Space of Maximal Filters in Distributive Semilattice Author: Tolesa Dekeba Bekele Department of Mathematics, Dire Dawa University, Dire Dawa, Ethiopia Email: toleyoobii2020@gmail.com Co-author: Tesfu Reta Gari Department of Mathematics, Dire Dawa University, Dire Dawa, Ethiopia Email: tesfur2013@gmail.com Abstract: In this paper, we contribute a topological representation of maximal filters of distributive meet semilattice with 0. We proved that the Compact Hausdorο¬ space is equivalent to T1- space in the case of maximal filters of pseudo complemented distributive semilattice. Finally, we show that in distributive semilattice with 0 the compact Hausdorff space of prime filters with the hull-kernel topology is equivalent to Regular space. Keywords: Prime filters of distributive semilattice; Space of maximal filters of distributive semilattice; Boolean space of distributive semilattice; Regular space distributive semilattice. Introduction The concept of distributivity in a lattice can be considered in several corresponding ways. In semilattices, we have a diverse situation because there are already several different concepts of distributivity. In this paper, we like to discuss the class of distributive meet-semilattices. A meet-semilattice is an idempotent commutative semigroup. i.e. an algebra (π, ∧) is said to be a semilattice if for any π, π, and π ∈ π; π ∧ π = π, π ∧ π = π ∧ π, and π ∧ (π ∧ π) = (π ∧ π) ∧ π The class of distributive semilattices is a significant subclass of semilattices. Several works are studying the class of distributive semilattices such as[1][3][4][12][13][16]and [18]. The class of 0-distributive semilattices is a good extension of the class of distributive semilattices. This extension is very important for the study of pseudocomplemented semilattices For 0distributive semilattices and pseudocomplemented semilattices we refer the readers to[2][3][5][6][7][10][12][15]. Filters and Ideals play an important character in the theory of semilattices, particularly in distributive semilattice, 0-distributive semilattice, and pseudocomplemented distributive semilattice. For the concept of filters in distributive semilattice, pseudo complement distributive semilattice, and 0-distributive semilattice we refer the readers to ΠΡΠΈΠ±ΠΊΠ°! ΠΡΡΠΎΡΠ½ΠΈΠΊ ΡΡΡΠ»ΠΊΠΈ Π½Π΅ Π½Π°ΠΉΠ΄Π΅Π½.[5][9][10][12]. Let (π, π) be a topological space, then the sub-collection β¬ of π is said to be the open base for π if each member of π can be expressed as a union of members of β¬. In other words let (π, π) be a topological space, then the sub-collection β¬ of π is said to be base, if for a point π₯ belonging to an open set π then there exist π ∈ β¬ such that π₯ ∈ π ⊆ π.[16]. The main objective of this paper is to address that, the space of prime (maximal) filters make some topological space.[16] Preliminaries: In this section, we state some important notations, definitions, and basic properties of semilattice. Let π = (π, ≤) be a semilattice; According to [1] A non-empty subset πΉ of π is called a filter if; π, π ∈ πΉimplies π ∧ π ∈ πΉ, and π ∈ πΉ, π ∈ π with π ≤ π implies π ∈ πΉ. A filter πΉ of π is said to be proper if πΉ ≠ π and πΉ ≠ ∅. A proper filter πΉ of π is called a prime filter if π − πΉ is a prime ideal. Dually, non-empty subset πΌ of π is called an ideal if; for any, π, π ∈ πΌ, there exist π ∈ πΌ such that π ≤ π, π ≤ π and π ∈ πΌ, π ∈ π with π ≤ π implies π ∈ πΌ. A prime ideal πΌ is a proper ideal of π, such that π ∧ π ∈ πΌ implies either π ∈ πΌ or π ∈ πΌ. Note that π is an ideal of itself if and only if π is directed above, π is always a filter of itself