In this section, we introduce very true filters of very true MTL-algebras. In particular, we focus on algebraic structures of *VF*(*L*) of all very true filters in the very true MTL-algebras and obtain that *VF*(*L*) forms a complete Heyting algebra. Moreover, we characterize subdirectly irreducible very true MTL-algebras and prove a representation theorem for very true MTL-algebras via very true filters.

* Let* (*L*, τ) *be a very true MTL*-*algebra and F be a filter of L*. *Then F is called a very true filter of* (*L*, τ) *if* *x* ∈ *F* *implies* τ (*x*) ∈ *F* *for all* *x* ∈ *L*.

We will denote the set of all very true filters of (*L*, τ) by *VF*[*L*].

*Considering* Example 3.2 (*c*), *one can easily check that the very true filters of* (*L*, τ) *are* {*a*, *b*, 1} *and* {1} *and L. However*, {*b*, 1} *is a filter of L but not a very true filter of* (*L*, τ).

Let (*L*, τ) be a very true MTL-algebra. For any nonempty set *X* of *L*, we denote by 〈*X*〉_{τ} the very true filter of (*L*, τ) generated by *X*, that is, 〈*X*〉_{τ} is the smallest very true filter of (*L*, τ) containing *X*. If *F* is a very true filter of (*L*, τ) and *x* ∉ *F*, we put 〈*F*, *x*〉_{τ} := 〈*F* ∪ {*x*}〉_{τ}.

The next theorem gives a concrete description of the very true filter generated by a subset of very true MTL-algebra (*L*, τ).

*Let* (*L*, τ) *be a very true MTL*-*algebra and X be a nonempty set of L*. *Then* 〈 *X* 〉_{τ} = {*x* ∈ *L*|*x*≥ τ(*y*_{1}) ⊙ … τ (*y*_{n}), *y*_{i} ∈ *X*, *n* ≥ 1}.

The proof is easy, and we hence omit the details.

*Considering* Example 3.2(*c*), *one can easily obtain that* 〈0, *a*〉_{τ} = 〈0, *b*〉_{τ} = 〈0, *a*, *b* 〉_{τ} = {*a*, *b*, 1}, 〈*L*〉_{τ} = 〈*a*, *b*, 1〉_{τ} = 〈0, *a*, 1〉_{τ} = 〈0, *b*, 1〉_{τ} = {1} and 〈0〉_{τ} = *L*.

*Let* *F*, *F*_{1}, *F*_{2} *be very true filters of* (*L*, τ) *and* *a* ∉ *F*. *Then*:

〈*a*〉_{τ} = {*x* ∈ *L*|*x* ≥ (τ*a*)^{n}, *n* ≥ 1},

〈*F* ∪ *a*〉_{τ} = {*x* ∈ *L*|*x* ≥ *f* ⊙(τ *a*)^{n}, *f* ∈ *F*} = *F* ∨ [τ *a*),

〈*F*_{1} ∪ *F*_{2} 〉_{τ} = {*x* ∈ *L*|*x* ≥ *f*_{1} ⊙ *f*_{2}, *f*_{1} ∈ *F*_{1}, *f*_{2} ∈ *F*_{2}},

*if* *a* ≤ *b*, *then* 〈*b*〉_{τ} ⊆ 〈*a*〉_{τ},

〈τ(*a*)〉_{τ} = 〈*a*〉_{τ},

〈*a*〉_{τ} ∨ 〈*b*〉_{τ} = 〈*a* ∧ *b*〉_{τ} = 〈*a* ⊙ *b*〉_{τ},

〈*a*〉_{τ} ∩ 〈_{b}〉_{τ} = 〈τ(*a*)∨ τ(*b*)〉_{τ}.

The proof of (1) – (5) are obvious.

(6) Since *a* ⊙ *b* ≤ *a* ∧ *b* ≤ *a*, *b*, we deduce that 〈*a*〉_{τ}, 〈*b*〉_{τ} ⊆ 〈*a*∧ *b*〉_{τ} ⊆ 〈*a* ⊙ *b*〉_{τ}. It follows from that 〈*a* 〉_{τ} ∨ 〈*b*〉_{τ}⊆ 〈*a*∧ *b*〉_{τ} ⊆ 〈*a* ⊙*b*〉_{τ}. Conversely, let *a* ∈ 〈*a* ⊙*b*〉_{τ}. Then for some natural number *n*≥ 1, *a* ≥(τ(*a* ⊙*b*))^{n} ≥( τ *a* ⊙ τ *b*)^{n} = (τ *a*)^{n} ⊙ (τ *b*)^{n}. Hence *a* ∈ 〈*a*〉_{τ} ∨ 〈*b*〉_{τ}, we deduce that 〈 *a* ⊙ *b*〉_{τ} ⊆ 〈*a*〉_{τ} ∨ 〈*b*〉_{τ}. Therefore 〈*a*〉_{τ} ∨ 〈*b*〉_{τ} = 〈*a* ∧ *b*〉_{τ} = 〈*a* ⊙ *b*〉_{τ}.

