Fuzzy Filter Spectrum of a BCK Algebra

Lemma 2.1. Let {η_α∣α ∈ Ω} be a family of fuzzy filters of X. Then, ⋂_α∈Ωη_α is a fuzzy filter of X.

Proof. Let x ∈ X. For any α ∈ Ω, η_α(1) ≥ η_α(x) since η_α <_FF X. Then, inf _α∈Ωη_α(1) ≥ inf _α∈Ωη_α(x) and so, ⋂_α∈Ωη_α(1) ≥ ⋂_α∈Ωη_α(x). (FF1) holds.

Moreover, for any ɛ > 0, there exists α(ɛ) ∈ Ω such that

Since ɛ is arbitrary, we get ⋂_α∈Ωη_α(x) ≥ min {⋂_α∈Ωη_α(x^*y^*) ^*, ⋂_α∈Ωη_α(y)}. So, (FF2) holds.

Therefore, ⋂_α∈Ωη_α is a fuzzy filter of X.

Lemma 2.2 (see, [16]). Let μ be a fuzzy filter of X. For any x, y ∈ X, if x ≤ y, then μ(x) ≤ μ(y).

Definition 2.3. Let μ be a fuzzy subset of X. Then the fuzzy filter generated by μ, which is denoted by 〈μ〉, is defined as

Obviously, we get μ⊆〈μ〉, and if μ <_FF X, then μ = 〈μ〉.

Lemma 2.4. If μ, η <_FFX, then μη = μ∩η.

Proof. Let x ∈ X, x = a ∨ b and μ, η be fuzzy filters. Then, by Lemma 2.2, μ(a) ≤ μ(a ∨ b) = μ(x) and η(b) ≤ η(a∨b) = η(x). Hence, min {μ(a), η(b)} ≤ μ∩η(x).

Therefore, μη ≤ μ∩η(x), or equivalently μη⊆μ∩η.

Conversely, μη(x) = sup _x=y∨z{min {μ(y), η(z)}} ≥ min {μ(x), η(x)} = μ∩η(x). So μη⊇μ∩η.

Thus, μη = μ∩η.

Corollary 2.5. If μ, η <_FFX, μη <_FFX.

Lemma 2.6. If  η <_FFX, μη⊆η.

Proof. Let   η <_FF X. If x = y∨z, then from Lemma 2.2 we know η(z) ≤ η(x). Thus, μη(x) = sup _x=y∨z{min {μ(y), μ(z)} ≤ sup _x=y∨z{η(z)} ≤ η(x). So, μη⊆η.

Definition 3.1. A nonconstant fuzzy filter μ of X is said to be s-prime if for all θ, σ <_FF X, θσ⊆μ implies θ⊆μ or σ⊆μ. In this case, we write μ <_FSP X.

Lemma 3.2. If σ is a fuzzy subset of X, then V(〈σ〉) = V(σ). So F(σ) = F(〈σ〉).

Proof. Let μ ∈ V(σ), then σ⊆μ and so 〈σ〉⊆μ. Hence, μ ∈ V(〈σ〉). Conversely, let μ ∈ V(〈σ〉), then 〈σ〉⊆μ. Note that σ⊆〈σ〉⊆μ, we get μ ∈ V(σ). Therefore, V(σ) = V(〈σ〉).

Theorem 3.3. Let ζ = {F(θ)∣θ <_FFX}. Then the pair (F(X), ζ) is a topological space.

Proof. Consider θ₀ = 0 and θ₁ = 1. Then θ₀, θ₁ <_FF X,  F(θ₀) = ∅ and F(θ₁) = F(X). Thus, F(X), ∅ ∈ ζ.

Then, we prove that ζ is closed under finite intersection.

Let η and θ be two fuzzy filters of X. We claim that V(θ) ∪ V(η) = V(θη). Let τ ∈ V(θη). Then, θη⊆τ. Since τ ∈ F(X), we have θ⊆τ or σ⊆τ. It follows that τ ∈ V(θ) ∪ V(η).

Conversely, let τ ∈ V(θ) ∪ V(η), then θ⊆τ or η⊆τ. By Lemma 2.6, θη⊆θ and θη⊆η. Thus, θη⊆τ and so τ ∈ V(θη). It follows that V(θ) ∪ V(η)⊆V(θη).