and a final segment πΉ of π is a filter of S if and only if πΉ is a sub-semilattice of π. Let πΉ(π) and πΌ(π) denote the set of all filters of π and the set of all ideals of S respectively. For any π ∈ π, let (a] denote the ideal {π₯ ∈ π: π₯ ≤ π} and [a) denotes the filter {π₯ ∈ π: π ≤ π₯}. Then, π β¦ [π) is an anti-monomorphism of S into πΉ(π) and hence we can consider S as a join subsemilattice of πΉ(π). Definition: 1. A homomorphism between two semilattices π and π is a map β: π → π with the property that β(π₯ ∧ π¦) = β(π₯) ∧ β(π¦). Theorem: 2. S has the largest (smallest) element if and only if the intersection of all non-empty filters (ideals) is again non-empty. Proof: Suppose that S has the smallest element, say π. For any ideal πΌ of π, if π₯, π¦ ∈ πΌ, then π ≤ π₯ and π ≤ π¦. Therefore π ∈ πΌ, for all πΌ ∈ πΌ(π). Hence π ∈ βπΌ∈πΌ(π) πΌ . Similarly, if S has the largest element says π‘, then π‘ ∈ βπΉ∈πΉ(π) πΉ . Conversely, suppose that the intersection of all non-empty filters (ideals) is again non-empty. Then π has the Largest (smallest) element. Proposition: 3. Let π be a semilattice, then πΌ(π) is closed under finite intersections. Proof: Suppose that πΌ1 , πΌ2 , … , πΌπ ∈ πΌ(π) for some n in a positive integer. Let π₯ ∈ βππ=1 πΌπ and let π ∈ π such that π ≤ π₯. Then π₯ ∈ πΌπ for all 1 ≤ π ≤ π and hence π ∈ πΌπ for all 1 ≤ π ≤ π. This implies that π ∈ βππ=1 πΌπ for all 1 ≤ π ≤ π. Let x and π¦ ∈ βππ=1 πΌπ then π₯, π¦ ∈ πΌπ for all π. Since each πΌπ is an ideal, π₯ ∨ π¦ ∈ πΌπ , for all i. Therefore π₯ ∨ π¦ ∈ βππ=1 πΌπ . Hence βππ=1 πΌπ is an ideal. Theorem: 4. [16] The following are equivalent in any semilattice π; a. (π, ≤) is a lattice. b. πΌ(π) is closed under an arbitrary intersection. c. For any πΉ ∈ πΉ(π),β π∈πΉ(π] is an ideal of π. d. πΌ(π) is a lattice under set inclusion. Proof: (a)⇒(b) Suppose(π, ≤) is the lattice. Let π be the smallest element in S, π ∈ πΌπ for all π, where {πΌπ }π∈π₯ ⊆ πΌ(π). This implies that π ∈ βπ∈π₯ πΌπ . Let π₯, π¦ ∈ βπ∈π₯ πΌπ then π₯, π¦ ∈ πΌπ , for all π. Therefore π₯ ∨ π¦ ∈ πΌπ , for all π ∈ π₯. Hence π₯ ∨ π¦ βπ∈π₯ πΌπ . (b) ⇒ (c) Suppose that for any πΉ ∈ πΉ(π), βπ ∈ πΉ (π] is an ideal of π. Let πΌ, πππ π½ ∈ πΌ(π) be ideals of π, then βπ∈πΌ {π₯ ∈ π: π₯ ≤ π¦, for all π¦ ∈ [π) ∩ [π) is the smallest ideal of π containing π∈π½ πΌ and π½. Therefore, πΌ(π) is closed under inclusion . (d) ⇒ (a) Suppose π πππ π ∈ π. Let πΌ be the least upper bound of (π] πππ (π] in πΌ(π), then there exist π ∈ πΌ such that π ≤ π and π ≤ π and hence πΌ = (π]. It follows that c is the least upper bound of a and b in S. Hence S is a lattice. Theorem: 5. Let π be a semilattice, π has the largest element if and only if the intersection of all non-empty filters is again nonempty. Proof: Suppose that π has the largest element, say π, then for any π ∈ π, π ≤ π. Let πΉ be a filter of π. Now π₯ ∈ πΉ implies π₯ ≤ π for all filter πΉ ∈ πΉ(π). Hence π ∈ βπΉ∈πΉ(π) πΉ . Therefore βπΉ∈πΉ(π) πΉ ≠ ∅. Conversely, suppose that the intersection of all non-empty filters is again non-empty. Then π has the largest element. Lemma: 6. Let π ⊂ π. Then π is a prime filter of π if and only if π − π is a prime ideal. Proof: Suppose π is a prime filter of π, then it is non-empty and proper. To show π − π is directed above, let π, π ∉ π − π. Suppose if possible, [π) ∩ [π) ∩ π − π = ∅, [π) ∩ [π) ⊆ π. This implies [π) ⊆ π or [π) ⊆ π. Since π is prime then π ∉ π − π or π ∉ π − π. This is a contradiction. Thus ([π) ∩ [π)) ∩ π − π ≠ ∅. Hence π − π is ideal. To show that π − π is the prime ideal of π, suppose that π ∧ π ∈ π − π ⇒ π ∧ π ∉ π ⇒ π ∉ π or π ∈ π. Therefore