By using *ij*-*gw*closed, *ij*-*gw*open and *ij*- *σ**g**w*closed sets, we introduce and study the notions of *ij*-\(wT_{\frac {1}{2}}\), *ij*-\(wT_{\frac {1}{2}}^{\sigma }\), *ij*-\(w^{\sigma }T_{\frac {1}{2}}\), *ij*-*w*normal, and *ij*-*gw*normal spaces.

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**Definition 6**

Let \(cl_{w_{j}}(\emptyset)\)= *∅*. A bi*w*ss (*X*,*w*_{1},*w*_{2}) is called

*ij*- *w**T*_{1} if for each distinct points *x*,*y*∈*X*, there exist a *w*_{i}-open set *U* and *w*_{j}-open set *V* s.t. *x*∈*U*,*y*∉*U* and *y*∈*V*,*x*∉*V*.

*ij*-\(wT_{\frac {1}{2}}\) if each *ij*-*gw*closed set *A* of *X*, \(cl_{w_{j}}(A)\)=*A*.

*ij*-\(wT_{\frac {1}{2}}^{\sigma }\) if each *ij*- *σ**g**w*closed set *A* of *X*, \(cl_{w_{j}}(A)\)=*A*.

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**Theorem 17**

A bi*w*ss (*X*,*w*_{1},*w*_{2}) is *ij*- *w**T*_{1} if every singleton in *X* is *ij*-*w*closed.

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*Proof*

Let *x*,*y*∈*X* and *x*≠*y*. Then, {*x*},{*y*} are *ij*-*w*closed sets. From Theorem 1, we have \(x \notin cl_{w_{i}}(\{y\})\) and \(y \notin cl_{w_{j}}(\{x\})\). Hence, there exist *w*_{i}-open set *U* containing *x* and *w*_{j}-open set *V* s.t. *x*∈*U*,*y*∉*U*, and *y*∈*V*,*x*∉*V*. Consequently, (*X*,*w*_{1},*w*_{2}) is a *ij*- *w**T*_{1} space. □

In view of Proposition 5, the class of *ij*-\(w T_{\frac {1}{2}}^{\sigma }\) spaces properly contains the class of *ij*-\(w T_{\frac {1}{2}}\) spaces.

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**Proposition 6**

Every *ij*-\(w T_{\frac {1}{2}}\) space is *ij*-\(w T_{\frac {1}{2}}^{\sigma }\).

The following example supports that the converse of the Proposition 6 is not true in general.

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**Example 12**

In Example 5, (*X*,*w*_{1},*w*_{2}) is a 21-\(wT^{\sigma }\frac {{~}_{1}}{2}\) space but not 21-\(wT_{\frac {1}{2}}\).

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**Theorem 18**

Let *X* be a *w*_{i}open set and \(int_{w_{j}}\{x\}\) is *w*_{j}open. A bi*w*ss (*X*,*w*_{1},*w*_{2}) is *ij*-\(wT_{\frac {1}{2}}\) iff {*x*} is *w*_{i}closed or {*x*}=\(int_{w_{j}}\{x\}\) for each *x*∈*X*.

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*Proof*

Suppose that {*x*} is not *w*_{i}closed for some *x*∈*X*. Then, by using Theorem 7, *X*∖{*x*} is *ij*-*gw*closed. Since (*X*,*w*_{1},*w*_{2}) is *ij*-\(wT_{\frac {1}{2}}\), then {*x*}=\(int_{w_{j}}\{x\}\). On the other hand, let *B* be an *ij*-*gw*closed set. By assumption, {*x*} is *w*_{i}closed or {*x*}=\(int_{w_{j}}\{x\}\) for any \(x{\in }cl_{w_{j}}B\).Case (I): Suppose {*x*} is *w*_{i}closed. If *x*∉*B*, then \(\{x\}{\subseteq }cl_{w_{j}}B {\setminus }B\), which is a contradiction to Theorem 8. Hence *x*∈*B*.Case (II): Suppose {*x*}=\(int_{w_{j}}\{x\}\) and \(x{\in }cl_{w_{j}}B\). Since {*x*}∩*B*≠*∅*, we have *x*∈*B*. Thus, in both cases, we conclude that \(cl_{w_{j}}B\)=*B*. Therefore, (*X*,*w*_{1},*w*_{2}) is *ij*-\(wT_{\frac {1}{2}}\) space. □

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**Theorem 19**

Suppose \(cl_{w_{i}}\emptyset \)= *∅*. If (*X*,*w*_{1},*w*_{2}) is an *ij*-\(wT_{\frac {1}{2}}^{\sigma }\) space, then {*x*} is *ji*-*gw*closed or {*x*}=\(int_{w_{j}}\{x\}\), for each *x*∈*X*.