(7) Since τ(*a*)≤ τ(*a*) ∨ τ(*b*), we deduce that 〈 τ(*a*) ∨ τ(*b*)〉_{τ} ⊆ 〈 τ(*a*)〉_{τ} = 〈*a* 〉_{τ}. Analogously, 〈 τ(*a*) ∨ τ(*b*) 〉_{τ}⊆ 〈 τ(*b*) 〉_{τ} = 〈*b* 〉_{τ}. It follows that 〈 τ(*a*) ∨ τ(*b*) 〉_{τ} ⊆ 〈*a*〉_{τ} ∩ 〈*b*〉_{τ}. Moreover, let *t* ∈ 〈*a*〉_{τ} ∩ 〈*b*〉_{τ}. Then for some natural number *n*, *m* ≥ 1, *t* ≥(τ(*a*))^{m} and *t* ≥(τ(*b*))^{n}. Hence *t*≥(τ(*a*))^{m}∨ (τ(*b*))^{n} ≥ (τ(*a*) ∨ τ(*b*))^{mn} = (τ(*a*∨ *b*))^{mn}, we deduce that *a*∈ 〈 τ(*a*)∨ τ(*b*)〉_{τ}, that is, 〈*a* 〉_{τ} ∩ 〈*b*〉_{τ} ⊆ 〈τ(*a*) ∨ τ(*b*) 〉_{τ}. Therefore 〈*a* 〉_{τ} ∩ 〈*b* 〉_{τ} = 〈 τ(*a*)∨ τ(*b*)〉_{τ}.

The next results shows that the algebraic structure of *VF*(*L*) of all very true filters in very true MTL-algebras forms a complete Heyting algebra.

*Let*(*L*, τ) *be a very true MTL*-*algebra. Define binary operations* ∧, ∨, ↦ *on VF*(*L*) *as follows*
: * for all* *F*_{1}, *F*_{2} ∈ *VF*(*L*), *F*_{1}∧ *F*_{2} = *F*_{1} ∩ *F*_{2}, *F*_{1} ∨ *F*_{2} = 〈 *F*_{1} ∪ *F*_{2} 〉_{τ}, *F*_{1}↦ *F*_{2} = {*x*∈ *L*|τ(*x*)∨ *f*_{1}∈ *F*_{2} *for any* *f*_{1}∈ *F*_{1}}. *Then* (*VF*(*L*), ∧, ∨, ↦, 1, *L*) *is a complete Heyting algebra*.

Suppose that {*F*_{i}}_{i∈ I} is a family of very true filters of (*L*, τ) . From Theorem 4.5, it is easy to check that the infimum of {*F*_{i}}_{i} ∈ *I* = ∩_{i∈ I}*F*_{i} and the supermum is ∨_{i ∈ I}*F*_{i}= {*x* ∈ *L*|*x* ≥ *f*_{i1} ⊙ *f*_{i2} ⊙ … *f*_{im}, *f*_{1j} ∈ *F*_{ij}, *i*_{j} ∈ *I*, 1 ≤ *j* ≤ *m*}. Therefore, (*VF*(*L*), ∧, ∨, 1, *L*) is a complete lattice under the inclusion order ⊆. Next, we define *F*_{1} ↦ *F*_{2} = {*x* ∈ *L*|τ(*x*)∨ *f*_{1} ∈ *F*_{2} for any *f*_{1} ∈ *F*_{1} 〉 for any *F*_{1}, *F*_{2} ∈ *VF*(*L*). And, we shall prove that *F*_{1}∩ *F*_{2}⊆ *F*_{3} if and only if *F*_{2}⊆ *F*_{1}↦ *F*_{3} for all *F*_{1}, *F*_{2}, *F*_{3} ∈ *VF*(*L*), that is, (*VF*(*L*), ∧, ∨, ↦, 1, *L*) is a complete Heyting algebra. In order to do this, we first show that *F*_{1} ↦ *F*_{2} is a very true filter of (*L*, τ).

Now, we will show that *F*_{1}↦ *F*_{2} is a very true filter of (*L*, τ). Clearly 1∈ *F*_{1}↦ *F*_{2}. Let *x*∈ *F*_{1} ↦ *F*_{2} and *x*≤ *y*, then for any *f*_{1} ∈ *F*_{1} such that τ(*x*)∨ *f*_{1} ∈ *F*_{2}. Since τ(*x*)∨ *f*_{1} ≤ τ(*y*)∨ *f*_{1} ∈ *F*_{2} and hence *y*∈ *F*_{1} ↦ *F*_{2}. Assume that *x*, *y*∈ *F*_{1} ↦ *F*_{2}, then for any *f*_{1}∈ *F*_{1}, τ(*x*)∨ *f*_{1}, τ(*y*)∨ *f*_{1} ∈ *F*_{2} and hence *f*_{1} ∨ τ(*x* ⊙*y*)∈ *F*_{2}. So *x* ⊙ *y* ∈ *F*_{1} ↦ *F*_{2}. Obviously, if *x*∈ *F*_{1} ↦ *F*_{2}, then τ(*x*)∈ *F*_{1} ↦ *F*_{2} and thus *F*_{1} ↦ *F*_{2} is a very true filter of (*L*, τ).