π ∈ π − π or π ∈ π − π. Conversely, suppose that π − π is a prime ideal, then π ≠ ∅, π. Since π − π is directed above, then π is directed below. If π, π ∈ π, then π, π ∉ π − π. Since π − π is prime, π ∧ π ∉ π − π. Thus π ∧ π ∈ π, and so π is a filter of π. To show that π is a prime filter, suppose if possible π ∩ π ⊆ π and π β π, π β π ⇒ π ∩ π − π ≠ ∅ , and π ∩ π − π ≠ ∅, there exists π ∈ π ∩ π − π and π ∈ π ∩ π − π, there exist π ∈ π − π with π, π ≤ π ⇒ π ∈ π ∩ π − π. This is a contradiction. Therefore π ⊆ π ππ π ⊆ π ⇒ π is prime. Lemma: 7. A filter πΉ of π is prime if and only if πΉ is proper and for any filters πΊ and π» of π, πΊ ∩ π» ⊂ π implies πΊ ⊂ π or π» ⊂ π. Proof: Let π be a prime filter of π, then π is proper (π ≠ π, ∅). Let πΊ and π» be filters of π. Suppose that πΊ ∩ π» ⊂ π ⇒ πΊ ∩ π» β π − π. Then πΊ β π − π or π» β π − π. Therefore πΊ ⊂ π or π» ⊂ π. Distributive Semilattice In this section, we recall the properties and definitions of ideals and filters of a semilattice, especially in distributive semilattice. Definition: 8. [3]A semilattice π is said to be distributive if, for any π, π,π ∈ π such that π ∧ π ≤ c, there existπ π₯ πππ π¦ ∈ π such that π ≤ π₯, π ≤ π¦, and π₯ ∧ π¦ = π. Proposition: 9. Let I be an ideal of S and J a filter of S such that πΌ ∩ π½ ≠ ∅. Then πΌ ∩ π½ is a distributive subsemilattice of S. Proof: Let πΌ be an ideal of π and π½ a filter of π, then πΌ ∩ π½ is a sub-semilattice of π. Let π, π, π ∈ πΌ ∩ π½ such that π ∧ π ≤ π. There exist π₯ and π¦ ∈ π such that π ≤ π₯, π ≤ π¦, and π₯ ∧ π¦ = π. Since πΌ is an ideal of π, there is π§ ∈ πΌ such that π ≤ π§, π ≤ π§, and π ≤ π§. Now π = π ∧ π§ = (π₯ ∧ π§) ∧ (π¦ ∧ π§), π ≤ π₯ ∧ π§ ∈ πΌ ∩ π½ and π ≤ π¦ ∧ π§ ∈ πΌ ∩ π½. Hence πΌ ∩ π½ is distributive. Theorem: 10. [16] Let (πΏ, ∧, ∨) be a lattice. Then the following are equivalent: a) (πΏ, ∧, ∨) is a distributive lattice. b) (πΏ, ∧) is a distributive meet semilattice. c) (πΏ, ∨) is a distributive join semilattice. Proof: (a)⇒(b): Suppose πΏ is distributive. Let π, π, and π ∈ π such that π ∧ π ≤ π, then put π₯ = π ∨ π and π¦ = π ∨ π. π ≤ π₯ and π ≤ π¦. Now π = (π ∧ π) ∨ π = (π ∨ π) ∧ (π ∨ π) = π₯ ∧ π¦. Hence (πΏ, ∧) is a distributive meet semilattice. (b)⇒(c): Let πΏ is distributive meet semilattice. Let π, π, πππ π ∈ πΏ such that π ≤ πΛ π, then put π₯ = π ∧ π, π¦ = π ∧ π then π ∧ π ≤ π, π ∧ π ≤ π. Consider π = π ∧ (π ∨ π) = (π ∧ π) ∨ (π ∧ π) = π₯ ∨ π¦. Hence (πΏ,∨) is a distributive join semilattice. (c) ⇒(a): Let (πΏ,∨) be a distributive join semilattice, then for any π, π and π ∈ πΏ, if π ≤ πΛ π, there exists π₯ πππ π¦ ∈ πΏ such that π = π₯ ∨ π¦. This implies that π ≤ π and π ≤ π. Now consider π ∧ (π ∨ π) = π = (π ∧ π) ∨ (π ∧ π). Hence (πΏ, ∧ ,∨) is a distributive lattice. Lemma: 11. If π is a distributive semilattice, π is directed above. Proof: Suppose π be a distributive semilattice, then for any π, π ∈ π, π ∧ π ∈ π, and π ∧ π ≤ π.There exists π₯, π¦ ∈ π such that π ≤ π₯, π ≤ π¦, and π₯ ∧ π¦ = π. Also π = π₯ ∧ π¦ ≤ π₯. Therefore for any π, π ∈ π, there existπ π₯ ∈ π such that π ≤ π₯ and π ≤ π₯. Hence π is directed above. Note: For an ideal, πΌ of π and π and π ∈ π, let ⟨π, πΌ⟩∗ = {π₯ ∈ π: π₯ ∧ π ∈ πΌ} and ⟨π, π⟩ = {π₯ ∈ π: π₯ ∧ π = π₯ ∧ π}. If πΌ = (π], we just write⟨ π, π⟩∗ for ⟨π, (π]⟩∗ . Theorem: 12. [16] The following are equivalent in any semilattice π; a. S is distributive. b. πΉ(π) is a distributive lattice. c. For an ideal, πΌ of π and π ∈ π, ⟨π, I⟩∗ is an ideal of π. d. For any π and π ∈ π, ⟨π, b⟩∗ is an ideal of π. e. For any a and π ∈ π, ⟨π, π⟩ is an ideal of π. f. Every filter which is maximal with respect to