the identity x ⊕ y = y ⊕ x, for any $x, y \in B$ . Let us note that the binary operation ⊕ is not necessarily commutative nor associative. In any basic algebra $(B; \oplus, \sim, 0)$ , the binary relation ≤ defined by x ≤ y ⇔ ∼ x ⊕ y = 1 is a partial order on $B$ .

The following theorem provides some properties of basic algebras.

Theorem 2.1. Let $(B; \oplus, \sim, 0)$ be a basic algebra. Then for all $x, y, z \in B$ , the following properties hold:

x ⊕ 0 =0 ⊕ x = x, ∼ x ⊕ x = x ⊕ ∼ x = 1, x ⊕ 1 =1 ⊕ x = 1,

x ⊙ 0 =0 ⊙ x = 0, ∼ x ⊙ x = x ⊙ ∼ x = 0, x ⊙ 1 =1 ⊙ x = x,

1 → x = 1 ⇝ x = x, x → 1 = x ⇝ 1 =1, x → x = x ⇝ x = 1, 0 → x = 0 ⇝ x = 1,

x ≤ y ⇔ x ø y = 0 ⇔ x ⊖ y = 0 ⇔ x → y = 1,

x ≤ y ⇒ y → z ≤ x → z. Particularly, ∼y ≤ ∼ x,

x ≤ y ⇒ x ∗ z ≤ y ∗ z, where ∗ ∈ {⊕ , ⊖, ⊙ , ∨ , ∧},

x ≤ y ⇒ z ø y ≤ z ø x, z ⇝ x ≤ z ⇝ y,

x ≤ y ⊕ x, x ≤ y → x, x ⊙ y ≤ y, x ⊖ ∼ y ≤ y, x ø y ≤ x,

∼ (x ∨ y) = ∼ x ∧ ∼ y.

An algebra $(IB; \to)$ of type (2) is called an implication basic algebra if for any $x, y, z \in IB$ , it satisfies the following axioms: (IB1) (x → x) → x = x, (IB2) (x → y) → y = (y → x) → x, (IB3) (((x → y) → y) → z) → (x → z) = x → x. In any implication basic algebra $(IB; \to)$ , there exists a constant $1 \in IB$ such that x → x = 1 and the binary operation ≤ defined by x ≤ y ⇔ x → y = 1 is a partial order and $(IB; \leq)$ is a join-semilattice with the greatest element 1, where (x → y) → y = (y → x) → x = x ∨ y.

Lemma 2.2. Let $(IB; \to)$ be an implication basic algebra. Then for each $x, y, z \in IB$ , the following properties hold:

1 → x = x, x → 1 =1,

y → (x → y) =1,

x ≤ y ⇔ y → z ≤ x → z for all $z \in IB$ .

Theorem 2.3. Let $(B; \oplus, \sim, 0)$ be a basic algebra. If we define x → y : = ∼ x ⊕ y, then $(B; \to)$ is an implication basic algebra. Conversely, every implication basic algebra $(B; \to)$ with least element 0, together with operations ∼ and ⊕ defined by ∼x : = x → 0 and x ⊕ y : = (x → 0) → y is a basic algebra.

A nonempty subset F of a basic algebra $B$ (implication basic algebra $IB$ ) containing 1 is called a weak filter of $B (IB)$ if for all $a, b, c \in B (IB)$ , (F1) a ∈ F and a → b ∈ F imply b ∈ F, (F2) a → b ∈ F and b ≤ a imply (b → c) → (a → c) ∈ F. A weak filter F is called a filter if (F3) a → b ∈ F and b → a ∈ F imply (c → a) → (c → b) ∈ F. A nonempty subset I of a basic algebra $B$ containing 0 is called a weak ideal of $B$ if for all $a, b, c \in B$ , (I1) b ∈ I and a ø b ∈ I imply a ∈ I, (I2) a ø b ∈ I and b ≤ a imply (c ø b) ø (c ø a) ∈ I. A weak ideal I is called an ideal, if, for any $a, b, c \in B$ , (I3) a ø b ∈ I and b ø a ∈ I imply (a ø c) ø (b ø c) ∈ I.

Notations. Let $B$ be a basic algebra. We define mappings I, P, D, M, Q, C from $B \times B$ to $B$ by, I (x, y) : = x → y, P (x, y) : = x ⊕ y, D (x, y) : = x ⊙ y, M (x, y) : = x ⊖ y, Q (x, y) : = x ⇝ y, C (x, y) : = x ø y.

In the following, we give some preliminary definitions resulting in topology.

It is worth noting that a set A with a family $T = {U_{α}}_{α \in I}$ of its subsets is called a topological space, denoted by $(A, T)$ , if $A, \emptyset \in T$ , any finite intersection, and any arbitrary union of members of $T$ is in $T$ . The members of $T$ are called "open sets" of A. The complement of $U \in T$ , that is, A \ U is considered as a closed set. A subset P of A is a neighborhood of x ∈ A, if there exists an open set U such that x ∈ U ⊆ P. In addition, a subfamily {U_α} _α∈I of $T$ is a base of $T$ if an α ∈ I is available for each $x \in U \in T$ such that x ∈ U_α ⊆ U. Or, equivalently, each U in $T$ is the union of members of {U_α} _α∈I. A subfamily {U_β} _β∈I of $T$ is a subbase for $T$ if the family of finite intersections of members of {U_β} _β∈I forms a base for $T$ . Let (A, ∗) be an algebra of type (2) and $T$ a topology on A. Then the operation ∗ is called left (right) continuous, if the map $l_{a}^{*} : A \to A (r_{a}^{*} : A \to A)$ defined by $l_{a}^{*} (x) = x * a (r_{a}^{*} (x) = a * x)$ is continuous for all a ∈ A. Or, equivalently, for any x in A and any open set U of x * a (a * x), there exists an open set V of x such that V * a ⊆ U (a * V ⊆ U) and ∗ is called semicontinuous, if the map $l_{a}^{*}$ and $r_{a}^{*}$ are continuous for all a ∈ A. Furthermore, ∗ is called continuous, if two open sets (neighborhoods) U and V are available for x and y, respectively, for any x, y in A and any open set (neighborhood) W of x * y such that U * V ⊆ W.

Let $(A, T_{1})$ and $(B, T_{2})$ be considered as two topological spaces. A mapping f : A → B is called a homeomorphism if f is a bijection and f and f^-1 are continuous. A topological space $(A, T)$ is a T₀-space if there exists a neighborhood containing one point but not the other for each pair of distinct points. a T₁-space if each point lies in a neighborhood not containing the other for each pair of distinct points, a T₂-space (Hausdorff space) if there exist two disjoint neighborhoods of x₁ and x₂, respectively, for each pair of distinct points x₁ and x₂. Further, it is considered as a regular space if every neighborhood of a point contains the closure of a neighborhood of that point, a normal space if there exist two disjoint open sets U and V for each disjoint closed sets C and D such that we have C ⊆ U and D ⊆ V, a T₃-space if it is a T₁-space and a regular space, a T₄-space if it is a T₁-space and a normal space or equivalently, if there exists an open set H for each closed set C and each open set O containing C, such that $C \subseteq H \subseteq \bar{H} \subseteq O$ , a zero dimensional space if it has clopen base (close and open base).

In the present study, $B = (B; \oplus, \sim, 0)$ and $IB = (IB; \to)$ denote a basic algebra and an implication basic algebra, respectively, and $T$ demonstrates a topology on $B$ ( $IB$ ), unless other cases are stated.

3 (Semi) Topological basic algebras

In this section, we introduce the notions of semitopological and topological (implication) basic algebras.

Definition 3.1. Let $T$ be a topology on $B$ ( $IB$ ). Then $(B, T)$ ( $(IB, T)$ ) is called a

left topological (implication) basic algebra if → is left continuous,

right topological (implication) basic algebra if → is right continuous,

semitopological (implication) basic algebra if → is semicontinuous,

topological (implication) basic algebra if → is continuous.

Example 3.2. Let $B = {0, a, b, 1}$ be a lattice whose Hasse diagram is shown in Fig. 1 and the operations ⊕ and ∼ on $B$ be defined as shown in Table 1. Routine calculations show that $(B; \oplus, \sim)$ is a basic algebra. It is not difficult to check that $T = {{0, a}, {b, 1}, B, \emptyset}$ is a topology on $B$ and $(B, T)$ is a topological basic algebra.