Next, we will prove that *F*_{1} ∧ *F*_{2} ≤ *F*_{3} if and only if *F*_{1} ≤ *F*_{2}↦ *F*_{3}. Assume that *F*_{1} ∧ *F*_{2} ≤*F*_{3}. Let *f*_{1} ∈ *F*_{1}. Then τ(*f*_{1})∈ *F*_{1} and for any *f*_{2} ∈ *F*_{2}, we have τ(*f*_{2})∨ *f*_{1}≥ *f*_{1}, τ(*f*_{2}) ∨ *f*_{1} ≥ τ(*f*_{2}). Hence τ(*f*_{2})∨ *f*_{1} ∈ *F*_{1} ∧ *F*_{2} ≤ *F*_{3} and hence *f*_{1} ∈ *F*_{1} ∧ *F*_{2}. Conversely, assume that *F*_{1} ≤ *F*_{2} ↦ *F*_{3}. Let *x* ∈ *F*_{2} ∧ *F*_{3}, then *x* ∈ *F*_{2} ↦ *F*_{3}. For any *f*_{3} ∈ *F*_{3}, we have τ(*x*) ∨ *f*_{3} ∈ *F*_{3}. Taking *f*_{3} = *x* ∈ *F*_{3}, we have *x* ∨ τ(*x*) = *x* ∈ *F*_{3}. Thus *F*_{1} ≤ *F*_{2} ↦ *F*_{3}.

Therefore, (*VF*(*L*), ∧, ∨, ↦, 1, *L*) is a complete Heyting algebra.

*Let* (*L*, τ) *be a very true MTL*-*algebra and F*∈ *VF*(*L*). *Then thefollowing conditions are equivalent*:

*F is a compact element of VF*(*L*),

*F is a principal very true filter of* (*L*, τ).

(1) ⇒ (2) Suppose that *F* is the compact element of *VF*(*L*). Since *F* = ∨_{x ∈ F} 〈*x* 〉_{τ}, then there exist *x*_{1}, *x*_{2} … *x*_{n} such that *F* = 〈*x*_{1} 〉_{τ} ∨ 〈*x**2* 〉_{τ} ∨…∨ 〈*x*_{n}〉_{τ}. By Proposition 4.5 (6), we have *F* = 〈*x*_{1} ⊙*x*_{2} ⊙ … ⊙*x*_{n}〉_{τ}. Therefore, *F* is a principal very true filter of (*L*, τ).

(2) ⇒(1) Let *F* be a principal very true filter of (*L*, τ) . Then there exists *x* ∈ *L* such that *F* = 〈*x*〉_{τ}. Suppose that {*F*_{i}}_{i ∈ I} ⊆ *VF*(*L*) and *F* = 〈*x*〉_{τ} ⊆ ∨_{i ∈ I} {*F*_{i}}. Then *x* ∈ ∨_{i ∈ I} *F*_{i}= 〈∪_{i ∈I} *F*_{i}〉_{τ}. It follows that there exist *i*_{j} ∈ *I*, *f*_{ij} ∈ *F*_{ij} for all 1 ≤ *j* ≤ *m* such that *x* ≥*f*_{i1} ⊙ *f*_{i2} ⊙ … *f*_{im}, that is, *x* ∈ 〈*F*_{i1} ∪ *F*_{i2} ∪ … ∪ *F*_{in} 〉_{τ} = *f*_{i1} ∨ *F*_{i2} ∨ … ∨ *F*_{in}. Hence *F*= 〈*x*〉_{τ} ⊆ *F*_{i1} ∨ *F*_{i2} ∨ … ∨ *F*_{in}. Therefore, *F* is a compact element of *VF*(*L*).

*Let* (*L*, τ) *be a very true MTL*-*algebra and* *θ be a congruence on L*. *Then θ is called a* very true congruence on (*L*, τ) *if* (*x*, *y*)∈ *θ* *implies* (τ(*x*), τ(*y*)) ∈ *θ*, *for any* *x*, *y* ∈ *L*.

*Considering* Example 3.2 (*c*), *one can see that* *R* = {{0,0}, {*a*, *a*}, {*b*, *b*}, {1,1}, {*a*, *b*}, {*b*, *a*}, {*a*, 1}, {1, *a*}, {*b*, 1}, {1, *b*}} *is a very true congruence on* (*L*,τ).

*For any very true MTL*-*algebra there exists a one to one correspondence between its very true filters and its very true congruences.*

The proof is easy, and we hence omit the details.

