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 Hausdorff 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 (𝑆, ∧) is said to be
a semilattice if for any 𝑎, 𝑏, and 𝑐 ∈ 𝑆;
𝑎 ∧ 𝑎 = 𝑎, 𝑎 ∧ 𝑏 = 𝑏 ∧ 𝑎, and 𝑎 ∧ (𝑏 ∧ 𝑐) = (𝑎 ∧ 𝑏) ∧ 𝑐
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 Ошибка!
Источник ссылки не найден.[5][9][10][12].
Let (𝑋, 𝒟) be a topological space, then the sub-collection ℬ of 𝒟 is said to be the open base
for 𝒟 if each member of 𝒟 can be expressed as a union of members of ℬ. In other words let
(𝑋, 𝒟) be a topological space, then the sub-collection ℬ of 𝒟 is said to be base, if for a
point 𝑥 belonging to an open set 𝔘 then there exist 𝔅 ∈ ℬ such that 𝑥 ∈ 𝔅 ⊆ 𝔘.[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 𝑆 = (𝑆, ≤) be a semilattice; According to [1] A non-empty subset 𝐹 of 𝑆 is
called a filter if; 𝑎, 𝑏 ∈ 𝐹implies 𝑎 ∧ 𝑏 ∈ 𝐹, and 𝑎 ∈ 𝐹, 𝑏 ∈ 𝑆 with 𝑎 ≤ 𝑏 implies 𝑏 ∈ 𝐹. A
filter 𝐹 of 𝑆 is said to be proper if 𝐹 ≠ 𝑆 and 𝐹 ≠ ∅. A proper filter 𝐹 of 𝑆 is called a prime
filter if 𝑆 − 𝐹 is a prime ideal. Dually, non-empty subset 𝐼 of 𝑆 is called an ideal if; for
any, 𝑎, 𝑏 ∈ 𝐼, there exist 𝑐 ∈ 𝐼 such that 𝑎 ≤ 𝑐, 𝑏 ≤ 𝑐 and 𝑎 ∈ 𝐼, 𝑏 ∈ 𝑆 with 𝑏 ≤ 𝑎 implies 𝑏 ∈
𝐼. A prime ideal 𝐼 is a proper ideal of 𝑆, such that 𝑎 ∧ 𝑏 ∈ 𝐼 implies either 𝑎 ∈ 𝐼 or 𝑏 ∈ 𝐼.
Note that 𝑆 is an ideal of itself if and only if 𝑆 is directed above, 𝑆 is always a filter of
itself and a final segment 𝐹 of 𝑆 is a filter of S if and only if 𝐹 is a sub-semilattice of 𝑆. Let
𝐹(𝑆) and 𝐼(𝑆) denote the set of all filters of 𝑆 and the set of all ideals of S respectively. For
any 𝑎 ∈ 𝑆, let (a] denote the ideal {𝑥 ∈ 𝑆: 𝑥 ≤ 𝑎} and [a) denotes the filter {𝑥 ∈ 𝑆: 𝑎 ≤ 𝑥}.
Then, 𝑎 ↦ [𝑎) is an anti-monomorphism of S into 𝐹(𝑆) and hence we can consider S as a join
subsemilattice of 𝐹(𝑆).
Definition: 1. A homomorphism between two semilattices 𝑆 and 𝑇 is a map ℎ: 𝑆 → 𝑇 with the
property that ℎ(𝑥 ∧ 𝑦) = ℎ(𝑥) ∧ ℎ(𝑦).
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 𝑚. For any ideal 𝐼 of 𝑆, if 𝑥, 𝑦 ∈ 𝐼, then
𝑚 ≤ 𝑥 and 𝑚 ≤ 𝑦. Therefore 𝑚 ∈ 𝐼, for all 𝐼 ∈ 𝐼(𝑆). Hence 𝑚 ∈ ⋂𝐼∈𝐼(𝑆) 𝐼 .
Similarly, if S has the largest element says 𝑡, then 𝑡 ∈ ⋂𝐹∈𝐹(𝑆) 𝐹 . Conversely, suppose that the
intersection of all non-empty filters (ideals) is again non-empty. Then 𝑆 has the Largest
(smallest) element.
Proposition: 3. Let 𝑆 be a semilattice, then 𝐼(𝑆) is closed under finite intersections.
Proof: Suppose that 𝐼1 , 𝐼2 , … , 𝐼𝑛 ∈ 𝐼(𝑆) for some n in a positive integer. Let 𝑥 ∈ ⋂𝑛𝑖=1 𝐼𝑖 and let
𝑎 ∈ 𝑆 such that 𝑎 ≤ 𝑥. Then 𝑥 ∈ 𝐼𝑖 for all 1 ≤ 𝑖 ≤ 𝑛 and hence 𝑎 ∈ 𝐼𝑖 for all 1 ≤ 𝑖 ≤ 𝑛. This
implies that 𝑎 ∈ ⋂𝑛𝑖=1 𝐼𝑖 for all 1 ≤ 𝑖 ≤ 𝑛. Let x and 𝑦 ∈ ⋂𝑛𝑖=1 𝐼𝑖 then 𝑥, 𝑦 ∈ 𝐼𝑖 for all 𝑖. Since
each 𝐼𝑖 is an ideal, 𝑥 ∨ 𝑦 ∈ 𝐼𝑖 , for all i. Therefore 𝑥 ∨ 𝑦 ∈ ⋂𝑛𝑖=1 𝐼𝑖 . Hence ⋂𝑛𝑖=1 𝐼𝑖 is an ideal.
Theorem: 4. [16] The following are equivalent in any semilattice 𝑆;
a. (𝑆, ≤) is a lattice.
b. 𝐼(𝑆) is closed under an arbitrary intersection.
c. For any 𝐹 ∈ 𝐹(𝑆),⋂ 𝑎∈𝐹(𝑎] is an ideal of 𝑆.
d. 𝐼(𝑆) is a lattice under set inclusion.
Proof: (a)⇒(b) Suppose(𝑆, ≤) is the lattice. Let 𝑎 be the smallest element in S, 𝑎 ∈ 𝐼𝑖 for all 𝑖,
where {𝐼𝑖 }𝑖∈𝛥 ⊆ 𝐼(𝑆). This implies that 𝑎 ∈ ⋂𝑖∈𝛥 𝐼𝑖 . Let 𝑥, 𝑦 ∈ ⋂𝑖∈𝛥 𝐼𝑖 then 𝑥, 𝑦 ∈ 𝐼𝑖 , for all 𝑖.
Therefore 𝑥 ∨ 𝑦 ∈ 𝐼𝑖 , for all 𝑖 ∈ 𝛥. Hence 𝑥 ∨ 𝑦 ⋂𝑖∈𝛥 𝐼𝑖 .