Fig. 1
Table 1

⊕ 0 a b 1 x ∼x → 0 a b 1

0 0 a b 1 0 1 0 1 1 1 1

a a a 1 1 a b a b 1 b 1

b b 1 b 1 b a b a a 1 1

1 1 1 1 1 1 0 1 0 a b 1

Example 3.3. Let $B = {0, a, b, c, d, e, f, 1}$ be a lattice whose Hasse diagram is shown in Fig. 2 and the operations ⊕ and ∼ on $B$ be defined as shown in Table 2. Routine calculations show that $(B, \oplus, \sim)$ is a basic algebra.
Fig. 2
Table 2

⊕ 0 a b c d e f 1 x ∼x → 0 a b c d e f 1

0 0 a b c d e f 1 0 1 0 1 1 1 1 1 1 1 1

a a a d f d e 1 1 a f a f 1 f e 1 1 f 1

b b d b e d 1 f 1 b e b e e 1 f 1 e 1 1

c c e f c 1 e f 1 c d c d d d 1 d 1 1 1

d d d d 1 d 1 1 1 d c d c e f c 1 e f 1

e e e 1 f 1 e 1 1 e b e b d b e d 1 f 1

f f 1 f e 1 1 f 1 f a f a a d f d e 1 1

1 1 1 1 1 1 1 1 1 1 0 1 0 a b c d e f 1

Let $T_{1} = {{0, c}, {a, e}, {b, f}, {d, 1}, {0, c, a, e},$ {0, c, b, f} , {0, c, d, 1} , {a, e, b, f} , {0, c, a, e, b, f} , {b, f, d, 1} , {a, e, d, 1} , {0, c, b, f, d, 1} , {0, c, a, e, d, $1}, {a, e, b, f, d, 1}, B, \emptyset}$ . It is not difficult to check that $(B, T)$ is a left topological basic algebra.

Proposition 3.4. Let $T$ be a topology on $B$ . If →(or ⇝) is left continuous, then for all $x \in B$ , the negation map S (x) =  ∼ x is a ∼-homeomorphism on $B$ .

Proof. Let $S (x) = \sim x \in U \in T$ , where $x \in B$ . Since ∼x = x → 0 and → is left continuous, there is a $V \in T$ such that x ∈ V and ∼V = V → 0 ⊆ U and so S is continuous. We observe that S is an involution, which implies that S^-1 is continuous. From ∼∼ x = x, it is obvious that S is onto and one to one, which implies that S is a ∼-homeomorphism.

The proof of the operation ⇝ is similar.□

Corollary 3.5. Let $T$ be a topology on $B$ . If ø(or ⊖) is right continuous, then for all $x \in B$ , the negation map S (x) =  ∼ x is a ∼-homeomorphism on $B$ .

Proof. By considering ∼x = 1 ø x (=1 ⊖ x) and right continuity of ø (⊖), the proof is similar to that of Proposition 3.4.□

Remark 3.6. If the negation map ∼ is continuous, then the maps n, n_l and n_r from $B \times B$ to $B \times B$ are defined by n (x, y) : = (∼ x, ∼ y) , n_l (x, y) : = (∼ x, y) and n_r (x, y) : = (x, ∼ y) are continuous.

Proposition 3.7. Let $T$ be a topology on $B$ . Then the following statements are equivalent:
$(B, T)$ is a topological basic algebra,

∼ is continuous and ∗ is continuous, where ∗ ∈ {⊕ , ⊙},

∗ is continuous, where ∗ ∈ {ø , ⊖ , ⇝}.

Proof. From (B5) - (B9) we have, $\begin{matrix} I (x, y) & = P \circ n_{l} (x, y) = S \circ D \circ n_{r} (x, y) \\ = S \circ M (x, y) = S \circ C \circ n (y, x) = Q \circ n (y, x) . \end{matrix}$ Now, the proof follows from Proposition 3.4, Corollary 3.5 and Remark 3.6.□

Proposition 3.8. Let $T$ be a topology on $B$ . Then the following statements are equivalent:
$(B, T)$ is a semitopological basic algebra,

∼ is continuous and ∗ is semicontinuous, where ∗ ∈ {⊕ , ⊙},

∗ is semicontinuous, where ∗ ∈ {ø , ⊖ , ⇝}.

Proof. (i) ⇒ (ii) Assume that $(B, T)$ is a semitopological basic algebra. By Proposition 3.4, ∼ is continuous. Let $x, a \in B$ and $U \in T$ such that x ⊕ a =  ∼ x → a ∈ U. Since → is left continuous, there is $V \in T$ such that ∼x ∈ V and V → a ⊆ U. By continuity of ∼, there is a W ∈ T such that x ∈ W and ∼W ⊆ V. Then W ⊕ a =  ∼ W → a ⊆ V → a ⊆ U, proving ⊕ is left continuous. Now, let $x, a \in B$ and $U \in T$ such that ∼a → x = a ⊕ x ∈ U. Since → is right continuous, there is a $V \in T$ such that x ∈ V and ∼a → V = a ⊕ V ⊆ U. Hence ⊕ is right continuous and so ⊕ is semicontinuous. By the similar way, we can prove that ⊙ and ⊖ are semicontinuous. (ii) ⇒ (iii) Assume that ⊕ is semicontinuous and let $x, a \in B$ and U ∈ T such that x ø a =  ∼ (a ⊕ ∼ x) ∈ U. Since ∼ is continuous there is V ∈ T such that a ⊕ ∼ x ∈ V and ∼V ⊆ U. Since ⊕ is right continuous there is V₁ ∈ T such that ∼x ∈ V₁ and a ⊕ V₁ ⊆ V. By continuity of ∼, there is V₂ ∈ T such that x ∈ V₂ and ∼V₂ ⊆ V₁. Then V₂ ø a =  ∼ (a ⊕ ∼ V₂) ⊆  ∼ (a ⊕ V₁) ⊆  ∼ V ⊆ U and so ø is left continuous. Now, let $x, a \in B$ and U ∈ T such that a ø x =  ∼ (x ⊕ ∼ a) ∈ U. Since ∼ is continuous, there is V ∈ T such that x ⊕ ∼ a ∈ V and ∼V ⊆ U. Since ⊕ is left continuous there is V₁ ∈ T such that x ∈ V₁ and V₁ ⊕ ∼ a ⊆ V. Then a ø V₁ = ∼ (V₁ ⊕ ∼ a) ⊆ ∼ V ⊆ U. Hence ø is right continuous and so ø is semicontinuous. If ⊙ is semicontinuous, then by (B5) - (B9), we can prove that ø, ⊖ and ⇝ are semicontinuous. (iii) ⇒ (i) Assume that ⇝ is semicontinuous. By Proposition 3.4, ∼ is continuous. Let $x, a \in B$ and $U \in T$ such that x → a = ∼ a ⇝ ∼ x ∈ U. By the right continuity of ⇝, there is $V \in T$ such that ∼x ∈ V and ∼a ⇝ V ⊆ U. By continuity of ∼, there is V₁ ∈ T such that x ∈ V₁ and ∼V₁ ⊆ V. Then V₁ → ∼ a = ∼ a ⇝ ∼ V₁ ⊆ ∼ a ⇝ V ⊆ U. Hence → is left continuous. By the similar way, we can prove that → is right continuous. Then $(B, T)$ is a semitopological basic algebra.□

Corollary 3.9. Let $T$ be a topology on $B$ . If ∗ is right (left) continuous, where ∗ ∈ {ø , ⊖ , ⇝}, then → is left (right) continuous.

Proof. The proof is similar to the proof of Proposition 3.8 (iii) ⇒ (i).□

Proposition 3.10. Let $(B, T)$ be a (semi) topological basic algebra. Then ∨ is (semi) continuous.

Proof. Suppose that $(B, T)$ is a topological basic algebra, $x, y \in B$ , $U \in T$ and x ∨ y = (x → y) → y ∈U. Since → is continuous, then there are two neighborhoods U₁ and U₂ of x → y and y respectively, such that U₁ → U₂ ⊆ U. Also, there are two neighborhoods V₁ and V₂ of x and y, respectively, such that V₁ → V₂ ⊆ U₁. Let V = V₂ ∩ U₂. Then V₁ → V ⊆ V₁ → V₂ ⊆ U₁ and so V₁ ∨ V = (V₁ → V) → V ⊆ U₁ → V ⊆ U₁ → U₂ ⊆ U. Thus, ∨ is continuous. Now, suppose that $(B, T)$ is a semitopological basic algebra. Since the operation ∨ is commutative, it is sufficient to prove that ∨ is left continuous. Let $x, y \in B, U \in T$ and x ∨ y ∈ U. Since → is right continuous, there is a neighborhood U₁ of x → y such that U₁ → y ⊆ U and there is a neighborhood V₁ of x such that V₁ → y ⊆ U₁. Then V₁ ∨ y = (V₁ → y) → y ⊆ U₁ → y ⊆ U. Hence, ∨ is left continuous on $B$ and so ∨ is semicontinuous.□

Proposition 3.11. Let $(B, T)$ be a (semi) topological basic algebra. Then ∧ is (semi) continuous.

Proof. Suppose that $(B, T)$ is a topological basic algebra. By Propositions 3.4 and 3.10, the mapping V (x, y) : = x ∨ y from $B \times B$ into $B$ and the negation map S are continuous. So, by (BA₉) and Remark 3.6, S∘ V ∘ n = ∧ is continuous.