Let (*L*, τ) be a very true MTL-algebra and *F* be a very true filter. We define the mapping τ_{F} : *L*/*F* → *L*/*F* such that τ_{F}([*x*])= [τ(*x*)] for any *x* ∈ *L*.

*Let* (*L*, τ) *be a very true MTL*-*algebra and F a very true filter of* (*L*, τ). *Then* (*L*/*F*, τ_{F}) *is a very true MTL*-*algebra*.

The proof is easy, and we hence omit the details.

*Let* (*L*, τ) *be a very true MTL*-*algebra*. *A proper very true filter F of* (*L*, τ) *is called a* prime very true filter *of* (*L*, τ), *if for all very true filter* *F*_{1}, *F*_{2} *of* (*L*, τ) *such that* *F*_{1} ∩ *F*_{2} ⊆ *F*, *then F*_{1} ⊆ *F or* *F*_{2} ⊆ *F*.

*Considering* Example 3.2(*c*), *one can easily obtain that* {*a*, *b*, 1} *is a prime very true filter of* (*L*, τ).

*Let* (*L*, τ) *be a very true MTL*-*algebra and* *F* *be a proper very true filter of* (*L*, τ). *Then the following are equivalent*:

*F is a prime very true filter of* (*L*, τ),

*if* τ(*x*) ∨ τ(*y*) ∈ *F* *for some* *x*, *y* ∈ *L*, *then* *x* ∈ *F* *or y* ∈ *F*,

(*L*/*F*, τ_{F}) *is a chain*.

(1) ⇒(2) Let τ(*x*) ∨ τ(*y*) ∈ *F* for some *x*, *y* ∈ *L*. Then 〈*x*〉_{τ} ∩ 〈*y*〉_{τ} = 〈 τ(*x*) ∨ τ(*y*) 〉_{τ} ∈ *F*. Since *F* is a prime very true filter of (*L*, τ), then 〈*x*〉_{τ} ⊆ *F* or 〈*y*〉_{τ} ⊆ *F*. Therefore, *x*∈ *F* or *y*∈ *F*.

(2) ⇒ (1) Suppose that *F*_{1},*F*_{2} ∈ *MF*[*L*] such that *F*_{1} ∩ *F*_{2} ⊆ *F* and *F*_{1} ⊈ *F* and *F*_{2} ⊈ *F*. Then there exist *x* ∈ *F*_{1} and *y* ∈ *F*_{2} such that *x*, *y* ∉ *F*. Since *F*_{1}, *F*_{2} are very true filters of (*L*, τ), then τ(*x*)∈ *F*_{1} and τ(*y*) ∈ *F*_{2}. From τ(*x*), τ(*y*) ≤ τ(*x*) ∨ τ(*y*), we obtain that τ(*x*)∨ τ (*y*) ∈ *F*_{1} ∩ *F*_{2} = *F*. By (2), we get *x* ∈ *F* or *y* ∈ *F*, which is a contradiction. Therefore, *F* is a prime very true filter of (*L*, τ).

(1) ⇔(3) From (2), one can obtain that every prime very true filter of (*L*, τ) must be a prime filter of *L*. Based on this, the equivalence of (1) and (3) is clear.

For proving the subdirect representation theorem of very true MTL-algebras we will need the following theorem.

*Let* (*L*, τ) *be a very true MTL*-*algebra and* *a*∈ *L*. *If a*≠ 1, *then there exists a prime very true filter* *P* *of* (*L*, τ) *such that* *a*∉ *P*.

Denote *F*_{a}= {*F*^{′}|*F*^{′} is a proper very true filter of (*L*, τ) such that *F* ⊆ *F*^{′}, *a* ∉ *F*^{′}}. Then *f*_{a} ≠ since *F* is a very true filter not containing *a* and *F*_{a} is a partially set under inclusion relation. Suppose that {*F*_{i}|*i* ∈ *I*} is a chain in *F*_{a}, then ∪ {*F*_{i}|*i* ∈ *I*} is a very true filter of (*L*, τ) and it is the upper bounded of this chain. By zorn’s Lemma, there exists a maximal element *P* in *F*_{a}. Now, we shall prove that *P* is the desire prime very true filter of ours. Since *P* ∈ *F*_{a}, then *P* is a proper very true filter and *a* ∉ *P*.

Let *x* ∨ *y* ∈ *P* for some *x*, *y* ∈ *L*. Suppose that *x* ∉ *P* and *y* ∉ *P*. Since *P* is strictly contained in 〈*P*, *x* 〉_{τ} and 〈*P*, *y*〉_{τ} and by the maximality of *P*, we deduce that 〈*P*, *x* 〉_{τ} ∉ *F*_{a} and 〈*P*, *y* 〉_{τ} ∉ *F*_{a}. Then *a*∈ 〈*P*, *x*〉_{τ} = *P* ∨[τ *x*) and *a* ∈ 〈*P*, *y*〉_{τ} = *P* ∨[τ(*y*)) . Then we have *a* ∈(*P* ∨[τ(*x*)))∧(*P* ∨[τ(*y*))) = *P* ∨([τ(*x*))∧[τ(*y*))) = *P*∨[τ(*x*)∨τ(*y*)) = *P* ∨[τ(*x*∨ *y*))∈ *P*, which implies that *a* ∈ *P*, a contradiction. Therefore, *P* is a prime very true filter such that *F*⊆ *P* and *a*∉ *P*. Put *F* = { 1 }, the result is easy to obtain.

