In general, the concept of topology represents the study of topological spaces. Important topological properties include *connectedness* and *compactness*.

A topology tells how elements of a set relate spatially to each other. The same set can have different topologies. For instance, the real line, the complex plane, and the Cantor set can be thought of as the same set with different topologies.

Let *X* be a set and let *τ* be a family of subsets of *X*. We denote by 𝒫 (*X*) the family of all subsets of *X*. Then *τ* is called a topology on *X* if:

**T**_{1}. Both the empty set ∅ and *X* are elements of *τ*;

**T**_{2}. Any union of elements of *τ* is an element of *τ*;

**T**_{3}. Any intersection of finitely many elements of *τ* is an element of *τ*.

If *τ* is a topology on *X*, then the pair (*X*, *τ*) is called a *topological space*. The notation *X*_{T} is used to denote a set *X* endowed with the particular topology *τ*.

The members of *τ* are called *open sets* in *X*. A subset of *X* is said to be *closed* if its complement is in *τ* (i.e., its complement is open). A subset of *X* may be open, closed, both (clopen set), or neither. The empty set ∅ and *X* itself are always both closed and open. An open set containing a point *x* is called a *neighborhood* of *x*.

A set with a topology is called a *topological space*. A topological space (*X*, *τ*) is called *connected* if {∅, *X*} is the set of all closed and open subsets of *X*.

A *base* (or *basis*) *β* for a topological space *X* with topology *τ* is a collection of open sets in *τ* such that every open set in *τ* can be written as a union of elements of *β*. We say that the base generates the topology *τ*.

#### Definition 3.1

([11]). *Let* (*X*, ≤) *be an ordered set. Then we define* ↑ : *𝒫*(*X*) → *𝒫*(*X*), *by* ↑ *S* = {*x* ∊ *X*|*a* ≤ *x*, *for some a* ∊ *S*}, *for any subset S of X. A subset F of X is called an upset if* ↑ *F* = *F*. *We denote by U*(*X*) *the set of all upsets of X. Anupset F is called finitely generated if there exists n* ∊ *N suchthat F* =↑ {*x*_{1}, *x*_{2}, *x*_{n}}, *for some x*_{1}, *x*_{2}, ... , *x*_{n} ∊ *X*.

In [11], O. Zahiri and R. A. Borzooei construct a topology on BL-algebras considering the notion of upsets, and inspired by their work we construct a topology in the case of residuated lattices and some of their results will become particular cases. We offer complete proofs for the results available in residuated lattices.

#### Definition 3.3

([11]). *Let* *τ* and *τ* *be two topologies on a given set X. If* *τ′* ⊆ *τ*, *then we say that* *τ* *is finer than* *τ*. *Let* (*X*, *τ*) *and* (*Y*, *τ* ) *be two topological spaces. A map f* : *X* → *Y is called continuous if the inverse image of each open set of Y is open in X. A homeomorphism is a continuous function, bijective and has a continuous inverse*.

#### Lemma 3.4

([11]). *Let β and β′ be two bases for topologies* *τ* *and* *τ′* , *respectively on X. Then the following are equivalent*:

(*i*) *τ′* *is iner than* *τ*;

(*ii*) *For any x* ∊ *X and each basis element B* ∊ *β containing x, there is a basis element B′* ∊ *β such that x* ∊ *B′* ⊆ *B*.

#### Definition 3.5

([11]). *Let* (*A*, *) *be an algebra of type 2 and* *τ* *be a topology on A. Then* (*A*, *, *τ*) *is called*:

(*i*) *Right* (*left*) *topological algebra, if for all a s A the map ** : *A* → *A defined by x* ↦ *a * x* (*x* ↦ *x * a*) *is continuous;*

(*ii*) *Semitopological algebra if A is a right and left topological algebra*.

*If* (*A*, *) *is a commutative algebra, then right and left topological algebras are equivalent*.

#### Definition 3.6

([11]). *Let A be a nonempty set*, {*_{i}}_{i∊I} *be a family of binary operations on A and* *τ* *be a topology on A. Then*:

(*i*) (*A*, {*_{i}}_{i∊I}, *τ*) *is a right* (*left*) *topological algebra, if for i s I*, (*A*, *_{i}, *τ*) *is a right* (*left*) *topological algebra;*

(*ii*) (*A*, {*_{i} , *τ*) *is a right (left) semitopological algebra, if for i* ∊ *I*, (*A*, *_{i}, *τ*) *is a right* (*left*) *semitopological algebra*.

On any residuated lattice *L* we define an operator ⊖ by setting for all *x, y* ∊ *L*,

$$x\ominus y={x}^{*}\oplus y.$$(3)

By (*r*_{5}), the identity (3) is equivalent with

$$x\ominus y={x}^{*}\oplus y={x}^{**}\to {y}^{**}={y}^{*}\to {x}^{*},\text{\hspace{0.17em}for\hspace{0.17em}all}\text{\hspace{0.17em}}x,y\in L.$$(4)

#### Definition 3.7

*Consider L a residuated lattice and a* ∊ *L. For any nonempty upset X of L we define the set*

$${D}_{a}(X)=\{x\in L|{a}^{n}\ominus x\in X,\text{\hspace{0.17em}}\mathit{\text{for all}}\text{\hspace{0.17em}}n\in \mathbb{N}\},$$

*where a*^{n} ⊖ *x* = *a* ⊖ (*a*^{n–1} ⊖ *x*), *for any n* ∊ {2, 3, 4,...}.

#### Lemma 3.8

*Let L be a residuated lattice and a, x* ∊ *L. Then*:

(*r*_{35}) *a*^{n} ⊖ *x* = (*a***)^{n} → *x***;

(*r*_{36}) (*a*^{n} ⊖ *x*)** = *a*^{n} ⊖ *x, for any n* ∊ ℕ.

#### Proof

(*r*_{35}). Mathematical induction relative to *n*.

If *n* = 2, by (*r*_{5}), (*r*_{3O}) and identity (4) we obtain successively ${a}^{2}\ominus x=a\ominus (a\ominus x)\stackrel{id.(4)}{=}{a}^{**}\to {({a}^{*}\oplus x)}^{**}\stackrel{({r}_{30})}{=}{a}^{**}\to ({a}^{*}\oplus x)\stackrel{id.(4)}{=}{a}^{**}\to ({a}^{**}\to {x}^{**})\stackrel{({r}_{5})}{=}({a}^{**}\odot {a}^{**})\to {x}^{**}={({a}^{**})}^{2}\to {x}^{**}$.