not containing an element is prime. g. All filters πΉ of S is the intersection of all prime filters of S covering πΉ. Proof: (a)⇒(b): Suppose π be a distributive semilattice. Let πΉ, πΊ, and π» ∈ πΉ(π), then; Consider (πΉ ∧ πΊ) ∨ (πΉ ∧ π») ⊆ πΉ ∧ (πΊ ∨ π») ⇒ π₯ ∈ πΉ ∧ (πΊ ∨ π»)⇒π₯ ∈ πΉ and π₯ ∈ (πΊ ∨ π»). π₯ ∈ πΉ and π₯ ∈ πΊ, π₯ ∈ π» ⇒ π₯ ∈ πΉ and π₯ ∈ πΊ, π₯ ∈ πΉ and π₯ ∈ π» ⇒ π₯ ∈ πΉ ∧ πΊ, π₯ ∈ πΉ ∧ π». Then π₯ ∈ (πΉ ∧ πΊ) ∨ (πΉ ∧ π»). Thus πΉ ∧ (πΊ ∨ π») ⊆ (πΉ ∧ πΊ) ∨ (πΉ ∧ π»). Hence πΉ ∧ (πΊ ∨ π») = (πΉ ∧ π») ∨ (πΉ ∧ π»), implies πΉ(π) is a distributive lattice. (b)⇒ (c): Suppose that π ∈ π and πΌ be an ideal of π, then ⟨π, I⟩∗ is an initial segment of π. Let π₯, π¦ ∈ π such that π₯ ∧ π and π¦ ∧ π ∈ πΌ, there is π ∈ πΌ such that π₯ ∧ π ≤ π and π¦ ∧ π ≤ π so that π ∈ ([π₯) ∨ [π) ∩ ([π¦) ∨ [π)) = ([π₯) ∩ [π¦)) ∨ [π). There exist π§ ∈ π such that π₯ ≤ π§, π¦ ≤ π§, and π§ ∧ π ≤ π. This implies that π§ ∈ ⟨π, I⟩∗ . Therefore ⟨π, I⟩∗ is an ideal of π. (c)⇒(d): Suppose that ⟨π, I⟩∗ be an ideal of π. Let π₯ ∈ ⟨π, b⟩∗ and π ∈ π such that π ≤ π₯, then π₯ ∧ π ∈ (π] and π ∧ π ≤ π₯ ∧ π ∈ (π]. Then π ∈ ⟨π, b⟩∗. Again let π₯, π¦ ∈ ⟨π, b⟩∗, then π₯ ∧ π and π¦ ∧ π ∈ (π]. There is π ∈ (π] such that π₯ ∧ π ≤ π and π¦ ∧ π ≤ π ⇒ π ∧ π ∈ (π]. Therefore π ∈ ⟨π, b⟩∗. (d)⇒(e): Suppose that ⟨π, b⟩∗ be an ideal of π, then⟨ π, a⟩∗ is also an ideal of π. Then ⟨π π⟩ = ⟨π, π⟩∗ ∩ ⟨π, π⟩∗ . Therefore ⟨π, π⟩ is ideal for π as πΌ(π) is closed under a finite intersection. (e)⇒(f): Suppose πΉ be a filter of π, which is maximal with respect to not containing an element π of π. Let π₯, π¦ ∈ π − π, then π ∈ (πΉΛ [π₯)) ∩ (πΉ Λ [π¦)). Therefore, there exists π ∈ πΉ such that π ∧ π₯ ≤ π and π ∧ π¦ ≤ π. Hence π₯ and π¦ ∈ ⟨π, π⟩ * which is an ideal of S. Thus π − πΉ is directed above, therefore πΉ is prime. (g)⇒(b): Let πΌ, π½, and πΎ be filters of π. For any prime filter πΉ of π; (πΌ ∩ π½) ∨ (πΌ ∩ πΎ) ⊂ πΉ ⇒ (πΌ ∩ π½) ⊂ πΉ or (πΌ ∩ πΎ) ⊂ πΉ ⇒ πΌ ⊂ πΉ and π½ ∨ πΎ ⊂ πΉ ⇒ πΌ ∩ (π½ ∨ πΎ) ⊂ πΉ ⇒ (πΌ ∩ π½) ∨ (πΌ ∩ πΎ) = πΌ ∩ (π½ ∨ πΎ). Hence πΉ(π) is distributive. Definition: 13. [10] A semilattice S with 0 is said to be pseudo complemented if, for each π ∈ π, there exists π∗ ∈ π such that π ∧ π = 0 if and only if π ≤ π∗ for all π ∈ π. An element π ∈ π is said to be dense if π∗ = 0 and is said to be closed if(π∗ )∗ = π. Theorem: 14. All pseudocomplemented semilattices are 0 − distributive. Proof: Let π be a pseudocomplemented semilattice and π₯1 , π₯2 , … , π₯π , πππ π ∈ π such that π₯ ∧ π1 = 0, π₯ ∧ π2 = 0, … , π₯ ∧ ππ = 0 and π₯1 ∨ π₯2 ∨ … ∨ π₯π exists. Since π₯π ≤ (π)∗ for all π′π , we get π₯1 ∨ π₯2 ∨ … ∨ π₯π ≤ (π)∗. Hence π ∧ (π₯1 ∨ π₯2 ∨ … ∨ π₯π ) = 0. Thus π is 0 − distributive. Ideals and Filters for Distributive Semilattice Theorem: 15. Let πΌ be an ideal of S and π½ a filter of S such that πΌ ∩ π½ ≠ ∅. then, πΌ ∩ π½ is a distributive sub-semilattice of π. Proof: Suppose πΌ be an ideal of π and π½ a filter of π such that πΌ ∩ π½ ≠ ∅. Let π, π ∈ πΌ ∩ π½, then π ∧ π ∈ πΌ. Since πΌ is the initial segment, there exist π ∈ π½ such that π ≤ π and π ≤ π. This implies π ≤ π ∧ π and π ∧ π ∈ π½ since π½ is the final segment. Therefore πΌ ∩ π½ is a subsemilattice of π. Let π, π, and π ∈ πΌ ∩ π½ such that π ∧ π ≤ π. There is π₯ and π¦ ∈ π such that π ≤ π₯, π ≤ π¦, and π₯ ∧ π¦ = π. Since πΌ is an ideal of π, there iπ π§ ∈ πΌ such that π ≤ π§, π ≤ π§, and π ≤ π§. now π = π ∧ π§ = (π₯ ∧ π§) ∧ (π¦ ∧ π§), π ≤ π₯ ∧ π§ ∈ πΌ ∩ π½ πnd π ≤ π¦ ∧ π§ ∈ πΌ ∩ π½. Therefore πΌ ∩ π½ is distributive. Corollary: 16. Every non-empty filter of π is a distributive sub-semilattice of π. Proof: Suppose that π½ is a nonempty