Now, we will prove that every very true MTL-algebra is a subdirect product of linearly ordered very true MTL-algebras.

*Each very true MTL*-*algebra is a subalgebra of the direct product of a system of linearly ordered very true MTL*-*algebras*.

The proof of this theorem is as usual and the only critical point is the above Theorem 4.15.

The next results shows that MTL-algebra is representable if and only if very true MTL-algebra is representable.

*Let* (*L*, τ) *be a very true MTL*-*algebra*. *Then the following conditions are equivalent*:

*L is representable*;

(*L*, τ) *is a subdirect product of linearly ordered very true MTL*-*algebras*.

(1) ⇒(2) Suppose that the MTL-algebra *L* is representable. Then by Theorem 2.9, there exists a system *S* of prime filter of *L* such that ∩ *S*= {1}. Since every prime filter of *L* contains a minimal prime filter, we get that in our case the intersection of all minimal prime filter is equal to {1}. Moreover, we will show that every minimal prime filter in (*L*, τ). Let *P* be a minimal prime filter of *L*. Then by Theorem 2.8, *P* = ∪ {a^{⊥}|*a* ∈ *P*}. If *x* ∈ *P*, then there is *a* ∉ *P* such that *x* ∨ *a* = 1, hence 1 = τ(*x*∨*a*) = τ(*x*)∨ τ(*a*). Since *a* ∉ *P*, we get τ(*a*)∉ *P*, therefore τ(*x*)∈ *P*, that means that *P* is a very true filter in (*L*, τ). Therefore, (*L*, τ) is a subdirect product of linearly ordered very true NM-algebras.

(2) ⇒ (1) The converse is trivial and we hence omit this.

*Let* *L* *be an MTL*-*algebra*. *Then the following condition are equivalent*:

*L is representable*,

*L can be embedded in a very true MTL*-*algebra* (*L*, τ).

(1) ⇒(2) If *L* is representable, then *L* is a subdirect product of MTL-chain. By Example 3.2 (b), any MTL-chain has a structure of very true MTL-algebra. Moreover, the class of very true MTL-algebras is a variety, so a direct product of very true MTL-algebra is still a very true MTL-algebra.

(2) ⇒(1) is straightforward, since any very true MTL-algebra is a subdirect product of very true MTL-chains.

*A very true MTL*-*algebra* (*L*, τ) *is said to be* *a subdirectly irreducible if it has the least nontrivial very true congruence*.

Let (*L*, τ) be subdirectly irreducible. Then by Theorem 4.10, there is a very true filter *F* of (*L*, τ) such that *θ*_{F} = *F*, that means, *F* is the least very true filter of (*L*, τ) such that *F* ≠ {1}. Thus, we can conclude that a very true MTL-algebra (*L*, τ) is said to be subdirectly irreducible if among the nontrivial very true filters of (*L*, τ) there exists the least one, i.e., ∩ {*F* ∈ *VF*(*L*)|*F* ≠ {1}} ≠ {1}.

*Considering* Example 3.2 (*c*), *one can easily check that the very true MTL*-* algebra* (*L*, τ) *is subdirectly irreducible*.

Next, we will show that every subdirectly irreducible very true MTL-algebra is linearly ordered. To prove this important result, we need the following several propositions and theorems.

*Let*(*L*, τ) *be a subdirectly irreducible very true MTL*-*algebra and* *F*_{1}, *F*_{2} ∈ *VF*(*L*). *If* *F*_{1} ∩ *F*_{2}= {1}, * then* *F*_{1}= {1} *or* *F*_{2}= {1}.

Suppose *F*_{1} ≠ {1} and *F*_{2} ≠ {1}, i.e., *F*_{1}, *F*_{2} ∈ ∩ {*F* ∈ *VF*(*L*)|*F* ≠ {1}}≠ {1}, then ∩{*F* ∈ *VF*(*L*)|*F* ≠ {1}} ≠ {1} ⊆ *F*_{1} ∩ *F*_{2}. By *F*_{1} ∩ *F*_{2} = {1}, we can get ∩ {*F* ∈ *VF*(*L*)|*F* ≠ {1}} = {1}, which contradicts the fact that (*L*, τ) is subdirectly irreducible. Hence, *F*_{1} = {1} or *F*_{2} = {1}.

*Let* (*L*, τ) *be a very true MTL*-*algebra*. *Then the following conditions are equivalent*:

(*L*, τ) *is a subdirectly irreducible very true MTL*-*algebra*,

*there exists an element* *a* ∈ *L*,*a* < 1, *such thatfor any* *x* ∈ *L*, *x* < 1 *and* *a* ∈ 〈*x*〉_{τ}.