Now, let $(B, T)$ be a semitopological basic algebra, $x, y \in B$ , $U \in T$ and x ∧ y = ∼ (∼ x ∨ ∼ y) ∈ U. By Proposition 3.4, the negation map ∼ is continuous. Then there is a neighborhood W of ∼x ∨ ∼ y such that ∼W ⊆ U. By Proposition 3.10, since ∨ is left continuous, there is a neighborhood U₁ of ∼x such that U₁ ∨ ∼ y ⊆ W. Moreover, there is an open neighborhood U₂ of x such that ∼U₂ ⊆ U₁. Now, for any z ∈ U₂, we have $\begin{matrix} z \land y = & \sim (\sim z \lor \sim y) \in \sim (\sim U_{2} \lor \sim y) \subseteq \\ \sim (U_{1} \lor \sim y) \subseteq \sim W \subseteq U \end{matrix}$ So, U₂ ∧ y ⊆ U. Hence ∧ is left continuous. Since ∧ is commutative, it is right continuous. Therefore, ∧ is semicontinuous.□

Theorem 3.12. There is a topology $T$ on $B$ such that ∗ is continuous for ∗ ∈ {∨ , ∧}.

Proof. For $a \in B$ , let $B_{a} = {x \in B ∣ a \leq x}$ . Obviously, $a \in B_{a}$ for all $a \in B$ and $B = \underset{a \in B}{\cup} B_{a}$ . Let $a, b, c \in B$ . If $a \in B_{b} \cap B_{c}$ , then $B_{a} \subseteq B_{b} \cap B_{c}$ . Hence $B = {B_{a} ∣ a \in B}$ is a base for the topology $T$ on $B$ where members of $T$ is the unions of members of B. For proof of continuity ∗ ∈ {∧ , ∨}, suppose that $a, b \in B$ and $U \in T$ such that a ∗ b ∈ U. Since for each $x \in B$ , $B_{x}$ is the smallest neighborhood of x, $a * b \in B_{a * b} \subseteq U$ . By (BA₆), it is easily to prove that $B_{a} * B_{b} \subseteq B_{a * b}$ . Therefore, ∗ is continuous.□

The following example shows that there is a topology $T$ on $B$ such that ∨, ∧ are continuous but $(B, T)$ is not a (semi) topological basic algebra.

Example 3.13. Let $B = {0, a, b, c, 1}$ be a lattice whose Hasse diagram is shown in Fig. 3. We define the operations ⊕ and ∼ on $B$ as shown in Table 3. Routine calculations show that $(B, \oplus, \sim)$ is a basic algebra. Moreover, $B_{a} = {a, c, 1}$ , $B_{b} = {b, 1}$ , $B_{c} = {c, 1}$ , $B_{1} = {1}$ and $B_{0} = B$ , then by Theorem 3.12, ${B_{x}}_{x \in B}$ is a base for the topology $T = {{1}, {b, 1}, {c, 1}, {a, c, 1}, {b, c, 1}, {a, b, c, 1}, B, \emptyset}$ on $B$ and ∨ and ∧ are continuous. Now, we show that $(B, T)$ is not a left topological basic algebra. For this, consider c → 0 = a ∈ {a, c, 1}. Then {c, 1} →0 = {0, a} ⊈ {a, c, 1}, {a, c, 1} →0 = {0, a, c} ⊈ {a, c, 1}, {b, c, 1} →0 = {0, a, b} ⊈ {a, c, 1} and {a, b, c, 1} →0 = {0, a, b, c} ⊈ {a, c, 1}, which imply that → is not left continuous. Hence $(B, T)$ is not a (semi) topological basic algebra.
Fig. 3
Table 3

⊕ 0 a b c 1 x ∼x → 0 a b c 1

0 0 a b c 1 0 1 0 1 1 1 1 1

a a c b 1 1 a c a c 1 b 1 1

b b a 1 c 1 b b b b a 1 c 1

c c 1 b 1 1 c a c a c b 1 1

1 1 1 1 1 1 1 0 1 0 a b c 1

4 Separation axioms on (semi) topological basic algebras

⊕	0	a	b	1	x	∼x	→	0	a	b	1
0	0	a	b	1	0	1	0	1	1	1	1
a	a	a	1	1	a	b	a	b	1	b	1
b	b	1	b	1	b	a	b	a	a	1	1
1	1	1	1	1	1	0	1	0	a	b	1

⊕	0	a	b	c	d	e	f	1	x	∼x	→	0	a	b	c	d	e	f	1
0	0	a	b	c	d	e	f	1	0	1	0	1	1	1	1	1	1	1	1
a	a	a	d	f	d	e	1	1	a	f	a	f	1	f	e	1	1	f	1
b	b	d	b	e	d	1	f	1	b	e	b	e	e	1	f	1	e	1	1
c	c	e	f	c	1	e	f	1	c	d	c	d	d	d	1	d	1	1	1
d	d	d	d	1	d	1	1	1	d	c	d	c	e	f	c	1	e	f	1
e	e	e	1	f	1	e	1	1	e	b	e	b	d	b	e	d	1	f	1
f	f	1	f	e	1	1	f	1	f	a	f	a	a	d	f	d	e	1	1
1	1	1	1	1	1	1	1	1	1	0	1	0	a	b	c	d	e	f	1

⊕	0	a	b	c	1	x	∼x	→	0	a	b	c	1
0	0	a	b	c	1	0	1	0	1	1	1	1	1
a	a	c	b	1	1	a	c	a	c	1	b	1	1
b	b	a	1	c	1	b	b	b	b	a	1	c	1
c	c	1	b	1	1	c	a	c	a	c	b	1	1
1	1	1	1	1	1	1	0	1	0	a	b	c	1

In this section, we get some conditions on (semi) topological basic algebras, which imply that they become a T_i-space.

Proposition 4.1. Let $(B, T)$ be a semitopological basic algebra. Then $(B, T)$ is a T₀-space if and only if it is a T₁-space.

Proof. Obviously, any T₁-space is a T₀-space. Let $(B, T)$ be a T₀-space and x ≠ y for $x, y \in B$ . Then either x → y ≠ 1 or y → x ≠ 1. Assume that x → y ≠ 1. Then there exists an open set U such that x → y ∈ U and 1 ∉ U or x → y ∉ U and 1 ∈ U. First, suppose that x → y ∉ U and 1 ∈ U. Since x → x = y → y = 1, there exist two neighborhoods V₁ and V₂ of x and y, respectively such that x → V₁ ⊆ U and V₂ → y ⊆ U. We claim that x ∉ V₂ and y ∉ V₁. If x or y belongs to V₁ ∩ V₂, then x → y ∈ U, which is a contradiction. By the similar way, the proof of other case is clear.□

Proposition 4.2. Let $(B, T)$ be a topological basic algebra. Then $(B, T)$ is a T₁-space if and only if it is a Hausdorff space.

Proof. Let $(B, T)$ be a T₁-space and x ≠ y for $x, y \in B$ . Then either x → y ≠ 1 or y → x ≠ 1. Assume that x → y ≠ 1. Since $B$ is a T₁-space, there exist two neighborhoods U and V of x → y and 1, respectively, such that x → y ∉ V and 1 ∉ U. Since → is continuous, there exist two neighborhoods V₁ and V₂ of x and y, respectively, such that V₁ → V₂ ⊆ U. We claim that V₁∩ V₂ = ∅. If z ∈ V₁ ∩ V₂, then 1 = z → z ∈ V₁ → V₂ ⊆ U, which is a contradiction. Thus $(B, T)$ is a Hausdorff space. The converse is obvious.□

Corollary 4.3. Let $(B, T)$ be a topological basic algebra. Then $(B, T)$ is a T_i-space if and only if it is a T_j-space, where i, j ∈ {0, 1, 2}.

Proof. It follows from Propositions 4.1 and 4.2.□

Proposition 4.4. Let F be a nonempty subset of $B$ . Then F is a (weak) filter of $B$ if and only if ∼F is an (a weak) ideal of $B$ .

Proof. Let F be a filter of $B$ . We show that ∼F satisfies (I1)-(I3). (I1) Let x ø y ∈ ∼ F and y ∈ ∼ F, for $x, y \in B$ . Then ∼y → ∼ x = ∼ (x ø y) ∈ F and ∼y ∈ F, thus, ∼x ∈ F. (I2) Let ∼ (∼ y → ∼ x) = x ø y ∈ ∼ F, y ≤ x and $c \in B$ . Since ∼y → ∼ x ∈ F, ∼x ≤ ∼ y and F is a filter, by (F2), for ∼c, we have (∼ x → ∼ c) → (∼ y → ∼ c) ∈ F. Thus we can conclude that (c ø y) ø (c ø x) = ∼ (∼ (c ø x) → ∼ (c ø y)) ∈ ∼ F. (I3) Let x ø y ∈ ∼ F, y ø x ∈ ∼ F and $c \in B$ . Since ∼ (∼ y → ∼ x) = x ø y ∈ ∼ F, ∼ (∼ x → ∼ y) = y ø x ∈ ∼ F, then ∼x → ∼ y ∈ F, ∼ y → ∼ x ∈ F, and so (∼ c → ∼ y) → (∼ c → ∼ x) ∈ F. Therefore, (x ø c) ø (y ø c) = ∼ (∼ (y ø c) → ∼ (x ø c)) ∈ ∼ F. Hence ∼F is an ideal of $B$ .

By the similar way, the proof of other side is clear.□

Theorem 4.5. Let $(B, T)$ be a semitopological basic algebra. If the ideal {0} (the filter {1}) is an open set, then $T$ is a discrete topology and so $(B, T)$ is a topological basic algebra.