filter of π. For any π, π ∈ π½, there exist π ∈ π½ such that π ≤ π and π ≤ π ⇒ π ≤ π ∧ π ⇒ π ∧ π ∈ π½, since π½ is the final segment. Then, π½ is a sub-semilattice of S. Let π, π, and π ∈ π½ such that π ∧ π ≤ π, there exists π₯, π¦ ∈ π such that π ≤ π₯, π ≤ π¦, and π₯ ∧ π¦ = π. Now as π, π, and π ∈ π½, then π₯, π¦ ∈ π½. Hence π₯ ∧ π¦ = π ∈ π½. Therefore π½ is distributive. Theorem: 17. Let πΌ be an ideal of π and πΏ is a sub-semilattice of π such that I ∩ πΏ = ∅, then, there exists a prime ideal P of S such that πΌ ⊂ π and π ∩ πΏ = ∅. Proof: Suppose that π = {π₯ ∈ π: π ≤ π₯ for some π ∈ πΏ}. Then π is the filter of π, π ∩ πΌ = ∅. Zorn's lemma, there exists a filter πΉ of π which is maximal among all filters containing πΏ and disjoint from πΌ. Now we have πΌ ⊂ π − πΉ and (π − πΉ) ∩ πΏ = ∅. Let us prove that πΉ is a prime filter. Let π₯, π¦ ∈ π − πΉ, since πΉ is maximal, there exists π ∈ πΌ ∩ (πΉΛ [π₯)) and π ∈ πΌ ∩ (πΉ ∨ [π¦)). There exist π ∈ πΌ such that π ≤ π and π ≤ π. This implies that π ∈ πΌ ∩ (πΉΛ [π₯)) ∩ (πΉ ∨ [π₯)) = πΌ ∩ (πΉ ∨ ([π₯) ∩ [π¦) ⇒ [π₯) ∩ [π¦) β πΉ. Thus πΉ is a prime filter of π. Corollary: 18. Let πΏ be sub-semilattice of π and π a prime ideal (filter) of πΏ. Then, there exists a prime ideal (filter) π of S such that π ∩ πΏ = π. Proof: Let π be a prime ideal of πΏ and put π1 = {π₯ ∈ π: π₯ ≤ π πππ π πππ π ∈ π}. Then π1 is an ideal of π and π − πΏ is a sub-semilattice of π and π1 ∩ (πΏ − π) = ∅. Then there exists a prime ideal π of S such that π1 ⊂ π and π ∩ (πΏ − π) = ∅. Hence π ∩ πΏ = π. Similarly, suppose π be a prime filter of πΏ and π2 = {π₯ ∈ π: π ≤ π₯ πππ π πππ π ∈ π}. Then π2 is a filter of π and π − πΏ is sub-semilattice of π and π2 ∩ (πΏ − π) = ∅. There exists a prime filter π of π such that π2 ⊂ π and π ∩ (πΏ − π) = ∅. Hence π ∩ πΏ = π. Corollary: 19. Let πΌ be an ideal of π, then the intersection of all prime ideals of π containing πΌ is again an ideal of S and is equal to πΌ. Proof: Let {ππ }π∈πΌ be the class of prime ideals containing I, then πΌ ⊆ ππ , ∀π ∈ πΌ. this implies πΌ ⊆∩ ππ, ∀π ∈ πΌ. Also if π ∈ π and π ∉ πΌ, then πΌ ∩ [π) = ∅. There exists a prime ideal ππ of π such that πΌ ⊂ ππ and ππ ∩ [π) = ∅ ⇒ π ∉ ππ . Therefore πΌ =∩ ππ , ∀π ∈ πΌ. Theorem: 20. Every maximal ideal (filter) of π is prime. Proof: Let π be distributive semilattice and π is a maximal filter of π. Then π − π is minimal ideal, there exists a prime ideal πΌ such that π − π ⊆ πΌ, then π ∩ πΌ = ∅. Let π₯, π¦ ∉ π, then π₯, π¦ ∈ π − π. There existπ π ∈ πΌ ∩ (π ∨ [π₯)) and π ∈ πΌ ∩ (π ∨ [π¦)). This implies there exists π ∈ πΌ such that π ≤ π and π ≤ π since πΌ directed above. Hence π ∈ πΌ ∩ (π ∨ [π₯)) ∩ (π ∨ [π¦)) ) = πΌ ∩ (π ∨ ([π₯) ∩ [π¦)) ⇒[π₯) ∩ [π¦) β π. Therefore π is a prime filter of π. Corollary: 21. Let π be distributive semilattice, π has the smallest (greatest) element if and only if the intersection of all prime ideals (filters) of π is non-empty. Proof: Suppose π has the largest element, ∅ ≠ βπΉ∈πΉ(π) πΉ ⊆ βπ∈πΉ(π) π, where πΉ any nonempty filters of π and π the prime filters of π. Therefore βπ∈πΉ(π) π ≠ ∅. Similarly, if π has a smallest element, ∅ ≠ βπΌ∈πΌ(π) πΌ ⊆ βπ∈πΌ(π) π, were πΌ any nonempty ideals of π and π the prime ideals of π. Therefore βπ∈πΉ(π) π ≠ ∅. Conversely, suppose that the intersection of all non-empty filters (ideals) is again nonempty, S has the largest (smallest) element. Definition: 22. Let π be a distributive semilattice with the smallest element 0 and πΏ is a subsemilattice of S, and then define 0(πΏ) = {π₯ ∈ π: π₯ ∧ π¦ = 0 for some π¦ ∈ πΏ}. Then 0(πΏ) is called the annihilator of πΏ. The annihilator of {π} is