Note that from Theorem 3.4 giving the ∧ family of filters of an *EQ*-algebra *E* which is closed under intersection. We can induce a uniform topology 𝒯_{Λ} on *E*. In this section we study topological properties on (*E*, 𝒯_{Λ}).

Let *E* be an *EQ*-algebra and *C*, *D* be subsets of *E*. Then we define *C* ∗ *D* as follows: *C* ∗ *D* = {*x* ∗ *y* : *x* ∈ *C*, *y* ∈ *D*}, where ∗ ∈ {∧, ⊗, ∼}.

#### Definition 4.1

*Let 𝓤 be a topology on* E. *Then* (E, 𝓤) *is called a* topological *EQ*-algebra(*TEQ*-algebra for short) *if the operations* ∧, ⊗ *and* ∼ *are continuous with respect to 𝓤*.

Recall that a topological space (*X*, 𝓤) is a discrete space if for any *x* ∈ *X*, {*x*} is an open set.

#### Example 4.2

*Every EQ-algebra with a discrete topology is a TEQ-algebra*.

#### Theorem 4.3

*The pair* (*E*, 𝒯_{Λ}) *is a TEQ-algebra*.

#### Proof

By Definition 4.1, it suffices to show that ∗ is continuous, where ∗ ∈ {∧, ⊗, ∼}. Indeed, assume that *x* ∗ *y* ∈ *G*, where *x*, *y* ∈ *E* and *G* is an open subset of *E*. Then there exist *U* ∈ 𝓚, *U*[*x* ∗ *y*] ⊆ *G*, and a filter F such that *U*_{F} ∈ 𝓚^{∗} and *U*_{F} ⊆ *U*. We claim that the following relation holds:
$${U}_{F}[x]\ast {U}_{F}[y]\subseteq {U}_{F}[x\ast y]\subseteq U[x\ast y].$$

Let *h*∗*k* ∈ *U*_{F} [*x*]∗*U*_{F} [*y*]. Then *h* ∈ *U*_{F} [*x*] and *k* ∈ *U*_{F} [*y*] we get that *x* ≡_{F} *h* and *y* ≡_{F} *k*. Hence *x* ∗ *y* ≡_{F} *h*∗*k*. From that we obtain (*x* ∗ *y*, *h* ∗ *k*) ∈ *U*_{F} ⊆ *U*. Hence *h* ∗ *k* ∈ *U*_{F} [*x* ∗ *y*] ⊆ *U*[*x* ∗ *y*]. Then *h* ∗ *k* ∈ *G*. Clearly, *U*_{F} [*x*] and *U*_{F} [*y*] are neighborhoods of *x* and *y*, respectively. Therefore, the operation ∗ is continuous. □

#### Example 4.4

*In Example 3.6, it is easy to check that* (*E*, 𝒯_{F}) *is a TEQ-algebra*.

#### Theorem 4.5

*Let* ∧ *be a family of filters of E which is closed under intersection*. *Any filter in the collection* ∧ *is a clopen subset of E for the topology* 𝒯_{Λ}.

#### Proof

Let *F* be a filter of *E* in ∧ and *y* ∈ *F*^{c}. Then *y* ∈ *U*_{F} [*y*] and we get *F*^{c} ⊆ ∪{*U*_{F}[*y*] : *y* ∈ *F*^{c}}. We claim that for all *y* ∈ *F*^{c}, *U*_{F} [*y*] ⊆ *F*^{c}. If *z* ∈ *U*_{F} [*y*], then *z* ≡_{F} *y*. Hence *z* ∼ *y* ∈ *F* .If *z* ∈ *F*, by Lemma 2.6 (i), we get that *y* ∈ *F*, which is a contradiction. So *z* ∈ *F*^{c} and we get ∪{*U*_{F} [*y*] : *y* ∈ *F*^{c}} ⊆ *F*^{c}. Hence *F*^{c} = ∪{*U*_{F} [*y*] : *y* ∈ *F*^{c}}. Since *U*_{F} [*y*] is open for all *y* ∈ *E*, it follows that *F* is a closed subset of *E*. We show that *F* = ∪{*U*_{F} [*y*] : *y* ∈ *F*}. If *y* ∈ *F*, then *y* ∈ *U*_{F} [*y*] and we get *F* ⊆ ∪{*U*_{F} [*y*] : *y* ∈ *F*}. Let *y* ∈ *F* .If *z* ∈ *U*_{F} [*y*], then *z* ≡_{F} *y* and so *y* ∼ *z* ∈ *F*. Since *y* ∈ *F*, by Lemma 2.6 (i), *z* ∈ *F*, and we get ∪{*U*_{F} [*y*] : *y* ∈ *F*} ⊆ *F* .So *F* is also an open subset of *E*. □

#### Theorem 4.6

*Let* ∧ *be a family of filters of E which is closed under intersection. For any x* ∈ *E and F* ∈ ∧, *U*_{F} [*x*] *is a clopen subset of E for the topology* 𝒯_{Λ}.