Proof. Let $x \in B$ . We know that x ø x = 0. Since ø is left and right continuous and {0} is an open set, there exist two neighborhoods U₁ and U₂ of x such that U₁ ø x = {0} = x ø U₂. For U = U₁ ∩ U₂ we have x ø U = U ø x = {0}. We claim that U = {x}. Suppose that y ∈ U. Then y ø x = x ø y = 0. Hence, by (BA₄), x = y and so $T$ is a discrete topology on $B$ , thus $(B, T)$ is a topological basic algebra. The proof for filter {1} follows from Proposition 4.4 and the first part of this theorem.□

Theorem 4.6. Let $T$ be a topology on $B$ , and for each $a \in B$ , $l_{a}^{\to}$ (resp. $r_{a}^{ø})$ be an open map. If there exists $U \in T$ containing 1 (resp. 0), then $(B, T)$ is a T₀-space.

Proof. Let x ≠ y and U be an open set containing 1. Since $l_{x}^{\to}$ and $l_{y}^{\to}$ are open maps, we have $l_{x}^{\to} (U) = U \to x$ and $l_{y}^{\to} (U) = U \to y$ are open sets containing x and y because x = 1 → x ∈ U → x and y = 1 → y ∈ U → y. We claim that x ∉ U → y or y ∉ U → x. Suppose that x ∈ U → y and y ∈ U → x. Then for some $a, b \in B$ , we have x = a → y and y = b → x, thus, by (BA₈) we get that x ≤ y and y ≤ x. Hence x = y, which is a contradiction. Then $(B, T)$ is a T₀-space. Now, let x ≠ y and U be an open set containing 0. Since $r_{x}^{ø}$ and $r_{y}^{ø}$ are open maps, we have $r_{x}^{ø} (U) = x ø U$ and $r_{y}^{ø} (U) = y ø U$ are open sets containing x and y, because x = x ø 0 ∈ x ø U and y = y ø 0 ∈ y ø U. We claim that x ∉ y ø U or y ∉ x ø U. Suppose that x ∈ y ø U and y ∈ x ø U. Then for some $a, b \in B$ , we have x = y ø a, y = x ø b, and by (BA₈), x ≤ y and y ≤ x. Then x = y, which is a contradiction. Hence $(B, T)$ is a T₀-space.□

Proposition 4.7. Let F be a filter (an ideal) of topological basic algebra $(B, T)$ . Then the following statements hold:

If 1 (resp. 0) is an interior point of F, then F is a closed set.

If F is an open set, then F is a closed set.

If $B$ is connected, then $B$ has no open proper filter (ideal).

Proof. (i) Let 1 be an interior point of filter F. It is sufficient to prove that F^c, the complement of F, is an open set. By assumption, there is a neighborhood U of 1 such that U ⊆ F. Since x → x = 1, for x ∈ F^c, there exist two neighborhoods V₁ and V₂ of x such that V₁ → V₂ ⊆ U. Let V = V₁ ∩ V₂. Then V → V ⊆ V₁ → V₂ ⊆ U. Now, we show that V ⊆ F^c. Suppose that V ⊈ F^c. Then for y ∈ V ∩ F and arbitrary element z ∈ V, we have y → z ∈ V → V ⊆ F, and so z ∈ F. This implies that V ⊆ F, which is a contradiction. Thus x ∈ V ⊆ F^c. Hence, F^c is an open set.

(ii) Let F be an open set of $B$ and x ∈ F^c. Since x → x = 1 ∈ F and $(B, T)$ is a topological basic algebra, there are two neighborhoods U₁, U₂ of x such that U₁ → U₂ ⊆ F. Consider U = U₁ ∩ U₂, then U → U ⊆ U₁ → U₂ ⊆ F. We claim that U ⊆ F^c. If U∩ F ≠ ∅, then there is y ∈ U ∩ F and for any z ∈ U, y → z ∈ U → U ⊆ F. Since y ∈ F, by (F1), z ∈ F, (U ⊆ F), which is a contradiction. Then U ⊆ F^c and F^c is an open set. Therefore, F is closed.

(iii) Let F be an open set. Then by (ii), F is closed set. Since $B$ is connected, $F = B$ and it is not proper filter of $B$ .

When F is an ideal of $B$ , the proof follows from Propositions 3.4 and 4.4.□

Proposition 4.8. Let $(B, T)$ be a topological basic algebra. If either of {1} or {0} is closed, then $(B, T)$ is a Hausdorff space.

Proof. Let $(B, T)$ be a topological basic algebra. Suppose that {1} is a closed set. Let $x, y \in B$ and x ≠ y. Then x → y ≠ 1 or y → x ≠ 1. Without the lost of generality, assume that x → y ≠ 1. Since {1} is closed, $B \ {1}$ is an open set containing x → y. Since $(B, T)$ is a topological basic algebra, there exist two neighborhoods U₁ and U₂ of x and y such that $U_{1} \to U_{2} \subseteq B \ {1}$ . We claim that U₁∩ U₂ = ∅. If y ∈ U₁ ∩ U₂, then $1 = y \to y \in U_{1} \to U_{2} \subseteq B \ {1}$ , which is a contradiction. Hence $B$ is a Hausdorff space.

Conversely, assume that $(B, T)$ is a Hausdorff space. We show that $B \ {1}$ is open. If x ≠ 1, then there exist two disjoint neighborhoods U₁, U₂ of x and 1, respectively. Thus 1 ∉ U₁ and $U_{1} \subseteq B \ {1}$ , which implies that $B \ {1}$ is an open set. In the above proof, if we replace {1} by {0} and → by ø, then, by a similar manner we can prove that if {0} is close. Hence, $(B, T)$ is a Hausdorff space.□

Proposition 4.9. Let $T$ be a topology on $B$ and assume that {0} is a closed set. Then the following statements hold:

if ø is semicontinuous, then $(B, T)$ is a T₁-space.

if ø is right or left continuous, then $(B, T)$ is a T₀-space.

Proof. (i) Let x ≠ y. Then x ø y ≠ 0 or y ø x ≠ 0. Suppose that x ø y ≠ 0. By assumption, there is a neighborhood U of x ø y such that 0 ∉ U. Since ø is semicontinuous, there exist two neighborhoods V₁ and V₂ of x and y, respectively, such that V₁ ø y ⊆ U and x ø V₂ ⊆ U. We claim that x ∉ V₂ and y ∉ V₁. Otherwise 0 = x ø x = y ø y ∈ U, which is a contradiction. Thus $(B, T)$ is a T₁-space.

(ii) Assume that ø is left continuous. Let $x, y \in B$ such that x ≠ y. Then x ø y ≠ 0 or y ø x ≠ 0. Suppose that x ø y ≠ 0. By assumption there is a neighborhood U of x ø y such that 0 ∉ U. On the other hand, there is a neighborhood V₁ of x such that V₁ ø y ⊆ U. We claim that y ∉ V₁. If y ∈ V, then 0 = y ø y ∈ U, which is a contradiction. Thus $(B, T)$ is a T₀-space.

The proof of the other case is similar.□

5 Induced (semi) topological basic algebras

In this section, we show that in every basic algebra $B$ , any arbitrary family of filters (ideals) induce a topology on $B$ , which is completely regular, normal and locally compact. In Theorem 5.31, it is shown that for any n ≥ 3, there exists a nontrivial topological implication basic algebra, which is normal and regular.

Definition 5.1. [12] Let Θ be an equivalence relation on $B$ .

Θ is called a weak congruence if for all $a, b, c \in B$ , (CW) (a, b) ∈ Θ ⇒ (c ø a, c ø b) ∈ Θ.

Θ is called a congruence if it is a weak congruence and satisfies (a, b) ∈ Θ ⇒ (a ø c, b ø c) ∈ Θ.

Theorem 5.2. Let Θ be an equivalence relation on $B$ . Then θ is a weak congruence if and only if it satisfies $(a, b) \in Θ \Rightarrow (a \to c, b \to c) \in Θ (CC)$ Proof. If Θ satisfies (CW) or (CC) and (a, b) ∈ Θ, then (∼ a, ∼ b) ∈ Θ. Now, suppose (CC) is satisfied. For $\sim c \in B$ , we get (∼ a → ∼ c, ∼ b → ∼ c) ∈ Θ. This means that (∼ (c ø a) , ∼ (c ø b)) ∈ Θ, which implies that (c ø a, c ø b) ∈ Θ. Conversely assume that (CW) is satisfied. Let (a, b) ∈ Θ, then (∼ a, ∼ b) ∈ Θ. For $\sim c \in B$ , we get (∼ c ø ∼ a, ∼ c ø ∼ b) = (∼ (a → c) , ∼ (b → c)) ∈ Θ, which implies that (a → c, b → c) ∈ Θ. □

Theorem 5.3. [7 , 12] (i) Let Θ be a (weak) congruence on $B$ . Then [1] _Θ and [0] _Θ are a (weak) filter and an (a weak) ideal of $B$ , respectively.

(ii) Let Θ be a congruence on $IB$ . Then [1] _Θ is a filter of $IB$ .

(iii) Let F be a (weak) filter of $B$ . Then the relation Θ_F is defined by $(a, b) \in Θ_{F} if and only if a \to b \in F and b \to a \in F$ is a (weak) congruence on $B$ such that [1] _{Θ
_F} = F.