Suppose *a*^{n–1} ⊖ *x* = (*a***)^{n–1} → *x***. By (*r*_{5}) we obtain successively ${a}^{n}\ominus x=a\ominus ({a}^{n-1}\ominus x)={a}^{**}\to ({({a}^{**})}^{n-1}\to {x}^{**})\stackrel{({r}_{5})}{=}({a}^{**}\odot {({a}^{**})}^{n-1})\to {x}^{**}={({a}^{**})}^{n}\to {x}^{**}$. Therefore, *a*^{n} ⊖ *x* = (*a***)^{n} → *x***, for any *n* ∊ ℕ.

(*r*_{36}). By (*r*_{30}) we have ${({a}^{n}\ominus x)}^{**}={[a\ominus ({a}^{n-1}\ominus x)]}^{**}\stackrel{id.(4)}{=}{[{a}^{*}\oplus ({a}^{n-1}\ominus x)]}^{**}\stackrel{({r}_{30})}{=}{a}^{*}\oplus ({a}^{n-1}\ominus x)\stackrel{id.(4)}{=}a\ominus ({a}^{n-1}\ominus x)={a}^{n}\ominus x$, for any *n* ∊ ℕ.

A directly consequence of (*r*_{35}) is the following result:

#### Proposition 3.9

*Let L be a residuated lattice and a* ∊ *L. For any nonempty upset X of L theset D*_{a} (*X*) = {*x* ∊ *L* : (*a***)^{n} → *x*** ∊ *X*, *for some n* ∊ ℕ}.

#### Corollary 3.11

*If L is a residuated lattice and a, x* ∊ *L, then*:

(*r*M_{37}) *a* ⊙ *x* ≤ *a* ∧ *x* ≤ *a* ∨ *x* ≤ *a*** ∨ *x*** ≤ *a* ⊕ *x*;

(*r*_{38}) *x*** ≤ *a* ⊖ *x* ≤ *a*^{n} ⊖ *x, for any n* ∊ ℕ.

#### Proof

(*r*_{37}). Indeed, *a* ⊙ *x* ≤ *a* ∧ *x* ≤ *a* ∨ *x* ≤ *a*** ∨ *x***. By (*r*_{27}) and identity (2) we obtain successively $a\oplus x\stackrel{id.(2)}{=}{a}^{*}\to {x}^{**}\ge {x}^{**}$ and $a\oplus x\stackrel{({r}_{27})}{=}x\oplus a\stackrel{id.(2)}{=}{x}^{*}\to {a}^{**}\ge {a}^{**}$, hence *a*** ∨ *x* ≤ *a* ⊕ *x*.

(*r*_{38}). Since ${a}^{n}\ominus x\stackrel{({r}_{35})}{=}{({a}^{**})}^{n}\to {x}^{**}\stackrel{({r}_{4})}{\ge}{a}^{**}\to {x}^{**}=a\ominus x\ge {x}^{**}$, hence *x*** ≤ *a* ⊖ *x* ≤ *a*^{n} ⊖ *x*, for any *n* ∊ ℕ.

The following properties hold for any residuated lattice:

(*r*_{39}) *z* → *y* ≤ (*x* → *z*) → (*x* → *y*) and *z* → *y* ≤ (*y* → *x*) → (*z* → *x*);

(*r*_{40} (*x* → *y*) ⊙ (*y* → *z*) ≤ *x* → *z*.

In the next result, we investigate some properties of the operator *D*_{a} (*X*) for nonempty upsets. Clearly, *D*_{a} (∅) = ∅.

In Proposition 3.3, [11], there are the properties of operator *D*_{a} for BL-algebras, we offer complete proofs for them in the case of residuated lattices:

#### Theorem 3.12

*If L is a residuated lattice and a, x* ∊ *L. Consider X, Y two nonempty upsets of L. Then*:

(*i*) *D*_{a}(*X*) *is an upset of L;*

(*ii*) 1 ∊ *D*_{a}(*X*), *a* ∊ *D*_{a}(*X*) and X ⊆ *D*_{a}(*X*);

(*iii*) *a*^{m} ⊖ (*a*^{n} ⊖ *x*) = *a*^{m+n} ⊖ *x, for any m, n* ∊ ℕ;

(*iv*) *if X* ⊆ *Y, then D*_{a} (*X*) ⊆ *D*_{a} (*Y*);

(*v*) *D*_{a} (*D*_{a} (*X*)) = *D*_{a} (*X*);

(*vi*) *if F is a filter of L, then D*_{a}(*F*) *is a filter of L;*

(*vii*) *if a* ≤ *s, then* *D*_{s} (*X*) ⊆ *D*_{a} (*X*);

(*viii*) *if* {*X*_{α}|*α* ∊ *I*} *is a family of upsets of L, then D*_{a}(∪{*X*_{α}|*α* ∊ *I*}) = ∪(*D*_{a} ({*X*_{α}|*α* ∊ *I*}));

(*ix*) *if* {*X*_{1}, *X*_{2},..., *X*_{n}} *is a finite set of upsets of L, then D*_{a} (∩{*X*_{i}| *i* = 1, 2,..., *n*}) = ∩{*D*_{a} (*X*_{i})| *i* = 1, 2,..., *n*};

(*x*) *D*_{a}(*D*_{x}(*X*)) = *D*_{x}(*D*_{a}(*X*));

(*xi*) *if X*_{1}, *X*_{2}, *X*_{3} *are upsets of L, then D*_{a}(*X*_{1}) ∩ [*D*_{a} (*X*_{2}) ∩ *D*_{a} (*X*_{3})] = [*D*_{a} (*X*_{1}) ∩ *D*_{a} (*X*_{2})] ∪ [*D*_{a} (*X*_{1}) ∩ *D*_{a} (*X*_{3})] *and* *D*_{a}(*X*_{1}) ∪ [*D*_{a}(*X*_{2}) ∩ *D*_{a}(*X*_{1})] = [*D*_{a}(*X*_{1}) ∪ *D*_{a}(*X*_{2})] ∩ [*D*_{a}(*X*_{1}) ∪ *D*_{a}(*X*_{3})].