#### Proof

First we show that (*U*_{F} [*x*])^{c} is open. If *y* ∈ (*U*_{F} [*x*])^{c}, then *y* ∼ *x* ∈ *F*^{c}. We claim that *U*_{F} [*y*] ⊆ (*U*_{F} [*x*])^{c}. If *z* ∈ *U*_{F} [*y*], then *z* ∈ (*U*_{F} [*x*])^{c}, otherwise *z* ∈ *U*_{F} [*x*], we get that *z* ∼ *y* ∈ *F* and *z* ∼ *x* ∈ *F*. Since *F* is a filter, we get that (*x* ∼ *z*) ⊗ (*z* ∼ *y*) ∈ *F*. By (*x* ∼ *z*) ⊗ (*z* ∼ *y*) < *x* ∼ *y* and *F* is a filter, it follows that *x* ∼ *y* ∈ *F*, which is a contradiction. Hence *U*_{F} [*y*] ⊆ (*U*_{F} [*x*])^{c} for all *y* ∈ (*U*_{F} [*x*])^{c}, and so *U*_{F} [*x*] is closed. It is clear that *U*_{F} [*x*] is open. So *U*_{F} [*x*] is a clopen subset of *E*. □

A topological space *X* is connected if and only if *X* has only *X* and ∅ as clopen subsets. Therefore we have the following corollary.

#### Corollary 4.7

*The space* (*E*, 𝒯_{Λ}) is not, in general, a connected space.

#### Theorem 4.8

𝒯_{∧} = 𝒯_{J}, where *J* = ∩{*F* : *F* ∈ ∧}.

#### Proof

Let 𝓚 and 𝓚^{∗} be as in Theorems 3.2 and 3.3. Now consider ∧_{0} = {*J*}, define (𝓚_{0})^{∗} = {*U*_{J}} and 𝓚_{0} = {*U* : *U*_{J} ⊆ *U*}. Let *G* ∈ 𝒯_{Λ}. So for each *x* ∈ *G*, there is *U* ∈ 𝓚 such that *U*[*x*] ⊆ *G*. From *J* ⊆ *F* we get that *U*_{J} ⊆ *U*_{F} for any filter *F* of ∧. Since *U* ∈ 𝓚, there exists *F* ∈ ∧ such that *U*_{F} ⊆ *U*. Hence *U*_{J} [*x*] ⊆ *U*_{F} [*x*] ⊆ *G*. Since *U*_{J} ∈ 𝓚_{0}, we get that *G* ∈ 𝒯_{J}. So 𝒯_{Λ} ⊆ 𝒯_{J}. Conversely, let *H* ∈ 𝒯_{J}. Then for any *x* ∈ *H* there is *U* ∈ 𝓚_{0} such that *U*[*x*] ⊆ *H*. Hence *U*_{J} [*x*] ⊆ *H*. Since ∧ is closed under intersection, so *J* ∈ ∧. Then we get *U*_{J} ∈ 𝓚 and so *H* ∈ 𝒯_{Λ}. Therefore, 𝒯_{J} ⊆ 𝒯_{Λ}. □

#### Corollary 4.9

*Let F and J be filters of E and F* ⊆ *J*. *Then J is clopen in the topological space* (*E*, 𝒯_{F}).

#### Proof

Consider ∧ = {*F*, *J*}. Then by Theorem 4.8, 𝒯_{Λ} = 𝒯_{F}. Hence by Theorem 4.5, *J* is clopen in the topological space (*E*, 𝒯_{F}). □

#### Lemma 4.11

*If F is a filter of E, then for all x* ∈ *E*, *U*_{F} [*x*] *is a clopen compact set in the topological space* (*E*, 𝒯_{F}).