Combining the above arguments we get V(θ) ∪ V(η) = V(θη), or equivalently, F(θ)∩F(η) = (F(X)∖V(θ))∩(F(X)∖V(η)) = (F(X)∖(V(θ) ∪ V(η)) = (F(X)∖V(θη)) = F(θη). By Corollary 2.5, θη <_FF X and so F(θ)∩F(η) = F(θη) ∈ ζ.

Finally, let {θ_α∣α ∈ Ω} be a family of fuzzy prime filters of X. We will prove that ⋂_α∈ΩV(θ_α) = V(⋃_α∈Ωθ_α).

Let μ ∈ ⋂_α∈ΩV(θ_α), then for any α ∈ Ω, μ ∈ V(θ_α) and so θ_α⊆μ. Hence, ⋃_α∈Ωθ_α⊆μ and thus μ ∈ V(⋃_α∈Ωθ_α).

Conversely, let μ ∈ V(⋃_α∈Ωθ_α), then ⋃_α∈Ωθ_α⊆μ. Thus, for any α ∈ Ω, θ_α⊆⋃_α∈Ωθ_α⊆μ. Hence, μ ∈ V(θ_α) for all α ∈ Ω and so μ ∈ ⋂_α∈ΩV(θ_α).

This shows that ⋂_α∈ΩV(θ_α) = V(⋃_α∈Ωθ_α).

By Lemma 3.2, we get V(⋃_α∈Ωθ_α) = V(〈⋃_α∈Ωθ_α〉) and so ⋂_α∈ΩV(θ_α) = V(〈⋃_α∈Ωθ_α〉).

Furthermore, we get ⋃_α∈ΩF(θ_α) = ⋃_α∈Ω(F(X)∖V(θ_α)) = F(X)∖⋂_α∈ΩV(θ_α) = F(X)∖V(〈⋃_α∈Ωθ_α〉) = F(〈⋃_α∈Ωθ_α〉) ∈ ζ.

It follows that (F(X), ζ) is a topological space.

Theorem 3.4. The collection

of ζ is a base of ζ where x_β <_FFX is defined by

Proof. By Lemma 3.2, for any x ∈ X, β ∈ (0,1], F(x_β) = F(〈x_β〉) and so F(x_β) ∈ ζ.

Now, we prove that ß is a base of ζ. It is sufficient to show that for all F(θ) ∈ ζ, and μ ∈ F(θ), there exists F(x_β) ∈ β such that μ ∈ F(x_β) and F(x_β)⊆F(θ).

Let F(θ) ∈ ζ and μ ∈ F(θ). Then, θ ⊈ μ and so there exists x ∈ X such that μ(x) < θ(x). Let θ(x) = β and then μ ∈ F(x_β). Moreover, for any σ ∈ V(θ), σ(x) ≥ θ(x) = β = x_β(x) and so x_β⊆σ. Thus σ ∈ V(x_β). This means V(θ)⊆V(x_β). It follows that F(x_β)⊆F(θ).

Therefore, ß is a base of ζ.

Theorem 3.5. FF-spec(X) is a T₀ space.

Proof. Let μ, η ∈ F(X) and μ ≠ η. Then, μ ⊈ η or η ⊈ μ.

If μ ⊈ η, then, η ∉ V(μ) but μ ∈ V(μ). Moreover, η ∈ F(μ) but μ ∉ F(μ).

If η ⊈ μ, similarly we can get μ ∈ F(η) but η ∉ F(η). It follows that FF-spec(X) is a T₀ space.

Lemma 3.6 (see [9].)Let μ be a fuzzy subset of X. Then, μ is a fuzzy filter of X if and only if μ_t is a filter of X for each t ∈ [0,1] wherever μ_t ≠ ∅.

Lemma 3.7. A non-constant fuzzy subset μ of X is a fuzzy prime filter if and only if μ_t is a prime filter of X for each t ∈ [0,1] whenever μ_t ≠ ∅.

Proof. Let μ be a fuzzy prime filter and t ∈ [0,1] such that μ_t ≠ ∅. Then by Lemma 3.6, μ_t is a filter of X.