#### Proof

(*i*). Clearly, *D*_{a}(*X*) ⊆↑ *D*_{a}(*X*). We consider *s* ∊↑ *D*_{a}(*X*), then there exists *x* ∊ *D*_{a}(*X*) such that *x* ≤ *s*. Since *x* ∊ *D*_{a}(*X*) we have *a*^{n} ⊖ *x* ∊ *X*, by (*r*_{35}) we deduce that (*a***)^{n} → *x*** ∊ *X*, for some *n* ∊ ℕ. By (*r*_{4}) and (*r*_{16}), since *x* ≤ *s* we obtain successively *x*** ≤ *s***, (*a***)^{n} → *x*** ≤ (*a***)^{n} → *s***. Since *X* is an upset of *L* and (*a***)^{n} → *x*** ∊ *X*, then (*a***)^{n} → *s*** ∊ *X*, hence *s* ∊ *D*_{a}(*X*) and so ↑ *D*_{a}(*X*) ⊆ *D*_{a}(*X*). We deduce that *D*_{a} (*X*) is an upset of *L*.

(*ii)*. Since *1* ∊ *X*, by (*r*_{35}) we have *a*^{n} ⊖ 1 = (*a***)^{n} → 1** = (*a***)^{n} → 1 = 1 ∊ *X*, for any *n* ∊ ℕ. Hence 1 ∊ *D*_{a} (*X*).

Since 1 ∊ *X*, by (*r*_{18} and (*r*_{35}) we have *a*^{n} ⊖ *a* = (*a***)^{n} → *a*** = 1 ∊ *X*, for any *n* ∊ ℕ. Hence *a* ∊ *D*_{a}(*X*).

Now, we consider *x* ∊ *X*. Then by and (*r*_{38}) we have *x* ≤ *x*** ≤ *a*^{n} ⊖ *x* ∊ *X*, hence *x* ∊ *D*_{a}(*X*). We deduce that *X* ⊆ *D*_{a} (*X*).

(*iii)*. Consider *m, n* ∊ ℕ and *a, x* ∊ *L*. By (*r*_{5}) and (*r*_{36}) we obtain successively ${a}^{m}\ominus ({a}^{n}\ominus x)\stackrel{id.(4)}{=}{({a}^{**})}^{m}\to {({a}^{n}\ominus x)}^{**}\stackrel{({r}_{36})}{=}{({a}^{**})}^{m}\to ({a}^{n}\ominus x)\stackrel{id.(4)}{=}{({a}^{**})}^{m}\to ({({a}^{**})}^{n}\to {x}^{**})\stackrel{({r}_{5})}{=}[{({a}^{**})}^{m}\odot {({a}^{**})}^{n}]\to {x}^{**}={({a}^{**})}^{m+n}\to {x}^{**}\stackrel{id.(4)}{=}{a}^{m+n}\ominus x$.

(*iv*). Consider *X* ⊆ *Y* and *x* ∊ *D*_{a}(*X*). Then there exists *n* ∊ ℕ such that *a*^{n} ⊖ *x* ∊ *X* ⊆ *Y* and so *x* ∊ *D*_{a}(*Y*). Hence *D*_{a}(*X*) ⊆ *D*_{a}(*Y*).

(*v*). Since *X* ⊆ *D*_{a} (*X*), by (*iv*) we have *D*_{a} (*X*) ⊆ *D*_{a} (*D*_{a} (*X*)). Consider *x* ∊ *D*_{a} (*D*_{a} (*X*)). Then there exists *m* ∊ ℕ such that *a*^{m} ⊖ *x* ∊ *D*_{a} (*X*) and so *a*^{n} ⊖ (*a*^{m} ⊖ *x*) ∊ *X*, for some *n* ∊ ℕ. By (*iii*) we have *a*^{n+m} ⊖ *x* ∊ *X* and so *x* ∊ *D*_{a} (*X*). Hence *D*_{a} (*X*) = *D*_{a} (*D*_{a} (*X*)).

(*iv*). Consider *X* ⊆ *Y* and *x* ∊ *D*_{a}(*X*). Then there exists *n* ∊ ℕ such that *a*^{n} ⊖ *x* ∊ *X* ⊆ *Y* and so *x* ∊ *D*_{a}(*Y*). Hence *D*_{a}(*X*) ⊆ *D*_{a}(*Y*).

(*v*). Since *X* ⊆ *D*_{a} (*X*), by (*iv*) we have *D*_{a} (*X*) ⊆ *D*_{a} (*D*_{a} (*X*)). Consider *x* ∊ *D*_{a} (*D*_{a} (*X*)). Then there exists *m* ∊ ℕ such that *a*^{m} ⊖ *x* ∊ *D*_{a} (*X*) and so *a*^{n} ⊖ (*a*^{m} ⊖ *x*) ∊ *X*, for some *n* ∊ ℕ. By (*iii*) we have *a*^{n+m} ⊖ *x* ∊ *X* and so *x* ∊ *D*_{a} (*X*). Hence *D*_{a} (*X*) = *D*_{a} (*D*_{a} (*X*)).

(*vi*). Consider *F* a filter of *L*. Then *F* is a nonempty upset and by (*ii*) we have 1 ∊ *D*_{a}(*F*). Let *x, x* → *y* ∊ *D*_{a}(*F*), then there are *m, n* ∊ ℕ such that (*a***)^{n} → *x*** ∊ *F* and (*a***)^{m} → (*x* → *y*)** ∊ *F*. By (*r*_{5}) and (*r*_{17}) we obtain successively $({({a}^{**})}^{n}\to {x}^{**})\to ({({a}^{**})}^{n+m}\to {y}^{**})=({({a}^{**})}^{n}\to {x}^{**})\to ({({a}^{**})}^{n}\odot {({a}^{**})}^{m}\to {y}^{**})\stackrel{({r}_{5})}{=}{({a}^{**})}^{m}\to [({({a}^{**})}^{n}\to {x}^{**})\to ({({a}^{**})}^{n}\to {y}^{**})]\stackrel{({r}_{39})}{\ge}{({a}^{**})}^{m}\to ({x}^{**}\to {y}^{**})\stackrel{({r}_{17})}{\ge}{({a}^{**})}^{m}\to {(x\to y)}^{**}\in F$.

Since *F* is a filter and (*a***)^{n} → *x*** ∊ *F*, then (*a***)^{n+m} → *y*** ∊ *F*, hence *y* ∊ *D*_{a}(*F*). We deduce that *D*_{a} (*F*) is a filter of *L*.

(*vii*). Consider *a* ≤ *s* and *x* ∊ *D*_{s}(*X*), then there exists *n* ∊ ℕ such that (*s***)^{n} → *x*** ∊ *X*. By (*r*_{4}) we obtain successively *a* ≤ *s*, *a*** ≤ *s***, (*a***)^{n} ≤ (*s***)^{n}, (*s***)^{n} → *x*** ≤ (*a***)^{n} → *x***. Since (*s***)^{n} → *x*** ∊ *X*, then (*a***)^{n} → *x*** ∊ *X*, hence *x* ∊ *D*_{a}(*X*). We deduce that *D*_{s}(*X*) ⊆ *D*_{a}(*X*).