(b) ⇒ (c) Suppose that for any 𝐹 ∈ 𝐹(𝑆), ⋂𝑎 ∈ 𝐹 (𝑎] is an ideal of 𝑆. Let 𝐼, 𝑎𝑛𝑑 𝐽 ∈ 𝐼(𝑆) be
ideals of 𝑆, then ⋃𝑎∈𝐼 {𝑥 ∈ 𝑆: 𝑥 ≤ 𝑦, for all 𝑦 ∈ [𝑎) ∩ [𝑏) is the smallest ideal of 𝑆 containing
𝑏∈𝐽
𝐼 and 𝐽. Therefore, 𝐼(𝑆) is closed under inclusion .
(d) ⇒ (a) Suppose 𝑎 𝑎𝑛𝑑 𝑏 ∈ 𝑆. Let 𝐼 be the least upper bound of (𝑎] 𝑎𝑛𝑑 (𝑏] in 𝐼(𝑆), then
there exist 𝑐 ∈ 𝐼 such that 𝑎 ≤ 𝑐 and 𝑏 ≤ 𝑐 and hence 𝐼 = (𝑐]. It follows that c is the least
upper bound of a and b in S. Hence S is a lattice.
Theorem: 5. Let 𝑆 be a semilattice, 𝑆 has the largest element if and only if the intersection of
all non-empty filters is again nonempty.
Proof: Suppose that 𝑆 has the largest element, say 𝑙, then for any 𝑎 ∈ 𝑆, 𝑎 ≤ 𝑙. Let 𝐹 be a filter
of 𝑆. Now 𝑥 ∈ 𝐹 implies 𝑥 ≤ 𝑙 for all filter 𝐹 ∈ 𝐹(𝑆). Hence 𝑙 ∈ ⋂𝐹∈𝐹(𝑆) 𝐹 .
Therefore ⋂𝐹∈𝐹(𝑆) 𝐹 ≠ ∅.
Conversely, suppose that the intersection of all non-empty filters is again non-empty. Then
𝑆 has the largest element.
Lemma: 6. Let 𝑃 ⊂ 𝑆. Then 𝑃 is a prime filter of 𝑆 if and only if 𝑆 − 𝑃 is a prime ideal.
Proof: Suppose 𝑃 is a prime filter of 𝑆, then it is non-empty and proper. To show 𝑆 − 𝑃 is
directed above, let 𝑎, 𝑏 ∉ 𝑆 − 𝑃. Suppose if possible, [𝑎) ∩ [𝑏) ∩ 𝑆 − 𝑃 = ∅, [𝑎) ∩ [𝑏) ⊆ 𝑃.
This implies [𝑎) ⊆ 𝑃 or [𝑏) ⊆ 𝑃. Since 𝑃 is prime then 𝑎 ∉ 𝑆 − 𝑃 or 𝑏 ∉ 𝑆 − 𝑃. This is a
contradiction. Thus ([𝑎) ∩ [𝑏)) ∩ 𝑆 − 𝑃 ≠ ∅. Hence 𝑆 − 𝑃 is ideal. To show that 𝑆 − 𝑃 is the
prime ideal of 𝑆, suppose that 𝑎 ∧ 𝑏 ∈ 𝑆 − 𝑃 ⇒ 𝑎 ∧ 𝑏 ∉ 𝑃 ⇒ 𝑎 ∉ 𝑃 or 𝑏 ∈ 𝑃. Therefore 𝑎 ∈
𝑆 − 𝑃 or 𝑏 ∈ 𝑆 − 𝑃.
Conversely, suppose that 𝑆 − 𝑃 is a prime ideal, then 𝑃 ≠ ∅, 𝑆. Since 𝑆 − 𝑃 is directed above,
then 𝑃 is directed below. If 𝑎, 𝑏 ∈ 𝑃, then 𝑎, 𝑏 ∉ 𝑆 − 𝑃. Since 𝑆 − 𝑃 is prime, 𝑎 ∧ 𝑏 ∉ 𝑆 − 𝑃.
Thus 𝑎 ∧ 𝑏 ∈ 𝑃, and so 𝑃 is a filter of 𝑆. To show that 𝑃 is a prime filter, suppose if
possible 𝑄 ∩ 𝑅 ⊆ 𝑃 and 𝑄 ⊈ 𝑃, 𝑅 ⊈ 𝑃 ⇒ 𝑄 ∩ 𝑆 − 𝑃 ≠ ∅ ,
and 𝑅 ∩ 𝑆 − 𝑃 ≠ ∅, there
exists 𝑎 ∈ 𝑄 ∩ 𝑆 − 𝑃 and 𝑏 ∈ 𝑅 ∩ 𝑆 − 𝑃, there exist 𝑐 ∈ 𝑆 − 𝑃 with 𝑎, 𝑏 ≤ 𝑐 ⇒ 𝑐 ∈ 𝑃 ∩ 𝑆 −
𝑃. This is a contradiction. Therefore 𝑄 ⊆ 𝑃 𝑜𝑟 𝑅 ⊆ 𝑃 ⇒ 𝑃 is prime.
Lemma: 7. A filter 𝐹 of 𝑆 is prime if and only if 𝐹 is proper and for any filters 𝐺 and 𝐻 of 𝑆,
𝐺 ∩ 𝐻 ⊂ 𝑃 implies 𝐺 ⊂ 𝑃 or 𝐻 ⊂ 𝑃.
Proof: Let 𝑃 be a prime filter of 𝑆, then 𝑃 is proper (𝑃 ≠ 𝑆, ∅). Let 𝐺 and 𝐻 be filters of 𝑆.
Suppose that 𝐺 ∩ 𝐻 ⊂ 𝑃 ⇒ 𝐺 ∩ 𝐻 ⊄ 𝑆 − 𝑃. Then 𝐺 ⊄ 𝑆 − 𝑃 or 𝐻 ⊄ 𝑆 − 𝑃. Therefore 𝐺 ⊂
𝑃 or 𝐻 ⊂ 𝑃.
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 𝑆 is said to be distributive if, for any 𝑎, 𝑏,𝑐 ∈ 𝑆 such that 𝑎 ∧
𝑏 ≤ c, there exist𝑠 𝑥 𝑎𝑛𝑑 𝑦 ∈ 𝑆 such that 𝑎 ≤ 𝑥, 𝑏 ≤ 𝑦, and 𝑥 ∧ 𝑦 = 𝑐.
Proposition: 9. Let I be an ideal of S and J a filter of S such that 𝐼 ∩ 𝐽 ≠ ∅. Then 𝐼 ∩ 𝐽 is a
distributive subsemilattice of S.
Proof: Let 𝐼 be an ideal of 𝑆 and 𝐽 a filter of 𝑆, then 𝐼 ∩ 𝐽 is a sub-semilattice of 𝑆. Let 𝑎, 𝑏, 𝑐 ∈
𝐼 ∩ 𝐽 such that 𝑎 ∧ 𝑏 ≤ 𝑐. There exist 𝑥 and 𝑦 ∈ 𝑆 such that 𝑎 ≤ 𝑥, 𝑏 ≤ 𝑦, and 𝑥 ∧ 𝑦 = 𝑐.