If F is a filter of $IB$ , then the relation Θ_F is a congruence relation on $IB$ such that [1] _{Θ
_F} = F.

(iv) Let I be an (a weak) ideal of $B$ . Then the relation Θ_I is defined by $(a, b) \in Θ_{I} if and only if a ø b \in I and b ø a \in I,$ is a (weak) congruence relation on $B$ such that [0] _{Θ
_I} = I.

Lemma 5.4. Let I (F) be an ideal (resp. filter) of $B$ . Then the following conditions hold:

If x ∈ I and $y \in B$ , then x ø y ∈ I,

If $x \in B$ and y ∈ F, then x → y ∈ F, x ∨ y ∈ F .

If x, y ∈ I (F), then x Θ_I y (x Θ_F y).

Proof. (i) Let x ∈ I and $y \in B$ . Then by (BA₇), (x ø y) ø x = 0 ∈ I, thus by (I1), we get that x ø y ∈ I. (ii) Let $x \in B$ and y ∈ F. Then by (BA₈), y → (x → y) =1 ∈ F, thus, by (F1) we get that x → y ∈ F. Also, since y ∈ F, by (F1), x ∨ y = (x → y) → y ∈ F. (iii) Let I be an ideal (filter) of $B$ and x, y ∈ I (resp. F). By (i) and (ii), x ø y ∈ I and y ø x ∈ I. Hence x Θ_I y. If F is a filter,then by the similar way, we can prove that x Θ_F y. □

Theorem 5.5. Let I be an (a weak) ideal of $B$ and F = ∼ I. Then, for all $a, b \in B$ , we get that, $(a, b) \in Θ_{I} if and only if (a, b) \in Θ_{F} .$

Proof. Let (a, b) ∈ Θ_I. Then ∼ (∼ b → ∼ a) = a ø b ∈ I and ∼ (∼ a → ∼ b) = b ø a ∈ I. Hence ∼a → ∼ b ∈ F and ∼b → ∼ a ∈ F. By Proposition 4.4, F is a (weak) filter. Then by (F2), if c = 0, then a → b ∈ F and b → a ∈ F. The converse is obvious. □

Definition 5.6. For an arbitrary element $x \in B (IB)$ and a nonempty subset U of $B (IB)$ , the set U [x] is defined by $U [x] = {y \in B ∣ x * y \in U, y * x \in U},$ for any ∗ ∈ {→ , ø} (resp. ∗ ∈ {→}).

Remark 5.7. In Definition 5.6, if I is a (weak)ideal of $B$ , then I [x] is obviously the equivalence class $[x]_{I} = {y \in B ∣ x Θ_{I} y}$ . It is satisfied for any (weak) filter of $B$ and $IB$ , too.

Theorem 5.8. Let $I$ be an arbitrary family of ideals of $B$ which is closed under finite intersections. Then $T_{I} = {U \subseteq B ∣ \forall x \in U, \exists I \in I, I [x] \subseteq U},$ is a topology on $B$ with the base $β = {I [x] ∣ x \in B, I \in I} = {[x]_{I} ∣ x \in B, I \in I}$ and $(B, T_{I})$ is a topological basic algebra and $T_{I}$ is called the induced topology by $I$ .

Proof. It is easy to prove that $T_{I}$ is closed under arbitrary unions. $T_{I}$ is a nontrivial topology because for any $x \in B$ and $I \in I, I [x] \in T_{I}$ (at least, there is a trivial ideal ${0} \in B$ ). Note that for any $x \in B$ , I [x] is the smallest neighborhood of x and $T_{I}$ is called the induced topology by $I$ .

Suppose that $I$ is a family of ideals of $B$ . Let $x, y \in B$ and $U \in T_{I}$ such that x ø y ∈ U. Then there exists an ideal I such that I [x ø y] ⊆ U. Since Θ_I is a congruence relation and I [x] is a class of Θ_I, we have I [x] ø I [y] ⊆ I [x ø y] ⊆ U. Hence by Proposition 3.7, $(B, T_{I})$ is a topological basic algebra. Clearly, β is a base for $T_{I}$ .□

Theorem 5.9. Let $F$ be an arbitrary family of filters of $B$ which is closed under finite intersections. Then $T_{F} = {U \subseteq B ∣ \forall x \in U, \exists F \in F, F [x] \subseteq U}$ is a topology on $B$ with the base $β = {F [x] ∣ x \in B, F \in F} = {[x]_{F} ∣ x \in B, F \in F}$ and $(B, T_{F})$ is a topological basic algebra and $T_{F}$ is called the induced topology by $F$ .

Proof. The proof is similar to the proof of Theorem 5.8. □

In the rest of paper, if $I = {I}$ and $F = {F}$ , then $T_{F}$ and $T_{I}$ are presented by $T_{F}$ and $T_{I}$ , respectively.

Example 5.10. Consider $B = {0, a, b, 1}$ be the basic algebra $(B; \oplus, \sim)$ as in Example 3.2. Routine calculations, show that the set of ideals of $B$ is ${{0, a}, {0, b}, {0}, B}$ .

Now, consider $I = {I_{0}, I_{1}, I_{2}}$ which I₀ = {0}, I₁ = {0, a} and I₂ = {0, b} are ideals of $B$ . Then I₀ [x] = {0}, for all $x \in B$ and $I_{1} [x] = {\begin{matrix} {0, a}, & x \in {0, a} \\ {x}, & x \in {b, 1} \end{matrix} and$ $I_{2} [x] = {\begin{matrix} {0, b}, & x \in {0, b} \\ {x}, & x \in {a, 1} \end{matrix}$

Thus β = {{0} , {1} , {a} , {b} , {a, 0} , {b, 0}} is a base for $T_{I}$ and $(B, T_{I})$ is a topological basic algebra.

Theorem 5.11. Let $I$ be an arbitrary family of weak ideals of $B$ which is closed under finite intersections. Then $(B, T_{I})$ is a left topological basic algebra which $β = {I [x] ∣ x \in B, I \in I}$ is the base of $T_{I}$ .

Proof. Suppose that $I$ is a family of weak ideals of $B$ . Let $a, y \in B$ and $U \in T_{I}$ such that a ø y ∈ U. Then there exists $I \in I$ such that I [a ø y] ⊆ U. We prove that a ø I [y] ⊆ I [a ø y]. Let b ∈ I [y]. Then b Θ_Iy, where Θ_I is a weak congruence. By (CW), a ø b Θ_I a ø y. Now, a ø b ∈ a ø I [y] ⊆ I [a ø y] ⊆ U. Thus ø is right continuous. Hence, by Corollary 3.9, the operation → is left continuous and $(B, T_{I})$ is a left topological basic algebra. □

Corollary 5.12. Let $F$ be an arbitrary family of weak filters of $B$ which is closed under finite intersections. Then ( $B, T$ ) is a left topological basic algebra which $β = {F [x] ∣ x \in B, F \in F}$ is the base of $T_{F}$ .

Theorem 5.13. If I is an (a weak) ideal of $B$ , then the induced topology by I is equal to induced topology by the (weak) filter F = ∼ I.

Proof. By Theorem 5.5, the proof is clear. □

Example 5.14. Consider $B = {0, a, b, c, d, e, f, 1}$ be the basic algebra $(B; \oplus, \sim)$ as in Example 3.3. Routine calculations, show that I = {0, c} is a weak ideal of $B$ . Then, $I_{1} [x] = {\begin{matrix} {0, c}, & x \in {0, c} \\ {a, e}, & x \in {a, e} \\ {b, f}, & x \in {b, f} \\ {d, 1}, & x \in {d, 1} \end{matrix}$

Thus β = {{0, c} , {a, e} , {b, f} , {d, 1}} is a base for $T_{I}$ and $(B, T_{I})$ is a left topological basic algebra. Easily, we can see that F = {1, d} is a weak filter of $B$ and the induced topology by weak filter F is equal to induced topology by weak ideal I.

Proposition 5.15. If F is a nontrivial filter of $B$ , then ∼F∩ F = ∅.

Proof. Let F be a filter of $B$ and x ∈ ∼ F ∩ F. Then ∼x = (x → 0) ∈ F and x ∈ F. Thus, by (F1), 0 ∈ F and $F = B$ , which is a contradiction. Now, let I be an ideal of $B$ . Then by Proposition 4.4, ∼I is a filter and then ∼I∩ I = F ∩ ∼ F = ∅. □

Corollary 5.16. If I is a nontrivial ideal of $B$ , then ∼I∩ I = ∅.

Proof. Let I be an ideal of $B$ . Then by Proposition 4.4, ∼I = F is a filter of $B$ and so ∼I ∩ I = ∼F∩ F = ∅. □

Definition 5.17. Let I be an ideal of $B$ . I is called supermaximal if $B = \sim I \cup I$ .

Remark 5.18. By Proposition 4.4, I is an ideal of $B$ , if and only if ∼I is a filter of $B$ . Then, the dual notion of supermaximal filters can be introduced such as supermaximal ideal.