(*viii*). Consider *x* ∊ *L*. Then we have the equivalences:

*x* ∊ *D*_{a}(∪{*X*_{α}|*α* ∊ *I*}) iff *a*^{n} ⊖ *x* ∊ ∪{*X*_{α}|*α* ∊ *I*} iff

*a*^{n} ⊖ *x* ∊ *X*_{α}, for some *n* ∊ ℕ, and α *s I* iff

*x* ∊ *D*_{a}(*X*_{α}), for some *n* ∊ ℕ and *α* ∊ *I* iff

*x* ∊ ∪(*D*_{a}({*X*_{α}|*α* ∊ *I*})). Hence *D*_{a}(∪{*X*_{α}|*α* ∊ *I*}) = ∪(*D*_{a}({*X*_{α}|*α* ∊ *I*})).

(*ix*). Following (*iv*), since ∩{*X*_{i}| *i* = 1, 2,..., *n*} ⊆ *X*_{i}, for any *i* ∊ {1, 2,..., *n*}, then *D*_{a}| *i* = 1, 2, *n*}) ⊆ *D*_{a}(*X*_{i}), for any *i* ∊ {1, 2, *n*}, hence *D*_{a}(∩{*X*_{i}| *i* = 1, 2, *n*}) ⊆ ∩{*X*_{i}| *i* = 1, 2, *n*}. Consider *x* ∊ ∩{*D*_{a}(*X*_{i})| *i* = 1, 2, *n*}. Then there exist *m*_{1}, *m*_{2}, *m*_{n} ∊ ℕ such that (*a***)^{mi} → *x*** ∊ *X*_{i}, for any *i* ∊ { 1 , 2 , ... , *n*}.

Consider now *p* = *max*{*m*_{1}, *m*_{2},...,*m*_{n}}. By (*r*_{4}), since *m*_{i} ≤ *p*, for any *i* ∊ {1, 2, *n*}, then (*a***)^{mi} → *x*** ≤ (*a***)^{p} → *x***. Since (*a***)^{mi} → *x*** ∊ *X*_{i}, then (*a***)^{p} → *x*** ∊ *X*_{i}, for any *i* ∊ {1, 2,..., *n*}. It follows that (*a***)^{p} → *x*** ∊ ∩{*X*_{i} = 1, 2, *n*}, hence *x* ∊ *D*_{a}(∩{*X*_{i} = 1, 2, *n*}). We deduce that *D*_{a}(∩{*X*_{i}| *i* = 1, 2,..., *n*}) = ∩{*D*_{a} (*X*_{i})| *i* = 1, 2,..., *n*}.

(*x*). We have successively:

*D*_{a} (*D*_{x} (*X*)) = {*t* ∊ *L*| *a*^{n} ⊖ *t* ∊ *D*_{x} (*X*), for some *n* ∊ ℕ} =

= {*t* ∊ *L*| *x*^{m} ⊖ (*a*^{n} ⊖ *t*) ∊ *X*, for some *m, n* ∊ ℕ}

$\stackrel{id.(4)}{=}\{t\in L|\text{\hspace{0.17em}}{({x}^{**})}^{m}\to {({a}^{n}\ominus t)}^{**}\in X,\text{\hspace{0.17em}}\text{for\hspace{0.17em}some}\text{\hspace{0.17em}}\text{m},n\in \mathbb{N}\}$

$\stackrel{({r}_{36})}{=}\{t\in L|\text{\hspace{0.17em}}{({x}^{**})}^{m}\to ({a}^{n}\ominus t)\in X,\text{\hspace{0.17em}}\text{for\hspace{0.17em}some}\text{\hspace{0.17em}}m,n\in \mathbb{N}\}$

$\stackrel{id.(4)}{=}\{t\in L|\text{\hspace{0.17em}}{({x}^{**})}^{m}\to [{({a}^{**})}^{n}\to {t}^{**}]\in X,\text{for\hspace{0.17em}some}\text{\hspace{0.17em}}m,n\in \mathbb{N}\}$

$\stackrel{({r}_{5})}{=}\{t\in L|\text{\hspace{0.17em}}{({a}^{**})}^{n}\to [{({x}^{**})}^{m}\to {t}^{**}]\in X,\text{\hspace{0.17em}for\hspace{0.17em}some\hspace{0.17em}}\mathit{\text{m}},\mathit{\text{n}}\in \mathbb{N}\}$

$\stackrel{id.(4)}{=}\{t\in L|\text{\hspace{0.17em}}{a}^{n}\ominus ({x}^{m}\ominus t)\in X,\text{\hspace{0.17em}}\text{for\hspace{0.17em}some}\text{\hspace{0.17em}}m,n\in \mathbb{N}\}$

= {*t* ∊ *L*|*x*^{m} ⊖ *t* ∊ *D*_{a} (*X*), for some *m* ∊ ℕ} = *D*_{x} (*D*_{a} (*X*)).

(*xi*). It follows from (*viii*) and (*ix*).

In universal algebra, for a nontrivial lattice *A*, a unary mapping *f* : *P*(*A*) → *P*(*A*) is called *laticeal modal operator* on *A* if it satisfies the conditions: For all *A*_{1}, *A*_{2} ⊆ *A*,

(1) *A*_{1} ⊆ *f*(*A*_{1});

(2) *f* (*f*(*A*_{1})) = *f* (*A*_{1});

(3) *f* (*A*_{1} ∩ *A*_{2}) = *f* (*A*_{1}) ∩ *f* (*A*_{2}).

A laticeal modal operator is called *monotone* ifit satisies:

(*m*) If *A*_{1} ⊆ *A*_{2}, then *f* (*A*_{1}) ⊆ *f* (*A*_{2}).

Following Theorem 3.12 (*ii*), (*iv*), (*v*), and (*viii*) we deduce:

#### Corollary 3.13

*Let a* ∊ *L. Then the map D*_{a} : *U*(*L*) → *U*(*L*) *is a laticeal monotone modal operator*.

#### Proof

Following Theorem 3.12 (*ii*), (*iv*), (*v*), and (*viii*) we deduce that the map *D*_{a} : *U*(*L*) → *U*(*L*) is a laticeal monotone modal operator.

