A new view of relationship between atomic posets and complete (algebraic) lattices

Theorem 3.1

Let P be an atomic poset. Then < ℬ(P); ≤ > is a complete lattice in which join and meet are given by

Proof

Define B(P0):={A⊆P0|A=D¯C¯A}. The map μ : (A, B) → A gives an order-isomorphism between ℬ(P) and ℬ(P₀).

We shall prove that ℬ(P₀) is a topped intersection structure. Let A_i ∈ ℬ(P₀) for i ∈ I. Then D¯C¯Ai=Ai for each i. By (5) in Proposition 2.16

Also ⋂i∈IAi⊆Ai for all i ∈ I, which, by (4) in Proposition 2.16, implies that

whence

Therefore D¯c¯⋂i∈IAi=⋂i∈IAi and hence ⋂i∈IAj∈B(P0).Also,P0⊆D¯C¯PO. So that P0=D¯C¯P0, which shows that ℬ(P₀) is topped.

By Theorem 2.7, ℬ(P₀) is a complete lattice in which meet is given by intersection. A formula for the join is given in Theorem 2.7 but we shall proceed more directly. We claim that

Let A=D¯⋂i∈IC¯Ai. Certainly A=D¯C¯A, by (7) in Proposition 2.16, and ⋃i∈IAi⊆A, by (2) in Proposition 2.16. Hence A is an upper bound for {A_i}_{i ∈ I} in ℬ(P₀). Also, if X is an upper bound in ℬ(P₀) for {A_i}_{i ∈ I}, then

Therefore A is indeed the required join. We may now appeal to Theorem 2.8 to deduce that ℬ(P) is a complete lattice in which joins and meets are given by

 □

Theorem 3.2

Let P be an atomic poset and ℬ(P) be the complete lattice by Theorem 3.1 Then φ(P₀) is join-dense in ℬ(P) and φ(𝒜(P)) is meet-dense in ℬ(P).

Proof

Let (A, B) ∈ ℬ(P). Then

However, according to Definition 2.15

Since (A, B) and ⋁ φ(A) are elements of ℬ(P) with the same second coordinate, ⋁ φ(A) = (A, B). Consequently φ(P₀) is join-dense in ℬ(P) and φ(𝒜(P)) is meet-dense in ℬ(P). □

Theorem 3.3

For every complete lattice L, there is an atomic poset P such that L is order-isomorphic to ℬ(P).

Proof

Suppose (L, ⊑) is a complete lattice. Define the atomic poset P= {0} ⋃ 𝒜(P)⋃ P₀, where 𝒜(P) = L and P₀ = L. Further, in P, let a ∥ b for ∀ a, b ∈ 𝒜(P) ; let a ≤ b iff a ⊑ b for ∀ a ∈ 𝒜(P), b ∈ P₀; let a ≤ b iff a ∥ b for ∀ a, b ∈ P₀. As L is a lattice, it is easy to see that P is an atomic poset. We want to show that (L, ⊑) is order-isomorphic to ℬ(P).

First note that for any X ⊆ P₀, we have

Among them, ↓_D(x) means the upper set of x in L. On the other hand, for any Y ⊆ 𝒜_P,

Therefore, X ∈ ℬ(P₀) iff D¯C¯P=X, or

Since ↑_L ⋁(↓_L ∧ X) = ↑_L(∧ X), hence, X ⊆ ℬ(P₀) iff X = ↑_L(∧ X). In other words, ℬ(P₀) are precisely the up-closed subsets of L generated by a single element. Hence, a subset of P₀ belongs to ℬ(P₀) if and only if it is a principal filter.

The mapping x ↦ ↑x provides an order-isomorphism between L and ℬ(P₀). Since ℬ(P₀) is isomorphic to ℬ(P), therefore (L, ⊑) is order-isomorphic to ℬ(P). □

Example 3.4

Let L = {a, b, c, d} be a complete lattice, the Hasse diagram of L is illustrated by Figure 1. We can get an atomic poset P by Theorem 3.3, whose Hasse diagram is illustrated by Figure 1, and can also get a complete lattice ℬ(P) which is isomorphic to L by Theorem 3.1. In Figure 1, a₁ and a₂ in P is a in L, b₁ and b₂ in P is b in L, c₁ and c₂ in P is C in L, d₁ and d₂ in P is d in L. In ℬ(P), A = ({d₂}, {a₁, b₁, c₁, d₁}), B = ({b₂, d₂}, {a₁, b₁}), C = ({c₂, d₂}, {a₁, c₁}), D = ({a₂, b₂, c₂, d₂}, {a₁, b₁, c₁, d₁}).

