Lattice and metric completions of the classical logic metric space and a comparison 1

Abstract

Based on the concept of truth degree for logical formulas, a pseudo-metric is constructed on the set of all classical propositional formulas, and a metric is naturally induced on the corresponding Lindenbaum algebra of Boolean type, which constitutes a metric space, called the classical logic metric space. We respectively study both lattice completions and metric completions of this space, and compare the two kinds of completions from the angle of lattice structure as well as metric structure. On one hand, it is proved that metric completions of the classical logic metric space are complete Boolean algebras, which also act as lattice completions of the Lindenbaum algebra. On the other hand, it is pointed out that the normal lattice completion of the Lindenbaum algebra constitutes a Boolean algebra as well, which is strictly smaller than metric completions of the classical logic metric space in the sense of order-embedding. Also, the normal lattice completion can be seen as a dense subspace of the metric completions in the sense of isometry.

Keywords

Truth degree classical logic metric space metric completion lattice completion D-poset

1 Introduction

It is known that, given a poset, there exist its lattice completions, especially the normal lattice completion, which are actually complete lattices such that the given poset can be embedded into them (see, e.g., [1 –4]). Meanwhile, given a metric space, there exist its metric completions, which are actually complete metric spaces taking the former as their dense subset in the sense of isometry (see, e.g., [8, 10]). Under this circumstance, if an object possesses both order structure and metric structure, that is, it is a poset as well as a metric space with respect to an order and a metric on it, respectively, then it seems interesting to consider both kinds of completions, lattice completions together with metric completions, for this object simultaneously. Generally speaking, metric completions of this kind of objects are probably no longer provided with order structure, and at the same time the lattice completions may possess no metric any more, which states that correspondence between the two kinds of completions is unclear. Therefore, a question naturally arises: Whether this result would be different for a special object? Does there exist any structure whose completions of both types possess order and metric structure simultaneously and can be consequently compared with each other?

In order to give a positive answer to this question, the present paper considers the classical propositional logic (see, e.g., [5 , 16]). In this logic, the set F (S) consisting of all formulas can be endowed with truth degree to evaluate how close a proposition is to always being true by use of evenly distributed probabilistic measure on the set of all valuations (see [12–14 , 16] in detail). Based on the concept of truth degree, a pseudo-metric is constructed on F (S), which naturally induces a metric d on the corresponding Lindenbaum algebra $B$ . On one hand, ( $B$ ,d) constitutes a metric space, called the classical logic metric space in [15, 17], and consequently we can consider the metric completions, say ( $\hat{B}$ , $\hat{d}$ ), of ( $B$ ,d). On the other hand, it was proved in [16] that $B$ is actually a Boolean algebra, whose lattice completions, especially the Dedekind-MacNeille completion (also known as the normal completion in [4]) denoted as DM $(B)$ , exist as a result (see [3, 4] in detail). In the present paper, it is firstly obtained that the metric completion ( $\hat{B}$ , $\hat{d}$ ) of ( $B$ ,d) is a complete Boolean algebra, all operators of which are uniformly continuous with respect to its metric $\hat{d}$ . Secondly, it is pointed out that the normal lattice completion DM $(B)$ of $B$ constitutes a Boolean algebra as well, which also can be considered as a metric space with respect to some metric extended from d. Furthermore, as a metric completion of $B$ , $\hat{B}$ is proved to be a lattice completion of $B$ as well, which is strictly larger than the normal completion DM $(B)$ in such a way that DM $(B)$ can be embedded into $\hat{B}$ . Meanwhile, as the normal lattice completion of $B$ , DM $(B)$ can also be considered as a metric space but not a metric completion of $B$ , which is dense in ( $\hat{B}$ , $\hat{d}$ ) in the sense of isometry.

2 The classical logic metric space

Let S = {p₁, p₂, …} be a countable set of propositional variables and F (S) the free algebra of type (¬ , →) generated by S, where ¬, → are unary and binary operators on S, respectively. Members of S are called atoms, and that of F (S) are called propositions or formulas. For A, B ∈ F (S), we use A ∨ B, A ∧ B to denote ¬A → B, ¬ (A → ¬ B), respectively. This is the language of classical propositional logic (see, e.g., [5 , 16]). Meanwhile, define a unary operator ¬ and a binary operator → on {0, 1} as $\begin{matrix} \neg 0 & = & 1, \neg 1 = 0, \\ a \to b = 0 iff a & = & 1 and b = 0, a, b \in {0, 1} . \end{matrix}$

Then {0, 1} becomes an algebra of type (¬ , →) as well. A mapping v : F (S) ⟶ {0, 1} is called a valuation of F (S) if v is a homomorphism of type (¬ , →), i.e., $\begin{matrix} v (\neg A) & = & \neg v (A), \\ v (A \to B) & = & v (A) \to v (B) \end{matrix}$ hold for every A, B ∈ F (S). The set consisting of all valuations of F (S) is denoted as Ω.

Note that a valuation v of F (S) is totally determined by its restriction v|_S since F (S) is the free algebra generated by S, thus v can be considered as a vector v = (v (p₁) , v (p₂) , …) ∈ {0, 1} ^ω, that is, an infinite binary sequence. In this sense, we do not distinguish Ω from {0, 1} ^ω in the following text.