We recall that, for a residuated lattice *L*, by *U*(*L*) we denote the set of all upsets of *L*.

#### Corollary 3.14

*For any a* ∊ *L, the map* *D*_{a} : *U*(*L*) → *U*(*L*) *is a closure operator and D*_{a} = *D*_{a**}.

#### Proof

Following Theorem 3.12 (*i*), (*ii*), (*iv*) and (*v*), we deduce that *D*_{a} is a closure operator.

Consider *X* an upset of *L*. Following Theorem 3.12 (*vii*), since by (*r*_{16}) we have *a* ≤ *a***, then *D*_{a**} (*X*) ⊆ *D*_{a}(*X*). Consider now *x* ∊ *D*_{a}(*X*). Then there exists *n* ∊ ℕ such that (*a***)^{n} → *x*** ∊ *X*. By (*r*_{15}) we have *a***** = *a***, then (*a*****)^{n} → *x*** ∊ *X*, hence *x* ∊ *D*_{a}**(*X*). We deduce that *D*_{a}(·) = *D*_{a**}(·).

#### Lemma 3.15

*Let a* ∊ *L and F be a filter of L. Then D*_{a} (*F*) = *D*_{1}(*F*) iff *a*** ∊ *F*.

#### Proof

” ⇐ ”. Consider *a* ∊ *L* and *F* a filter of *L* such that *D*_{a} (*F*) = *D*_{1}(*F*). Following Theorem 3.12 (*ii*), we have *a* ∊ *D*_{a}(*F*) = *D*_{1}(*F*), then *a* ∊ *D*_{1} (*F*), and we have successively *a* ∊ *D*_{1}(*F*), (1**)^{n} → *a*** ∊ *F*, 1^{n} → *a*** ∊ *F*, 1 → *a*^{**} ∊ *F*, *a*^{**} ∊ *F*.

” ⇐ ”. Suppose *a*** ∊ *F*. By Theorem 3.12 (*vii*), since *a* ≤ 1, then *D*_{1}(*F*) ⊆ *D*_{a}(*F*). Consider now *x* ∊ *D*_{a} (*F*). Then there exists *n* ∊ ℕ such that (*a***)^{n} → *x*** ∊ *F*. Since *a*** ∊ *F* and *F* is a filter of *L*, then (*a***)^{n} ∊ *F*, for any *n* ∊ ℕ. By (*a*^{**})^{n} ∊ *F*, (*a*^{**})^{n} → *x*^{**} ∊ *F*, we deduce *x*^{**} ∊ *F*. Since *x*^{**} = (1^{**})^{n} → *x*^{**} ∊ *F*, then *x* ∊ *D*_{1}(*F*), hence *D*_{a} (*F*) ⊆ *D*_{1}(*F*). We deduce that *D*_{a} (*F*) = *D*_{1}(*F*).

#### Corollary 3.16

*Let L be an involutive residuated lattice, a* ∊ *L and F be a filter of L. Then D*_{a} (*F*) = *F* *iff a* ∊ *F*.

#### Proof

” ⇐ ”. Consider *a* ∊ *L* and *F* a filter of *L* such that *D*_{a}(*F*) = *F*. Following Theorem 3.12 (*ii*), we have *a* ∊ *D*_{a} (*F*) = *F*, then *a* ∊ *F*.

” ⇐ ”. Suppose *a* ∊ *F*. By Theorem 3.12 (*ii*), *a* ∊ *F* ⊆ *D*_{a}(*F*). Consider now *x* ∊ *D*_{a}(*F*). Then there exists *n* ∊ ℕ such that (*a***)^{n} → *x*** ∊ *F*. Since *a* ≤ *a*** ∊ *F* and *F* is a filter of *L*, then (*a***)^{n} ∊ *F*, for any *n* ∊ ℕ. By (*a***)^{n} ∊ *F*, (*a***)^{n} → *x*** ∊ *F*, we deduce *x*** ∊ *F*, hence *x* ∊ *F*. Therefore, *D*_{a}(*F*) ⊆ *F*. We deduce that *D*_{a} (*F*) = *F*.

*There are residuated lattices L such that D*_{a} (*M*) *is a maximal filter, but M* ∉ *Max*_{i} (*L*). *We consider L* = {0, *n, a, b, c, d*, 1} *with* 0 < *n* < *a* < *b* < *c, d* < 1, *but c and d are incomparable*.

*Then ([12]) L becomes a distributive residuated lattice relative to the followingoperations*:

*Clearly*, 〈*a*〉 = {*a, b, c, d*, 1} *is the unique maximal filter of L and* 〈*b*〉 = {*b, c, d*, 1} *is a filter of L. Since D*_{c}(〈*b*〉) = {*x* ∊ *L*|(*c***)^{n} → *x*** ∊ 〈*b*〉, *for some n* ∊ ℕ} = {*x* ∊ *L*|1 → *x*** ∊ 〈*b*〉} = {*x* ∊ *L*|*x*** ∊ 〈*b*〉} = {*a, b, c, d*, 1} = 〈*a*〉. *Therefore, D*_{c}(〈*b*〉) = 〈*a*〉 *is the unique maximal filter of L, but* 〈*b*〉 *is not maximal*.

#### Proposition 3.18

*Let a* ∊ *L and F be a filter of L. Then*:

(*i*) *if M is a maximal filter of L and a*** ∊ *M*, *then D*_{a} (*M*) = *M;*

(*ii*) *D*_{1}(*F*) *is a prime filter iff D*_{a} (*F*) *is prime;*

(*iii*) (*a*^{n})* ∊ *F, for some n* ∊ ℕ *iff D*_{a} (*F*) = *L;*

(*iv*) *if M is a maximal filter of L, then D*_{a} (*M*) = *L iff a* ∊ *L* \ *M*.

*Moreover, if M is a maximal filter of L, then D*_{a}(*M*) = *M iff a* ∊ *M*.