Theorem 5.19. Let I be an ideal of $B$ . Then the following statements are equivalent:

I is a supermaximal ideal of $B$ ,

$T_{I} = {\emptyset, I, \sim I, B}$ ,

$| T_{I} | = 4$ .

Proof. (i) ⇒ (ii) Let I be a supermaximal ideal of $B$ and $a \in B = I \cup \sim I$ . If a ∈ I, a/I = 0/I = I and if a ∈ ∼ I, a/(∼ I) =1/(∼ I) = (∼ I). So, $B / F = {I, \sim I}$ is a base for topology $T_{I}$ . Hence $T_{I} = {\emptyset, I, \sim I, B}$ . (ii) ⇒ (iii) It is straightforward. (iii) ⇒ (i) Let I be an ideal of $B$ . Then $\emptyset \neq [0]_{I} = I \in T_{I}$ . By Proposition 4.4 and Corollary 5.16, β = {I, ∼ I} is a base of $T_{I}$ and $B = \sim I \cup I$ . Hence, I is a supermaximal ideal of $B$ . □

Example 5.20. Let $B = {0, a, b, c, d, e, f, g, h, i,$ j, 1} be the lattice that Hasse diagram is shown in Fig. 4 and the operations ⊕, ∼ and ø on $B$ are defined as shown in Table 4. Routine calculations show that (B ; ⊕ , ∼) is a basic algebra. Moreover, it is not difficult to check that I₁ = {0, a} is a weak ideal and I = {0, b, d, f, h, j} is a supermaximal ideal of $B$ . Now, let $I = {I}$ . $I [x] = {\begin{matrix} {0, b, d, f, h, j}, & x \in {0, b, d, f, h, j} \\ {a, c, e, g, i, 1}, & x \in {a, c, e, g, i, 1} \end{matrix}$ Then {{0, b, d, f, h, j} , {a, c, e, g, i, 1}} is a base for $T_{I}$ . Hence $(B, T_{I})$ is a topological basic algebra.

Fig. 4

$(B; \oplus, \sim)$ .

Table 4

The Cayley tables of the operations on $B$

⊕	0	a	b	c	d	e	f	g	h	i	j	1	x	∼x
0	0	a	b	c	d	e	f	g	h	i	j	1	0	1
a	a	a	1	c	g	e	i	g	c	i	e	1	a	b
b	b	1	b	1	b	1	b	1	b	1	b	1	b	a
c	c	i	1	c	1	e	i	1	c	i	e	1	c	d
d	d	e	b	1	d	e	f	g	b	i	j	1	d	c
e	e	g	1	c	g	e	1	g	c	1	e	1	e	f
f	f	c	b	c	d	1	f	g	h	i	b	1	f	e
g	g	e	1	1	g	e	i	g	1	i	e	1	g	h
h	h	i	b	c	b	e	f	1	h	i	j	1	h	g
i	i	c	1	c	g	1	i	g	c	i	1	1	i	j
j	j	g	b	c	d	e	b	g	h	1	j	1	j	i
1	1	1	1	1	1	1	1	1	1	1	1	1	1	0
(a) $(B; \oplus; \sim)$

ø	0	a	b	c	d	e	f	g	h	i	j	1
0	0	0	0	0	0	0	0	0	0	0	0	0
a	a	0	a	0	a	0	a	0	a	0	a	0
b	b	b	0	j	f	h	d	f	j	d	h	0
c	c	h	a	0	c	h	c	h	a	h	c	0
d	d	d	0	d	0	d	d	0	d	d	d	0
e	e	j	a	j	e	0	e	j	e	j	a	0
f	f	f	0	f	f	f	0	f	f	0	f	0
g	g	d	a	d	a	d	g	0	g	d	g	0
h	h	h	0	0	h	h	h	h	0	h	h	0
i	i	f	a	f	i	f	a	f	i	0	i	0
j	j	j	0	j	j	0	j	j	j	j	0	0
1	1	b	a	d	c	f	e	h	g	j	i	0
(b) $(B; ø, \sim)$

Now, consider the weak ideal I₁ = {0, a}. Then $I_{1} = {\begin{matrix} {0, a}, & x \in {0, a} \\ {1, b}, & x \in {1, b} \\ {c, h}, & x \in {c, h} \\ {d, g}, & x \in {d, g} \\ {e, j}, & x \in {e, j} \\ {f, i}, & x \in {f, i} \end{matrix}$ Hence {{0, a} , {1, b} , {c, h} , {d, g} , {e, j} , {f, i}} is a base for $T_{I_{1}}$ . Therefore, $(B, T_{I_{1}})$ is a left topological basic algebra.

Proposition 5.21. Let I be a proper ideal (filter) of $B$ . In the topological space $(B, T_{I})$ , the following properties hold:

For each x ∈ I, I [x] = I.

For each $x \in B$ , I and I [x] are clopen subsets of $B$ .

For each $x \in B$ , I and I [x] are compact subsets of $B$ .

$(B, T_{I})$ is zero dimensional, disconnected, locally compact, completely regular and normal.

Proof. (i) The proof is clear.

(ii) Let $x \in B$ . Then I [x] is an open set. It is sufficient to prove that (I [x]) ^c is an open set, too. Let y ∈ I [x] ^c. Then I [x]∩ I [y] = ∅, which yields I [y] ⊆ (I [x]) ^c. Since I [y] is an open set, I [x] ^c is an open set. Therefore, I [x] (and so I = I [0]) is a clopen subset of $B$ .

(iii) Suppose that $O$ is an arbitrary open covering of I [x]. Then there exists an open set $V \in O$ containing x and thus I [x] ⊆ V, because I [x] is the smallest neighborhood of x in this topology. Therefore, I [x] and I [0] = I (I [1] = I) are compact.

(iv) By (ii), I is a proper clopen subset of $B$ , then $(B, T_{I})$ is disconnected and it is zero dimensional, because $β = {I [x] ∣ x \in B}$ is a clopen base for topology $T_{I}$ . Since for any $x \in B$ , I [x] is a compact neighborhood of x, $(B, T_{I})$ is a locally compact space. Now, let $x \in B$ and V be a neighborhood of x. Then by (ii), there is a closed neighborhood I [x] of x which I [x] ⊆ V. Therefore $(B, T_{I})$ is regular. Since $(B, T_{I})$ is locally compact, $(B, T_{I})$ is completely regular. Now, let C and O be a closed and open set, respectively, such that C ⊆ O. Since for x ∈ C, I [x] is the smallest neighborhood of x, then I [x] ⊆ O. By (ii), I [x] is a clopen subset of $B$ . Then H = ⋃ _x∈CI [x] is a clopen subset of $B$ . Therefore $C \subseteq H \subseteq \bar{H} \subseteq O$ , which implies that $(B, T_{I})$ is a normal space. □

Now, in the following, we present the notion of quotient basic algebras and study the interaction of topological basic algebras and topological quotient basic algebras and find their properties.

Let $B$ be a basic algebra and F be a filter of $B$ . We define ⊕, → and ∼ on $B / F$ by $\begin{matrix} x / F \oplus y / F = (x \oplus y) / F, x / F \to y / F = (x \to y) / \\ F and \sim (x / F) = (\sim x) / F . \end{matrix}$

It is easy to check that quotient algebra $(B / F; \oplus, \sim)$ is a basic algebra. Let $T$ be a topology on $B$ and $π_{F} : B ⟼ B / F$ be a canonical projection. Suppose $T$ is a topology on $B$ and U is a subset of $B / F$ . Then U is an open subset of $B / F$ if and only if π^-1 (U) is an open subset of $T$ . Now, consider $T_{π_{F}} = {U \subseteq B / F ∣ π_{F}^{- 1} (U) \in T} .$

$T_{π_{F}}$ is the topology on $B / F$ which is called the quotient topology induced by π_F. It is well known that it is the greatest topology on $B / F$ making π_F continuous.

If we consider $l_{a}^{\to} (x) = L_{a} (x) = x \to a$ and $r_{a}^{\to} (x) = R_{a} (x) = a \to x$ , then it is easy to show that for each $a \in B$ , π_F ∘ L_a = L_a/F ∘ π_F and π_F ∘ R_a = R_a/F ∘ π_F.

In the following, we give some conditions that $B / F$ becomes a topological basic algebra.

Proposition 5.22. Let $(B, T)$ be a topological basic algebra and F be a filter of $B$ . If natural projection $π_{F} : B ⟼ B / F$ is an open map, then the quotient basic algebra $B / F$ with quotient topology is a topological basic algebra.

Proof. It is sufficient to prove that → is continuous to quotient topology. Let V be an open neighborhoods of (x → y)/F = x/F → y/F. Then $π_{F}^{- 1} (V)$ is an open subset of $B$ and $x \to y \in π_{F}^{- 1} (V) = U$ . Since $(B, T)$ is a topological basic algebra, there exist open neighborhoods U₁, U₂ of x and y, respectively, such that $U_{1} \to U_{2} \in π_{F}^{- 1} (V)$ . Then V₁ = π_F (U₁) and V₂ = π_F (U₂) are open sets in $B / F$ and x/F ∈ V₁ and y/F ∈ V₂, because π_F is an open map. Hence V₁ → V₂ ⊆ π_F (U₁ → U₂) ⊆ U. Therefore → is continuous. □

Proposition 5.23. Let $(B, T_{F})$ be a topological basic algebra induced by $F$ and F be a filter of $B$ . If $F \subseteq \cap {F_{i} ∣ F_{i} \in F}$ , then $π_{F} : B ⟼ B / F$ is an open map.