#### Proof

By Theorem 4.6, it is enough to show that *U*_{F} [*x*] is a compact set. Let *U*_{F} [*x*] ⊆ ∪_{α∈I} *O*_{α}, where each *O*_{α} is an open set of *E*. Since *x* ∈ *U*_{F} [*x*], there exists α ∈ *I* such that *x* ∈ *O*_{α}. Then *U*_{F} [*x*] ⊆ *O*_{α}. Hence *U*_{F} [*x*] is compact. Therefore *U*_{F} [*x*] is a clopen compact set in the topological space (*E*, 𝒯_{F}). □

#### Theorem 4.12

*Let* ∧ *be a family of filters of E which is closed under intersection*. *Then* (*E*, 𝒯_{Λ}) *is a first-countable, zero-dimensional, disconnected and completely regular space*.

#### Proof

By Theorem 4.8, it is suffices to show that (*E*, 𝒯_{J}) is a first-countable, zero-dimensional, disconnected and completely regular space. Let *x* ∈ *E*. By Remark 4.10 (iv), {*U*_{J} [*x*]} is a denumerable fundamental system of neighborhoods of *x*, so (*E*, 𝒯_{J}) is first-countable. Let 𝓑_{J} = {*U*_{J} [*x*] : *x* ∈ *E*}. By Remark 4.10 (iii) and Theorem 4.6, we get that *B*_{J} is a clopen basis of (*E*, 𝒯_{J}), hence (*E*, 𝒯_{J}) is a zero-dimensional space. By Corollary 4.7, we get that (*E*, 𝒯_{J}) is a disconnected space. By Lemma 4.11 and Remark 4.10 (ii), *U*_{J} [*x*] is a compact neighborhood of *x*. Hence (*E*, 𝒯_{J}) is a locally compact space. Let *x* ∈ *E* and *V* be a neighborhood of *x*. By Remark 4.10 (ii) and Lemma 4.11, there exists closed neighborhood *U*_{J} [*x*] of *x* such that *U*_{J} [*x*] ⊆ *V*. Therefore, (*E*, 𝒯_{J}) is a regular space. Since (*E*, 𝒯_{J}) is a locally compact space, we get that it is completely regular. □

#### Theorem 4.13

*Let* ∧ *be a family of filters of E which is closed under intersection*. *Then* (*E*, 𝒯_{Λ}) *is a discrete space if and only if there exists F* ∈ ∧ *such that U*_{F} [*x*] = {*x*} *for all x* ∈ *E*.

#### Proof

Let 𝒯_{Λ} be a discrete topology on *E*. If for any *F* ∈ ∧, there exists *x* ∈ *E* such that *U*_{F} [*x*] ≠ {*x*}. Let *J* = ∩∧. Then *J* ∈ ∧, there exists *x*_{0} ∈ *E* such that *U*_{J} [*x*_{0}] ≠ {*x*_{0}}. It follows that there exists *y*_{0} ∈ *U*_{F} [*x*_{0}] and *x*_{0} ≠ *y*_{0}. By Remark 4.10 (ii), *U*_{J} [*x*_{0}] is the smallest neighborhood of *x*_{0}. Hence {*x*_{0}} is not an open subset of *E*, which is a contradiction. Conversely, for any *x* ∈ *E*, there exists *F* ∈ ∧ such that *U*_{F} [*x*] = {*x*}. Hence {*x*} is an open set of *E*. Therefore, (*E*, 𝒯_{Λ}) is a discrete space. □

#### Theorem 4.14

*Let* ∧ *be a family of filters of E which is closed under intersection*, *J* = ∩∧ *and E be a separated EQ-algebra. Then the following conditions are equivalent*:

(*E*, 𝒯_{J}) *is a discrete space*;

*J* = {1}.

#### Proof

⇒ (*ii*): By Theorem 4.13, we have *U*_{J} [1] = {1}. We show that *J* ⊆ *U*_{J}[1]. Let *x* ∈ *J*. By Proposition 2.2 (vii), we get that *x* ≤ *x* ∼ 1. Since *J* is a filter and *x* ∈ *J*, hence *x* ∼ 1 ∈ *J*. So *x* ∈ *U*_{J} [1]. It follows that *J* ⊆ *U*_{J} [1]. Since *U*_{J} [1] = {1} and 1 ∈ *J*. Therefore, *J* = {1}.