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*Proof*

Follows directly from Theorem 15 and Definition 6. □

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**Lemma 1**

If {*x*} is *ji*-*gw*closed, then (*X*,*w*_{1},*w*_{2}) is an *ij*-*w*-\(T_{\frac {1}{2}}^{\sigma }\) space, for each *x*∈*X*.

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*Proof*

Straightforward. □

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**Definition 7**

A bi*w*ss (*X*,*w*_{1},*w*_{2}) is called

Pairwise \(wT_{\frac {1}{2}}\) if it is both *ij*-\(wT_{\frac {1}{2}}\) and *ji*-\(wT_{\frac {1}{2}}\).

Pairwise \(wT^{\sigma }\frac {{~}_{1}}{2}\) if it is both *ij*-\(wT_{\frac {1}{2}}^{\sigma }\) and *ji*-\(wT_{\frac {1}{2}}^{\sigma }\).

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**Proposition 7**

If (*X*,*w*_{1},*w*_{2}) is a pairwise \(wT_{\frac {1}{2}}\) space, then it is pairwise \(wT_{\frac {1}{2}}^{\sigma }\).

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*Proof*

Uncomplicated. □

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**Remark 10**

The converse of Proposition *7* is not true as can be seen from the next example.

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**Example 13**

Let *X*,*w*_{1},*w*_{2} be as in Example *12*. Then, (*X*,*w*_{1},*w*_{2}) is also a 21-\(wT_{\frac {1}{2}}^{\sigma }\) space, and therefore, it is a pairwise \(wT_{\frac {1}{2}}^{\sigma }\) space. But (*X*,*w*_{1},*w*_{2}) is not a pairwise \(wT_{\frac {1}{2}}\) space.

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**Definition 8**

A bi*w*ss (*X*,*w*_{1},*w*_{2}) is called an *ij*-\(w^{\sigma }T_{\frac {1}{2}}\) if *ij*- *G**W**C*(*X*)=*ij*- *σ**G**W**C*(*X*).

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**Proposition 8**

Every *ij*-\(wT_{\frac {1}{2}}\) space is *ij*-\(w^{\sigma }T_{\frac {1}{2}}\).

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*Proof*

Obvious. □

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**Remark 11**

The converse of Proposition *8* may not be applicable as we see in the next example.

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**Example 14**

Let *X*= {1,2,3,4}. Define weak structures *w*_{1},*w*_{2} on *X* as follows: *w*_{1}= {*∅*,{1,3},{1,4},{2,3,4}} and *w*_{2}= {*∅*,{2},{1,2},{3,4},{1,3,4}}. Then, (*X*,*w*_{1},*w*_{2}) is an 12-\(w^{\sigma }T_{\frac {1}{2}}\) space but not 12-\(wT_{\frac {1}{2}}\).

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**Remark 12**

*ij*-\(w^{\sigma }T_{\frac {1}{2}}\) and *ij*-\(wT_{\frac {1}{2}}^{\sigma }\) spaces are independent as may be seen from Example 15 and Example *16*.

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**Example 15**

Let *w*_{1}= {*∅*,{1},{1,2}},*w*_{2}={*∅*,{3},*X*} be weak structures on *X*= {1,2,3}, then (*X*,*w*_{1},*w*_{2}) is a 12-\(wT^{\sigma }\frac {{~}_{1}}{2}\) space but not 12-\(w^{\sigma }T_{\frac {1}{2}}\).

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**Example 16**

In Example *14*, (*X*,*w*_{1},*w*_{2}) is an 12-\(w^{\sigma }T_{\frac {1}{2}}\), but it is not 12-*w*-\(T_{\frac {1}{2}}^{\sigma }\).

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**Theorem 20**

Let \(cl_{w_{j}}(\emptyset)\)= *∅*. A bi*w*ss (*X*,*w*_{1},*w*_{2}) is *ij*-\(wT_{\frac {1}{2}}\) if and only if it is both *ij*-\(wT_{\frac {1}{2}}^{\sigma }\) and *ij*-\(w^{\sigma }T_{\frac {1}{2}}\) space.