Suppose that X_n = {0, 1} and μ_n is the evenly distributed probabilistic measure on X_n (n = 1, 2, ⋯), i.e., $μ_{n} (\emptyset) = 0, μ_{n} (X_{n}) = 1, μ_{n} ({0}) = μ_{n} ({1}) = \frac{1}{2} .$

Let $X = \prod_{n = 1}^{\infty} X_{n}$ and μ = μ₁× μ₂ × ⋯ be the infinite product measure. Then it was proved in [7] that μ is actually an evenly distributed probabilistic measure on X = {0, 1} ^ω = Ω for which $E \times \prod_{n = m + 1}^{\infty} X_{n}$ is measurable and $μ (E \times \prod_{n = m + 1}^{\infty} X_{n}) = (μ_{1} \times \dots \times μ_{m}) (E),$ where $E \subseteq \prod_{n = 1}^{m} X_{n}$ (m = 1, 2, ⋯).

Definition 2.1. [13 , 17] Let A ∈ F (S). Define $[A] = {v \in Ω ∣ v (A) = 1}, τ (A) = μ ([A]) .$ (1)

Then τ (A) is called the truth degree of A.

Note that if A ∈ F (S) and A = A (p₁, ⋯ , p_n) contains n atoms p₁, ⋯ , p_n, then $\begin{matrix} [A] & = & {(v (p_{1}), \dots, v (p_{n})) \in \prod_{i = 1}^{n} X_{i} ∣ v (A) = 1} \\ \times \prod_{i = n + 1}^{\infty} X_{i} \end{matrix}$ is certainly measurable. Consequently the truth degree τ (A) of A exists and Equation. (1) is well-defined. For example, it is easy to calculate that $τ (p_{1}) = \frac{1}{2}$ , $τ (p_{1} \to p_{2}) = \frac{3}{4}$ , $τ (p_{1} \land p_{2} \land p_{3}) = \frac{1}{8}$ , etc.

Proposition 2.2. [13 , 17] LetA, B ∈ F (S). Then

τ (A) =1 if and only ifAis a tautology;τ (A) =0 if and only ifAis a contradiction.

τ (A) + τ (¬ A) =1.

τ (A ∨ B) = τ (A) + τ (B) - τ (A ∧ B).

IfA → Bis a tautology, thenτ (A) ≤ τ (B).

Define a mapping ρ : F (S) × F (S) ⟶ [0, 1] for A, B ∈ F (S) as $ρ (A, B) = 1 - τ ((A \to B) \land (B \to A)) .$ (2)

Then the following proposition can be obtained.

Proposition 2.3. [13 , 17] The mappingρdefined by Equation. (2) is a pseudo-metric onF (S). That is, forA, B, C ∈ F (S),

ρ (A, B) =0 if and only ifA ≈ B; ρ (A, B) =1 if and only ifA ≈ ¬ B, where ≈ is the logical equivalence relation onF (S), i.e., $A \approx B iff v (A) = v (B) holds for every v \in Ω .$

ρ (A, B) = ρ (B, A).

ρ (A, B) + ρ (B, C) ≥ ρ (A, C).

Besides, the following conditions also hold for every A, B ∈ F (S):

ρ (A, ⊤) =1 - τ (A), ρ (A, ¬ ⊤) = τ (A), where ⊤ stands for any tautology in F (S).

ρ (A, B) = μ ({v ∈ Ω ∣ v (A) ≠ v (B)}).

Note that it was proved in [16] that the logical equivalence relation ≈ is a congruence relation on F (S), and consequently we obtain a quotient class of F (S) by ≈, denoted as $B$ , that is, $B$ = F (S)/≈ $= {\bar{A} ∣ A \in F (S)}$ , where $\bar{A} = {B \in F (S) ∣ A \approx B}$ . Define a binary relation ≤ and operators ∨, ∧ , ′, → on $B$ as follows: $\bar{A} \leq \bar{B} if and only if A \to B is a taulology,$ (3) $\begin{matrix} \bar{A} \lor \bar{B} & = & \bar{A \lor B}, \bar{A} \land \bar{B} = \bar{A \land B}, \\ (\bar{A})^{'} & = & \bar{\neg A}, \bar{A} \to \bar{B} = \bar{A \to B} . \end{matrix}$ (4)

Then it was proved in [16] that ≤ is a partial order on $B$ , and $(B, \lor, \land,^{'}, \bar{\neg ⊤}, \bar{⊤})$ constitutes a Boolean algebra with respect to ≤, taking $\bar{\neg ⊤}, \bar{⊤}$ as the least and greatest element in $B$ , respectively. This Boolean algebra $B$ is known as the Lindenbaum algebra (see [16] in detail).

Since the mapping ρ in Equation. (2) is indeed a pseudo-metric on F (S) satisfying that ρ (A, B) =0 if and only if A ≈ B, then ρ naturally induces a metric d on the Lindenbaum algebra $B$ , i.e., $d : B \times B ⟶ [0, 1]$ , as follows: $d (\bar{A}, \bar{B}) = ρ (A, B), \bar{A}, \bar{B} \in B .$ (5)

As a result, $(B, d)$ constitutes a metric space, called the classical logic metric space (also the logic metric space in [13 , 17]).

3 Metric completions of the classical logic metric space

It was proved in [15] that the classical logic metric space $(B, d)$ , as a metric space, is not complete. In other words, one can find a Cauchy sequence in $B$ not converging to any equivalence class of formulas in F (S). However, there exist metric completions of $(B, d)$ , which are certainly complete metric spaces as shown in the following proposition.