⇒ (*i*): Let *J* = {1}. Since *E* is separated, we can get that *U*_{J} [*x*] = {*x*}. It follows that (*E*, 𝒯_{J}) is discrete. □

#### Corollary 4.15

*Let* ∧ *be a family of filters of E which is closed under intersection*, *J* = ∩∧ *and E be a separated EQ-algebra*. *Then* (*E*, 𝒯_{J}) *is a Hausdorff space if and only if J* = {1}.

#### Proof

Let (*E*, 𝒯_{J}) be a Hausdorff space. First we show that for any *x* ∈ *E*, *U*_{J} [*x*] = {*x*}. If there exists *x* ≠ *y* ∈ *U*_{J} [*x*], then *y* ∈ *U*_{J} [*x*] ∩ *U*_{J} [*y*]. By Remark 4.10 (ii), *U*_{J} [*x*] and *U*_{J} [*y*] are the smallest neighborhoods of *x* and *y*, respectively. Hence for any neighborhood *U* of *x* and neighborhood *V* of *y*, we have that *y* ∈ *U*_{J} [*x*] ∩ *U*_{J} [*y*] ⊆ *U* ∩ *V* ≠ ∅, which is a contradiction. Hence by Theorems 4.13 and 4.14, *J* = {1}. The other side of the proof directly follows from Theorem 4.14. □

#### Definition 4.16

*Let E*_{1} *and E*_{2} *be EQ-algebras*. *A mapping φ* : *E*_{1} → *E*_{2} *is called an EQ-morphism from E*_{1} *to E*_{2} *if*
$$\phi (x\ast y)=\phi (x)\ast \phi (y)$$ *for any* ∗ ∈ {∧, ⊗, ∼}. *If, in addition, the mapping φ is bijective, then we call φ an EQ*-isomorphism. *Note that φ*(1) = 1 *when φ is an EQ-morphism*.

#### Proposition 4.17

*Let φ* : *E*_{1} → *E*_{2} *be an EQ-morphism. Then the following properties hold*:

*if F is a filter of E*_{2}, *then the set φ*^{−1}(*F*) *is a filter of E*_{1};

*if φ is surjective and F is a filter of E*_{1}, *then φ*(*F*) *is a filter of E*_{2}.

#### Proof

It is easy to prove by definition of filters. □

#### Lemma 4.18

*Let E*_{1} *and E*_{2} *be EQ-algebras and F be a filter of E*_{2}. *If φ* : *E*_{1} → *E*_{2} *is an EQ-isomorphism, then*
$$(a,b)\in {U}_{{\phi}^{-1}(F)}\text{\hspace{0.17em}}if\text{\hspace{0.17em}}and\text{\hspace{0.17em}}only\text{\hspace{0.17em}}if\text{\hspace{0.17em}}(\phi (a),\phi (b))\in {U}_{F},for\text{\hspace{0.17em}}any\text{\hspace{0.17em}}a,b\in E.$$

#### Proof

For any (*a, b*) ∈ *U*_{φ−1(F)} ⇔ *a* ∼ *b* ∈ *φ*^{−1}(*F*) ⇔ *φ*(*a*) ∼ *φ*(*b*) ∈ *F* ⇔ (*φ*(*a*), *φ*(*b*)) ∈ *U*_{F}. □

#### Theorem 4.19

*Let E*_{1} *and E*_{2} *be EQ-algebras and F be a filter of E*_{2}. *If φ* : *E*_{1} → *E*_{2} *is an EQ-isomorphism, then the following properties hold*:

*for any a* ∈ *E*_{1}, *φ*(*U*_{φ−1(F)} [*a*]) = *U*_{F} [*φ*(*a*)];

*for any b* ∈ *E*_{2}, *φ*^{−1}(*U*_{F}[*b*]) = *U*_{φ−1(F)}[*φ*^{−1}(*b*)].

#### Proof

(i) Let *b* ∈ *φ**U*_{φ−1(F)}[*a*]). Then there exists *c* ∈ *U*_{φ−1(F)}[*a*] such that *b* = *φ*(*c*). It follows that *a* ∼ *c* ∈ *φ*^{−1}(*F*) ⇒ *φ*(*a*) ∼ *φ*(*c*) ∈ *F* ⇒ *φ*(*a*) ∼ *b* ∈ *F* ⇒ *b* ∈ *U*_{F} [*φ*(*a*)].