#### Proof

(*i*). Let *M* be a maximal filter of *L* and *a*** ∊ *M*. Following Theorem 3.12 (*ii*) and (*vi*) we have *D*_{a}(*M*) is a filter and *M* ⊆ *D*_{a}(*M*). We must to prove that *D*_{a}(*M*) is a proper filter of *L*. If 0 ∊ *D*_{a}(*M*), then by (*r*_{5}) and (*r*_{15}) we obtain successively 0 ∊ *D*_{a}(*M*), (*a***)^{n} → 0** ∊ *M*, (*a***)^{n} → 0 ∊ *M*, (*a***)^{n-1} ⊙ *a*** → 0 ∊ *M*, (*a***)^{n-1} → [*a*** → 0] ∊ *M*, (*a***)^{n-1} → *a**** ∊ *M*, (*a***)^{n-1} → *a** ∊ *M*. Since *a*** ∊ *M* and *M* is a filter of *L*, then (*a***)^{n-1} ∊ *M*. Since (*a***)^{n-1} ∊ *M*, (*a*** )^{n-1} → *a** ∊ *M*, then *a** ∊ *M*. Since *a**, *a*** ∊ *M* and *M* is a filter of *L*, then 0 = *a** ⊙ *a*** ∊ *M*, a contradiction. Hence *D*_{a}(*M*) ≠ *L*.

By hypothesis, *M* is a maximal filter and *M* ⊆ *D*_{a}(*M*) ≠ *L*, then *D*_{a}(*M*) = *M*.

(*ii*). Let *F* beafilter of *L*. Following Theorem 3.12 (*vi*), *D*_{1}(*F*) and *D*_{a} (*F*) are filters of *L*. Suppose *D*_{1}(*F*) is a prime filter of *L* and *x* ∨ *y* ∊ *D*_{1}(*F*). If *x* ∨ *y* ∊ *D*_{1}(*F*), then (1**)^{n} → (*x* ∨ *y*)** = (*x* ∨ *y*)** ∊ *F*, for any *n* ∊ ℕ. Since *F* is a filter and (*x* ∨ *y*)** ≤ (*a***)^{n} → (*x* ∨ *y*)**, then (*a***)^{n} → (*x* ∨ *y*)** ∊ *F*, hence *x* ∨ *y* ∊ *D*_{a}(*F*).

Since *x* ∨ *y* ∊ *D*_{1}(*F*) and *D*_{1}(*F*) is prime, then *x* ∊ *D*_{1}(*F*) or *y* ∊ *D*_{1}(*F*). It follows that *x*** ∊ *F* or *y*** ∊ *F*. We have successively *x*** ∊ *F*, *x*** ≤ (*a***)^{n} → *x*** ∊ *F* or *y*** ∊ *F*, *y*** ≤ (*a***)^{m} → *y*** ∊ *F*, then *x* ∊ *D*_{a}(*F*) or *y* ∊ *D*_{a}(*F*), for some *m, n* ∊ ℕ. Hence *D*_{a}(*F*) is prime.

Now, suppose *D*_{a}(*F*) is prime and for *a* = 1 we deduce *D*_{i}(*F*) is prime, too.

(*iii*). Suppose (*a*^{n})* ∊ *F*, for some *n* ∊ ℕ. Following Theorem 3.12 (*ii*), F ⊆ *D*_{a}(*F*), then (*a*^{n})* ∊ *D*_{a}(*F*). By Theorem 3.12 (*ii*) and (*vi*), we obtain *a* ∊ *D*_{a} (*F*) and *D*_{a} (*F*) is a filter, then *a*^{n} ∊ *D*_{a} (*F*), for any *n* ∊ ℕ. Since *a*^{n} ∊ *D*_{a}(*F*), (*a*^{n})* ∊ *D*_{a}(*F*) and *D*_{a}(*F*) is a filter, then 0 = *a*^{n} ⊙ (*a*^{n})* ∊ *D*_{a}(*F*). Hence *D*_{a}(*F*) = *L*.

Now, suppose *D*_{a}(F) = *L*. Then *0 ∊ D*_{a}(F), (*a*^{**})→ 0^{**} ∊ *F*, (*a*^{**})^{n} → 0 ∊ *F*, for some *n ∊ ℕ*.

We prove by induction that (*a*^{**})^{n+1} → 0 = *a*^{n+1} → 0, for *n ∊ ℕ*. Clear, for *n* = 0. Suppose (*a*^{**})^{n} → 0 = *a*^{n} → 0, for *n* ∊ ℕ. By (*r*_{5}), (*a*^{**})^{n+1} → 0 = ((*a***)^{n} ⊙ *a*^{**}) → 0 = *a*^{**} → ((*a*∊**)^{n} → 0) = *a*^{**} → (*a*^{n} → 0) = *a*^{n} → *a*∊^{***} = *a*^{n} → (*a* → 0) = *a*^{n+1} → 0.

Since (*a*^{**})^{n} → 0 ∊ *F*, then *a*^{n} → 0 ∊ F, that is (*a*^{n})* ∊ *F*.

(*iv*) Consider *M* a maximal filter of *L*.

”(⇒)”. Consider *D*_{a}(M) = *L*. If *a* ∊ *M*, then we get *a*^{**} ∊ *M*, and by (*i*) we obtain *L* = *D*_{a}(M) = *M*, a contradiction. We deduce *a* ∉ *M*, that is, *a* ∊ *L* \ *M*.

” ⇐ ”. Consider *a ∊ L \ M*. Since *M* is a maximal filter, then following Proposition 2.8 there is *n* ∊ ⊙ such that (*a*^{n})^{*} ∊ *M*, and by (*iii*) we obtain *D*_{a}(M) = *L*.

Now, the fact that *D*_{a}(M) = *M* iff *a* ∊ *M* is routine.

Georgescu et al. (2015)[15] called *Gelfand residuated lattices* those residuated lattices in which any prime filter is included in a unique maximal ilter. They are also called *normal residuated lattices*. Examples of Gelfand residuated lattices are Boolean algebras, BL-algebras and Stonean residuated lattices (see [8]).

#### Proposition 3.19

*Let a ∊ L and P ∊ Spec*_{i} (L) be aprime filter. If P = D_{a} (P), then L is Gelfand (normal) residuated lattice. The converse does not hold.

#### Proof

Let *P* be a prime filter of *L* such that *P = D*_{a} (P). Using Zorn’s Lemma we deduce that *P* is contained in a maximal filter. Suppose that there are two distinct maximal filters *M*_{1} and *M*_{2} such that *P* ⊆ *M*_{1} and *P* ⊆ *M*_{2}. Since *M*_{1} ≠ M_{2}, there is *a* ∊ *M*_{1} such that *a* ∌ *M*_{2}. By Theorem 3.12 (*ii*) and (*vi*) we have that *a* ∊ *D*_{a}(*P*) and *D*_{a}(*P*) is a filter, then (*a*^{n})^{**} ∊ *D*_{a}(*P*) = *P*, for any *n* ∊ ℕ. Hence (*a*^{n})^{**} ∊ *P*, for any *n* ∊ ℕ.