Proposition 3.1. [10] Each metric space (X, r) can be mapped by a one-to-one isometryfonto a dense subset of a complete metric space $(\hat{X}, \hat{r})$ .

Under this circumstance, the pair $〈 f, (\hat{X}, \hat{r}) 〉$ is called a metric completion of (X, r), and we also say that $(\hat{X}, \hat{r})$ is a metric completion of (X, r) whenever there is no confusion (see [10] in detail). Note that f is a one-to-one isometry, i.e., $r (x, y) = \hat{r} (f (x), f (y)), x, y \in X .$

Therefore, (X, r) can be considered as a subspace of $(\hat{X}, \hat{r})$ in the sense of isometry and consequently is dense in $(\hat{X}, \hat{r})$ .

As to the classical logic metric space $(B, d)$ , there certainly exist its metric completions. Suppose that $(\hat{B}, \hat{d})$ is one such metric completion, then $(\hat{B}, \hat{d})$ is a complete metric space taking $(B, d)$ as its dense subset. In order to study further structure of $(\hat{B}, \hat{d})$ , the following concepts are needed.

Definition 3.2. [9, 11] A difference poset (briefly, D-poset) is a quadruple (D, ≤ , ⊖ , 1), where D is a nonempty set partially ordered by ≤, 1 is the greatest element of D, and ⊖ is the difference operator which defines for every a, b ∈ D with a ≤ b an element b ⊖ a in such a way that the following conditions hold:

b ⊖ a ≤ b.

b ⊖ (b ⊖ a) = a.

a ≤ b ≤ c implies c ⊖ b ≤ c ⊖ a, and (c ⊖ a) ⊖ (c ⊖ b) = b ⊖ a.

It can be easily inferred that in any D-poset,0 = 1 ⊖1 is the least element.

Definition 3.3. [11] Let D be a D-poset. If D is a lattice and $T$ is a topology on D, then $(D, T)$ is called a topological lattice D-poset if the following conditions hold:

The mapping $⊖ : (G, T_{0}) ⟶ (D, T)$ is continuous, where G = {(a, b) ∈ D × D ∣ a ≤ b} and $T_{0}$ is the relative topology on G induced by the product topology $T \times T$ , i.e., $T_{0} = (T \times T) |_{G}$ .

The mappings $\lor, \land : (D \times D, T \times T) ⟶ (D, T)$ are continuous.

Furthermore, if $T$ is uniform and the mappings ⊖, ∨ , ∧ are uniformly continuous, then $(D, T)$ is called a uniform lattice D-poset.

Example 3.4. Consider the classical logic metric space $(B, d)$ . Note that $(B, \lor, \land,^{'}, \bar{\neg ⊤}, \bar{⊤})$ is a Boolean algebra, where the operators ∨, ∧ , ′ are defined by Equation. (4). Let $G = {(a, b) \in B \times B ∣ a \leq b}$ , and define $⊖ : G ⟶ B$ as $b ⊖ a = b \land a^{'}, a, b \in B, a \leq b .$ (6)

Then b ⊖ a ≤ b is obviously true, and b ⊖ (b ⊖ a) = b ∧ (b ∧ a′) ′ = b ∧ (b′ ∨ a) = (b ∧ b′)∨ (b ∧ a) = b ∧ a = a since a ≤ b. Assume that a ≤ b ≤ c, then c ⊖ b = c ∧ b′ ≤ c ∧ a′ = c ⊖ a. Moreover, (c ⊖ a) ⊖ (c ⊖ b) = (c ∧ a′) ∧ (c ∧ b′) ′ = (c ∧ a′) ∧ (c′ ∨ b) = (c ∧ a′ ∧ c′) ∨ (c ∧ a′ ∧ b) = c ∧ a′ ∧ b = b ∧ a′ = b ⊖ a. Therefore, it follows from Definition. 3.2 that $(B, \leq, ⊖, 1)$ is a D-poset, where $1 = \bar{⊤}$ and ≤ is the partial order on $B$ defined by Equation. (3).

Let $T$ be the metric topology on $B$ induced by d. Then it was proved in [10] that $T$ is same with some uniform topology on $B$ , and consequently $T$ is uniform. Assume that $a, b \in B$ , and d (a, b) < ɛ holds for ɛ > 0 with $a = \bar{A}, b = \bar{B}$ (A, B ∈ F (S)). Then it follows from Equation. (5), Proposition. 2.3 and $a^{'} = \bar{\neg A}, b^{'} = \bar{\neg B}$ that $\begin{matrix} d (a^{'}, b^{'}) & = & μ ({v \in Ω ∣ v (\neg A) \neq v (\neg B)}) \\ = & μ ({v \in Ω ∣ 1 - v (A) \neq 1 - v (B)}) \\ = & μ ({v \in Ω ∣ v (A) \neq v (B)}) \\ = & d (a, b) < ɛ . \end{matrix}$