Fig. 1

The figure in Example 3.4

Theorem 3.5

Let P be an atomic poset. Then < ℬ^o(P);≤> is a complete lattice in which join and meet are given by

Proof

We shall prove that ℬ^o(P) is a topped intersection structure. Let A_i ∈ ℬ^o(P) for i ∈ I. Then DCAio=Ai for each i. By (6) in Proposition 2.16

Also ⋂i∈IAi⊆Ai for all i ∈ I, which, by (3) in Proposition 2.16, implies that

whence

Therefore Dc⋂i∈IAio=⋂i∈IAi and hence ⋂i∈IAi∈Bo(P).Also,P0⊆DCP0o. So taht P0=DCP0o, which shows that ℬ^o(P) is topped.

By Theorem 2.7, ℬ^o(P) is a complete lattice in which meet is given by intersection. A formula for the join is given in Theorem 2.7 but we shall proceed more directly. We claim that

Let A=D⋃i∈ICAio. Certainly A=DCAo, by (8) in Proposition 2.16, and ⋃i∈IAi⊆A, by (2) in Proposition 2.16.

Hence A is an upper bound for {A_i}_{i ∈ I} in ℬ^o(P). Also, if X is an upper bound in ℬ^o(P) for {A_i}_{i ∈ I}, then

Therefore A is indeed the required join. We may now appeal to Theorem 2.8 to deduce that ℬ^o(P) is a complete lattice in which joins and meets are given by

 □

Theorem 3.6

For every complete lattice L, there is an atomic poset P such that L is order-isomorphic to ℬ^o(P).

Proof

Suppose (L, ≤) is a complete lattice. Define the atomic poset P = 𝒜(P)∪ P₀, where 𝒜(P) = L and P₀ = L. Further, in P,

1. let a ∥ b for ∀ a, b ∈ 𝒜(P);

2. let a ≤ b for ∀ a ∈ 𝒜(P), b ∈ P₀ iff a = b in case that a is the least element in L, a ≤ b in case that b is the largest element in L, in other cases a ≱ b in L;

3. let a ≤ b for ∀ a, b ∈ P₀ iff a ≤ b in L.

Under the order relation defined in P, it is easy to see that P is an atomic poset. We want to show that (L, ≤) is order-isomorphic to ℬ(P).

First note that for any X ⊆ P₀ which does not contain the least and largest elements, we have

Among them, ↑_L(x) means the upper set of x in L. On the other hand,

If X(⊆ P₀) which contains the least or largest elements, we can easily check that DCXo=↓(⋁X). Therefore, we have {DCXo|X⊆P0}={↓x|x∈L}. Hence, X ⊆ ℬ^o(P) iff X = ↓_L(⋁ X). In other words, ℬ^o(P) are precisely the lower sets of L generated by a single element. It is obvious that ℬ^o(P) is isomorphic to L. □

Example 3.7

Let L = {a, b, c, d} be a complete lattice, the Hasse diagram of L is illustrated by Figure 2. We can get an atomic poset P by Theorem 3.6, whose Hasse diagram is illustrated by Figure 2, and can also get a complete lattice ℬ^o(P) which is isomorphic to L by Theorem 3.5. In Figure 2, a₁ and a₂ in P is a in L, b₁ and b₂ in P is b in L, c₁ and c₂ in P is C in L, d₁ and d₂ in P is d in L. In ℬ^o(P), A = {a₂}, B = {a₂, b₂}, C = {a₂, c₂}, D = {a₂, b₂, c₂, d₂}.

Fig. 2

The figure in Example 3.7

Theorem 4.1

Let P be an atomic poset. Then < ℱ(P);≤> forms an algebraic lattice.

Proof

We first show that < ℱ(P);≤> is a complete lattice. To show that ℱ(P);≤> is a complete lattice it suffices to show that ℱ(P);≤>is a topped intersection structure by Lemma 2.7. Given any subset T ⊆ ℱ(P), it suffices to show that ⋂ T ∈ ℱ(P). Suppose X is a finite subset of ⋂ T. Then X ⊆ t for each t ∈ T. Since each t ∈ ℱ(P), we have D¯C¯X⊆t for each t ⊆ T. This implies D¯C¯X⊆⋂T and so ⋂ T ∈ ℱ(P). It is easy to see L₀ ∈ ℱ(P) and so < ℱ(P);≤ > is a topped intersection structure.