Suppose x∨y ∈ μ_t. It follows that μ(x∨y) ≥ t. Since μ is prime, we have max {μ(x), μ(y)} ≥ μ(x∨y) ≥ t and thus μ(x) ≥ t or μ(y) ≥ t. It follows that x ∈ μ_t or y ∈ μ_t. Therefore μ_t is a prime filter.

Conversely, suppose that for each t ∈ [0,1], μ_t is a prime filter whenever μ_t ≠ ∅. If μ is not a fuzzy prime filter, then there exist x, y ∈ X such that μ(x∨y) > max {μ(x), μ(y)}. Take t satisfying μ(x∨y) > t > max {μ(x), μ(y)}. Then x∨y ∈ μ_t. Since μ_t is a prime filter of X, then x∨y ∈ μ_t implies x ∈ μ_t or y ∈ μ_t. But on the other hand, μ(x) ≤ max {μ(x), μ(y)} < t and μ(y) ≤ max {μ(x), μ(y)} < t imply x ∉ μ_t and y ∉ μ_t, a contradiction. It follows that μ is indeed a fuzzy prime filter.

Lemma 3.8 (see [13].)Let X be a bounded commutative BCK-algebra and F be a BCK-filter of X. Then, F is prime if and only if, for any filters A, B, F = A∩B implies F = A or F = B.

Theorem 3.9. Let X be a bounded commutative BCK-algebra and μ be a fuzzy s-prime filter. Then for each t ∈ [0,1], μ_t is a prime filter of X whenever μ_t ≠ ∅ and μ_t ≠ X.

Proof. Let μ be a fuzzy s-prime filter and t ∈ [0,1], μ_t ≠ ∅. Then by Lemma 3.6, μ_t is a filter.

Let A, B be two filters such that μ_t = A ∩ B. Define the fuzzy subset θ = tχ_A and σ = tχ_B. It is easy to see that θ and σ are fuzzy filters of X. Note that

Since μ_t = A∩B, then for any x ∈ A∩B = μ_t, μ(x) ≥ t = θ∩σ(x) and so μ(x) ≥ θ∩σ(x) for all x ∈ X. Thus μ⊇θ∩σ. It follows from μ being a fuzzy s-prime filter that θ⊆μ or σ⊆μ. Without loss of generality let θ⊆μ. Then, for any x ∈ A, θ(x) = tχ_A(x) = t ≤ μ(x) and so x ∈ μ(t). This means that A⊆μ_t. But μ_t = A∩B implies μ_t⊆A and thus μ_t = A. Therefore, μ_t is a prime filter by Lemma 3.8.

Theorem 3.10. Let X be a bounded commutative BCK-algebra. If μ is a fuzzy s-prime filter, then it is a fuzzy prime filter.

Proof. The proof follows from Lemma 3.7 and Theorem 3.9.

Example 3.11. Let X = {0,1}. Define the operation * on X as follows: 0 * 0 = 0, 0 * 1 = 0, 1 * 0 = 1 and 1*1 = 0. It is easy to see that 〈X; *, 0〉 is a bounded commutative BCK-algebra. Define a fuzzy subset μ of X by μ(0) = 0, μ(1) = 1/2. Clearly μ is a fuzzy prime filter of X.

Moreover, we define the fuzzy filters σ and θ by σ(x) = 1/2 for all x ∈ X and θ(1) = 1, θ(0) = 0. Then, we get θσ⊆μ but σ ⊈ μ and θ ⊈ μ. Therefore, μ is not a s-prime fuzzy filter.

Lemma 3.12. F is a prime filter of X if and only if $$ is a fuzzy s-prime filter, where α ∈ [0,1), and $$ is defined by

Proof. Let F be a prime filter. Then, by Lemma 3.6, we can easily see that $$ is a fuzzy filter.

Let θ, σ be two fuzzy prime filters such that $$, we will prove $$ or $$. If it is not true, then there exist x, y ∈ X∖F such that θ(x) > α and σ(y) > α. Since F is prime, then x∨y ∉ F. Note that $$, then

Thus, x∨y ∈ F, a contradiction. It follows that $$ or $$, and so $$ is a fuzzy s-prime filter.