Hence the operator’ $: (B, d) ⟶ (B, d)$ is uniformly continuous with respect to d. Moreover, assume that $a_{1}, a_{2}, b_{1}, b_{2} \in B$ , and d (a₁, a₂) < ɛ, d (b₁, b₂) < ɛ hold for ɛ > 0 with $a_{1} = \bar{A_{1}}, a_{2} = \bar{A_{2}}, b_{1} = \bar{B_{1}}, b_{2} = \bar{B_{2}}$ (A₁, A₂, B₁, B₂ ∈ F (S)). Then $\begin{matrix} d (a_{1} \to b_{1}, a_{2} \to b_{2}) \\ = & μ ({v \in Ω ∣ v (A_{1} \to B_{1}) \neq v (A_{2} \to B_{2})}) \\ = & μ ({v \in Ω ∣ v (A_{1}) \to v (B_{1}) \neq v (A_{2}) \to v (B_{2})}) \\ \leq & μ ({v \in Ω ∣ v (A_{1}) \neq v (A_{2}) or v (B_{1}) \neq v (B_{2})}) \\ \leq & μ ({v \in Ω ∣ v (A_{1}) \neq v (A_{2})}) \\ + μ ({v \in Ω ∣ v (B_{1}) \neq v (B_{2})}) \\ = & d (a_{1}, a_{2}) + d (b_{1}, b_{2}) < 2 ɛ . \end{matrix}$

Hence the operator $\to : (B, d) \times (B, d) ⟶ (B, d)$ is also uniformly continuous with respect to d. Since for every $a, b \in B$ , we have a ∨ b = a′ → b, a ∧ b = (a → b′) ′, and b ⊖ a = b ∧ a′ (a ≤ b), hence the operators ⊖, ∨ , ∧ are uniformly continuous with respect to d as well. This proves that $(B, d)$ is a uniform lattice D-poset.

Definition 3.5. Let (B, ∨ , ∧ , ′, 0, 1) be a Boolean algebra and r a metric on B. If the operators ∨, ∧ , ′ on B are uniformly continuous with respect to r, then the pair (B, r) is called a metric Boolean algebra.

It can be directly obtained the following corollary from above.

Corollary 3.6. The classical logic metric space $(B, d)$ is a metric Boolean algebra.

The following proposition was proved in [11]. And for the definition of net, please refer to [8, 10] in detail.

Proposition 3.7. [11] If $(D, U)$ is a Hausdorff uniform lattice D-poset in which all monotone nets are Cauchy, then there exist extensions of ≤ and ⊖ on $\hat{D}$ such that $(\hat{D}, \hat{U})$ is also a Hausdorff uniform lattice D-poset, and $\hat{D}$ is a complete lattice, where $(\hat{D}, \hat{U})$ is a completion of the uniform space $(D, U)$ .

We see in the above that the classical logic metric space $(B, d)$ is a uniform lattice D-poset, which is clearly Hausdorff. In the following, we are going to prove that the metric completion $(\hat{B}, \hat{d})$ of $(B, d)$ is also a uniform lattice D-poset as well as a metric Boolean algebra. First of all, we need a lemma.

Lemma 3.8. Every monotone sequence in $(B, d)$ is Cauchy.

Proof. We only prove the case for increasing sequences in $(B, d)$ , since the proof for decreasing ones is similar. Let {a_n} be an increasing sequence in $(B, d)$ , i.e., a₁≤ a₂ ≤ ⋯. Suppose that $a_{n} = \bar{A_{n}}$ (A_n ∈ F (S), n = 1, 2, ⋯), then it follows from Eq. (3) that v (A₁)≤ v (A₂) ≤ ⋯ holds for every v ∈ Ω. Hence it can be obtained that τ (A₁)≤ τ (A₂) ≤ ⋯ by Equation. (1), that is, {τ (A_n)} is an increasing sequence in [0,1] and consequently is Cauchy. Let ɛ be any given positive number, then there exists an m ∈ Z⁺ such that |τ (A_k) - τ (A_k+l) | < ɛ holds whenever k ≥ m (l = 1, 2, ⋯). Note that {v ∈ Ω ∣ v (A_k+l) =1} = {v ∈ Ω ∣ v (A_k) =1} ∪ {v ∈ Ω ∣ v (A_k) =0, v (A_k+l) =1}, and {v∈ Ω ∣ v (A_k) =1} ∩ {v ∈ Ω ∣ v (A_k) =0, v (A_k+l) =1} = ∅. As a result, $\begin{matrix} τ (A_{k + l}) & = & μ ({v \in Ω ∣ v (A_{k + l}) = 1} \\ = & μ ({v \in Ω ∣ v (A_{k}) = 1}) \\ + μ ({v \in Ω ∣ v (A_{k}) = 0, v (A_{k + l}) = 1}) \\ = & τ (A_{k}) + μ ({v \in Ω ∣ v (A_{k}) \neq v (A_{k + l})}) \\ = & τ (A_{k}) + ρ (A_{k}, A_{k + l}) . \end{matrix}$

Hence d (a_k, a_k+l) = ρ (A_k, A_k+l) = τ (A_k+l) - τ (A_k)<ɛ holds whenever k ≥ m (l = 1, 2, ⋯). This proves that {a_n} is a Cauchy sequence in $(B, d)$ , and the proof is completed.□

Since the classical logic metric space $(B, d)$ is a Hausdorff uniform lattice D-poset in which all monotone sequences are Cauchy as shown in Lemma. 3.8, it follows from Proposition. 3.7 that the metric completion $(\hat{B}, \hat{d})$ of $(B, d)$ is a Hausdorff uniform lattice D-poset as well, and $\hat{B}$ is a complete lattice. Next, we are going to prove that $(\hat{B}, \hat{d})$ is also a metric Boolean algebra.