To show that < ℱ(P);≤> is algebraic, note that D¯C¯X is a compact element for each finite X in L₀. To see this, we first show D¯C¯X∈F(P). Let X₁ be a finite subset of D¯C¯X,thenD¯C¯X1⊆D¯C¯(D¯C¯X)=D¯C¯X by Proposition 2.16, which implies D¯C¯X∈F(P). Then let {A_i|i ∈ I} be a directed subset of ℱ(P) such that

By Proposition 2.16, X⊆D¯C¯X. Therefore X⊆⋃i∈IAi. Since X is finite and {A_i|i ∈ I} is directed, X ⊆ A_k for some k ∈ I. But A_k ∈ ℱ(P), therefore D¯C¯X⊆Ak. By Definition 2.9 and Lemma 2.10, D¯C¯X is a compact element for each finite X.

Next we will show that for any T ∈ ℱ(P), T=⋃{D¯C¯X|X⊆finT}. For any X ⊆^fin T, as T ∈ ℱ(P), we have D¯C¯X⊆T.Then⋃{D¯C¯X|X⊆finT}⊆T.AsX⊆D¯C¯X,SoT=⋃X⊆⋃{D¯C¯X|X⊆finT}. Therefore T=⋃{D¯C¯X|X⊆finT}. Therefore < ℱ(P);≤> forms an algebraic lattice by Definition 2.11. □

Corollary 4.2

Let P be a finite atomic poset. Then F(P)={D¯C¯X|X⊆L}.

Proof

First we will show F(P)⊆{D¯C¯X|X⊆P}.∀A∈F(P), we have A ⊆ P₀ and for every finite subset X⊆A,D¯C¯X⊆A. As P is finite, we have that A is finite and D¯C¯A⊆A.SinceA⊆D¯C¯A by Proposition 2.16, therefore A=D¯C¯A.SoF(P)⊆{D¯C¯X|X⊆P}. Then we will show {D¯C¯X|X⊆P}⊆F(P). Since we show in Theorem 4.1, D¯C¯X is a compact element in < ℱ(P);≤> for each finite X in P₀. Since P is finite, so {D¯C¯X|X⊆L}⊆F(P).ThereforeF(P)={D¯C¯X|X⊆P}.  □

Theorem 4.3

For every algebraic lattice D, there is an atomic poset P such that D is order-isomorphic to ℱ(P).

Proof

Suppose (D, ⊑) is an algebraic lattice. Define a poset (P, ≤) = {0} ⋃ 𝒜(P)⋃ P₀ with 𝒜(P) = D, P₀ = K(D), where K(D) stands for the set of compact elements of D. Further, in P, let a∥ b for ∀a, b ∈ 𝒜(P); let a ≤ b iff b ⊑ a for ∀ a ∈ 𝒜(P), b ∈ P₀; let a ≤ b iff b ⊑ a for ∀ a, b ∈ P₀. As D is an algebraic lattice, it is easy to see that P is an atomic poset. We want to show that (D, ⊑) is order-isomorphic to ℱ(P).

First note that for any X ⊆ P₀, we have

Among them, ↑_D(⋁ X) means the upper set of ⋁ X in D. On the other hand,

Therefore, I ∈ ℱ(P) iff D¯C¯X⊆I for any finite subset X ⊆^fin I, or

for each X ⊆^fin I. Since

this is equivalent to say that I is a downward closed, directed subset of compact elements of D. A downward closed, directed subset is called an ideal. Hence, a subset of P₀ belongs to ℱ(P) if and only if it is an ideal.

At last, by the classical result about algebraic domains [3]: an algebraic domain is isomorphic to the ideal completion of the poset of its compact elements through the isomorphism

that is, ℱ(P) is isomorphic to D. □

Example 4.4

Let D ={a, b, c, d, e} be an alegraic lattice, the Hasse diagram of D is illustrated by Figure 3. We can get an atomic poset P by Theorem 4.3, whose Hasse diagram is illustrated by Figure 3, and can also get an alegraic lattice ℱ(P) which is isomorphic to D by Theorem 4.1. In Figure 3, a₁ and a₂ in P is a in D, b₁ and b₂ in P is b in D, c₁ and c₂ in P is C in D, d₁ and d₂ in P is d in D. In ℱ(P), A = {a₂}, B = {a₂, b₂}, C = {a₂, c₂}, D = {a₂, d₂}, E = {a₂, b₂, c₂, d₂, e₂}.

Fig. 3

The figure in Example 4.4

Theorem 4.5

Let P be an atomic poset. Then < ℱ^o(P);≤> forms an algebraic lattice.

Proof

We first show that > ℱ^o(P);≤> is a complete lattice. To show that ℱ^o(P);≤> is a complete lattice it suffices to show that <ℱ^o(P);≤> is a topped intersection structure by Lemma 2.7. Given any subset T ⊆ ℱ^o(P), it suffices to show that ⋂ T ∈ ℱ^o(P). Suppose X is a finite subset of ⋂T. Then X ⊆ t for each t ∈ T. Since each t ∈ ℱ^o(P), we have DCXo⊆t for each t ⊆ T. This implies DCXo⊆⋂T and so ⋂T ∈ ℱ^o(P). It is easy to see P₀∈ ℱ^o(P) and so <ℱ^o(P);≤> is a topped intersection structure.