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*Proof*

Suppose that (*X*,*w*_{1},*w*_{2}) is an *ij*-\(wT_{\frac {1}{2}}\) space. Then, by Propositions 6 and 8, (*X*,*w*_{1},*w*_{2}) is both *ij*-\(wT_{\frac {1}{2}}^{\sigma }\) and *ij*-\(w^{\sigma }T_{\frac {1}{2}}\) space. Conversely, suppose that (*X*,*w*_{1},*w*_{2}) is both *ij*-\(wT_{\frac {1}{2}}^{\sigma }\) and *ij*-\(w^{\sigma }T_{\frac {1}{2}}\). Let *A*∈*i**j*- *G**W**C*(*X*). Since (*X*,*w*_{1},*w*_{2}) is an *ij*-\(w^{\sigma }T_{\frac {1}{2}}\) space, *A*∈*i**j*- *σ**G**W**C*(*X*). Since (*X*,*w*_{1},*w*_{2}) is an *ij*-\(wT_{\frac {1}{2}}^{\sigma }\) space, then \(cl_{w_{j}}(A)\)=*A*. Therefore, (*X*,*w*_{1},*w*_{2}) is *ij*-\(wT_{\frac {1}{2}}\). □

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**Definition 9**

A bi*w*ss (*X*,*w*_{1},*w*_{2}) is called *ij*-*w*normal if for each *w*_{i}closed set *A* and *w*_{j}closed set *B* s.t. *A*∩*B*= *∅*, there are *w*_{j}open set *U* and *w*_{i}open set *V* s.t. *A*⊆*U*,*B*⊆*V*, and *U*∩*V*= *∅*.

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**Theorem 21**

Let (*X*,*w*_{1},*w*_{2}) be a bi*w*ss. Consider the following statements:

(*X*,*w*_{1},*w*_{2}) is *ij*-*w*normal,

For each *w*_{i}closed set *A* and *w*_{j}open set *N* with *A*⊆*N*, there exists *w*_{j}open set *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\),

For each *w*_{i}closed set *A* and each *ij*-*gw*closed set *H* with *A*∩*H*= *∅*, there exist *w*_{j}open set *U* and *w*_{i}open set *V* s.t. *A*⊆*U*,*H*⊆*V* and *U*∩*V*= *∅*,

For each *w*_{i}closed set *A* and *ij*-*gw*open *N* with *A*⊆*N*, there exists *w*_{j}open set *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\).

Then, the implications (1)⇒(2) and (3)⇒(4)⇒(2) are hold.

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*Proof*

Obvious. □

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**Theorem 22**

Let (*X*,*w*_{1},*w*_{2}) be a bi*w*ss. If \(cl_{w_{i}}(A)\) is *w*_{i}closed for each *w*_{j}open or *ij*-*gw*closed, then the statements in Theorem *21* are equivalent.

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*Proof*

According to Theorem *21*, we need to prove (2)⇒(1) and (1)⇒(3) only. (2)⇒(1): Let *A* be a *w*_{i}closed set and *B* be a *w*_{j}closed set with *A*∩*B*= *∅*. Then, *X*∖*B* is a *w*_{j}open set with *A*⊆*X*∖*B*. Thus, by (2) there exists *w*_{j} open set *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }X{\setminus }B\). Hence *A*⊆*U* and \(B{\subseteq }X{\setminus }cl_{w_{i}}(U)\). Since \(cl_{w_{i}}(U)\) is *w*_{i}closed for each *w*_{j}open *U*, then \(X{\setminus }cl_{w_{i}}(U)\)=*V* is *w*_{i}open and *U*∩*V*= *∅*. Hence (*X*,*w*_{1},*w*_{2}) is *ij*-*w*normal. (1)⇒(3): Let *A* be a *w*_{i}closed set and *H* be an *ij*-*gw*closed set with *A*∩*H*= *∅*. Then, *H*⊆*X*∖*A*. From Definition 3, we have \(cl_{w_{j}}(H){\subseteq }X{\setminus }A\). Since *H* is *ij*-*gw*closed, then \(cl_{w_{j}}(H)\) is *w*_{j}closed. Since \(A{\cap }cl_{w_{j}}(H)\)= *∅*, then from (1) there exist *w*_{j}open set *U* and *w*_{i}open set *V* s.t. \(A{\subseteq }U, H{\subseteq }cl_{w_{j}}(H){\subseteq }V\) and *U*∩*V*= *∅*. □