Theorem 3.9. The metric completion $(\hat{B}, \hat{d})$ of the classical logic metric space $(B, d)$ is a metric complete Boolean algebra.

Proof. First of all, it follows from Proposition. 3.7 that there exist extensions of ≤ and ⊖ (defined by Equations. (3) and (6), respectively) on $\hat{B}$ such that $\hat{B}$ is a complete lattice, and the extended operators ⊖, ∨ , ∧ on $\hat{B}$ (we use the same notations when there is no confusion) are uniformly continuous with respect to $\hat{d}$ . Note that the greatest (or least) element of $\hat{B}$ is the same greatest (or least) element of $B$ (as proved in [11]).

Let $a, b, c \in \hat{B}$ . Since $(B, d)$ is a dense subset of $(\hat{B}, \hat{d})$ , then there exist sequences {a_n} , {b_n} , {c_n} in $B$ such that $lim_{n \to \infty} a_{n} = a$ , $lim_{n \to \infty} b_{n} = b$ , and $lim_{n \to \infty}$ c_n = c. For every n (n = 1, 2, ⋯), a_n ∧ (b_n ∨ c_n) = (a_n ∧ b_n) ∨ (a_n ∧ c_n) holds since $B$ is a Boolean algebra. Then a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) holds since ∨, ∧ are uniformly continuous with respect to $\hat{d}$ . As a result, $(\hat{B}, \lor, \land, 0, 1)$ is a distributive complete lattice.

Define a unary operator $^{c} : \hat{B} ⟶ \hat{B}$ as $x^{c} = 1 ⊖ x, x \in \hat{B} .$ (7)

Then for every $a \in B$ , a^c = 1 ⊖ a = 1 ∧ a′ = a′ by Equation. (6), which states that the operator ^c on $\hat{B}$ is an extension of the complementary operator ^′ on $B$ . Besides, the operator ^c is uniformly continuous with respect to $\hat{d}$ since ⊖ is uniformly continuous with respect to $\hat{d}$ .

Assume $x \in \hat{B}$ , then there exists a sequence {x_n} in $B$ such that $lim_{n \to \infty} x_{n} = x$ . It follows from the continuity of ⊖ that $x^{c} = 1 ⊖ x = 1 ⊖ lim_{n \to \infty} x_{n} = lim_{n \to \infty} (1 ⊖ x_{n}) = lim_{n \to \infty} x_{n}^{'} .$ Hence, it follows from the continuity of ∨ that $x \lor x^{c} = (lim_{n \to \infty} x_{n}) \lor (lim_{n \to \infty} x_{n}^{'}) = lim_{n \to \infty} (x_{n} \lor x_{n}^{'}) = 1 .$

Similarly, it can be proved that x ∧ x^c = 0, which states that ^c is a complementary operator on $\hat{B}$ . Therefore, $(\hat{B}, \lor, \land,^{c}, 0, 1)$ is a complete Boolean algebra, where the operators ∨, ∧ , ^c are uniformly continuous with respect to $\hat{d}$ , and consequently $(\hat{B}, \hat{d})$ is a metric complete Boolean algebra.□

4 Lattice completions of the classical logic metric space and comparison with metric completions

From above we see that the metric completion $(\hat{B}, \hat{d})$ of the classical logic metric space $(B, d)$ is a complete metric space, and $\hat{B}$ is a complete Boolean algebra with all its operators uniformly continuous with respect to $\hat{d}$ . A question naturally arises: Whether $\hat{B}$ is also a lattice completion, especially the normal completion, of $B$ ?

Definition 4.1. [4] Let P be a poset. If φ : P ⟶ L is an order-embedding, i.e., $x \leq_{P} y if and only if φ (x) \leq_{L} φ (y), x, y \in P,$ and L is a complete lattice, then we say that L is a lattice completion of P (via the order-embedding φ).

Since $(B, d)$ can be considered as a dense subset of $(\hat{B}, \hat{d})$ , then define $φ : B ⟶ \hat{B}$ as $φ (a) = a, a \in B .$

Note that the order on $\hat{B}$ is an extension of the order ≤ on $B$ as shown in the proof of Theorem. 3.9, hence it is easy to obtain that $a \leq b if and only if φ (a) \leq φ (b), a, b \in B,$ that is, φ is an order-embedding. Since $\hat{B}$ is a complete lattice by Theorem. 3.9, then $\hat{B}$ is actually a lattice completion of $B$ .

Given a poset P, there exist many lattice completions of P, some of which are unnecessarily large (see [4] in detail). However, among all the lattice completions of P, there is a smallest one in the sense of order-isomorphism. In fact, define $DM (P) = {A \subseteq P ∣ (A^{u})^{l} = A},$ (8) where the operators ^u and ^l are defined by $\begin{matrix} A^{u} & = & {x \in P ∣ \forall a \in A, a \leq x}, \\ A^{l} & = & {x \in P ∣ \forall a \in A, a \geq x}, \end{matrix} A \subseteq P .$

Then it was proved in [4] that (DM (P) , ⊆) is a complete lattice, and the mapping φ : P ⟶ DM (P) defined by $φ (x) = ↓ x, x \in P$ is an order-embedding, where ↓x = {x} ^l = {y ∈ P ∣ y ≤ x}. Therefore, DM (P) is a lattice completion of P, known as the Dedekind-MacNeille completion (also the normal completion) of P (see [1 –4] for more details).