Conversely, *b* ∈ *U*_{F} [*φ*(*a*)] ⇒ *φ*(*a*) ∼ *b* ∈ *F* ⇒ *φ*^{−1}(*φ*(*a*) ∼ *b*) ∈ *φ*^{−1}(*F*) ⇒ *a* ∼ *φ*^{−1}(*b*) ∈ *φ*^{−1}(*F*) ⇒ *φ*^{−1}(*b*) ∈ *U*_{φ−1(F)}[*a*] ⇒ *b* ∈ *φ*(*U*_{φ−1(F)}[*a*]).

(ii) *a* ∈ *φ*^{−1}(*U*_{F}[*b*]) ⇔ *φ*(*a*) ∈ *U*_{F}[*b*] ⇔ *φ*(*a*) ∼ *b* ∈ *F* ⇔ *φ*^{−1}(*φ*(*a*) ∼ *b*) ∈ *φ*^{−1}(*F*) ⇔ *a* ∼ *φ*^{−1}(*b*) ∈ *φ*^{−1}(*F*) ⇔ *a* ∈ *U*_{φ−1(F)}[*φ*^{−1}(*b*)]. □

#### Theorem 4.20

*Let E*_{1} *and E*_{2} *be EQ-algebras and F be a filter of E*_{2}. *If φ* : *E*_{1} → *E*_{2} *is an EQ-isomorphism, then φ is a continuous map from* (*E*_{1}, 𝒯_{φ−1(F)}) to (*E*_{2}, 𝒯_{F}).

#### Proof

Let *A* ∈ 𝒯_{F}. By Remark 4.10 (i), we can get that *A* = ∪_{a∈A} *U*_{F} [*a*]. It follows that *φ*^{−1}(*A*) = *φ*^{−1}(∪_{a∈A} *U*_{F} [*a*]) = ∪_{a∈A} *φ*^{−1}(*U*_{F} [*a*]). We claim that if *b* ∈ *φ*^{−1} (*U*_{F} [*a*]), then *U*_{φ−1(F)}[*b*] ⊆ *φ*^{−1}(*U*_{F} [*a*]). Indeed, let *c* ∈ *U*_{φ−1(F)}[*b*], we get that *c* ∼ *b* ∈ *φ*^{−1} (*F*), so *φ*(*c*) ∼ *φ*(*b*) ∈ *F*. Since *φ*(*b*) ∈ *U*_{F} [*a*], we get that *φ*(*b*) ∼ *a* ∈ *F*. It follows that *φ*(*c*) ∼ *a* ∈ *F*. Thus we have that *φ*(*c*) ∈ *U*_{F} [*a*]. So *c* ∈ *φ*^{−1}(*U*_{F} [*a*]). Hence *φ*^{−1}(*U*_{F} [*a*]) = ∪_{b∈φ−1(UF[a])} *U*_{φ−1(F)} [*b*] ∈ 𝒯_{φ−1(F)}. Therefore *φ*^{−1}(*A*) = ∪_{a∈A} *φ*^{−1} (*U*_{F} [*a*]) ∈ 𝒯_{φ−1(F)}. So *φ* is a continuous map. □

#### Theorem 4.21

*Let E*_{1} *and E*_{2} *be EQ-algebras and F be a filter of E*_{2}. *If φ* : *E*_{1} → *E*_{2} *is an EQ-isomorphism, then φ is a quotient map from* (*E*_{1}, 𝒯_{φ−1(F)}) to (*E*_{2}, 𝒯_{F}).

#### Proof

From Theorem 4.20 we get that *φ* is a continuous surjective map. It is enough to show that *φ* is an open map. Let *A* be an open set of (*E*_{1}, 𝒯_{φ−1(F)}). We claim that *φ*(*A*) is an open set of (*E*_{2}, 𝒯_{F}). Let *a* ∈ *φ*(*A*). We shall show that *U*_{F} [*a*] ⊆ *φ*(*A*). Indeed, for any *b* ∈ *U*_{F} [*a*], we get that *b* ∼ *a* ∈ *F*. By Lemma 4.15, we have *φ*^{−1}(*a*) ∼ *φ*^{−1}(*b*) ∈ *φ*^{−1}(*F*). Hence *φ*^{−1}(*b*) ∈ *U*_{φ−1(F)}[*φ*(*a*)]. Since *a* ∈ *φ*(*A*) and *φ* is injective we get that *φ*^{−1}(*a*) ∈ *A*. By Remark 4.10 (i), it follows that *U*_{φ−1(F)}[*φ*^{−1}(*a*)] ⊆ *A*. So *φ*^{−1}(*b*) ∈ *A*, we get that *b* ∈ *φ*(*A*). Therefore, *U*_{F} [*a*] ⊆ *φ*(*A*). So *φ* is a quotient map. □

#### Corollary 4.22

*Let E*_{1} *and E*_{2} *be EQ-algebras and F be a filter of E*_{2}. *If φ* : *E*_{1} → *E*_{2} *is an EQ-isomorphism, then φ is a homeomorphism map from* (*E*_{1}, 𝒯_{φ−1(F)}) to (*E*_{2}, 𝒯_{F}).