Conversely, let $$ be a fuzzy s-prime filter. By Theorem 3.10, $$ is also a fuzzy prime filter. Then, by Lemma 3.7, (χ_F) _t = F is a prime filter, where α ≤ t ≤ 1.

Corollary 3.13. F is a prime filter of X if and only if χ_F is a fuzzy s-prime filter.

Lemma 3.14 (see [13].)Let X be a bounded implicative BCK-algebra, then x ∧ x^* = 0 and x ∨ x^* = 1.

Lemma 3.15. Let μ be a fuzzy filter of a bounded commutative BCK-algebra X. Then, μ(0) = min {μ(x), μ(x^*)} for all x ∈ X.

Proof. Since μ is a fuzzy filter, we have μ(0) ≥ min {μ(x*0) ^*, μ(x)} = min {μ(x^*), μ(x)} for all x ∈ X. On the other hand, μ(0) ≤ min {μ(x^*), μ(x)}, since any fuzzy filter is order preserving. Thus, μ(0) = min {μ(x), μ(x^*)}.

Lemma 3.16. If μ is a fuzzy filter of a bounded BCK-algebra X, then μ₁ = {x ∈ X∣μ(x) = μ(1)} is a filter of X and $$ is a fuzzy filter of X.

Proof. Let μ be a fuzzy filter and take t = μ(1). Then, μ_t = μ₁ and so μ_t = μ₁ is a filter of X by Lemma 3.6. Clearly $$ is a fuzzy filter.

Lemma 3.17. Let X be a bounded commutative BCK-algebra and μ be a fuzzy s-prime filter of X. Then μ(1) = 1.

Proof. Suppose that μ(1) < 1. Since μ is non-constant, there exists a ∈ X such that μ(a) < μ(1). Define fuzzy subset θ and σ of X by

and σ(x) = μ(1) for all x ∈ X. By Lemma 3.16, $$is a fuzzy filter and clearly σ is a fuzzy filter. Note that θ(1) = 1 > μ(1) and σ(a) = μ(1) > μ(a), we get θσ ⊈ μ. But note that for any x, y ∈ X

Thus, θσ(x) = sup _x=y∨z{min {θ(y), σ(z)}} ≤ sup _x=y∨z{μ(y∨z)} = sup {μ(x)} = μ(x) for any x ∈ X, a contradiction. Therefore, μ(1) = 1.

Lemma 3.18. Let X be a bounded implicative BCK-algebra and μ be a fuzzy s-prime filter of X. Then for any x ∈ X, μ(x) = 1 or μ(x^*) = 1.

Proof. By Lemma 3.14, x∨x^* = 1, for all x ∈ X. Since μ is a fuzzy s-prime filter, we get that μ₁ is a prime filter of X by Theorem 3.9. Hence, x∨x^* = 1 ∈ μ₁ implies x ∈ μ₁ or x^* ∈ μ₁. Therefore, μ(x) = μ(1) = 1 or μ(x^*) = μ(1) = 1 by Lemma 3.17.

Theorem 3.19. Let X be a bounded implicative BCK-algebra and μ be a fuzzy s-prime filter of X. Then, for x ∈ X, μ(x) = μ(1) = 1 or μ(x) = μ(0).

Proof. By Lemma 3.14, x∨x^* = 1 and then μ(1) = μ(x∨x^*) = μ(x) or μ(x^*) since μ is a fuzzy s-prime filter. By Lemma 3.18, we get μ(x) = 1 or μ(x^*) = 1. If μ(x^*) = 1, then μ(0) = min {μ(x), μ(x^*)} = μ(x) by Lemma 3.15. If μ(x^*) ≠ 1, then μ(x) = 1 = μ(1).

Lemma 3.20. Let X be a bounded BCK-algebra. Then, a filter F of X is proper if and only if 0 ∉ F.

Proof. If 0 ∉ F, then clearly F is proper.

Conversely, let F be proper. If 0 ∈ F, then for any x ∈ X, (x^*0^*) ^* = (x^*1) ^* = 0^* = 1 ∈ F and so x ∈ F. It follows that F = X, a contradiction. Therefore, 0 ∉ F.

Lemma 3.21 (see [13].)Let X be a bounded commutative BCK-algebra and F be a filter of X. If x ∈ X∖F, then there is a prime filter A of X such that F⊆A and x ∉ A.