Following Proposition 2.8, there is *n* ≥ 1 such that (*a*^{n})^{*} ∊ *M*_{2}. Then (*a*^{n})^{**} ∉ *M*_{2}, hence (*a*^{n})^{**} ∉ *P*, a contradiction.

For the converse we consider the residuated lattice *L* from Remark 3.17 (*v*). The prime filter of *L* are (*a*) = {*a, b, c, d, 1*}, 〈*b*〉 = {*b, c, d, 1*}, 〈c〉 = {*c, 1*} and 〈*d*〉 {*d, 1*}. The maximal filter of *L* is 〈*a*〉, which include the all other prime filters. Hence *L* is Gelfand, but *D*_{c}(〈b〉) = 〈*a*〉 ≠ 〈*b*〉.

#### Proposition 3.20

*Let F ∊ ℱ*_{i} (L) and a ∊ F. For x ∊ D_{a}(F) the following assertions are equivalent:

(*i*)*D*_{a}(F) = *F*;

(*ii*)*x*^{**} ∊ *F, then x ∊ F*.

#### Proof

(*i*) ⇒ (*ii*). Consider *D*_{a}(F) = F. By hypothesis *F* is a filter and *a* ∊ *F*, then *a*^{**} ∊ *F*, (*a*^{**})^{n} ∊ *F*, for every *n* ∊ ℕ. Since *F* is a ilter and (*a*^{**})^{n} ∊ *F*, (*a*^{**})^{n} → *x* ∊ *F*, then *x* ∊ *F*. We obtain successively *F* = *D*_{a}(F) = {*x* ∊ *L* | (*a*^{**})^{n} → *x*^{**} ∊ *F*, for some *n* ∊ ℕ} = {*x ∊ L* | *x*^{**} ∊ *F*}, that is, if *x*^{**} ∊ *F*, then *x* ∊ *F*, for all *x* ∊ *L*.

(*ii*) ⇒ (*i*). By hypothesis *F* is a filter and *a* ∊ *F*, then *a*^{**} ∊ *F*, (*a*^{**})^{n} ∊ *F*, for every *n* ∊ ℕ. Since *F* is a filter and (*a*^{**})^{n} ∊ *F*, (*a*^{**})^{n} → *x*^{**} ∊ *F*, then *x*^{**} ∊ *F*. We obtain successively *D*_{a}(F) = {*x* ∊ *L* | (*a*^{**})^{n} → *x*^{**} ∊ *F*, for some *n* ∊ ℕ} = {*x ∊ L* | *x*^{**} ∊ *F*} = *F*.

#### Theorem 3.21

*Let a ∊ L and F be a filter of L. For x ∊ D*_{a} (*F*) *the following assertions are equivalent*:

(*i*)*D*_{a}(*F*) = *F*

(*ii*)*x*^{**} ∊ F iff x ∊ *F*.

#### Proof

(*i*) ⇒ (*ii*). Consider *D*_{a}(F) = *F*. By hypothesis *F* is a filter and *x* ∊ *D*_{a}(F) = *F*, by (*r*_{16}) we obtain *x* ≤ *x*^{**} ∊ *F*, so, if *x* ∊ *F*, then *x*^{**} ∊ *F*. Now, we prove that if *x*^{**} ∊ *F*, then *x* ∊ *F*. Since *F* is a filter and *x*^{**} ∊ *F*, *x*^{**} ≤ (*a*^{**})^{n} — *x*^{**} ∊ *F*, for any *n ∊ N*, then *x* ∊ *D*_{a}(*F*) = *F*.

(*ii*) ⇒ (*i*). By Theorem 3.12, (*ii*) we have *F* ⊆ *D*_{a}(*F*). Now, we consider *x* ∊ *D*_{a}(*F*) such that *x*** ∊ *F* iff *x* ∊ *F*. Since *x*** ∊ *F* and *x*** ≤ (*a***)^{n} → *x*** ∊ *F*, for any *n* ∊ ℕ, then we obtain successively *D*_{a}(*F*) = {*x* ∊ *L* | (*a***)^{n} → *x*** ∊ *F*, for some *n* ∊ ℕ} ⊆ {*x* ∊ *L* | *x*** ∊ *F*} ⊆ *F*. Therefore, *D*_{a}(*F*) = *F*.

#### Corollary 3.23

*The set τ*_{a} = {D_{a}(X)|X ϵ U(L)} is a topology on L and (*L*, τ_{a}) *is a topological space*.

#### Proof

Let *a ϵ L*. Clearly, *D*_{a}(ϕ) = *ϕ* and *D*_{a}(*L*) = *L*. Following Theorem 3.12 (*viii)* and (*ix*) the set *τ*_{a} = {*D*_{a}(*X*)|*X* ϵ *U*(*L*)} is a topology on *L* and (*L, τ*_{a}) is a topological space.

#### Proposition 3.24

*Theset* β_{a} = {*D*_{a} (↑ *x*)|*x* ϵ *L*)} *is a base for the topology τa on L*.

#### Proof

Let *Z* be an open subset of (*L, τ*_{a}). Then there is *X* ϵ *U*(*L*) such that *Z* = *D*_{a}(*X*). Since *X* is an upset, then *X* = *u*{↑ *x*|*x* ϵ *X*} and by Theorem 3.12 (*viii*), we deduce that *D*_{a}(*X*) = ∪{*D*_{a} (↑ *x*)| *x* ϵ *X*}. Hence β_{a} is a base for the topology *τa* on *L*.

#### Corollary 3.25

*Every open set X relative to the topology τ*_{a} is an upset. But the converse does not hold.

#### Proof

Following Corollary 3.23 and Proposition 3.24 we have (*L, τ*_{a}) is a topological space with *β *_{a} a base for the topology τ_{a}. Since every union of elements of *β*_{a} is an upset and every open set *X* relative to τ_{a} can be written as a union of elements of *β*_{a}, then *X* is an upset.

For the converse we consider the residuated lattice *L* from Remark 3.17 (*v*), where *↑ n* = {*n, a, b, c, d*, 1} is an upset of *L*. The upsets of *L* are ↑ 0 = *L*, ↑ *n* = {*n, a, b, c, d*, 1}, ↑ *a* = {*a, b, c, d*, 1}, ↑ *b* = {*b, c, d*, 1}, ↑ *c* = {*c*, 1}, ↑ *d* = {*d*, 1} and *↑ 1 =* {1}, so *U* (*L*) = {↑ 0, ↑ *n*, ↑ *a*, ↑ *b*, ↑ *c*, ↑ *d*, ↑ 1}. Since the base of topology *τ*_{p} on *L* is the set *β*_{p} = {*D*_{p}(↑ *x*) | *p ϵ L* and ↑ *x* ϵ *U*(*L*)} = {*z* ϵ *L*| (*p*^{**})^{n} → *z*^{**} ϵ↑ *x* for some *n* ϵ ℕ}, then it is easy to verify that *↑ n* can not be written as a union of elements of *βp*. Therefore, *↑ n* is not an open set relative to the topology *τ*_{p}.