#### Proof

It clearly follows from Theorem 4.21. □

Recall that a uniform space (*X*, 𝓚) is totally bounded if for each *U* ∈ 𝓚, there exist *x*_{1},..., *x*_{1} ∈ *X* such that $X={\cup}_{i=1}^{n}U[{x}_{i}]$.

#### Theorem 4.23

*Let F be a filter of E*. *Then the following conditions are equivalent*:

*the topological space* (*E*, 𝒯_{F}) *is compact*;

*the topological space* (*E*, 𝒯_{F}) *is totally bounded*;

*there exists P* = {*x*_{1}, ... , *x*_{n}} ⊆ *E such that for all a* ∈ *E there exists x*_{i} ∈ *P such that a* ≡_{F} *x*_{i}.

#### Proof

⇒ (2): The proof is straightforward.

⇒ (3): Since (*E*, 𝒯_{F}) is totally bounded, there exist *x*_{1},..., *x*_{n} ∈ *E* such that $E={\cup}_{i=1}^{n}{U}_{F}[{x}_{i}]$. Now let *a* ∈ *E*. Then there exists *x*_{i}, such that *a* ∈ *U*_{F} [*x*_{i}], therefore *a* ∼ *x*_{i}, ∈ *F* i.e. *a* ≡_{F} *x*_{i}.

⇒ (1): For any *a* ∈ *E*, by hypothesis, there exists *x*_{i} ∈ *P* such that *a* ∼ *x*_{i} ∈ *F*. We can get that *a* ∈ ∪_{F} [*x*_{i}], hence $E={\cup}_{i=1}^{n}{U}_{F}[{x}_{i}]$. Now let *E* = ∪_{α∈I} *O*_{α}, where each *O*_{α} is an open set of *E*. Then for any *x*_{i} ∈ *E* there exists α_{i} ∈ *I* such that *x*_{i}, ∈ *O*_{αi}. Since *O*_{αi} is an open set, *U*_{F} [*x*_{i}] ⊆ *O*_{αi}, so we have that
$$E=\underset{i\in I}{\cup}{U}_{F}[{x}_{i}]\subseteq \underset{i=1}{\overset{n}{\cup}}{O}_{{\alpha}_{i}}.$$

Therefore $E={\cup}_{i=1}^{n}{O}_{{\alpha}_{i}}$, whence (*E*, 𝒯_{F}) is compact.

□

#### Theorem 4.24

*If F is a filter of E such that F*^{c} *is a finite set, then the topological space* (*E*, 𝒯_{F}) *is compact*.

#### Proof

Let *E* = ∪_{α∈I} *O*_{α}, where each *O*_{α} is an open subset of *E*. Let *F*^{c} = {*x*_{1},... ,*x*_{n}}. Then there exist α, α_{1},...,α_{n} ∈ *I* such that 1 ∈ *O*_{α}, *x*_{1} ∈ *O*_{α1},...,*x*_{n} ∈ *O*_{αn}. Then *U*_{F} [1] ⊆ *O*_{α}, but *U*_{F} [1] = *F*. Hence $E={\cup}_{i=1}^{n}{O}_{{\alpha}_{i}}\cup {O}_{\alpha}$. □

#### Theorem 4.25

*If F is a filter of E, then F is a compact set in the topological space* (*E*, 𝒯_{F}).

#### Proof

Let *F* ⊆ ∪_{α∈I} *O*_{α}, where each *O*_{α} is open set of *E*. Since 1 ∈ *F*, there is α ∈ I such that 1 ∈ *O*_{α}. Then *F* = *U*_{F} [1] ⊆ *O*_{α}. Hence F is a compact set in the topological space (*E*, 𝒯_{F}). □

Our next target is to establish the convergence of *EQ*-algebras using the convergence of nets.

#### Definition 4.26

*Let E be an EQ-algebra and* (*D*, ≤) *be an upward directed set. If for any* α ∈ *D we have a*_{α} ∈ *E*, *then we call* {*a*_{α}}_{α∈D} *a* net *of E*.