Since 𝐼 is an ideal of 𝑆, there is 𝑧 ∈ 𝐼 such that 𝑎 ≤ 𝑧, 𝑏 ≤ 𝑧, and 𝑐 ≤ 𝑧. Now 𝑐 = 𝑐 ∧ 𝑧 =
(𝑥 ∧ 𝑧) ∧ (𝑦 ∧ 𝑧), 𝑎 ≤ 𝑥 ∧ 𝑧 ∈ 𝐼 ∩ 𝐽 and 𝑏 ≤ 𝑦 ∧ 𝑧 ∈ 𝐼 ∩ 𝐽. Hence 𝐼 ∩ 𝐽 is distributive.
Theorem: 10. [16] Let (𝐿, ∧, ∨) be a lattice. Then the following are equivalent:
a) (𝐿, ∧, ∨) is a distributive lattice.
b) (𝐿, ∧) is a distributive meet semilattice.
c) (𝐿, ∨) is a distributive join semilattice.
Proof: (a)⇒(b): Suppose 𝐿 is distributive. Let 𝑎, 𝑏, and 𝑐 ∈ 𝑆 such that 𝑎 ∧ 𝑏 ≤ 𝑐, then put 𝑥 =
𝑎 ∨ 𝑐 and 𝑦 = 𝑏 ∨ 𝑐. 𝑎 ≤ 𝑥 and 𝑏 ≤ 𝑦. Now 𝑐 = (𝑎 ∧ 𝑏) ∨ 𝑐 = (𝑎 ∨ 𝑐) ∧ (𝑏 ∨ 𝑐) = 𝑥 ∧ 𝑦.
Hence (𝐿, ∧) is a distributive meet semilattice.
(b)⇒(c): Let 𝐿 is distributive meet semilattice. Let 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 ∈ 𝐿 such that 𝑐 ≤ 𝑎˅𝑏, then put
𝑥 = 𝑎 ∧ 𝑐, 𝑦 = 𝑏 ∧ 𝑐 then 𝑎 ∧ 𝑐 ≤ 𝑎, 𝑏 ∧ 𝑐 ≤ 𝑏. Consider 𝑐 = 𝑐 ∧ (𝑎 ∨ 𝑏) = (𝑐 ∧ 𝑎) ∨ (𝑐 ∧
𝑏) = 𝑥 ∨ 𝑦. Hence (𝐿,∨) is a distributive join semilattice.
(c) ⇒(a): Let (𝐿,∨) be a distributive join semilattice, then for any 𝑎, 𝑏 and 𝑐 ∈ 𝐿, if 𝑐 ≤ 𝑎˅𝑏,
there exists 𝑥 𝑎𝑛𝑑 𝑦 ∈ 𝐿 such that 𝑐 = 𝑥 ∨ 𝑦. This implies that 𝑐 ≤ 𝑎 and 𝑐 ≤ 𝑏. Now consider
𝑐 ∧ (𝑎 ∨ 𝑏) = 𝑐 = (𝑐 ∧ 𝑎) ∨ (𝑐 ∧ 𝑏). Hence (𝐿, ∧ ,∨) is a distributive lattice.
Lemma: 11. If 𝑆 is a distributive semilattice, 𝑆 is directed above.
Proof: Suppose 𝑆 be a distributive semilattice, then for any 𝑎, 𝑏 ∈ 𝑆, 𝑎 ∧ 𝑏 ∈ 𝑆, and 𝑎 ∧ 𝑏 ≤
𝑏.There exists 𝑥, 𝑦 ∈ 𝑆 such that 𝑎 ≤ 𝑥, 𝑏 ≤ 𝑦, and 𝑥 ∧ 𝑦 = 𝑏. Also 𝑏 = 𝑥 ∧ 𝑦 ≤ 𝑥. Therefore
for any 𝑎, 𝑏 ∈ 𝑆, there exist𝑠 𝑥 ∈ 𝑆 such that 𝑎 ≤ 𝑥 and 𝑏 ≤ 𝑥. Hence 𝑆 is directed above.
Note: For an ideal, 𝐼 of 𝑆 and 𝑎 and 𝑏 ∈ 𝑆, let ⟨𝑎, 𝐼⟩∗ = {𝑥 ∈ 𝑆: 𝑥 ∧ 𝑎 ∈ 𝐼} and ⟨𝑎, 𝑏⟩ = {𝑥 ∈
𝑆: 𝑥 ∧ 𝑎 = 𝑥 ∧ 𝑏}. If 𝐼 = (𝑏], we just write⟨ 𝑎, 𝑏⟩∗ for ⟨𝑎, (𝑏]⟩∗ .
Theorem: 12. [16] The following are equivalent in any semilattice 𝑆;
a.
S is distributive.
b.
𝐹(𝑆) is a distributive lattice.
c.
For an ideal, 𝐼 of 𝑆 and 𝑎 ∈ 𝑆, ⟨𝑎, I⟩∗ is an ideal of 𝑆.
d.
For any 𝑎 and 𝑏 ∈ 𝑆, ⟨𝑎, b⟩∗ is an ideal of 𝑆.
e.
For any a and 𝑏 ∈ 𝑆, ⟨𝑎, 𝑏⟩ is an ideal of 𝑆.
f.
Every filter which is maximal with respect to not containing an element is
prime.
g.
All filters 𝐹 of S is the intersection of all prime filters of S covering 𝐹.
Proof: (a)⇒(b): Suppose 𝑆 be a distributive semilattice. Let 𝐹, 𝐺, and 𝐻 ∈ 𝐹(𝑆), then;
Consider (𝐹 ∧ 𝐺) ∨ (𝐹 ∧ 𝐻) ⊆ 𝐹 ∧ (𝐺 ∨ 𝐻) ⇒ 𝑥 ∈ 𝐹 ∧ (𝐺 ∨ 𝐻)⇒𝑥 ∈ 𝐹 and 𝑥 ∈ (𝐺 ∨ 𝐻).
𝑥 ∈ 𝐹 and 𝑥 ∈ 𝐺, 𝑥 ∈ 𝐻 ⇒ 𝑥 ∈ 𝐹 and 𝑥 ∈ 𝐺, 𝑥 ∈ 𝐹 and 𝑥 ∈ 𝐻 ⇒ 𝑥 ∈ 𝐹 ∧ 𝐺, 𝑥 ∈ 𝐹 ∧ 𝐻.
Then 𝑥 ∈ (𝐹 ∧ 𝐺) ∨ (𝐹 ∧ 𝐻). Thus 𝐹 ∧ (𝐺 ∨ 𝐻) ⊆ (𝐹 ∧ 𝐺) ∨ (𝐹 ∧ 𝐻).
Hence 𝐹 ∧ (𝐺 ∨ 𝐻) = (𝐹 ∧ 𝐻) ∨ (𝐹 ∧ 𝐻), implies 𝐹(𝑆) is a distributive lattice.