Lemma 3.22 (see [13].)Let X be a bounded implicative BCK-algebra. Then, for any a ∈ X, the filter 〈a〉, generated by a, is a set of elements x in X satisfying a ≤ x.

Lemma 3.23. Let X be a bounded implicative BCK-algebra and a ≠ 0. Then, 〈a〉 ≠ X.

Proof. By Lemma 3.22, 0 ∉ 〈a〉 and thus 〈a〉 ≠ X.

Lemma 3.24 (see [16].)For a bounded commutative BCK-algebra X, one gets

• (1)

  x^** = x for all x ∈ X.

• (2)

  x^*∧y^* = (x∨y) ^*, x^*∨y^* = (x∧y) ^* for all x, y ∈ X.

• (3)

  x^*y^* = yx for all x, y ∈ X.

Theorem 3.25. Let X be a bounded implicative BCK-algebra. Then,

• (i)

  if β₁, β₂ ∈ (0,1], β = min {β₁, β₂} and x, y ∈ X, then $$.

• (ii)

  if β ∈ (μ(0), 1] and x, y ∈ X, then F(x_β) ∪ F(y_β) = F((x∧y) _β).

Moreover, F(x_β) is both open or closed, where μ ∈ F(X).

• (iii)

  if F(x_β) = F(X), where x ∈ X and β ∈ (0,1], then x = 0.

Proof. (i) If $$, then μ(x) < β₁ and μ(y) < β₂. By Theorem 3.10, μ is a fuzzy prime filter, and then μ(x∨y) ≤ max {μ(x), μ(y)}. Since μ(x) < β₁ and μ(y) < β₂, then μ(x) ≠ μ(1) and μ(y) ≠ μ(1). It follows from Theorem 3.19 that μ(x) = μ(0) and μ(y) = μ(0). Thus, μ(x∨y) ≤ max {μ(x), μ(y)} = μ(0) = min {μ(x), μ(y)} < min {β₁, β₂} = β. Therefore, μ ∈ F((x∨y) _β).

Conversely, if μ ∈ F(x∨y) _β, then μ(x∨y) < β and so μ(x∨y) < β₁, μ(x∨y < β₂). Note that x ≤ x∨y and y ≤ x∨y, we get μ(x) ≤ μ(x∨y) < β₁, and μ(y) ≤ μ(x∨y) < β₂, since μ is order preserving. Thus, $$ and $$, or equivalently, $$.

Therefore, (i) holds.

(ii) Let μ ∈ F(x_β)  ∪  F(y_β). Then, μ(x) < β or μ(y) < β. By Lemma 3.17, we have μ(1) = 1 ≥ β > μ(x), μ(1) = 1 ≥ β > μ(y). Therefore, x ∉ μ₁ or y ∉ μ₁. On the other hand, by Lemma 3.14, x∨x^* = 1 ∈ μ₁, y∨y^* = 1 ∈ μ₁. Note that μ₁ = μ₁ is a prime filter of X by Theorem 3.9. If x ∉ μ₁, then x∨x^* ∈ μ₁ implies x^* ∈ μ₁. If y ∉ μ₁, then y∨y^* ∈ μ₁ implies y^* ∈ μ₁. Therefore, we get that x^* ∈ μ₁ or y^* ∈ μ₁. Note that x∧y ≤ x, y, we get x^* ≤ (x∧y) ^* and y^* ≤ (x∧y) ^*. Thus, μ(x^*) ≤ μ((x∧y) ^*) and μ(y^*) ≤ μ((x∧y) ^*), and so max {μ(x^*), μ(y^*)} ≤ μ((x∧y) ^*). But μ(x^*) = μ(1) or μ(y^*) = μ(1) implies that 1 = max {μ(x^*), μ(y^*)} ≤ μ((x∧y) ^*) or μ((x∧y) ^*) = μ(1). This means, (x∧y) ^* ∈ μ₁.