(1) ⇒ (2) Suppose that (*L*, τ) is a subdirectly irreducible very true MTL-algebra, i.e., ∩ {*F* ∈ *VF*(*L*)|*F* ≠ {1}} ≠ {1}, then ∩ {〈*x*〉_{τ}|*x* <1} ≠ {1}. Take *a* ∈ ∩ {〈*x*〉_{τ}|*x*<1} satisfying *a* ≠ 1, then for any *x* ∈ *L*, *x* ≠ 1, *a* ∈ 〈*x*〉_{τ}, by Theorem 4.5 (1), there exists *m* ∈ *N*, such that *a* ≥(τ(*x*))^{m}. Clearly, *a* is the element that we need.

(2) ⇒(1) Conversely, we need to prove that for any *F* ∈ *VF*(*L*), if *F* ≠ {1}, then *a* ∈ *F*. In fact, by *F* ≠ {1}, it follows that there exists *x* ∈ *F*, *x* <1, and then, by the known condition, *a*∈ 〈*x*〉_{τ}, furthermore, *a* ∈ *F*, so *a* ∈ ∩ {*F* ∈ *VF*(*L*)|*F*≠ {1}}. Hence, ∩ {*F* ∈ *VF*(*L*)|*F* ≠ {1}} ≠ {1}, i.e., (*L*, τ) is a subdirectly irreducible very true MTL-algebra.

We recall that a non-unit element *a* ∈ *L* is said to be a *coatom* of *L* if *a* ≤ *b*, then *b* ∈ {*a*, 1}, i.e. *b* = *a* or *b* = 1 ([25]). In the following proposition, we will show that every subdirectly irreducible very true MTL-algebra has at most one coatom.

*Let* (*L*, τ) *be a subdirectly irreducible very true MTL*-*algebra*. *For any* *x*,*y* ∈ *L*, * if* *x* ∨ *y* = 1, *then* *x* = 1 * or* *y* = 1.

For any *x*, *y* ∈ *L*, if *x*∨*y* = 1, then by Theorem 4.5, we have 〈*x*〉_{τ} ∩ 〈*y*〉_{τ} = 〈τ(*x*)∨ τ(*y*)〉_{τ} = 〈τ(*x* ∨ *y*)〉_{τ}= 〈1〉_{τ}= {1}. By Proposition 4.21, we have 〈*x*〉_{τ} = {1} or 〈*y*〉_{τ= {1}, hence x = 1 or y = 1}.

The following Theorem shows that the subdirectly irreducible very true MTL-algebra (*L*, τ) is linearly ordered, that is to say, the fuzzy truth value of all propositions in very true MTL logic are comparable. This is of key importance from the logical point of view.

*Let* (*L*, τ) *be a very true MTL*-*algebra*. *Then the following conditions are equivalent*:

(*L*, τ) *is a subdirectly irreducible very true MTL*-*algebra*,

(*L*, τ) *is a chain*.

(1) ⇒(2) Suppose (*L*, τ) is a subdirectly irreducible very true MTL-algebra. Applying Definition 2.1, we have (*x* → *y*)∨(*y* → *x*) = 1 for any *x*, *y* ∈ *L*, then by Proposition 4.23, we have *x*→ *y* = 1 or *y* → *x* = 1, i.e., *x* ≤ *y* or *y* ≤ *x*, so (*L*, τ) is a chain.

(2) ⇒ (1) Since (*L*, τ) is a chain and nontrivial, there exists a unique dual atom, denoted as *a*. Suppose *F* is any very true filter of (*L*, τ) satisfying *F* ≠ {1}, then *a* ∈ *F*. Since *F* is chosen arbitrarily from *VF*(*L*), then *a* ∈ ∩ {*F* ∈ *VF*(*L*)|*F*≠ {1}}. Hence ∩ {*F* ∈ *VF*(*L*)|*F* ≠ {1}} ≠ {1}, i.e., (*L*, τ) is a subdirectly irreducible very true MTL-algebra.

In what follows, we introduce the stabilizer of a nonempty subset set *X* with respect to a very true operator τ and study some properties of them. Let (*L*, τ) be a very tue MTL-algebra. Given a nonempty subset *X* of *L*, we put *R*_{τ}(*X*) = {*a* ∈ *L*|τ(*a*)→ *x* = *x*, ∀ *x* ∈ *X*}, and *L*_{τ}(*X*) = {*a* ∈ *L*|*x* → τ (*a*) = τ(*a*), ∀ *x* ∈ *X*}, which are called *right and left stabilizer* of *X* with respect to τ. Clearly, *R*_{τ} (*X*), *L*_{τ} (*X*) ≠ Ø. In fact, 1 ∈ *R*_{τ} (*X*) ∩ *L*_{τ} (*X*). In particular, if τ = *id*_{L}, which is a right and left stabilizer of *X* (see [26]).