Proposition 4.2. [1 –3] LetPbe a poset. Ifh : P ⟶ Lis an order-embedding, andLis a complete lattice, then there exists an order-embeddingh^* : DM (P) ⟶ L, which is an extension ofh.

It follows from Proposition. 4.2 that among all the lattice completions of a given poset, the normal completion is the smallest one in the sense of order-isomorphism.

As to the classical logic metric space $(B, d)$ , since $B$ is a Boolean algebra, then $DM (B)$ defined by Equation. (8) is the normal lattice completion of $B$ . Meanwhile, we have already obtained that $\hat{B}$ is also a lattice completion of $B$ . Therefore, it follows from Proposition. 4.2 that $\hat{B}$ is not less than $DM (B)$ in such a way that $DM (B)$ can be embedded into $\hat{B}$ . In the following, we will further prove that $\hat{B}$ is strictly larger than $DM (B)$ .

Proposition 4.3.Let $(\hat{B}, \hat{d})$ be the metric completion of the classical logic metric space $(B, d)$ . Define $B^{*} = {x \in \hat{B} ∣ \exists E \subseteq B s . t . x = \sup_{\hat{B}} E} .$ (9)

Then $B^{*}$ is a lattice completion of $B$ .

Proof. Note that $B$ can be considered as a subset of $\hat{B}$ . Suppose that $x \in B$ . Let E = {x}, then it is obvious that $E \subseteq B$ and $x = \sup_{\hat{B}} E$ . Therefore, $x \in B^{*}$ , and consequently $B \subseteq B^{*}$ . Define the mapping $φ : B ⟶ B^{*}$ as $φ (a) = a, a \in B .$

Since $B^{*} \subseteq \hat{B}$ and the order on $\hat{B}$ is an extension of the order ≤ on $B$ , hence φ is clearly an order-embedding.

It only needs to prove that $B^{*}$ is a complete lattice. In fact, since $0 = \sup_{\hat{B}} \emptyset$ , then $0 \in B^{*}$ . Suppose that $\emptyset \neq Q \subseteq B^{*}$ . For every x ∈ Q, there exists a set $E_{x} \subseteq B$ such that $x = \sup_{\hat{B}} E_{x}$ . Let E = ∪ _x∈QE_x, then $E \subseteq B$ . For every x ∈ Q, E = ∪ _x∈QE_x ⊇ E_x, and consequently $\sup_{\hat{B}} E \geq \sup_{\hat{B}} E_{x} = x$ . Therefore, $\sup_{\hat{B}} E$ is an upper bound of Q in $\hat{B}$ . Furthermore, suppose that y is also an upper bound of Q in $\hat{B}$ . Then for every x ∈ Q and every z ∈ E_x, $y \geq x = \sup_{\hat{B}} E_{x} \geq z$ . As a result, y is also an upper bound of E = ∪ _x∈QE_x, and consequently $y \geq \sup_{\hat{B}} E$ . This proves that $\sup_{\hat{B}} E$ is the least upper bound of Q in $\hat{B}$ , i.e., $\sup_{\hat{B}} Q = \sup_{\hat{B}} E \in B^{*}$ . Therefore, $B^{*}$ is closed under (nonempty) $\sup_{\hat{B}}$ and contains the least element 0, hence $B^{*}$ is a complete lattice (as proved in [4]). The proof is completed.□

Lemma 4.4.Let (B, r) be a metric Boolean algebra as well as a complete metric space. If {a_n} is an increasing Cauchy sequence in (B, r) and $a = sup_{n} {a_{n}}$ exists, then {a_n} converges toa.

Proof. Since {a_n} is an increasing Cauchy sequence in (B, r) and (B, r) is metric complete, then {a_n} converges to an element a^* ∈ B. On one hand, it follows from Definition. 3.5 and a_n ∨ a = a (n = 1, 2, ⋯) that $a = lim_{n \to \infty} (a_{n} \lor a) = (lim_{n \to \infty} a_{n}) \lor a = a^{*} \lor a$ . Hence a^* ≤ a. On the other hand, fix an a_n. Then it follows from Definition. 3.5 and a_n = a_n ∧ a_n+k (k = 1, 2, ⋯) that $a_{n} = lim_{k \to \infty} (a_{n} \land a_{n + k}) = a_{n} \land (lim_{k \to \infty} a_{n + k}) = a_{n} \land a^{*}$ . Hence a_n ≤ a^* holds for every n (n = 1, 2, ⋯). This means that a^* is an upper bound of {a_n}, and consequently a ≤ a^*. As a result, {a_n} converges to a^* = a.□

Proposition 4.5. $B^{*}$ is a sublattice of $\hat{B}$ .