If x∧y ∈ μ₁, then μ(0) ≥ min {μ(((x∧y)*0) ^*), μ(x∧y)} = min {μ((x∧y) ^*), μ(x∧y)} = μ(1). It follows that μ(x) = μ(1) for all x ∈ X, a contradiction. Thus, x∧y ∉ μ₁. By Lemma 3.18, μ(x∧y) = μ(0). Hence, μ(x∧y) < β and so μ ∈ F((x∧y) _β). It follows that F(x_β) ∪ F(y_β)⊆F((x∧y) _β).

Conversely, let μ ∈ F((x∧y) _β). Then, μ(x∧y) < β ≤ 1 = μ(1). Thus, x∧y ∉ μ₁. Since (x ∧ y) ∨ (x ∧ y) ^* = 1 ∈ μ₁, then (x ∧ y) ^* ∈ μ₁. By Lemma 3.24, (x ∧ y) ^* = x^* ∨ y^* and so x^* ∈ μ₁ or y^* ∈ μ₁. If x^* ∈ μ₁ (or y^* ∈ μ₁), then μ(0) ≥ min {μ((x*0) ^*), μ(x)} = min {μ(x^*), μ(x)} = μ(x) (or μ(0) ≥ μ(y)). Thus, x ∉ μ₁ or y ∉ μ₁. It follows that F((x ∧ y) _β)⊆F(x_β) ∪ F(y_β).

Combining the above arguments, we get F(x_β) ∪ F(y_β) = F((x∧y) _β).

In order to prove F(x_β) is closed, we will show F(x_β) = V((x^*) _β).

Let μ ∈ F(x_β). Then, μ ∈ F(〈x_β〉) by Lemma 3.2. Thus, 〈x_β〉 ⊈ μ and so μ(x) < β ≤ 1 = μ(1). Hence, x ∉ μ₁. Note that x ∨ x^* = 1 ∈ μ₁ we get x^* ∈ μ₁, which implies that μ(x^*) = μ(1) = 1 ≥ β and so (x^*) _β⊆μ. It follows that μ ∈ V((x^*) _β) and thus F(x_β)⊆V((x^*) _β).

Conversely, let μ ∈ V((x^*) _β). Then, (x^*) _β⊆μ and so μ(x^*) ≥ β > μ(0). By Theorem 3.19, μ(x^*) = μ(1). Note that μ(0) ≥ min {((x*0) ^*), μ(x)} = min {μ(x^*), μ(x)} = μ(x), we get that μ(x) = μ(0) < β, or equivalently, 〈x_β〉 ⊈ μ. Thus, μ ∈ F(x_β) and so V((x^*) _β)⊆F(x_β).

Combining the above two sides, we get F(x_β) = V((x^*) _β).

(iii) Let F(x_β) = F(X). We claim that x = 0. If this is not true, by Lemma 3.23,〈x〉 ≠ X. By Lemma 3.21, there exists a prime filter P of X such that 〈x〉⊆P. On the other hand, by Lemma 3.12, χ_P ∈ F(X) = F(x_β).

Therefore, 〈x〉 ⊈ P, a contradiction. It follows that x = 0.

Example 3.26. Let X = {0,1, 2,3} and *-table and ∨-table be given as follows.

Then (x; *, 0) is a bounded implicative BCK-algebra and 3 is a unit. It is easy to see that P = {1,3} is a filter of X. From ∨-table, we can see easily that P is prime. So, χ_P is a fuzzy s-prime filter by Lemma 3.12. Let β = 1/2. Then, 3_β⊆μ and so 〈3_β〉⊆μ. Hence, μ ∉ F(3_β). Therefore F(3_β) ≠ F(X).

Theorem 3.27. Let X be a bounded implicative BCK-algebra and X_α = {μ ∈ F(X)∣μ(0) = α} for α ∈ [0,1). Then, X_α is a Hausdorff space.