*Considering* Example 3.2 (*c*). *Let* *X* = {*a*, *b*}. *Then* *L*_{τ}(*X*) = {*a*, *b*, 1} *and* *R*_{τ}(*X*) = {1}.

*Let* *L* *be a MTL*-*algebra and* *X*, *Y* ⊆ *L*. *Then the following conditions hold*:

1 ∈ *R*_{τ}(*X*)∩ *L*_{τ}(*X*),

*If* *X* ⊆ *Y*, *then* *L*_{τ}(*X*) ⊆ *L*_{τ}(*Y*) *and **R*_{τ}(*X*) ⊆ *R*_{τ}(*Y*),

*X* ⊆ *L*_{τ} (*R*_{τ}(*X*)) ∩ *R*_{τ}(*L*_{τ}(*X*)),

*R*_{τ}(*X*) = *R*_{τ}(*L*_{τ} (*R*_{τ}(*X*))) *and* *L*_{τ} (*X*) = *L*_{τ} (*R*_{τ} (*L*_{τ}(*X*))),

*L*_{τ}(*L*) = *R*_{τ} (*L*) = {1} *and* *L*_{τ} (1) = *R*_{τ} (1) = *L*,

*If* Ø ≠ *X* ⊆ *L*, *then* *R*_{τ} (*X*) *is a very true filter of* (*L*, τ).

The proof of (1)-(5) is easy by Proposition 2.2.

(6) Let *a*, *b* ∈ *R*_{τ}(*X*). Then τ(*a*)→ *x* = *x* and τ(*b*) → *x* = *x* for all *x* ∈ *X*. Hence by Proposition 2.2 (12), (τ(*a*) ⊙ τ(*b*))→ *x* = τ(*a*) → (τ(*b*) → *x*) = τ(*a*) → *x* = *x* for all *x* ∈ *X* and so τ(*a* ⊙ *b*) → *x* ≤(τ(*a*) ⊙ τ(*b*)) → *x* = *x*. On the other hand, we have *x* ≤ τ(*a* ⊙ *b*) → *x*. Therefore *x* = τ(*a* ⊙ *b*) → *x* for all *x* ∈ *X* and hence *a* ⊙ *b* ∈ *R*_{τ}(*X*). Now, let *a* ≤ *b* and *a* ∈ *R*_{τ} (*X*). Then τ(*a*) → *x* = *x*, for all *x* ∈ *X*. Hence by Proposition 3.3, τ(*b*) → *x* ≤ τ(*a*) → *x* = *x*. Since by *x* ≤ τ(*b*) → *x*, then τ(*b*)→ *x* = *x* and *b* ∈ *R*_{τ}(*X*). Finally, one can easy check that if *a* ∈ *R*_{τ}(*X*) then τ(*a*)∈ *R*_{τ}(*X*). Therefore *R*_{τ}(*X*) is a very true filter of (*L*, τ).

*Let* *F* *be a very true filter of* (*L*, τ). *Then* *R*_{τ}(*F*) *is a pseudocomplemented of* *F* *in the complete Heyting algebra* (*VF*(*L*), ∧, ∨, ↦, 1, *L*).

First, we prove that *F* ∩ *R*_{τ} (*F*) = {1}. Let *x* ∈ *F* ∩ *R*_{τ} (*F*). Since *x* ∈ *R*_{τ} (*F*), then for any *a* ∈ *F*, τ(*x*) → *a* = *a*. Now, since *x* ∈ *F*, put *a* = *x* we have *x* = 1. Therefore, *F* ∩ *R*_{τ} (*F*) = {1}. Now, let *G* be a very true filter of (*L*, τ) such that *F* ∩ *G* = {1}. Let *a* ∈ *G*, then τ(*a*) ∈ *G*. Then for any *x* ∈ *F*, since τ(*a*), *x* ≤ τ(*a*) ∨ *x*, *x* ∈ *F*, then τ(*a*) ∨ *x* ∈ *F* and τ(*a*) ∨ *x* ∈ *G* and so τ(*a*) ∨ *x* ∈ {1}. Hence τ(*a*) ∨ *x* = 1, that is ((τ(*a*) → *x*) → *x*) ∧ ((*x* → τ(*a*)) → τ(*a*)) = 1 and so ((τ(*a*) → *x*) → *x*) = 1. Hence, τ(*a*) → *x* ≤ *x*. On the other hand, *x* ≤ τ(*a*) → *x*, then τ(*a*) → *x* = *x* and *a* ∈ *R*_{τ} (*F*). Thus *G* ⊆ *R*_{τ} (*F*). Therefore, *R*_{τ} (*F*) is a pseudocomplemented of *F* in the complete Heyting algebra (*VF*(*L*), ∧, ∨, ↦, 1, *L*).

*Let* (*L*, τ) *be a very true MTL*-*algebra*. *Then* (*VF*(*L*), ∧, ∨, ↦, 1, *L*) *is a complete pseudocomple*-*mented Heyting lattice*.