Proof. Let $F$ be an arbitrary family of filters of $B$ which is closed under finite intersection and F be a filter of $B$ . By Theorem 5.9, $β = {F_{i} [x] ∣ x \in B, F_{i} \in F} = {[x]_{F_{i}} ∣ x \in B, F_{i} \in F}$ is a base for $T_{F}$ . So, it is sufficient to show that for all [x] _{F
_i} ∈ β, π_F ([x] _{F
_i}) is an open set or $π_{F}^{- 1} π_{F} ([x]_{F_{i}}) \in T_{F}$ . Let [x] _{F
_i} ∈ β, for $x \in B$ and $F_{i} \in F$ . Suppose $a \in π_{F}^{- 1} π_{F} ([x]_{F_{i}})$ . Then a/F = π_F (a) ∈ π_F ([x] _{F
_i}). Hence, there exists b ∈ [x] _{F
_i} such that a/F ∈ π_F (b) = b/F and aΘ_Fb. Since aΘ_Fb and F ⊆ F_i, we get that aΘ_{F
_i}b and bΘ_{F
_i}x. By transitivity condition, aΘ_{F
_i}x i.e, a ∈ [x] _{F
_i}. So $π_{F}^{- 1} π_{F} ([x]_{F_{i}}) \subseteq [x]_{F_{i}}$ . Clearly, $[x]_{F_{i}} \subseteq π_{F}^{- 1} π_{F} ([x]_{F_{i}})$ . Therefore, $π_{F}^{- 1} π_{F} ([x]_{F_{i}}) = [x]_{F_{i}} \in T_{F}$ and π is an open map. □

Corollary 5.24. Let $(B, T_{F})$ be a topological basic algebra induced by $F$ and F be a filter of $B$ . If $F \subseteq \cap {F_{i} ∣ F_{i} \in F}$ , then the quotient basic algebra $B / F$ with quotient topology is a topological basic algebra.

Proof. By Propositions 5.22 and 5.23, the proof is clear. □

Proposition 5.25. Let $(B, T_{F})$ be a semitopological basic algebra induced by $F$ and F be a filter of $B$ . Then the following statements hold:

F is open if and only if x/F is open for any $x \in B$

F is open if and only if π_F is open and the quotient topology on $B / F$ is discrete.

If F is open, then $B / F$ is a Hausdorff topological basic algebra.

Proof. (i) Let F be an open set. It is easy to prove that for each $x \in B$ , $x / F = L_{x}^{- 1} (F) \cap R_{x}^{- 1} (F)$ . Since F is open, then by continuity L_x and R_x, x/F is open. Now, if for all $x \in B$ , x/F is open in $B$ , then 1/F = F is open in $B$ . (ii) Let F be an open set. Then by (i), for each $x \in B$ , x/F is open. It is easy to see that for each $U \in T_{F}$ , $π_{F}^{- 1} \circ π_{F} (U) = ⋃_{x \in U} x / F$ is open in $B$ . Hence π_F is open in $B / F$ and for each $x \in B$ , π_F (x/F) = {x/F} is open in $B / F$ . Conversely, let the quotient topology on $B / F$ be a discrete topology and π_F be an open map. Since {x/F} is an open in $B / F$ , $π_{F}^{- 1} (F) = F$ is an open subset of $B$ . (iii) Let F be an open set. Then by (ii), $B / F$ is discrete and it is clear that $B / F$ is a Hausdorff topological basic algebra. □

Now, we investigate the topological implication basic algebras and prove that, there exists at least one nontrivial regular and normal topological implication basic algebra of cardinality n.

Theorem 5.26. Let $F$ be an arbitrary family of (weak) filters of $IB$ which is closed under finite intersections. Then $T_{F} = {U \subseteq IB ∣ \forall x \in U, \exists F \in F, F [x] \subseteq U}$ is a topology on $IB$ with the base $β = {F [x] ∣ x \in IB, F \in F}$ and $(IB, T_{F})$ is a (left) topological implication basic algebra and $T_{F}$ is called the induced topology by $F$ .

Proof. The proof is similar to the proof of Theorem 5.9 and Corollary 5.12. □

Example 5.27. Let $IB = {a, b, c, d, e, 1}$ be the join semilattice that Hasse diagram is shown in Fig. 5 and the operation → on $IB$ is defined as shown in Table 6. Routine calculations show that $(IB, \to)$ is an implication basic algebra. Moreover, it is not difficult to check that {{1} , {b, 1} , {b, d, 1} , ${d, 1}, {a, e, 1}, {a, b, c, e, 1}, {a, d, e, 1}, {1}, B}$ is the set of filters of $IB$ . Now, by setting $F = {f_{1}, f_{2}, f_{3}}$ which f₁ = {a, e, 1}, f₂ = {a, b, c, e, 1} and f₃ = {a, d, e, 1} are filters of $IB$ , we have $f_{1} [x] = {\begin{matrix} {a, e, 1}, & x \in {a, e, 1} \\ {b, c}, & x \in {b, c} \\ {d}, & x \in {d} \end{matrix},$ $f_{2} [x] = {\begin{matrix} {a, b, c, e, 1}, & x \in {a, b, c, e, 1} \\ {d}, & x \in {d} \end{matrix}$ and $f_{3} [x] = {\begin{matrix} {a, d, e, 1}, & x \in {a, d, e, 1} \\ {b, c}, & x \in {b, c} \end{matrix}$ Then β = {{d} , {b, c} , {a, e, 1} , {a, b, c, e, 1} , {a, d, e, 1}} is a base for $T_{F}$ and $(IB, T_{F})$ is a topological implication basic algebra.

Fig. 5

Table 5

$(IB, \to)$

→	a	b	c	d	e	1
a	1	b	b	d	1	1
b	a	1	e	d	e	1
c	e	1	1	d	1	1
d	a	b	c	1	e	1
e	e	b	b	d	1	1
1	a	b	c	d	e	1

Table 6

Induced implication basic algebra

→	a	b	c	d	1	⇀	z ^∗	a	b	c	d	1
a	1	d	1	d	1	z ^∗	1	a	b	c	d	1
b	c	1	1	1	1	a	z ^∗	1	d	1	d	1
c	c	d	1	d	1	b	z ^∗	c	1	1	1	1
d	a	c	c	1	1	c	z ^∗	c	d	1	d	1
1	a	b	c	d	1	d	z ^∗	a	c	c	1	1
						1	z ^∗	a	b	c	d	1
(a) $({IB}_{5}; \to)$						(b) ( ${IB}^{*}; ⇀)$

Theorem 5.28. Let F be a proper (weak) filter of $IB$ . Then, the topological space $(IB, T_{F})$ is zero-dimensional, disconnected, locally compact, completely regular and normal and for each $x \in IB$ , F and F [x] are clopen and compact subsets of $(IB, T_{F})$ .

Proof. The proof is similar to the proof of Proposition 5.21.□

Theorem 5.29. Let $(IB, \to)$ be an implication basic algebra and $z^{*} \notin IB$ . Then there is an operation ⇀ on ${IB}^{*} = IB \cup {z^{*}}$ such that $({IB}^{*}, ⇀)$ is an implication basic algebra. Moreover, if F is a filter of $IB$ , then F^∗ = F ∪ {z^∗} is a filter of ${IB}^{*}$ .

Proof. We define ⇀ on ${IB}^{*} = IB \cup {z^{*}}$ as follows: $x ⇀ y = {\begin{matrix} x \to y, & x, y \in IB \\ y, & x = z^{*}, y \in IB \\ z^{*}, & x \in IB, y = z^{*} \\ 1, & x = z^{*}, y = z^{*} \end{matrix}$ It is not difficult to see that all conditions (IB1)-(IB3) are satisfied. Let F be a filter of $IB$ . We prove that F^∗ = F ∪ {z^∗} satisfies (F1)-(F3). (F1) Suppose that a ∈ F^∗, a ⇀ b ∈ F^∗. If a ∈ F, a ⇀ b ∈ F or a ⇀ b = z^∗, then b ∈ F or b = z^∗, respectively and so b ∈ F^∗. If a = z^∗ and a ⇀ b ∈ F, then b = z^∗ ∈ F^∗ or b = z^∗ ⇀ b ∈ F ⊆ F^∗. (F2) Suppose a ⇀ b ∈ F^∗, b ≤ a and $c \in {IB}^{*}$ . We will have following cases:

Case 1. Let a ⇀ b ∈ F, b ≤ a and $a, b \in IB$ . If $c \in IB$ , then by (F2), (b ⇀ c) ⇀ (a ⇀ c) ∈ F ⊆ F^∗ and if c = z^∗, then (b ⇀ z^∗) ⇀ (a ⇀ z^∗) = z^∗ ⇀ z^∗ = 1 ∈ F ⊆ F^∗.