denoted by 0(π) = {π₯ ∈ π: π₯ ∧ π = 0}. An ideal πΌ of π is called annihilator ideal if πΌ = 0(πΏ), for some sub-semilattice πΏ ππ π. Proposition: 23. Suppose S has the smallest element 0 and πΉ is a proper filter of π, then πΉ is a maximal filter if and only if π − πΉ is a minimal prime ideal. Proof: Let πΉ be a maximal filter of S. Since π is distributive, πΉ is the prime filter. Thus π − πΉ is a prime ideal. To prove the minimalism of π − πΉ, assume π − πΉ is not minimal ideal. Then there exists a prime ideal πΌ such that πΌ ⊆ π − πΉ which implies πΉ ⊆ π − πΌ which contradicts the maximalist of πΉ. Hence π − πΉ is a minimal prime ideal. Conversely, suppose that π − πΉ is a minimal prime ideal, then 0 ∈ π − πΉ ⊆ 0(πΉ). Let π₯ ∈ 0(πΉ), then π₯ ∧ π¦ = 0, for some π¦ ∈ πΉ. Since π − πΉ is prime and π¦ ∈ πΉ, then π₯ ∈ π − πΉ. Therefore 0(πΉ) ⊆ π − πΉ and hence 0(πΉ) = π − πΉ. Since 0(πΉ) = π − πΉ, then π − πΉ is a prime ideal, and so πΉ is a prime filter. Suppose if possible, πΉ is not maximal filter, then πΉ ⊂ π, for some maximal filter π ππ π. then π − π ⊂ π − πΉ = 0(πΉ), which contradicts the minimalist of π − πΉ. Therefore πΉ is a maximal filter. Corollary: 24. If π is a distributive semilattice with the smallest element 0, then for any π ∈ π, (π)∗ = ⟨π, 0⟩∗ = {π₯ ∈ π: π ∧ π₯ = 0} is an ideal of π. Proof: Suppose π is distributive semilattice with 0. For π ∈ π, (π)∗ = {π₯ ∈ π: π ∧ π₯ = 0}. Let π₯ ∈ (π)∗ , π ∈ π such that π ≤ π₯,then π = π₯ ∧ π. Now π ∧ π = π ∧ (π₯ ∧ π) = (π ∧ π₯) ∧ π = 0 ∧ π = 0. Therefore π ∈ (π)∗ . Let π₯, π¦ ∈ (π)∗ , ∧ π₯ = 0 and π ∧ π¦ = 0. Consider π ∧ (π₯ ∨ π¦) = (π ∧ π₯) ∨ ( π ∧ π¦) = 0 ∨ 0 = 0. Therefore π₯ ∧ π¦ ∈ (π)∗ . Hence (π)∗ is an ideal of S. Space of Maximal Filters in Distributive Semilattice A topological space (π, π) is said to be compact if every open set cover for π has a finite subcover. Let π be distributive semilattice and π the set of all prime filters of π, for any π ∈ π, let ππ = {πΉ ∈ π: π ∈ πΉ}. Then for any π, π ∈ π, ππ ∩ ππ = ππΛπ and {ππ : π ∈ π} forms a base for some topology π· on π, which is called the hull-kernel topology on π. Theorem: 25. [2] Let π be distributive semilattice, then π is isomorphic to meet subsemilattices {ππ : π ∈ π} of the lattice of all open subsets of (π, π). Proof: Let π be the class of subsemilattices of {ππ : π ∈ π}. Define π: π → π by π(π) = ππ , for any π ∈ π. Let π, π ∈ π, and π ∈ π, π ∉ π(π ∧ π) ⇔ π ∧ π ∉ π ⇔ π ∉ π or π ∉ π ⇔ π ∉ π(π ) or π ∉ π( π)⇔π ∉ π(π) ∩ π( π). Therefore π(π ∧ π) = π(π) ∩ π( π). Hence π is a homomorphism of a lattice. Let π ≠ π, then there exists a prime filter π such that π ∈ π and from π ∉ π and vice versa. Then ππ ≠ ππ . Hence π is injective. And since every element of π has the form ππ , for any π ∈ π, then π is surjective. Therefore π is an isomorphism. Deο¬nition: 26. A topological space π is said to be Hausdorο¬ space if for any points π₯, π¦ ∈ π such that π₯ ≠ π¦ there exist open sets π, π ⊆ π such that π₯ ∈ π, π¦ ∈ π, and π ∩ π = ∅. Theorem: 27. Let S has the smallest element 0 and X be the set of all maximal filters of S. Then X is Hausdorff and a totally disconnected subspace of (π, π·). Proof: Let π, π ∈ π be maximal filters such that π ≠ π. There existπ π ∈ π and π ∉ π. This implies π ∈ ππ and π ∉ ππ , then π ∈ ππ and π ∈ π − ππ and ππ ∩ π − ππ = ∅. Therefore π is Hausdorff. Let π ≠ π ∈ π, then we can suppose that π β° π. There existπ π ∈ π such that π ∉ π. Therefore, ππ is clopen and π ∈ ππ and π ∉ ππ . Therefore, π is totally disconnected. Theorem: 28. Let S be a pseudo-complemented distributive semilattice. So space π for all maximal