Proof. Let $x \in \hat{B}$ and $E \subseteq B$ . Define $x \land E = {\begin{matrix} {x \land e ∣ e \in E}, E \neq \emptyset, \\ \emptyset, E = \emptyset . \end{matrix}$

Firstly, we are going to prove that $x \land \sup_{\hat{B}} E = \sup_{\hat{B}} (x \land E) .$ (10)

In fact, if E =∅, then Equation. (10) holds since $\sup_{\hat{B}} \emptyset = 0$ . If E≠ ∅, then elements of E can be listed as b₁, b₂, ⋯ (in case E is finite, repeat an element infinitely) since $| B | = ω$ . Let a_n = b₁ ∨ ⋯ ∨ b_n (n = 1, 2, ⋯). Then {a_n} is an increasing sequence in $(B, d)$ , hence it follows from Lemma. 3.8 that {a_n} is Cauchy with respect to d. Since $(\hat{B}, \hat{d})$ is a metric completion of $(B, d)$ , {a_n} is also an increasing Cauchy sequence in $(\hat{B}, \hat{d})$ . Since $(\hat{B}, \hat{d})$ is a metric Boolean algebra and $\hat{B}$ is a complete lattice by Theorem. 3.9, then it follows from Lemma. 4.4 that $lim_{n \to \infty} a_{n} = \sup_{\hat{B}} {a_{n} ∣ n = 1, 2, \dots} = \sup_{\hat{B}} E$ holds in $(\hat{B}, \hat{d})$ . Since the operators ∨, ∧ on $\hat{B}$ are uniformly continuous with respect to $\hat{d}$ , then $lim_{n \to \infty} (x \land a_{n}) = x \land lim_{n \to \infty} a_{n} = x \land \sup_{\hat{B}} E .$

Let $y = \sup_{\hat{B}} (x \land E)$ , then it is easy to obtain that $y \leq x \land \sup_{\hat{B}} E$ . Meanwhile, it follows from (x ∧ a_n) ∨ y = y (n = 1, 2, ⋯) that $\begin{matrix} y & = & lim_{n \to \infty} [(x \land a_{n}) \lor y] = [lim_{n \to \infty} (x \land a_{n})] \lor y \\ = & (x \land \sup_{\hat{B}} E) \lor y . \end{matrix}$

Hence $y \geq x \land \sup_{\hat{B}} E$ , and Equation. (10) follows.

Next, we will prove that $B^{*}$ is a sublattice of $\hat{B}$ based on Equation. (10). It is obvious that $\emptyset \neq B^{*} \subseteq \hat{B}$ . Assume that $u, v \in B^{*}$ , then it follows from the proof of Proposition. 4.3 that $u \lor_{\hat{B}} v \in B^{*}$ . Thus, it only needs to prove that $u \land_{\hat{B}} v \in B^{*}$ , and the index in $\land_{\hat{B}}$ is omitted in the sequel. In fact, there exist $E, F \subseteq B$ such that $u = \sup_{\hat{B}} E$ , $v = \sup_{\hat{B}} F$ . We only consider the case that E≠ ∅, F≠ ∅. It follows from Equation. (10) that $\begin{matrix} u \land v & = & u \land \sup_{\hat{B}} F \\ = & \sup_{\hat{B}} (u \land F) \\ = & \sup_{\hat{B}} [(\sup_{\hat{B}} E) \land F] \\ = & \sup_{\hat{B}} {f \land \sup_{\hat{B}} E ∣ f \in F} \\ = & \sup_{\hat{B}} {\sup_{\hat{B}} (f \land E) ∣ f \in F} \\ = & \sup_{\hat{B}} {e \land f ∣ e \in E, f \in F} \in B^{*} . \end{matrix}$

This proves that $B^{*}$ is a sublattice of $\hat{B}$ .□

Proposition 4.6. $\hat{B}$ is strictly larger than $B^{*}$ .

Proof. Define A₁, A₂, ⋯ in F (S) as follows: $\begin{matrix} A_{1} & = & p_{1} \lor p_{2}, \\ A_{k} & = & A_{k - 1} \land (p_{n} \lor \dots \lor p_{n + k}), k = 2, 3, \dots, \end{matrix}$ where $n = \sum_{i = 1}^{k} i$ . Let $a_{n} = \bar{A_{n}}$ (n = 1, 2, ⋯). Then it is easy to infer that {a_n} is a decreasing sequence in $B$ , and it follows from Lemma. 3.8 that {a_n} converges to an element e of $\hat{B}$ . Since the operator ∧ on $\hat{B}$ is uniformly continuous with respect to $\hat{d}$ , then $\begin{matrix} e \land a_{n} & = & (lim_{k \to \infty} a_{n + k}) \land a_{n} = lim_{k \to \infty} (a_{n + k} \land a_{n}) \\ = & lim_{k \to \infty} a_{n + k} = e . \end{matrix}$

Hence e ≤ a_n holds for every n (n = 1, 2, ⋯). It only needs to prove that $e \notin B^{*}$ .

In fact, let b be an element of $B$ with b ≠ 0. Then there exist a formula B ∈ F (S) such that $b = \bar{B}$ , and B = B (p_{i
₁}, ⋯ , p_{i
_l}) is not a contradiction. Hence there exists a valuation v ∈ Ω such that v (B) =1. Define v^* ∈ Ω as $v^{*} (p_{j}) = {\begin{matrix} v (p_{j}), & j = i_{1}, \dots, i_{l}, \\ 0, & j \notin {i_{1}, \dots, i_{l}} . \end{matrix}$