(b)⇒ (c): Suppose that 𝑎 ∈ 𝑆 and 𝐼 be an ideal of 𝑆, then ⟨𝑎, I⟩∗ is an initial segment of 𝑆. Let
𝑥, 𝑦 ∈ 𝑆 such that 𝑥 ∧ 𝑎 and 𝑦 ∧ 𝑎 ∈ 𝐼, there is 𝑏 ∈ 𝐼 such that 𝑥 ∧ 𝑎 ≤ 𝑏 and 𝑦 ∧ 𝑎 ≤ 𝑏 so that
𝑏 ∈ ([𝑥) ∨ [𝑎) ∩ ([𝑦) ∨ [𝑎)) = ([𝑥) ∩ [𝑦)) ∨ [𝑎). There exist 𝑧 ∈ 𝑆 such that 𝑥 ≤ 𝑧, 𝑦 ≤
𝑧, and 𝑧 ∧ 𝑎 ≤ 𝑏. This implies that 𝑧 ∈ ⟨𝑎, I⟩∗ . Therefore ⟨𝑎, I⟩∗ is an ideal of 𝑆.
(c)⇒(d): Suppose that ⟨𝑎, I⟩∗ be an ideal of 𝑆. Let 𝑥 ∈ ⟨𝑎, b⟩∗ and 𝑐 ∈ 𝑆 such that 𝑐 ≤ 𝑥, then
𝑥 ∧ 𝑎 ∈ (𝑏] and 𝑐 ∧ 𝑎 ≤ 𝑥 ∧ 𝑎 ∈ (𝑏]. Then 𝑐 ∈ ⟨𝑎, b⟩∗.
Again let 𝑥, 𝑦 ∈ ⟨𝑎, b⟩∗, then 𝑥 ∧ 𝑎 and 𝑦 ∧ 𝑎 ∈ (𝑏]. There is 𝑐 ∈ (𝑏] such that 𝑥 ∧ 𝑎 ≤ 𝑐 and
𝑦 ∧ 𝑎 ≤ 𝑐 ⇒ 𝑎 ∧ 𝑐 ∈ (𝑏]. Therefore 𝑐 ∈ ⟨𝑎, b⟩∗.
(d)⇒(e): Suppose that ⟨𝑎, b⟩∗ be an ideal of 𝑆, then⟨ 𝑏, a⟩∗ is also an ideal of 𝑆.
Then ⟨𝑎 𝑏⟩ = ⟨𝑎, 𝑏⟩∗ ∩ ⟨𝑏, 𝑎⟩∗ . Therefore ⟨𝑎, 𝑏⟩ is ideal for 𝑆 as 𝐼(𝑆) is closed under a finite
intersection.
(e)⇒(f): Suppose 𝐹 be a filter of 𝑆, which is maximal with respect to not containing an element
𝑎 of 𝑆. Let 𝑥, 𝑦 ∈ 𝑆 − 𝑃, then 𝑎 ∈ (𝐹˅[𝑥)) ∩ (𝐹 ˅[𝑦)). Therefore, there exists 𝑏 ∈ 𝐹 such
that 𝑏 ∧ 𝑥 ≤ 𝑎 and 𝑏 ∧ 𝑦 ≤ 𝑎. Hence 𝑥 and 𝑦 ∈ ⟨𝑎, 𝑏⟩ * which is an ideal of S. Thus 𝑆 − 𝐹 is
directed above, therefore 𝐹 is prime.
(g)⇒(b): Let 𝐼, 𝐽, and 𝐾 be filters of 𝑆. For any prime filter 𝐹 of 𝑆; (𝐼 ∩ 𝐽) ∨ (𝐼 ∩ 𝐾) ⊂ 𝐹 ⇒
(𝐼 ∩ 𝐽) ⊂ 𝐹 or (𝐼 ∩ 𝐾) ⊂ 𝐹 ⇒ 𝐼 ⊂ 𝐹 and 𝐽 ∨ 𝐾 ⊂ 𝐹 ⇒ 𝐼 ∩ (𝐽 ∨ 𝐾) ⊂ 𝐹 ⇒ (𝐼 ∩ 𝐽) ∨ (𝐼 ∩
𝐾) = 𝐼 ∩ (𝐽 ∨ 𝐾). Hence 𝐹(𝑆) is distributive.
Definition: 13. [10] A semilattice S with 0 is said to be pseudo complemented if, for each 𝑎 ∈
𝑆, there exists 𝑎∗ ∈ 𝑆 such that 𝑎 ∧ 𝑏 = 0 if and only if 𝑏 ≤ 𝑎∗ for all 𝑏 ∈ 𝑆. An element 𝑎 ∈ 𝑆
is said to be dense if 𝑎∗ = 0 and is said to be closed if(𝑎∗ )∗ = 𝑎.
Theorem: 14. All pseudocomplemented semilattices are 0 − distributive.
Proof: Let 𝑆 be a pseudocomplemented semilattice and 𝑥1 , 𝑥2 , … , 𝑥𝑛 , 𝑎𝑛𝑑 𝑎 ∈ 𝑆 such that 𝑥 ∧
𝑎1 = 0, 𝑥 ∧ 𝑎2 = 0, … , 𝑥 ∧ 𝑎𝑛 = 0 and 𝑥1 ∨ 𝑥2 ∨ … ∨ 𝑥𝑛 exists. Since 𝑥𝑖 ≤ (𝑎)∗ for all 𝑖′𝑠, we
get 𝑥1 ∨ 𝑥2 ∨ … ∨ 𝑥𝑛 ≤ (𝑎)∗. Hence 𝑎 ∧ (𝑥1 ∨ 𝑥2 ∨ … ∨ 𝑥𝑛 ) = 0. Thus 𝑆 is 0 − distributive.
Ideals and Filters for Distributive Semilattice
Theorem: 15. Let 𝐼 be an ideal of S and 𝐽 a filter of S such that 𝐼 ∩ 𝐽 ≠ ∅. then, 𝐼 ∩ 𝐽 is a
distributive sub-semilattice of 𝑆.
Proof: Suppose 𝐼 be an ideal of 𝑆 and 𝐽 a filter of 𝑆 such that 𝐼 ∩ 𝐽 ≠ ∅. Let 𝑎, 𝑏 ∈ 𝐼 ∩ 𝐽, then
𝑎 ∧ 𝑏 ∈ 𝐼. Since 𝐼 is the initial segment, there exist 𝑐 ∈ 𝐽 such that 𝑐 ≤ 𝑎 and 𝑏 ≤ 𝑐. This
implies 𝑐 ≤ 𝑎 ∧ 𝑏 and 𝑎 ∧ 𝑏 ∈ 𝐽 since 𝐽 is the final segment. Therefore 𝐼 ∩ 𝐽 is a subsemilattice of 𝑆. Let 𝑎, 𝑏, and 𝑐 ∈ 𝐼 ∩ 𝐽 such that 𝑎 ∧ 𝑏 ≤ 𝑐. There is 𝑥 and 𝑦 ∈ 𝑆 such that 𝑎 ≤
𝑥, 𝑏 ≤ 𝑦, and 𝑥 ∧ 𝑦 = 𝑐. Since 𝐼 is an ideal of 𝑆, there i𝑠 𝑧 ∈ 𝐼 such that 𝑎 ≤ 𝑧, 𝑏 ≤ 𝑧, and 𝑐 ≤
𝑧. now 𝑐 = 𝑐 ∧ 𝑧 = (𝑥 ∧ 𝑧) ∧ (𝑦 ∧ 𝑧), 𝑎 ≤ 𝑥 ∧ 𝑧 ∈ 𝐼 ∩ 𝐽 𝑎nd 𝑏 ≤ 𝑦 ∧ 𝑧 ∈ 𝐼 ∩ 𝐽.