Case 2. Let a ⇀ b ∈ F, b ≤ a and $a = z^{*}, b \in IB$ . Then a ⇀ b = z^∗ ⇀ b = b ∈ F. If $c \in IB$ , then by Lemma 5.4 (ii), (b ⇀ c) ⇀ (z^∗ ⇀ c) = (b ⇀ c) ⇀ c = b ∨ c ∈ F ⊆ F^∗ and if c = z^∗, (b ⇀ z^∗) ⇀ (z^∗ ⇀ z^∗) = z^∗ ⇀ 1 =1 ∈ F^∗.

Case 3. Let a ⇀ b ∈ F, b ≤ a and a = b = z^∗. If $c \in IB$ or c = z^∗, then (z^∗ ⇀ c) ⇀ (z^∗ ⇀ c) =1 ∈ F^∗.

Case 4. Let a ⇀ b = z^∗. Then b = z^∗ and $a \in IB$ . For $c \in IB$ , by (IBA₃), (z^∗ ⇀ c) ⇀ (a ⇀ c) = c ⇀ (a ⇀ c) =1 ∈ F^∗ and for c = z^∗, (z^∗ ⇀ z^∗) ⇀ (a ⇀ z^∗) = z^∗ ⇀ z^∗ = 1 ∈ F^∗. (F3) Suppose a ⇀ b ∈ F^∗, b ⇀ a ∈ F^∗ and c ∈ IB^∗. We consider the following cases:

Case 1. Let a ⇀ b ∈ F, b ⇀ a ∈ F. If $c \in IB$ , then by (F3), (c ⇀ a) ⇀ (c ⇀ b) ∈ F ⊆ F^∗. Now, if c = z^∗, then (z^∗ ⇀ a) ⇀ (z^∗ ⇀ b) = a ⇀ b ∈ F ⊆ F^∗.

Case 2. Let a ⇀ b ∈ F, b ⇀ a = z^∗. Then a = z^∗ and b ∈ F. If $c \in IB$ , then (c ⇀ z^∗) ⇀ (c ⇀ b) = z^∗ ⇀ (c ⇀ b) = c ⇀ b. Since b ∈ F, by Lemma 5.4(ii), we have c ⇀ b ∈ F ⊆ F^∗. Now, if c = z^∗, then (z^∗ ⇀ z^∗) ⇀ (z^∗ ⇀ b) =1 ⇀ b = b ∈ F ⊆ F^∗.

Case 3. Let a ⇀ b = z^∗, b ⇀ a ∈ F. In this case, proof is similar to the proof of Case 2. We observe that if a ⇀ b = b ⇀ a = z^∗, then a = b = z^∗, and so 1 = z^∗ ⇀ z^∗ = a ⇀ b = z^∗, which is a contradiction. □

Example 5.30. Let ${IB}_{5} = {a, b, c, d, 1}$ be the join-semilattice whose Hasse diagram is below (see Fig. 6). We define an operation → on ${IB}_{5}$ as shown in Table 6(a). Then $({IB}_{5}, \to)$ is an implication basic algebra in which F = {a, c, 1} is a filter. Now, by Theorem 5.29, ${IB}^{*} = {IB}_{5} \cup {z^{*}}$ (see Fig. 7) together with the operation ⇀ which is shown in Table 6(b) is an implication basic algebra of order 6 with filter F^∗ = {z^∗, a, c, 1}.

Fig. 6

${IB}_{5}$ .

Fig. 7

${IB}^{*}$ .

Theorem 5.31. For any n ≥ 3, there is a nontrivial topological implication basic algebra $(IB, T)$ of cardinality n such that $(IB, T)$ is regular and normal.

Proof. Let ${IB}_{3} = {a, b, 1}$ be the join-semilattice which Hasse diagram is shown in Fig. 8 and let the operation → be defined by Table 7(a). Then $({IB}_{3}, \to)$ is an implication basic algebra and F = {a, 1} is a filter of ${IB}_{3}$ . By Corollary 5.28, $T_{F} = {{a, 1}, {b}, {IB}_{3}, \emptyset}$ is a topology on ${IB}_{3}$ and $({IB}_{3}, T_{F})$ is a normal and regular topological implication basic algebra. Now, suppose that ${IB}^{*} = {z^{*}} \cup {IB}_{3}$ , where $z^{*} \notin {IB}_{3}$ . By Theorem 5.29, $({IB}^{*}, ⇀)$ is an implication basic algebra (see Fig. 9 and Table 7(b)) in which F^∗ = {z^∗, a, 1} is a filter. By Corollary 5.28, $T_{F^{*}} = {{z^{*}, a, 1}, {b}, {IB}^{*}, \emptyset}$ is a topology on ${IB}^{*}$ and $({IB}^{*}, T_{F^{*}})$ is a normal and regular topological implication basic algebra. By the similar way, for each n ≥ 3, we can construct a normal and regular topological implication basic algebra of order n.□

Fig. 8

${IB}_{3}$ .

Fig. 9

${IB}^{*}$ .

Table 7

Induced topological implication basic algebra of any order

→	a	b	1	⇀	z ^∗	a	b	1
a	1	b	1	z ^∗	1	a	b	1
b	a	1	1	a	z ^∗	1	b	1
1	a	b	1	b	z ^∗	a	1	1
				1	z ^∗	a	b	1
(a) $({IB}_{3}, \to)$				(b) $({IB}^{*}, ⇀)$

In the following example, we construct an implication basic algebra of order 9, by the basic algebra $B$ of order 8, as shown in Example 3.3 and by using the it’s weak filter, we make the left topological implication basic algebra.

Example 5.32. Consider $B = {0, a, b, c, d, e, f, 1}$ be the basic algebra $(B; \oplus, \sim)$ as in Example 3.3. By Theorem 5.29, we construct the implication basic algebra $(B, \to)$ of order 9 which Hasse diagram is shown in Fig. 10 and the operation → is defined as shown in Table 8. Since F = {d, 1} is a weak filter of $B$ , F^* = {d, z^*, 1} is a weak filter of $B$ . Now, let $F = {F^{*}}$ .

Fig. 10

Table 8

Table8

→	0	a	b	c	d	e	f	z ^*	1
0	1	1	1	1	1	1	1	z ^*	1
a	f	1	f	e	1	1	f	z ^*	1
b	e	e	1	f	1	e	1	z ^*	1
c	d	d	d	1	d	1	1	z ^*	1
d	c	e	f	c	1	e	f	z ^*	1
e	b	d	b	e	d	1	f	z ^*	1
f	a	a	d	f	d	e	1	z ^*	1
z ^*	0	a	b	c	d	e	f	1	1
1	0	a	b	c	d	e	f	z ^*	1

Then, $F^{*} = {\begin{matrix} {0, c}, & x \in {0, c} \\ {a, e}, & x \in {a, e} \\ {b, f}, & x \in {b, f} \\ {d, z^{*}, 1}, & x \in {d, z^{*}, 1} \end{matrix}$

Hence, β = {{0, c} , {a, e} , {b, f} , {d, z^*, 1}} is a base of $T_{F^{*}}$ and $(B, T_{F^{*}})$ is the left topological implication basic algebra.

6 Conclusions and future works

In this paper, we defined the notions of (semi) topological basic algebra and (semi) topological implication basic algebra by (semi) continuity → on (implication) basic algebras. We found the relations between (semi) continuity operations due to the existence of some various operations on basic algebras. Further, any operations * ∈ {⊕ , ⊙ , ⊖ , ø , ⇝ , ∧ , ∨} are (semi) continuous if → is (semi) continuous. In addition, there is a topology $T$ on $B$ such that ∨, ∧ are continuous, but $(B, T)$ is not a topological basic algebra.

Then, we evaluated separation axioms on (semi) topological basic algebras and proved that topological basic algebra $(B, T)$ is a T_i-space if and only if $(B, T)$ is a T_j-space, for i, j ∈ {0, 1, 2}. In addition, some conditions were provided to make $(B, T)$ to a T_i-space, for i ∈ {0, 1, 2}.

In the sequel, by using a family of ideals (filters) of basic algebras, we introduced topological basic algebras and investigated some topological properties of topological basic algebras and topological quotient basic algebras.

Currently, studying topologies of logic algebras based on MV-algebras, are mainly through (filters) ideals of corresponding algebra, which are (pre-filters) pre-ideals corresponding to basic algebras. In basic algebras, pre-ideals of $B$ can not induce a topology on $B$ such that every operation on $B$ is continuous with respect to this topology. In basic algebras, the family of weak ideals(weak filters) makes a weak congruence(not congruence) and we used them to induce left topological basic algebras $(B, T)$ , which are zero dimensional, normal, disconnected, locally compact and completely regular left topological space.

Further, by using an implication basic algebra $IB$ of cardinality n with filter F, we constructed an implication basic algebra ${IB}^{*}$ of cardinality n + 1 with filter F^* = F ∪ {z^*}, which $z^{*} \notin IB$ and proved that, there exists at least one nontrivial regular and normal topological implication basic algebra of cardinality n. In future research, the notions of prime filters and spectrum topology on implication basic algebras can be considered, along with the conditions these topological spaces can be a T_i-space, for i ∈ {0, 1, 2}.

Footnotes

Acknowledgments

The authors would like to express their sincere gratitude to the referees for their valuable suggestions and comments.

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