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**Theorem 23**

Let (*X*,*w*_{1},*w*_{2}) be a bi*w*ss. Consider the following statements:

(*X*,*w*_{1},*w*_{2}) is *ij*-*w*normal,

For each *w*_{i}closed set *A* and *w*_{j}closed set *B* s.t. *A*∩*B*= *∅*, there exist *ij*-*gw*open *U* and *ji*-*gw*open *V* s.t. *A*⊆*U*,*B*⊆*V* and *U*∩*V*= *∅*,

For each *w*_{i}closed set *A* and *w*_{j}open *N* with *A*⊆*N*, there exists *ij*-*gw*open *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\).

Then, the implication (1)⇒(2)⇒(3) is hold.

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*Proof*

(1)⇒(2): Let *A* be a *w*_{i}closed set and *B* be a *w*_{j}closed set with *A*∩*B*= *∅*. Since (*X*,*w*_{1},*w*_{2}) is *ij*-*w*normal, then there exist *w*_{j}open set *U* and *w*_{i}open set *V* s.t. *A*⊆*U*,*B*⊆*V* and *U*∩*V*= *∅*. From Corollary 1, there exist *ij*-*gw*open *U* and *ji*-*gw*open *V* s.t. *A*⊆*U*,*B*⊆*V* and *U*∩*V*= *∅*. (2)⇒(3): Let *A* be a *w*_{i}closed set and *N* be a *w*_{j}open set with *A*⊆*N*. Then, *A*∩*X*∖*N*= *∅*. From (2), there exist *ij*-*gw*open *U* and *ji*-*gw*open *V* s.t. *A*⊆*U*,*X*∖*N*⊆*V*, and *U*∩*V*= *∅*. Since *X*∖*V* is *ji*-*gw*closed, *N* is *w*_{j}open, and *X*∖*V*⊆*N*, then from Definition 3, we have \(cl_{w_{i}}(X{\setminus }V){\subseteq }N\). Since *U*⊆*X*∖*V*, hence \(U{\subseteq }cl_{w_{i}}(U){\subseteq }cl_{w_{i}}(X {\setminus } V)\). Consequently, \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\). □

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**Theorem 24**

Let (*X*,*w*_{1},*w*_{2}) be an *ij*-\(wT_{\frac {1}{2}}\). If \(cl_{w_{i}}(U)\) is *w*_{i}closed for each *ij*-*gw*closed and \(int_{w_{j}}(U)\) is *w*_{j}open for each *ij*-*gw*closed *U*, then the statements in Theorem *23* are equivalent.

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*Proof*

According to Theorem 23, we need to prove (3)⇒(1). (3)⇒(1): Let *A* be a *w*_{i}closed set and *B* be a *w*_{j}closed set with *A*∩*B*= *∅*. Take *N*= *X*∖*B*, then by using (3) there exists *ij*-*gw*open *U* s.t. \(A \subseteq U \subseteq cl_{w_{i}}(U) \subseteq N\). Since (*X*,*w*_{1},*w*_{2}) is an *ij*-\(wT_{\frac {1}{2}}\) space, then, \(int_{w_{j}}(U)\)=*U*. By assumption *U* is *w*_{j}open. Also, \(X{\setminus }cl_{w_{i}}(U)\) is *w*_{i}open and \(B{\subseteq }X{\setminus }cl_{w_{i}}(U)\). □

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**Definition 10**

A bi*w*ss (*X*,*w*_{1},*w*_{2}) is called *ij*-*gw*normal if for each *ji*-*gw*closed set *A* and *ij*-*gw*closed set *B* s.t. *A*∩*B*= *∅*, there are *w*_{j}open set *U* and *w*_{i}open set *V* s.t. *A*⊆*U*,*B*⊆*V* and *U*∩*V*= *∅*.

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**Remark 13**

It is clear that every *ij*-*gw*normal space is *ij*-*w*normal. It can be checked that the converse is not true by the following example.