Then v^* (B) =1. Let m = max {i₁, …, i_l}, then v^* (A_k) =0 whenever $\sum_{i = 1}^{k} i \geq m + 1$ . Hence $\bar{B} \leq \bar{A_{k}}$ does not hold whenever k is large enough, that is, b ≤ a_k does not hold for some a_k. Since we have proved that e ≤ a_n holds for every n (n = 1, 2, ⋯), then b ≤ e does not hold for every $b \in B$ unless b = 0. Suppose on the contrary that $e \in B^{*}$ , then there exist a set $E \subseteq B$ such that $e = \sup_{\hat{B}} E$ . Thus it follows from Equation. (10) that $e = e \land \sup_{\hat{B}} E = \sup_{\hat{B}} (e \land E) \leq \sup_{\hat{B}} {b \in B ∣ b \leq e} = 0$ , and consequently $lim_{n \to \infty} a_{n} = e = 0$ . Since the operator $^{c} : \hat{B} \to \hat{B}$ defined by Equation. (7) is uniformly continuous with respect to $\hat{d}$ , then $lim_{n \to \infty} a_{n}^{c} = (lim_{n \to \infty} a_{n})^{c} = 0^{c} = 0^{'} = 1$ . Note that $a_{n}^{c} = \bar{\neg A_{n}}$ , and $\begin{matrix} \neg A_{1} & = & \neg p_{1} \land \neg p_{2}, \\ \neg A_{k} & = & \neg A_{k - 1} \lor (\neg p_{n} \land \dots \land \neg p_{n + k}), k = 2, 3, \dots, \end{matrix}$ where $n = \sum_{i = 1}^{k} i$ . It is easy to verify that $\begin{matrix} τ (\neg A_{1}) & = & \frac{1}{2^{2}}, \\ τ (\neg A_{k}) & \leq & \frac{1}{2^{2}} + \dots + \frac{1}{2^{k + 1}}, k = 2, 3, \dots . \end{matrix}$

Hence $\begin{matrix} \hat{d} (a_{k}^{c}, 1) & = & ρ (\neg A_{k}, ⊤) = 1 - τ (\neg A_{k}) \\ \geq & 1 - (\frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots) = \frac{1}{2} . \end{matrix}$ This contradicts with $lim_{n \to \infty} a_{n}^{c} = 1$ , and the proof is completed.□

Theorem 4.7. $\hat{B}$ is a lattice completion, but not the normal completion, of $B$ .

Proof. It follows from Propositions. 4.3–4.6 that there exists a lattice completion, $B^{*}$ , of $B$ such that $\hat{B}$ is strictly larger than $B^{*}$ . Since $\hat{B}$ is a lattice completion of $B$ as shown in the above, then it follows from Prop. 4.2 that $\hat{B}$ is strictly larger than the normal completion $DM (B)$ of $B$ in such a way that $DM (B)$ can be embedded into $\hat{B}$ . This completes the proof.□

As we see in the above, $\hat{B}$ is strictly larger than the normal completion $DM (B)$ of $B$ in such a way that $DM (B)$ can be embedded into $\hat{B}$ . If we consider $DM (B)$ as a subset of $\hat{B}$ , then the mapping $\hat{d} |_{DM (B)} : DM (B) \times DM (B) ⟶ [0, 1]$ is naturally a metric on $DM (B)$ . In this sense, $(DM (B), \hat{d} |_{DM (B)})$ can be considered as a dense subspace of $(\hat{B}, \hat{d})$ , since $B$ is dense in $\hat{B}$ and $B$ can be considered as a subset of $DM (B)$ . However, as a metric space, $(DM (B), \hat{d} |_{DM (B)})$ is not complete, since by Proposition. 4.6 one can find a sequence {a_n} in $B \subseteq DM (B)$ converging to $e \notin B^{*}$ , then of course $e \notin DM (B)$ .

Note that since $B$ is a Boolean algebra, then its normal completion $DM (B)$ is also a Boolean algebra as proved in [3]. However, as a subset of $\hat{B}$ in the sense of order-embedding, $DM (B)$ is not a sub-Boolean algebra of the Boolean algebra $\hat{B}$ . In fact, it can be proved that $DM (B)$ is not closed under the operator ^c defined by Equation. (7).

5 Conclusions

In this paper, we have studied lattice completions and metric completions, respectively, of the classical logic metric space ( $B$ ,d), and compared the two kinds of completions from the angle of lattice structure as well as metric structure. We proved that the metric completion ( $\hat{B}$ , $\hat{d}$ ) of ( $B$ ,d) is a complete Boolean algebra, all operators of which are uniformly continuous with respect to its metric $\hat{d}$ . We also pointed out that the normal lattice completion DM $(B)$ of $B$ constitutes a Boolean algebra as well, which also can be considered as a metric space with respect to some metric extended from d. Furthermore, $\hat{B}$ was proved to be a lattice completion of $B$ as well, which is strictly larger than the normal completion DM $(B)$ in such a way that $DM (B)$ can be embedded into $\hat{B}$ . Meanwhile, DM $(B)$ can also be considered as a metric space but not a metric completion of $B$ , which is dense in ( $\hat{B}$ , $\hat{d}$ ) in the sense of isometry. Besides, if $B$ is a complete lattice, then DM $(B)$ can be proved to be isomorphic to $B$ in [3, 4]. That is, $B$ is the normal completion of itself in the sense of order-isomorphism. Otherwise, $B$ is strictly less than DM $(B)$ . The further comparison among $B$ , DM $(B)$ and $\hat{B}$ will be investigated in a forthcoming paper. Relevant references such as [18] will help to improve our future work.

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