Therefore 𝐼 ∩ 𝐽 is distributive.
Corollary: 16. Every non-empty filter of 𝑆 is a distributive sub-semilattice of 𝑆.
Proof: Suppose that 𝐽 is a nonempty filter of 𝑆. For any 𝑎, 𝑏 ∈ 𝐽, there exist 𝑐 ∈ 𝐽 such that 𝑐 ≤
𝑎 and 𝑐 ≤ 𝑏 ⇒ 𝑐 ≤ 𝑎 ∧ 𝑏 ⇒ 𝑎 ∧ 𝑏 ∈ 𝐽, since 𝐽 is the final segment. Then, 𝐽 is a sub-semilattice
of S. Let 𝑎, 𝑏, and 𝑐 ∈ 𝐽 such that 𝑎 ∧ 𝑏 ≤ 𝑐, there exists 𝑥, 𝑦 ∈ 𝑆 such that 𝑎 ≤ 𝑥, 𝑏 ≤ 𝑦,
and 𝑥 ∧ 𝑦 = 𝑐. Now as 𝑎, 𝑏, and 𝑐 ∈ 𝐽, then 𝑥, 𝑦 ∈ 𝐽. Hence 𝑥 ∧ 𝑦 = 𝑐 ∈ 𝐽. Therefore 𝐽 is
distributive.
Theorem: 17. Let 𝐼 be an ideal of 𝑆 and 𝐿 is a sub-semilattice of 𝑆 such that I ∩ 𝐿 = ∅, then,
there exists a prime ideal P of S such that 𝐼 ⊂ 𝑃 and 𝑃 ∩ 𝐿 = ∅.
Proof: Suppose that 𝑇 = {𝑥 ∈ 𝑆: 𝑎 ≤ 𝑥 for some 𝑎 ∈ 𝐿}. Then 𝑇 is the filter of 𝑆, 𝑇 ∩ 𝐼 = ∅.
Zorn's lemma, there exists a filter 𝐹 of 𝑆 which is maximal among all filters containing 𝐿 and
disjoint from 𝐼. Now we have 𝐼 ⊂ 𝑆 − 𝐹 and (𝑆 − 𝐹) ∩ 𝐿 = ∅. Let us prove that 𝐹 is a prime
filter. Let 𝑥, 𝑦 ∈ 𝑆 − 𝐹, since 𝐹 is maximal, there exists 𝑎 ∈ 𝐼 ∩ (𝐹˅[𝑥)) and 𝑏 ∈ 𝐼 ∩ (𝐹 ∨
[𝑦)). There exist 𝑐 ∈ 𝐼 such that 𝑎 ≤ 𝑐 and 𝑏 ≤ 𝑐. This implies that 𝑐 ∈ 𝐼 ∩ (𝐹˅[𝑥)) ∩
(𝐹 ∨ [𝑥)) = 𝐼 ∩ (𝐹 ∨ ([𝑥) ∩ [𝑦) ⇒ [𝑥) ∩ [𝑦) ⊄ 𝐹. Thus 𝐹 is a prime filter of 𝑆.
Corollary: 18. Let 𝐿 be sub-semilattice of 𝑆 and 𝑃 a prime ideal (filter) of 𝐿. Then, there exists
a prime ideal (filter) 𝑄 of S such that 𝑄 ∩ 𝐿 = 𝑃.
Proof: Let 𝑃 be a prime ideal of 𝐿 and put 𝑃1 = {𝑥 ∈ 𝑆: 𝑥 ≤ 𝑎 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑎 ∈ 𝑃}. Then 𝑃1 is
an ideal of 𝑆 and 𝑆 − 𝐿 is a sub-semilattice of 𝑆 and 𝑃1 ∩ (𝐿 − 𝑃) = ∅. Then there exists a
prime ideal 𝑄 of S such that 𝑃1 ⊂ 𝑄 and 𝑄 ∩ (𝐿 − 𝑃) = ∅. Hence 𝑄 ∩ 𝐿 = 𝑃.
Similarly, suppose 𝑃 be a prime filter of 𝐿 and 𝑃2 = {𝑥 ∈ 𝑆: 𝑎 ≤ 𝑥 𝑓𝑜𝑟 𝑠𝑜𝑚𝑎 𝑎 ∈ 𝑃}. Then
𝑃2 is a filter of 𝑆 and 𝑆 − 𝐿 is sub-semilattice of 𝑆 and 𝑃2 ∩ (𝐿 − 𝑃) = ∅. There exists a prime
filter 𝑄 of 𝑆 such that 𝑃2 ⊂ 𝑄 and 𝑄 ∩ (𝐿 − 𝑃) = ∅. Hence 𝑄 ∩ 𝐿 = 𝑃.
Corollary: 19. Let 𝐼 be an ideal of 𝑆, then the intersection of all prime ideals of 𝑆 containing 𝐼
is again an ideal of S and is equal to 𝐼.
Proof: Let {𝑃𝑖 }𝑖∈𝛼 be the class of prime ideals containing I, then 𝐼 ⊆ 𝑃𝑖 , ∀𝑖 ∈ 𝛼. this implies
𝐼 ⊆∩ 𝑃𝑖, ∀𝑖 ∈ 𝛼. Also if 𝑎 ∈ 𝑆 and 𝑎 ∉ 𝐼, then 𝐼 ∩ [𝑎) = ∅. There exists a prime ideal 𝑃𝑖 of 𝑆
such that 𝐼 ⊂ 𝑃𝑖 and 𝑃𝑖 ∩ [𝑎) = ∅ ⇒ 𝑎 ∉ 𝑃𝑖 . Therefore 𝐼 =∩ 𝑃𝑖 , ∀𝑖 ∈ 𝛼.
Theorem: 20. Every maximal ideal (filter) of 𝑆 is prime.