#### Definition 4.27

*Let* {*a*_{α}}_{α∈D} *be a net of E*. *In the topological space* (*E*, 𝒯_{F}), *say that* {*a*_{α}}_{α∈D}

converges to the point *a* of *E if for any neighborhood U of a, there exists d*_{0} ∈ *D such that a*_{α} ∈ *U for any* α ≥ *d*_{0};

Cauchy sequence *if there exists d*_{0} ∈ *D such a*_{α} ≡_{F} *a*_{β} *for any* α, *β* ≥ *d*_{0}.

A net {*a*_{α}}_{i∈D}, which converges to *a* is said to be convergent. For simplicity, we write lim*a*_{α} = *a* and we say that *a* is a limit of {*a*_{α}}_{α∈D}.

#### Example 4.28

*Consider the TEQ-algebra* (*E*, 𝒯_{F}) *in Example 4.4. Clearly*, (ℕ, ≤) *is an upward directed set, where* ℕ *is a natural number set*. *We define* {*a*_{n}}_{n∈ℕ} *as a*_{0} = 0, *a*_{1} = *a*, *a*_{2} = *b*, *a*_{n} = 1, *n* ≥ 3. *It is easy to check that* {*a*_{n}}_{n∈ℕ} *is a net of E*. *Let n*_{0} = 3. *For any neighborhood U of 1, if n* ≥ 3, *then* 1 ∈ *U*. *Therefore*, lim*a*_{n} = *1*.

#### Theorem 4.29

*Let* {*a*_{α}}_{α∈D} *and* {*b*_{α}}_{α∈D} *be nets of E and F be a filter of E. Then in the topological space* (*E*, 𝒯_{F}) *we have*:

*if* lim*b*_{α} = *b and* lim*a*_{α} = *a, for some a, b* ∈ *E, then the sequence* {*a*_{α} ∗ *b*_{α}}_{α∈D} *is convergent and* lim*a*_{α} ∗ *b*_{α} = *a* ∗ *b, for any operation* ∗ ∈ {∧, ⊗, ∼};

*any convergent sequence of E is a Cauchy sequence*.

#### Proof

Let lim*a*_{α} = *a*, lim*b*_{α} = *b* and ∗ ∈ {∧, ⊗, ∼}, for some *a, b* ∈ *E*. For any neighborhood *W* of *a* ∗ *b* we get that *U*_{F} [*a* ∗ *b*] ⊆ *W*. Clearly, *U*_{F} [*a*] and *U*_{F} [*b*] are neighborhoods of *a* and *b*, respectively. By hypothesis, there exist *d*_{1}, *d*_{2} ∈ *D* such that *a*_{α} ∈ *U*_{F} [*a*] and *b*_{α} ∈ *U*_{F} [*b*], for any α ≥ *d*_{1} and α ≥ *d*_{2}. Since *D* is an upward directed set, then there exists *d*_{0} ∈ *D* such that *d*_{0} ≥ *d*_{1} and *d*_{0} ≥ *d*_{2}. By Theorem 4.3, we get that *U*_{F} [*a*] ∗ *U*_{F} [*a*] ⊆ *U*_{F} [*a* ∗ *b*]. So *a*_{α} ∈ *U*_{F} [*a*] and *b*_{α} ∈ *U*_{F} [*b*], for any α ≥ *d*_{0}. It follows that *a* ∗ *b* ∈ *U*_{F} [*a*] ∗ *U*_{F} [*a*] ⊆ *U*_{F} [*a* ∗ *b*] ⊆ *U*, for any α ≥ *d*_{0} ∈ *D*. Therefore, lim *a*_{α} ∗ *b*_{α} = *a* ∗ *b*.

Let {*a*_{α}}_{α∈D} be a net of *E* and lim*a*_{α} = *a*. For the neighborhood *U*_{F} [*a*] of *a*, there exists *d* ∈ *D* such that *a*_{α} ∈ *D*, for any α ≥ *d*. So if α, *β* ≥ *d*, then *a*_{α}, *a*_{β} ∈ *U*_{F} [*a*] that is *a*_{α} ≡_{F} *a* and *a*_{β} ≡_{F} *a*. It follows that *a*_{α} ≡_{F} *a*_{β}. Therefore, {*a*_{α}}_{α∈D} is a Cauchy sequence.

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