It follows from Theorem 4.6 and Theorem 4.27

Now, we use of the right and left stabilizers of a very true MTL-algebra to produce a basis for a topology on it. Then we show that the generated topology by this basis is Baire, connected, locally connected and separable.

* Let* (*L*, τ) *be a very true MTL*-*algebra and* *X* ⊆ *L*. *Define a mapping* α :
$\mathcal{P}$(
$\mathcal{L}$,≪) → $\mathcal{P}$(
$\mathcal{L}$,≪)*such that* α(*X*) = *R*_{τ}(*L*_{τ}(*X*)) *for all* *X* ∈ $\mathcal{P}$(
$\mathcal{L}$). *Then the following conditions hold*,

α * is a closure map on* (*L*, τ),

*X* ⊆ α(*Y*) *if and only if* α(*X*) ⊆ α(*Y*), *for all* *Y* ⊆ *L*,

β_{α} = {*X* ⊆
$\mathcal{P}$($\mathcal{L}$)|α(*X*) = *X*} * is a basis for a topology on* (*L*, τ).

It follows from Proposition 4.26 (2),(3),(4).

The proof is easy.

Let β_{α} = {*X* ∈ $\mathcal{P}$($\mathcal{L}$) |α(*X*) = *X*}. It is clear that Ø ∈ β_{α}. Also, by Proposition 4.26 (5), α(*L*)= *R*_{τ} (*L*_{τ}(*L*)) = *R*_{τ} ({1}). Thus, α(*L*) = *L*, and *L* ∈ β_{α}. Now, suppose that *X*, *Y* ∈ β_{α}. Then α(*X*) = *X* and α(*Y*) = *Y*. We prove that *X* ∩ *Y* ∈ β_{α}. Since *X* ∩ *Y* ⊆ *X*, *Y*, by (1), α(*X* ∩ *Y*) ⊆ α(*X*), α (*Y*). Thus, α(*X* ∩ *Y*) ⊆ α(*X*) ∩ α(*Y*). Also, since *X*, *Y* ∈ β_{α}, we have α(*X* ∩ *Y*) ⊆ *X* ∩ *Y*. Moreover, by Proposition 4.26 (3), *X* ∩ *Y* ⊆ α(*X* ∩ *Y*). Then α(*X* ∩ *Y*) = *X* ∩ *Y*, and so *X* ∩ *Y* ∈ β_{α}. Therefore, β_{α} is a basis.

*Based on* Theorem 4.29, *we introduce the topological space*, (*L*,$\mathcal{T}$_{⊘}) *is called* stabilizer topology *on very true MTL*-*algebras*.

From Proposition 4.26(6), one can obtain that *R*_{τ}(*X*) ∈ *VF*(*L*), for any *X* ⊆ *L* and hence every element of β_{α} is a very true filter of (*L*, τ).

*The stabilizer topology* (*L*,$\mathcal{T}$_{⊘}) *is connected and locally connected*.

The proof is easy, and we hence omit the details.

*The stabilizer topology* (*L*,$\mathcal{T}$_{⊘}) *is Hausdorffspace if and only if* *L* = {1}.

The proof is easy, so we hence omit the details.

* The stabilizer topology* (*L*,$\mathcal{T}$_{⊘}) *is separable*.

First, we show that if Ø ≠ *X* ⊆ *L* such that 1 ∈ *X*, then
$\overline{X}$ = *L*. Let Ø ≠ *X* ⊆ *L* such that 1 ∈ *X*. We only show that *L* ⊆ $\overline{X}$. Let *x* ∈ *L*. If *x* = 1, then *x* ∈ $\overline{X}$. Hence $\overline{X}$ = *L*. Now, suppose that 1 ≠ *x* ∈ *L*. Then there exists an open subset *U* ∈ β_{α} such that *x* ∈ *U*. Since *U* ∈ *VF*(*L*) and 1 ∈ *U*, we have *U* ∩(*X* − {*x*}) ≠ Ø. Hence *x* ∈ $\overline{X}$, and so $\overline{X}$ = *L*. Since 1 ∈ β_{α}, thus $\overline{1}$ = *L*. Hence (*L*,$\mathcal{T}$_{⊘}) is separable.

*The stabilizer topology* (*L*,$\mathcal{T}$_{⊘}) *is a Baire space*.

Let *U* ∈ $\mathcal{T}$_{⊘}). Since *U* ∈ *VF*(*L*), we have 1 ∈ *U*. Then by Theorem 4.32, $\overline{U}$ = *L*. Thus, every open set of (*L*,$\mathcal{T}$_{⊘}) is dense. On the other hand, for each collection of open set *U*_{n}, ∩ *U*_{n} ∈ *VF*(*L*), so 1 ∈ ∩ *U*_{n}. Thus, by Theorem 4.32, ∩ *U*_{n} is dense. Therefore, (*L*,$\mathcal{T}$_{⊘}) is a Baire space.

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