Proof: Let 𝑆 be distributive semilattice and 𝑄 is a maximal filter of 𝑆. Then 𝑆 − 𝑄 is minimal
ideal, there exists a prime ideal 𝐼 such that 𝑆 − 𝑄 ⊆ 𝐼, then 𝑄 ∩ 𝐼 = ∅. Let 𝑥, 𝑦 ∉ 𝑄,
then 𝑥, 𝑦 ∈ 𝑆 − 𝑄. There exist𝑠 𝑎 ∈ 𝐼 ∩ (𝑄 ∨ [𝑥)) and 𝑏 ∈ 𝐼 ∩ (𝑄 ∨ [𝑦)). This implies there
exists 𝑐 ∈ 𝐼 such that 𝑎 ≤ 𝑐 and 𝑏 ≤ 𝑐 since 𝐼 directed above. Hence 𝑐 ∈ 𝐼 ∩ (𝑄 ∨ [𝑥)) ∩
(𝑄 ∨ [𝑦)) ) = 𝐼 ∩ (𝑄 ∨ ([𝑥) ∩ [𝑦)) ⇒[𝑥) ∩ [𝑦) ⊈ 𝑄. Therefore 𝑄 is a prime filter of 𝑆.
Corollary: 21. Let 𝑆 be distributive semilattice, 𝑆 has the smallest (greatest) element if and
only if the intersection of all prime ideals (filters) of 𝑆 is non-empty.
Proof: Suppose 𝑆 has the largest element, ∅ ≠ ⋂𝐹∈𝐹(𝑆) 𝐹 ⊆ ⋂𝑃∈𝐹(𝑆) 𝑃, where 𝐹 any nonempty
filters of 𝑆 and 𝑃 the prime filters of 𝑆. Therefore ⋂𝑃∈𝐹(𝑆) 𝑃 ≠ ∅. Similarly, if 𝑆 has a smallest
element, ∅ ≠ ⋂𝐼∈𝐼(𝑆) 𝐼 ⊆ ⋂𝑃∈𝐼(𝑆) 𝑃, were 𝐼 any nonempty ideals of 𝑆 and 𝑃 the prime ideals
of 𝑆. Therefore ⋂𝑃∈𝐹(𝑆) 𝑃 ≠ ∅.
Conversely, suppose that the intersection of all non-empty filters (ideals) is again nonempty, S
has the largest (smallest) element.
Definition: 22. Let 𝑆 be a distributive semilattice with the smallest element 0 and 𝐿 is a subsemilattice of S, and then define 0(𝐿) = {𝑥 ∈ 𝑆: 𝑥 ∧ 𝑦 = 0 for some 𝑦 ∈ 𝐿}. Then 0(𝐿) is
called the annihilator of 𝐿. The annihilator of {𝑎} is denoted by 0(𝑎) = {𝑥 ∈ 𝑆: 𝑥 ∧ 𝑎 = 0}. An
ideal 𝐼 of 𝑆 is called annihilator ideal if 𝐼 = 0(𝐿), for some sub-semilattice 𝐿 𝑜𝑓 𝑆.
Proposition: 23. Suppose S has the smallest element 0 and 𝐹 is a proper filter of 𝑆, then 𝐹 is a
maximal filter if and only if 𝑆 − 𝐹 is a minimal prime ideal.
Proof: Let 𝐹 be a maximal filter of S. Since 𝑆 is distributive, 𝐹 is the prime filter. Thus 𝑆 − 𝐹
is a prime ideal. To prove the minimalism of 𝑆 − 𝐹, assume 𝑆 − 𝐹 is not minimal ideal. Then
there exists a prime ideal 𝐼 such that 𝐼 ⊆ 𝑆 − 𝐹 which implies 𝐹 ⊆ 𝑆 − 𝐼 which contradicts the
maximalist of 𝐹. Hence 𝑆 − 𝐹 is a minimal prime ideal.
Conversely, suppose that 𝑆 − 𝐹 is a minimal prime ideal, then 0 ∈ 𝑆 − 𝐹 ⊆ 0(𝐹). Let 𝑥 ∈
0(𝐹), then 𝑥 ∧ 𝑦 = 0, for some 𝑦 ∈ 𝐹. Since 𝑆 − 𝐹 is prime and 𝑦 ∈ 𝐹, then 𝑥 ∈ 𝑆 − 𝐹.
Therefore 0(𝐹) ⊆ 𝑆 − 𝐹 and hence 0(𝐹) = 𝑆 − 𝐹. Since 0(𝐹) = 𝑆 − 𝐹, then 𝑆 − 𝐹 is a prime
ideal, and so 𝐹 is a prime filter. Suppose if possible, 𝐹 is not maximal filter, then 𝐹 ⊂ 𝑀, for
some maximal filter 𝑀 𝑜𝑓 𝑆. then 𝑆 − 𝑀 ⊂ 𝑆 − 𝐹 = 0(𝐹), which contradicts the minimalist
of 𝑆 − 𝐹. Therefore 𝐹 is a maximal filter.
Corollary: 24. If 𝑆 is a distributive semilattice with the smallest element 0, then for any 𝑎 ∈ 𝑆,
(𝑎)∗ = ⟨𝑎, 0⟩∗ = {𝑥 ∈ 𝑆: 𝑎 ∧ 𝑥 = 0} is an ideal of 𝑆.
Proof: Suppose 𝑆 is distributive semilattice with 0. For 𝑎 ∈ 𝑆, (𝑎)∗ = {𝑥 ∈ 𝑆: 𝑎 ∧ 𝑥 = 0}. Let
𝑥 ∈ (𝑎)∗ , 𝑏 ∈ 𝑆 such that 𝑏 ≤ 𝑥,then 𝑏 = 𝑥 ∧ 𝑏. Now 𝑎 ∧ 𝑏 = 𝑎 ∧ (𝑥 ∧ 𝑏) = (𝑎 ∧ 𝑥) ∧ 𝑏 =
0 ∧ 𝑏 = 0. Therefore 𝑏 ∈ (𝑎)∗ . Let 𝑥, 𝑦 ∈ (𝑎)∗ , ∧ 𝑥 = 0 and 𝑎 ∧ 𝑦 = 0. Consider 𝑎 ∧ (𝑥 ∨
𝑦) = (𝑎 ∧ 𝑥) ∨ ( 𝑎 ∧ 𝑦) = 0 ∨ 0 = 0. Therefore 𝑥 ∧ 𝑦 ∈ (𝑎)∗ . Hence (𝑎)∗ is an ideal of S.
Space of Maximal Filters in Distributive Semilattice
A topological space (𝑋, 𝒟) is said to be compact if every open set cover for 𝑋 has a finite
subcover. Let 𝑆 be distributive semilattice and 𝑋 the set of all prime filters of 𝑆, for any 𝑎 ∈ 𝑆,
let 𝑋𝑎 = {𝐹 ∈ 𝑋: 𝑎 ∈ 𝐹}. Then for any 𝑎, 𝑏 ∈ 𝑆, 𝑋𝑎 ∩ 𝑋𝑏 = 𝑋𝑎˄𝑏 and {𝑋𝑎 : 𝑎 ∈ 𝑆} forms a
base for some topology 𝐷 on 𝑋, which is called the hull-kernel topology on 𝑋.