#### Proposition 3.26

*If u, v ϵ L such that u ≤ v, then the topology τ*_{v} is finer than topology τ_{u}.

#### Proof

We denote by *β*_{u}, β_{v} basis of the topology *τ*_{u}, respectively *τ*_{v}.

Consider *t ϵ L* and *D*_{u} (↑ x) ⊆ β_{u} an element of the basis *β*_{u} such that *t ϵ D*_{u}(↑ x). Then there exists *n ϵ N* such that (*u*^{**})^{n} → t^{**} ϵ↑ x, that is, *x ≤ (u*^{**})^{n} → t^{**}.

Following Theorem 3.12 (*ii)* we have *t ϵ D*_{v}(↑ t). Following Theorem 3.12 (*vii)*, since *u ≤ v*, then *D*_{v} (↑ t) ⊆ D_{u}(↑ t). We prove that *D*_{u} (↑ t) ⊆ D_{u}(↑ x). Now, we consider s *ϵ D*_{u} (↑ t). Then there exists *m ϵ N* such that (*u*^{**})^{m} → s^{**} ϵ ↑ t and so *t ≤ (u*^{**})^{m} → s^{**}. By *(r*_{4}), (r_{35}) and (r_{36}) we obtain successively t ≤ (u^{**})^{m} → s^{**}, ${t}^{**}\stackrel{({r}_{4})}{\le}{[{({u}^{**})}^{m}\to {s}^{**}]}^{**}\stackrel{({r}_{35}),({r}_{36})}{=}{({u}^{**})}^{m}\to {s}^{**}$. Since $x\le {({u}^{**})}^{n}\to {t}^{**}\le {({u}^{**})}^{n}\to [{({u}^{**})}^{m}\to {s}^{**}]\stackrel{({r}_{5})}{=}{({u}^{**})}^{m+n}\to {s}^{**}$, then *x ≤ (u*^{**})^{m+n} → s^{**}, hence (*u*^{**})^{m+n}→ s^{**} ϵ ↑ x and so s *ϵ Du (↑ x)*.

Since *D*_{u} (↑ t) ⊆ D_{u} (↑ x) and *D*_{v} (↑ t) ⊆ D_{u} (↑ t), it follows that *D*_{v} (↑ t) ⊆ D_{u} (↑ t) ⊆ D_{u} (↑ x).

Clearly, *D*_{v} (↑ *t*) ⊆ β_{v} is an element of the basis β_{v}. We deduce that for any *t* ϵ *D*_{u} (↑ *x*) ⊆ β_{u}, there is a basis element *D*_{v} (↑ *t*) ⊆ β_{v} such that *t* ϵ *D*_{v} (↑ *t*) ⊆ *D*_{u}(↑ *t*) ⊆ *D*_{u} (↑ *x*). Following Lemma 3.4 we deduce that the topology *τv* is finer than topology *τu*.

#### Lemma 3.27

*If X is a nonempty subset of L and a ϵ L, then X is a compact subset of* (*L*, τ*a*) *iff X* ⊆ *D*_{a} (↑ {*x*_{i1}, x_{i2}, x_{in}}), for some *x*_{i1}, *x*_{i2}, *x*_{in} ϵ *X and i* ϵ *I*.

#### Proof

“ ⇐ “. Suppose*X ⊆ D*_{a}(↑ {*x*_{i1}, x_{i2},..., x_{in}}), for some *x*_{i1}, x_{i2},..., x_{in} ϵ X and {*D*_{a}(*X*_{i})|*i* ϵ *I*} be a family of open subsets of *L* whose union contains *X*. For any *j* ϵ {1, 2,..., *n*}, there is *i*_{j} ϵ *I* such that *x*_{ij} ϵ *D*_{a} (*X*_{ij}). Following Theorem 3.12 (*i*) and (*iv*), we have *D*_{a}(*x*_{ij}) ⊆ *D*_{a}(*X*_{ij}), for any *j* ϵ {1,2,..., *n*}. By Theorem 3.12 (*viii*) weobtain *X ⊆ D*_{a} (↑ {*xi*_{1}, *x*_{i2}*x*_{in}}) = *D*_{a} (*x*_{i1} ) ∪ *D*_{a} (*x*_{i2})∪...∪*D*_{a} (*x*_{in}) ⊆ D_{a} (*X*_{i1})∪*D*_{a} (*X*_{i2})∪...∪*D*_{a} (*X*_{in}). Hence *X* is compact.

*“ ⇒ “*. Now, suppose *X* be a compact subset of *L*. Since *X* ⊆ ∪{↑ *x*|*x* ϵ *X*}, then by Theorem 3.12(ii), *X* ⊆ *u*{*D*_{a}(↑ *x*)|*x* ϵ *X*}. Hence {*D*_{a}(↑ *x*)|*x* ϵ *X*} is a family of open subsets of *L* whose union contains *X*.

By hypothesis, there are *x*_{1}, x_{2},..., x_{n} ∊ X such that *X ⊆ D*_{a}(↑ x_{1}) ∪ D_{a}(↑ x_{2}) ∪ ... ∪ D_{a}(↑ x_{n}). By Theorem 3.12(*viii*) we obtain *X* ⊆ D_{a}(↑ {x_{1}, x_{2},..., x_{n}}).

#### Theorem 3.28

*The topological space* (*L*, *τ*_{a}) *is connected*.

#### Proof

Consider *X* a non-empty subset of *L* such that is both closed and open relative to the topology *T*_{a}. If *X* is an open set, by Corollary 3.25, then *X* is an upset. If 0 ∊ *X* and *X* is an upset, then *X = L*. If 0 ∊ *L -X*, since *X* is closed, we get *L - X* is open, by Corollary 3.25, *L - X* is an upset of *L*, since *0 ∊ L - X*, then *L - X = L*. Hence *X* = ∅, a contradiction. We conclude that {∅, *L*} is the set of all subsets of *L* which are both closed and open. That is, (*L, τ*_{a}) is connected.

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