filters of S with the hull-kernel topology is a Boolean space (i.e. a compact Hausdorff and a totally disconnected space). Proof: Suppose that π is the space of all maximal filters of π with the hull-kernel topology. Let π, π ∈ π such that π ≠ π. There exists π ∈ π and π ∉ π ⇒ π ∈ ππ and π ∉ ππ . This implies π ∈ ππ and π ∈ π − ππ and ππ ∩ π − ππ = ∅. Therefore π is Hausdorff. Again let π ≠ π ∈ π, we can suppose that π β° π. There existπ π ∈ π such that π ∉ π. Hence ππ is clopen and π ∈ ππ and π ∉ ππ . Therefore π is totally disconnected. Also let {ππ } π ∈ πΏ be a basic open cover for π, for some πΏ ⊆ π, then π ⊆ βπ∈πΏ ππ . Let πΉ = [πΏ) = {π₯ ∈ π: βππ ≤ π₯, where ππ ∈ πΏ}. Suppose if possible, πΉ is a proper filter of π, then there exists a maximal filter π of π such that πΉ ⊆ π. Then πΏ ⊆ π as πΏ ⊆ πΉ. Therefore π ∉ π which is a contradiction and hence πΉ = π. In particular, 0 ∈ πΉ = [πΏ) ⇒0 = βππ=1 ππ , for some ππ ∈ πΏ. If π is a maximal filter, then ππ ∉ π for some π. Therefore π = βππ=1 πππ , and hence {ππ1 , ππ2 , … , πππ } is a finite subcover of {ππ }π ∈ πΏ. Hence ππ is compact. This implies π is compact. Therefore π Boolean space. Theorem: 29. Let π be a distributive semilattice with 0 and π be the space of prime filters with the hull-kernel topology. Then the following are equivalent: a. π is Hausdorff space, b. π is a T1-space, c. Every prime filter of S is maximal, d. πΉ is the only prime filter of π such that 0(πΉ) ∩ πΉ = ∅. Proof: (a) ⇒(b): Suppose that π be Hausdorff space; for any π ≠ π ∈ π there exists prime filters π and π in π such that π ∈ ππ , and π ∈ ππ , with ππ ∩ ππ = ∅. Therefore π ∈ ππ − ππ , π ∈ ππ − ππ . Hence π is a T1-space. (b)⇒(c): Suppose that π be a prime filter of π.Let π ∈ π − ππ , and π ∈ π − ππ , then π ∈ π and π ∉ π. This implies π ⊂ [π ∪ {π}) = {π₯ ∈ π: β(ππ ∪ π) ≤ π₯, where1 ≤ π ≤ π, for some, ππ ∈ π} = {π₯ ∈ π: (βππ ) ∪ π ≤ π₯ for some ππ ∈ π}= {π₯ ∈ π: βππ ≤ π₯ and π ≤ π₯ for some ππ ∈ π} = π. Hence π is a maximal filter. (c)⇒(d): Suppose πΉ be a prime filter of π, then πΉ is the maximal filter. Hence π − πΉ = 0(πΉ). Therefore 0(πΉ) ∩ πΉ = ∅. (d)⇒ (a): Suppose πΉ be the only prime filter of π such that 0(πΉ) ∩ πΉ = ∅. Let π ≠ π ∈ π such that π ∈ πΉ and π ∉ πΉ, then πΉ ∈ ππ and πΉ ∉ ππ . Therefore ππ ∩ ππ = ∅. Hence π is Hausdorff. Deο¬nition: 30. A topological space π is said to be a Regular space if for any points π₯ ∈ π and for every closed set π ⊆ π such that π₯ ∉ π, there exist open sets π, π ⊆ π such that π₯ ∈ π, π ⊆ π, and π ∩ π = ∅. Theorem: 31. Let π be a pseudo-complemented distributive semilattice and π be the space of prime filters with the hull-kernel topology. π is a compact hausdorff space if and only if π is a regular space. Proof: Supposed that π be a compact Hausdorff space and let ππ ⊂ π be a closed subspace. Let π be a prime filter of π such that π ∈ π − ππ . We need to contract open sets π and π ⊆ π with ππ ⊆ π and π ∈ π such that π ∩ π = ∅. Since π is Hausdorff space, for each prime filter π ∈ ππ there exists a neighborhood πππ of π and πππ of ππ , where ππ ∈ π and ππ ∈ π such that π ∈ πππ and π ∈ πππ and πππ ∩ πππ = ∅. Let ππ = {πππ : ππ ∈ π}. Then βππ ∈π πππ is an open cover of ππ , and since π is compact, ππ is a compact subspace of π. Therefore there exist finitely many elements π1 , π2 , … , ππ ∈ π such that ππ1 , ππ2 , . . . , πππ form subcover ππ . Then ππ ⊆ βππ=1 πππ . Let and π = βππ=1 πππ ,and π = βππ=1 πππ . Then π and π are open sets and ππ ⊆ π and π ∈ π, such that π ∩ π = ∅. To show that π and π are really disjointed, assume that π ∈ π ∩ π ≠ ∅. Then π ∈ πππ , for each π, and