Theorem: 25. [2] Let 𝑆 be distributive semilattice, then 𝑆 is isomorphic to meet
subsemilattices {𝑋𝑎 : 𝑎 ∈ 𝑆} of the lattice of all open subsets of (𝑋, 𝒟).
Proof: Let 𝑋 be the class of subsemilattices of {𝑋𝑎 : 𝑎 ∈ 𝑆}. Define 𝑓: 𝑆 → 𝑋 by 𝑓(𝑎) = 𝑋𝑎 , for
any 𝑎 ∈ 𝑆. Let 𝑎, 𝑏 ∈ 𝑆, and 𝑃 ∈ 𝑋, 𝑃 ∉ 𝑓(𝑎 ∧ 𝑏) ⇔ 𝑎 ∧ 𝑏 ∉ 𝑃 ⇔ 𝑎 ∉ 𝑃 or 𝑏 ∉ 𝑃 ⇔ 𝑃 ∉
𝑓(𝑎 ) or 𝑃 ∉ 𝑓( 𝑏)⇔𝑃 ∉ 𝑓(𝑎) ∩ 𝑓( 𝑏).
Therefore 𝑓(𝑎 ∧ 𝑏) = 𝑓(𝑎) ∩ 𝑓( 𝑏). Hence 𝑓 is a homomorphism of a lattice. Let 𝑎 ≠ 𝑏, then
there exists a prime filter 𝑃 such that 𝑎 ∈ 𝑃 and from 𝑏 ∉ 𝑃 and vice versa. Then 𝑋𝑎 ≠ 𝑋𝑏 .
Hence 𝑓 is injective. And since every element of 𝑌 has the form 𝑋𝑎 , for any 𝑎 ∈ 𝑆, then 𝑓 is
surjective. Therefore 𝑓 is an isomorphism.
Definition: 26. A topological space 𝑋 is said to be Hausdorff space if for any points 𝑥, 𝑦 ∈ 𝑋
such that 𝑥 ≠ 𝑦 there exist open sets 𝑈, 𝑉 ⊆ 𝑋 such that 𝑥 ∈ 𝑈, 𝑦 ∈ 𝑉, and 𝑈 ∩ 𝑉 = ∅.
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 (𝑋, 𝐷).
Proof: Let 𝑃, 𝑄 ∈ 𝑌 be maximal filters such that 𝑃 ≠ 𝑄. There exist𝑠 𝑎 ∈ 𝑃 and 𝑎 ∉ 𝑄. This
implies 𝑃 ∈ 𝑌𝑎 and 𝑄 ∉ 𝑌𝑎 , then 𝑃 ∈ 𝑌𝑎 and 𝑄 ∈ 𝑌 − 𝑌𝑎 and 𝑌𝑎 ∩ 𝑌 − 𝑌𝑎 = ∅. Therefore 𝑌 is
Hausdorff. Let 𝑃 ≠ 𝑄 ∈ 𝑌, then we can suppose that 𝑃 ≰ 𝑄. There exist𝑠 𝑎 ∈ 𝑃 such that 𝑎 ∉
𝑄. Therefore, 𝑌𝑎 is clopen and 𝑃 ∈ 𝑌𝑎 and 𝑄 ∉ 𝑌𝑎 . Therefore, 𝑌 is totally disconnected.
Theorem: 28. Let S be a pseudo-complemented distributive semilattice. So space 𝑋 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 𝑋 is the space of all maximal filters of 𝑆 with the hull-kernel topology.
Let 𝑃, 𝑄 ∈ 𝑋 such that 𝑃 ≠ 𝑄. There exists 𝑎 ∈ 𝑃 and 𝑎 ∉ 𝑄 ⇒ 𝑃 ∈ 𝑋𝑎 and 𝑄 ∉ 𝑋𝑎 . This
implies 𝑃 ∈ 𝑋𝑎 and 𝑄 ∈ 𝑋 − 𝑋𝑎 and 𝑋𝑎 ∩ 𝑋 − 𝑋𝑎 = ∅. Therefore 𝑋 is Hausdorff. Again
let 𝑃 ≠ 𝑄 ∈ 𝑋, we can suppose that 𝑃 ≰ 𝑄. There exist𝑠 𝑎 ∈ 𝑃 such that 𝑎 ∉ 𝑄. Hence 𝑋𝑎 is
clopen and 𝑃 ∈ 𝑋𝑎 and 𝑄 ∉ 𝑋𝑎 . Therefore 𝑋 is totally disconnected.
Also let {𝑋𝑎 } 𝑎 ∈ 𝐿 be a basic open cover for 𝑋, for some 𝐿 ⊆ 𝑆, then 𝑋 ⊆ ⋃𝑎∈𝐿 𝑋𝑎 . Let 𝐹 =
[𝐿) = {𝑥 ∈ 𝑆: ⋀𝑎𝑖 ≤ 𝑥, where 𝑎𝑖 ∈ 𝐿}. Suppose if possible, 𝐹 is a proper filter of 𝑆, then there
exists a maximal filter 𝑀 of 𝑆 such that 𝐹 ⊆ 𝑀. Then 𝐿 ⊆ 𝑀 as 𝐿 ⊆ 𝐹. Therefore 𝑀 ∉ 𝑌 which
is a contradiction and hence 𝐹 = 𝑆. In particular, 0 ∈ 𝐹 = [𝐿) ⇒0 = ⋀𝑛𝑖=1 𝑎𝑖 , for some 𝑎𝑖 ∈ 𝐿.
If 𝑀 is a maximal filter, then 𝑎𝑖 ∉ 𝑀 for some 𝑖. Therefore 𝑋 = ⋃𝑛𝑖=1 𝑌𝑎𝑖 , and hence
{𝑋𝑎1 , 𝑋𝑎2 , … , 𝑋𝑎𝑛 } is a finite subcover of {𝑋𝑎 }𝑎 ∈ 𝐿. Hence 𝑋𝑎 is compact. This implies 𝑋 is
compact. Therefore 𝑋 Boolean space.
Theorem: 29. Let 𝑆 be a distributive semilattice with 0 and 𝑋 be the space of prime filters with
the hull-kernel topology. Then the following are equivalent:
a. 𝑋 is Hausdorff space,
b. 𝑋 is a T1-space,
c. Every prime filter of S is maximal,
d. 𝐹 is the only prime filter of 𝑆 such that 0(𝐹) ∩ 𝐹 = ∅.
Proof: (a) ⇒(b): Suppose that 𝑋 be Hausdorff space; for any 𝑎 ≠ 𝑏 ∈ 𝑆 there exists prime filters
𝑃 and 𝑄 in 𝑆 such that 𝑃 ∈ 𝑋𝑎 , and 𝑄 ∈ 𝑋𝑏 , with 𝑋𝑎 ∩ 𝑋𝑏 = ∅.
Therefore 𝑃 ∈ 𝑋𝑎 − 𝑋𝑏 , 𝑄 ∈ 𝑋𝑏 − 𝑋𝑎 . Hence 𝑋 is a T1-space.