Proof. Let μ, σ ∈ X_α and μ ≠ σ. We claim that μ₁ ≠ σ₁. Otherwise, if μ₁ = σ₁, then for x ∈ μ₁ = σ₁, μ(x) = μ(1) = 1 = σ(1) = σ(x) and for x ∉ μ₁ = σ₁, μ(x) = μ(0) = α = σ(0) = σ(x) by Theorem 3.19, a contradiction. Thus, μ₁ ⊈ σ₁ or σ₁ ⊈ μ₁. Let μ₁ ⊈ σ₁. Then, x ∈ μ₁∖σ₁ implies x^* ∈ σ₁. Moreover, μ(0) ≥ min {μ(x * 0) ^*, μ(x)} = min {μ(x^*), μ(x)} = μ(x^*) since μ(x) = μ(1) = 1. Thus μ(x^*) = μ(0) ≠ μ(1) and so x^* ∉ μ₁. Therefore, x^* ∈ σ₁∖μ₁. Hence,

Let t ∈ (α, 1]. Then, (x_t)(x) = t > σ(x), and so σ ∈ F(x_t). Note that ((x^*) _t)(x^*) = t > α = μ(x^*), we get μ ∈ F((x^*) _t). Moreover, we get

It follows that X_α is a Hausdorff space.

Corollary 3.28. Let X be a bounded implicative BCK-algebra. Then X₀ = {μ ∈ F(X)∣μ(0) = 0} is a Hausdorff space.

Lemma 3.29. S(A) = S(〈A〉) and if A⊆B, then S(A)⊆S(B).

Proof. Since A⊆〈A〉, then A ⊈ P implies 〈A〉 ⊈ P. Then, P ∈ S(A) implies P ∈ S(〈A〉).

Conversely, if P ∈ S(〈A〉), then 〈A〉 ⊈ P. Hence, A ⊈ P since A⊆P implies 〈A〉⊆P. Therefore, P ∈ S(A).

Thus, S(A) = S(〈A〉). Similarly, we can prove that A⊆B implies S(A)⊆S(B).

Proposition 3.30. The family T(X) = {S(F)∣F ∈ L(X)} forms a topology on F-spec(X).

Proof. First, we get

Then, for any family {S(F_α)} _α∈Ω,

Finally,

Therefore, T(X) is a topology on F-spec(X).

Theorem 3.31. Let X be a bounded implicative BCK-algebra and the map f : F-spec(X) → X_α is defined by $$ where $$ is defined in Lemma 3.12. Then, f is a homeomorphism.

Proof. (a) f is well defined.

By Lemma 3.12, $$ is a fuzzy s-prime filter for any P ∈ F-spec(X). Note that 0 ∉ P, then $$ and so $$. Thus, f is well defined.

• (b)

  Clearly f is injective.

• (c)

  f is surjective.

For any μ ∈ X_α, by Theorem 3.19, μ(x) = μ(1) = 1 or μ(x) = μ(0) = α. Hence, $$. By Theorem 3.9, we get μ₁ is a prime filter of X. Thus, μ₁ ∈ F-spec(X) and so $$. It follows that f is surjective.

• (d)

  f is continuous.

Let F_α(θ) = F(θ)∩X_α be an open set of X_α. We will prove that f⁻¹(F_α(θ)) is an open set of F-spec(X). It is sufficient to prove that f⁻¹(F_α(θ)) = ⋃_α≤t≤1S(θ_t), since θ_t is a filter of X by Lemma 3.6.

First, let P ∈ ⋃_α<t<1S(θ_t), then there exists some t such that α < t < 1 and P ∈ S(θ_t). Thus, θ_t ⊈ P and there exists x ∈ θ_t∖P. Hence $$. Therefore, $$ and so $$. This shows that f(P) ∈ F_α(θ). It follows that P ∈ f⁻¹(F_α(θ)).

Conversely, let P ∈ f⁻¹(F_α(θ)), then $$. Hence, $$, and thus there exists x ∈ X such that $$. Therefore, x ∉ P and so $$. We can take t such that α < t₁ < θ(x). Then, $$. It follows that θ_t1 ⊈ P and so $$.

Combining the above two hands, we get f⁻¹(F_α(θ)) = ⋃_α<t<1S(θ_t). So f is continuous.

• (e)

  f⁻¹ is continuous. It is sufficient to prove that f(S(F)) is an open set of X_α for any F ∈ L(X).

We will prove that $$.

Thus, $$.

Conversely, we get

So that $$.

Therefore, $$. By Lemma 3.6, we can easily see that $$ is a fuzzy filter of X and so $$ is an open set of X_α. It follows that f⁻¹ is continuous.

Corollary 3.32. Let X be a bounded implicative BCK-algebra. Then, F-spec(X) is a Hausdorff space.