To show that < ℱ^o(P); ≤> is algebraic, note that DCXo is a compact element for each finite X in P₀. To see this, we first show DCXo∈Fo(P). Let X₁ be a finite subset of DCXo. Then DCX1o⊆Dc(DCX∘)o=DCXo by Proposition 2.16, which implies DCXo∈Fo(P). Then let {A_i|i ∈ I} be a directed subset of ℱ^o(P) such that

By Proposition 2.16, X⊆DCXo. Therefore X⊆⋃i∈IAi. Since X is finite and {A_i|i ∈ I} is directed, X ⊆ A_k for some k ∈ I. But A_k ∈ ℱ^o(P), therefore DCXo⊆Ak. By Definition 2.9 and Lemma 2.10, DCXo is a compact element for each finite X.

Next we will show that for any T∈Fo(P),T=⋃{DCXo|X⊆finT}. For any X ⊆^finT, as T ∈ ℱ^o(P), we have DCXo⊆T.Then⋃{DCXo|X⊆finT}⊆T.AsX⊆DCXo,SoT=⋃X⊆⋃{DCXo|X⊆finT}. Therefore T=⋃{DCXo|X⊆finT}. Therefore < ℱ^o(P);≤> forms an algebraic lattice by Definition 2.11. □

Corollary 4.6

Let P be a finite atomic poset. Then Fo(P)={DCXo|X⊆L}.

Proof

First we will show Fo(P)⊆{DCXo|X⊆P}.∀A∈Fo(P), we have A ⊆ P₀ and for every finite subset X⊆A,DCXo⊆A. As P is finite, we have that A is finite and DCAo⊆A.AsA⊆DCAo by Proposition 2.16. Therefore A=DCAo.SoFo(P)⊆{DCXo|X⊆P}. Then we will show {DCXo|X⊆P}⊆Fo(P). As we show in Theorem 4.5, DCXo is a compact element in < ℱ^o(P);≤> for each finite X in P₀. As P is finite, so {DCXo|X⊆L}⊆Fo(P).ThereforeFo(P)={DCXo|X⊆P}.  □

Theorem 4.7

For every algebraic lattice D, there is an atomic poset P such that D is order-isomorphic to ℱ^o(P).

Proof

Suppose (D, ≤) is an algebraic lattice. Define a poset (P, ≤) = 𝒜(P) ⋃ P₀ with 𝒜(P) = D, P₀ = K(D), where K(D) stands for the set of compact elements of D. Further, in P,

1. let a ∥ b for ∀a, b ∈ 𝒜(P);

2. let a ≤ b for ∀ a ∈ 𝒜(P), b ∈ P₀ iff a = b in the case that a is the least in D, a ≤ b in the case that b is the largest element in L, in other cases a ≱ b in L;

3. let a ≤ b for ∀a, b ∈ P₀ iff a ≤ b in D.

As D is an algebraic lattice, it is easy to see that P is an atomic poset. We want to show that (D, ≤) is order-isomorphic to ℱ^o(P).

First note that for any X(⊆ P₀) which does not contain the least and largest elements, we have

Among them, ↑D(x) means the upper set of x in D. On the other hand,

If X(⊆ P₀) which contains the least or largest element, we can easily check that DCXo=↓(⋁X). Therefore, I∈Fo(P)iffDCXo⊆I for any finite subset X ⊆^finI, or

this is equivalent to say that I is a downward closed, directed subset of compact elements of D. A downward closed, directed subset is called an ideal. Hence, a subset of P₀ belongs to ℱ^o(P) if and only if it is an ideal.

At last, by the classical result about algebraic domains [3]: an algebraic domain is isomorphic to the ideal completion of the poset of its compact elements through the isomorphism

that is, ℱ^o(P) is isomorphic to D. □

Example 4.8

Let D= {a, b, c, d, e} be an alegraic lattice, the Hasse diagram of D is illustrated by Figure 4. We can get an atomic poset P by Theorem 4.7, whose Hasse diagram is illustrated by Figure 4, and can also get an alegraic lattice ℱ^o(P) which is isomorphic to L by Theorem 4.5. In Figure 4, a₁ and a₂ in P is a in D, b₁ and b₂ in P is b in D, c₁ and c₂ in P is C in D, d₁ and d₂ in P is d in D. In ℱ^o(P), A = {a₂}, B = {a₂, b₂}, C = {a₂, c₂}, D = {a₂, d₂}, E = {a₂, b₂, c₂, d₂, e₂}.