(b)⇒(c): Suppose that 𝑃 be a prime filter of 𝑆.Let 𝑃 ∈ 𝑋 − 𝑋𝑏 , and 𝑄 ∈ 𝑋 − 𝑋𝑎 , then 𝑎 ∈
𝑃 and 𝑏 ∉ 𝑃. This implies 𝑃 ⊂ [𝑃 ∪ {𝑏}) = {𝑥 ∈ 𝑆: ⋀(𝑎𝑖 ∪ 𝑏) ≤ 𝑥, where1 ≤ 𝑖 ≤ 𝑛, for some,
𝑎𝑖 ∈ 𝑃} = {𝑥 ∈ 𝑆: (⋀𝑎𝑖 ) ∪ 𝑏 ≤ 𝑥 for some 𝑎𝑖 ∈ 𝑃}= {𝑥 ∈ 𝑆: ⋀𝑎𝑖 ≤ 𝑥 and 𝑏 ≤ 𝑥 for some
𝑎𝑖 ∈ 𝑃} = 𝑆. Hence 𝑃 is a maximal filter.
(c)⇒(d): Suppose 𝐹 be a prime filter of 𝑆, then 𝐹 is the maximal filter. Hence 𝑆 − 𝐹 = 0(𝐹).
Therefore 0(𝐹) ∩ 𝐹 = ∅.
(d)⇒ (a): Suppose 𝐹 be the only prime filter of 𝑆 such that 0(𝐹) ∩ 𝐹 = ∅. Let 𝑎 ≠ 𝑏 ∈ 𝑆 such
that 𝑎 ∈ 𝐹 and 𝑏 ∉ 𝐹, then 𝐹 ∈ 𝑋𝑎 and 𝐹 ∉ 𝑋𝑏 . Therefore 𝑋𝑎 ∩ 𝑋𝑏 = ∅. Hence 𝑋 is Hausdorff.
Definition: 30. A topological space 𝑋 is said to be a Regular space if for any points 𝑥 ∈ 𝑋 and
for every closed set 𝑃 ⊆ 𝑋 such that 𝑥 ∉ 𝑃, there exist open sets 𝑈, 𝑉 ⊆ 𝑋 such that 𝑥 ∈ 𝑈, 𝑃 ⊆
𝑉, and 𝑈 ∩ 𝑉 = ∅.
Theorem: 31. Let 𝑆 be a pseudo-complemented distributive semilattice and 𝑋 be the space of
prime filters with the hull-kernel topology. 𝑋 is a compact hausdorff space if and only if 𝑋 is a
regular space.
Proof: Supposed that 𝑋 be a compact Hausdorff space and let 𝑋𝑎 ⊂ 𝑋 be a closed subspace.
Let 𝑃 be a prime filter of 𝑆 such that 𝑃 ∈ 𝑋 − 𝑋𝑎 . We need to contract open sets 𝑈 and 𝑉 ⊆ 𝑋
with 𝑋𝑎 ⊆ 𝑈 and 𝑃 ∈ 𝑉 such that 𝑈 ∩ 𝑉 = ∅. Since 𝑋 is Hausdorff space, for each prime
filter 𝑄 ∈ 𝑋𝑎 there exists a neighborhood 𝑋𝑎𝑖 of 𝑄 and 𝑋𝑏𝑖 of 𝑋𝑎 , where 𝑎𝑖 ∈ 𝑄 and 𝑏𝑖 ∈ 𝑃 such
that 𝑄 ∈ 𝑋𝑎𝑖 and 𝑃 ∈ 𝑋𝑏𝑖 and 𝑋𝑎𝑖 ∩ 𝑋𝑏𝑖 = ∅.
Let 𝑋𝑎 = {𝑋𝑎𝑖 : 𝑎𝑖 ∈ 𝑄}. Then ⋃𝑎𝑖 ∈𝑄 𝑋𝑎𝑖 is an open cover of 𝑋𝑎 , and since 𝑋 is compact, 𝑋𝑎 is
a compact subspace of 𝑋. Therefore there exist finitely many elements 𝑎1 , 𝑎2 , … , 𝑎𝑛 ∈ 𝑄 such
that 𝑋𝑎1 , 𝑋𝑎2 , . . . , 𝑋𝑎𝑛 form subcover 𝑋𝑎 . Then 𝑋𝑎 ⊆ ⋃𝑛𝑖=1 𝑋𝑎𝑖 .
Let and 𝑈 = ⋃𝑛𝑖=1 𝑋𝑎𝑖 ,and 𝑉 = ⋂𝑛𝑖=1 𝑋𝑏𝑖 . Then 𝑈 and 𝑉 are open sets and 𝑋𝑎 ⊆ 𝑈 and 𝑃 ∈ 𝑉,
such that 𝑈 ∩ 𝑉 = ∅. To show that 𝑈 and 𝑉 are really disjointed, assume that 𝑃 ∈ 𝑈 ∩ 𝑉 ≠ ∅.
Then 𝑃 ∈ 𝑋𝑎𝑖 , for each 𝑖, and 𝑃 ∈ 𝑋𝑏𝑖 , for some 𝑖. That is 𝑃 ∈ 𝑈 ∩ 𝑉 ⇒ 𝑃 ∈ (⋂𝑛𝑖=1 𝑋𝑎𝑖 ) ∩
(⋃𝑛𝑖=1 𝑋𝑏𝑖 ) ⇒ 𝑃 ∈ 𝑋𝑎𝑖 for all 𝑖 and 𝑃 ∈ 𝑋𝑏𝑖 for some 𝑖. Then 𝑃 ∈ 𝑋𝑎𝑖 ∩ 𝑋𝑏𝑖 for some 𝑖. This is
in contradiction. Therefore 𝑈 ∩ 𝑉 = ∅. Hence 𝑋 is a regular space.
Conversely, suppose 𝑋 is a regular space. Let 𝑃, 𝑄 ∈ 𝑋 such that 𝑃 ≠ 𝑄. Then without loss of
generality, let 𝑃 ≰ 𝑄, then there exists a prime filter 𝑃 containing 𝑎 ∈ 𝑃 but not 𝑄. Then there
exists a clopen set 𝑋𝑎 such that 𝑃 ∈ 𝑋𝑎 and 𝑋 − 𝑋𝑎 is a closed set. Let 𝑄 ∈ 𝑋 − 𝑋𝑎 , since 𝑋 is
a regular space, there exist an open sets 𝑈, 𝑎𝑛𝑑 𝑉 ⊆ 𝑋 such that ∈ 𝑃 ∈ 𝑋𝑎 ⊆ 𝑈, 𝑄 ∈ 𝑋 − 𝑋𝑎 ⊆
𝑉 such that 𝑈 ∩ 𝑉 = ∅. Therefore 𝑋 is Hausdorff space.
Conclusion
Generally, when we consider the class of maximal (prime) filters 𝑋 of distributive
semilattice with 0 and pseudo complemented semilattice;
𝑋 is a T1-space ⇔ 𝑋 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.
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