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**Theorem 25**

Let (*X*,*w*_{1},*w*_{2}) be a bi*w*ss. Consider the following statements:

(*X*,*w*_{1},*w*_{2}) is *ij*-*gw*normal,

For each *ji*-*gw*closed set *A* and *ij*-*gw*open set *N* with *A*⊆*N*, there exists *w*_{j}open set *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\),

For each *ji*-*gw*closed set *A* and *ij*-*gw*closed set *B* s.t. *A*∩*B*= *∅*, there exist *w*_{j}open set *U* s.t. *A*⊆*U* and \(cl_{w_{i}}(U){\cap }B\)= *∅*.

Then, the implication (1)⇒(2)⇒(3) is hold.

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*Proof*

Obvious. □

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**Remark 14**

If \(cl_{w_{i}}(U)\) is *w*_{i}closed for each *w*_{i}open set *U*, then the statements in Theorem *25* are equivalent.

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**Theorem 26**

Let (*X*,*w*_{1},*w*_{2}) be a bi*w*ss. Consider the following statements:

(*X*,*w*_{1},*w*_{2}) is *ij*-*gw*normal,

For each *ji*-*gw*closed set *A* and *ij*-*gw*closed set *B* s.t. *A*∩*B*= *∅*, there exist *ij*- *σ**g**w*open set *U*, *ji*- *σ**g**w*open set *V* s.t. *A*⊆*U*,*B*⊆*V* and *U*∩*V*= *∅*,

For each *ji*-*gw*closed set *A* and *ij*-*gw*open set *N* with *A*⊆*N*, there exists *ij*- *σ**g**w*open set *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\).

Then, the implication (1)⇒(2)⇒(3) is hold.

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*Proof*

(1)⇒(2) Follows directly from Proposition 5. (2)⇒(3) Let *A* be a *ji*-*gw*closed set and *N* be an *ij*-*gw*open set with *A*⊆*N*. Take *B*= *X*∖*N*. Then, by assumption, there exist *ij*- *σ**g**w*open set *U*, *ji*- *σ**g**w*open set *V* s.t. *A*⊆*U*,*B*⊆*V* and *U*∩*V*= *∅*. Hence, *U*⊆*X*∖*V*,*X*∖*V*⊆*N*. Since *X*∖*V* is *ji*- *σ**g**w*closed, then \(cl_{w_{i}}(X{\setminus }V){\subseteq }N\) and so \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\). □

The question that comes to our mind, under what conditions can be achieved parity in Theorem 26.

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**Theorem 27**

Let (*X*,*w*_{1},*w*_{2}) be an *ij*-*w*-\(T_{\frac {1}{2}}^{\sigma }\) space. If \(int_{w_{j}}(U)\) is *w*_{j}open and \(int_{w_{i}}(U)\) is *w*_{i}open for each *ij*- *σ**g**w*open set *U*, then the statements in Theorem *26* are equivalent.

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*Proof*

Straightforward. □

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**Corollary 4**

If a bi*w*ss (*X*,*w*_{1},*w*_{2}) is *ij*-*gw*normal, then for each *ji*-*gw*closed set *A* and *ij*- *σ**g**w*open set *N* with *A*⊆*N*, there exists *ij*- *σ**g**w*open set *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\).

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*Proof*

Obvious from Proposition 5. □

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**Theorem 28**

If a bi*w*ss (*X*,*w*_{1},*w*_{2}) is *ij*-*gw*normal, then for each *ji*-*gw*closed set *A* and *ij*-*gw*closed set *B* s.t. *A*∩*B*= *∅*, there exist *ij*-*gw*open set *U* and *ji*-*gw*open set *V* s.t. *A*⊆*U*,*B*⊆*V* and *U*∩*V*= *∅*.

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*Proof*

Clear. □

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**Theorem 29**

If a bi*w*ss (*X*,*w*_{1},*w*_{2}) is *ji*-*w*-\(T_{\frac {1}{2}}^{\sigma }\) and \(cl_{w_{i}}(\emptyset)\)= *∅*. Consider the following statements:

(*X*,*w*_{1},*w*_{2}) is *ij*-*gw*normal,

For each *ji*-*gw*closed set *A* and *ij*-*gw*open set *N* with *A*⊆*N*, there exists *ij*-*gw*open set *U* s.t. \(A{\subseteq }U{\subseteq }cl_{w_{i}}(U){\subseteq }N\).

Then, the implication (1)⇒(2) is hold.

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*Proof*

Obvious. □