π ∈ πππ , for some π. That is π ∈ π ∩ π ⇒ π ∈ (βππ=1 πππ ) ∩ (βππ=1 πππ ) ⇒ π ∈ πππ for all π and π ∈ πππ for some π. Then π ∈ πππ ∩ πππ for some π. This is in contradiction. Therefore π ∩ π = ∅. Hence π is a regular space. Conversely, suppose π is a regular space. Let π, π ∈ π such that π ≠ π. Then without loss of generality, let π β° π, then there exists a prime filter π containing π ∈ π but not π. Then there exists a clopen set ππ such that π ∈ ππ and π − ππ is a closed set. Let π ∈ π − ππ , since π is a regular space, there exist an open sets π, πππ π ⊆ π such that ∈ π ∈ ππ ⊆ π, π ∈ π − ππ ⊆ π such that π ∩ π = ∅. Therefore π is Hausdorff space. Conclusion Generally, when we consider the class of maximal (prime) filters π of distributive semilattice with 0 and pseudo complemented semilattice; π is a T1-space ⇔ π is compact hausdorff space ⇔Regular space. Acknowledgment : The correspondent author is very thankful to the anonymous referee for the comments and suggestions on this paper. Especially, his wife Ms. Tigist Tamirat for her support and commitment during doing this paper. References [1] Balbes, Raymond. "A Representation Theory for Prime and Implicative Semilattices." Transactions of the American Mathematical Society 136 (1969): 261-67. Accessed December 3, 2020. doi:10.2307/1994713. [2] Blyth, T. S.: Ideals and filters of pseudo-complemented semilattices. Proc. Edinb. Math. Soc., II. Ser. 23 (1980), 301-316 [3] Celani, Sergio Arturo, and Luciano Javier González. " Notes on mildly distributive semilattices", Mathematica Slovaca 67, 5 (2017): 1073-1084, doi: https://doi.org/10.1515/ms-2017-0033 [4] Celani, Sergio Arturo, and Kiyomitsu Horiuchi. "topological representation of distributive semilattices." scientiae Mathematicae japonicae 58, no. 1 (2003): 55-65. [5] Cornish, W. (1973). Congruences on distributive pseudocomplemented lattices. Bulletin of the Australian Mathematical Society, 8(2), 161-179. doi:10.1017/S0004972700042404 [6] Grätzer, G.: Lattice Theory: First Concepts and Distributive Lattices. Freeman (1971). [7] Kiss, S. A. Semilattices and a ternary operation in modular lattices. Bull. Amer. Math. Soc. 54 (1948), no. 12, 1176—1179 [8] MacLane, S. (1938). M. H. Stone. Topological representations of distributive lattices and Brouwerian logics, vol. 67 (1937–1938), pp. 1–25. Journal of Symbolic Logic, 3(2), 90-91. doi:10.2307/2267630 [9] Murty, P.V.R., Rao, V.V.R. Characterization of certain classes of pseudo complemented semi-lattices. Algebra Universalist 4, 289–300 (1974). [10] O. Frink, Pseudo complements in semilattices, Duke Math. J. Vol.29 (1962), 505–514. [11] P. Balasubramani and P. V. Venkatanarasimhan, Characterizations of the 0-Distributive Lattices, Indian J. Pure Appl. Math., 32(3) (2001) 315-324. [12] Pawar, Y., & Thakare, A. (1978). O-Distributive Semilattices. Canadian Mathematical Bulletin, 21(4), 469-475. doi:10.4153/CMB-1978-080-6 [13] Rhodes Modular and distributive semilattice. Trans. Amer. Math. Society. Vol. 201 (1975), 31-41 [14] Stone, M.H.: The theory of representations for Boolean algebras. Trans. Amer. Math. Soc. 40, 37–111 (1936) [15] Swamy, U. Maddana Representation of Universal Algebra by Sheaves. Proc. Amer. Math. Soc. 45 (1974), 55–58. [16] Swamy U.Maddana. Distributive Semilattices, Mathematics seminar notes Kobe University, 1979, Vol 7, 211-233. [17] Talukder, M. R. and Noor, A. S. A., Modular ideals of a join semilattice directed below, Southeast Asian Bull. of Math. 23, 18- 37 (1998). [18] W.H. Cornish, Characterization of Distributive and Modular Semilattices, Math. Japonica, 22 (1977), 159–174.