Consider E as a bornological vector space (bvs). A subset A of E is said to be bornological semi-open (bs-op, in short) if for every sequence {xn}n∈ℕ⊆E, and xn⁢→𝑠⁢x then x∈A then xn∈A∀n>no.

Consider E as a bounded vector space (bvs). A subset A contained within E is termed bornological semi-closed or abbreviated as bs-cl if the following condition is satisfied: If you have a sequence {xn} indexed by natural numbers ℕ, where all the elements are part of A and these elements xn tend to approach a point x in E according to the bornological semi convergence, then it must also be the case that x belongs to the set A.

In a (bvs) every b-op (b-cl) set is bs-op (bs-cl), but the converse is not true in general.

Take A be a b-op (b-cl) set in a (bvs) E. Then:

If B is bs-op (bs-cl) set in E , then B ⁢ ⋂ A is bs-op (bs-cl) set in E .

(i) Take A to be a b-op and take {xn}n∈ℕ⊆E semi converging bornological to a point x∈A∩B, then x∈A and x∈B. If x∈A and by Remark 1A is bs-op then xn∈A∀n>no. If x∈B and B is bs-op then xn∈B∀n>no, we have xn∈A∩B∀n>no then A∩B is bs-op.

Now take A be a b-cl {xn}n∈ℕ⊂A∩B and xn⁢→𝑠⁢x in E then {xn}n∈ℕ⊆B and {xn}n∈ℕ⊆A. If {xn}n∈ℕ⊆B and since B is bs-cl we have xn⁢→𝑠⁢x then x∈B. If {xn}n∈ℕ⊆A and since A is b-cl then by Remark 1A is bs-cl we have xn⁢→𝑠⁢x then x∈A, then x∈A∩B.

(ii) Same approach as above proof.

Take E be a (bvs) and B⊆A⊆E. Then:

If B is bs-op (bs-cl) set in A and A bs-op (bs-cl) set in E then B is bs-op (bs-cl) set in E .

(i) Let {xn}n∈ℕ⊆E and xn⁢→𝑠⁢x such that x∈B, thus x∈B⊆A we have x∈A, since A is bs-op set in E and {xn}n∈ℕ⊆A, xn⁢→𝑠⁢x such that x∈B, since B is bs-op in A then {xn}n∈ℕ⊆B, we have B is bs-op in E.

If we take E to be a bounded vector space (bvs), then the bornological semi-closure of a subset A in E, denoted as bs-clr⁢(A), is the intersection of all bs-cl subsets of E that contain the set A.

Let E, F are (bvs) and let f be a mapping from E into F. We say that f is a seq bs-con map at a point x if for any sequence {xn}n∈ℕ in E, xn⁢→𝑠⁢x then f⁢(xn)⁢→𝑠⁢f⁢(x) in F. If f is a seq bs-con map at every point x in E, then f is called seq bs-con map.

Inclusion map i:A→F in a bornological vector space (bvs) is sequential bornological semi continuous map (bs-con) map if and only if A is closed set in E.

Let E and F are (bvs)s, a map f:E→F is called:

bs-cl map if ∀ b-cl set A in X , f ⁢ ( A ) is bs-cl set in F

Let E and F are bornological vector spaces (bvs)s, a map f:E→F is called:

bs-op map if ∀ b-op set A in E , f ⁢ ( A ) is bs-op set in F

Every b-cl (b-op) map is bs-cl (bs-op) map [10].

Let E,F and H are (bvs), f:E→F, g:F→H, then, if f is b-cl map and g is bs-cl map, then g∘f is bs-cl.

Let A⊆E is a b-cl set, since f b-cl map then f⁢(A) b-cl set in F, by g bs-cl map we have g⁢(f⁢(A)) is bs-cl set in H, then g∘f is bs-cl map.

Let E,F and H are (bvs)s, f:E→F, g:F→H

If f is bss-cl map and g is bss-cl map then g ∘ f is bss-cl map.

(i) Let A⊆E is a bs-cl set, since f bss-cl map then f⁢(A) is b-cl set by Remark 2 then f⁢(A) is bs-cl in F, by g bss-cl map we have g⁢(f⁢(A)) is b-cl set in H, then g∘f is bss-cl map.

(ii) Let A⊆E is a bs-cl set, since f bsi-cl map, then f⁢(A) is bsi-cl set in F. On the other hand, g is bsi-cl map, we have g⁢(f⁢(A)) is bsi-cl set in H, then g∘f is bsi-cl map.

Let A be subset of a (bvs) E, then the inclusion map i:A→E is b-cl (bs-cl) iff A is b-cl set in E.

Let E,F are (bvs)s. Assuming f:E→F and g:E→F seq bs-con map then:

c ⁢ f where c ⁢ f : E → F seq bs-con map ∀ c ∈ K

(i) Let x∈E and {xn} be a sequence in (bvs) E, such that xn⁢→𝑠⁢x. Since f is a seq bs-con at x, then f⁢(xn)⁢→𝑠⁢f⁢(x) in F implies c⁢f⁢(xn)⁢→𝑠⁢c⁢f⁢(x)∀c∈K. Then ∀x we have c⁢f seq bs-con.

(ii) Let x∈E and {xn} be a sequence in (bvs) E, such that xn⁢→𝑠⁢x. Since f, g seq bs-con map at x then f⁢(xn)⁢→𝑠⁢f⁢(x) in F and g⁢(xn)⁢→𝑠⁢g⁢(x) in F, then f⁢(xn)+g⁢(xn)⁢→𝑠⁢f⁢(x)+g⁢(x) in F. We have (f+g)⁢(xn)⁢→𝑠⁢(f+g)⁢(x) in F, then f+g is a seq bs-con map at every point x in E. Then f is seq bs-con map.

(iii) Let x∈E and {xn} be a sequence in (bvs) E, such that xn⁢→𝑠⁢x. Since f, g seq bs-con map at x then f⁢(xn)⁢→𝑠⁢f⁢(x) in F and g⁢(x)⁢→𝑠⁢g⁢(x) in F then f⁢(xn)⋅g⁢(xn)⁢→𝑠⁢f⁢(x)⋅g⁢(x) in F, hence (f⋅g)⁢(xn)⁢→𝑠⁢(f⁢g)⁢(x) in F at every x in E then f⋅g is seq bs-con map.

Let E and F are (bvs)s and A a non-empty subset of E if f:E⁢→𝑠⁢F is a seq bs-con map then the restriction fA is a seq bs-con map, where A has the relative bornology BA.

Let x∈A and {xn}n∈ℕ⊆A such that xn⁢→𝑠⁢x in A. Since A⊆E then xn⁢→𝑠⁢x in E. Since f:E→F is a seq bs-con map then if xn⁢→𝑠⁢x we have f⁢(xn)⁢→𝑠⁢f⁢(x) hence fA is a seq bs-con map.

A mapping f where f:E→F from a bornological vector spaces (bvs) E into (bvs) F is seq bs-con map, for every B is bs-cl set in F then f−1⁢(B) is bs-cl set in E.

Let B is bs-cl set in F, if f−1⁢(B)=ϕ and thus the proof is complete. If f−1⁢(B)≠ϕ. Let xn∈f−1⁢(B) such that xn⁢→𝑠⁢x in f−1⁢(B)⊆E, we have f⁢(xn)∈B. Since f is seq bs-con map then f⁢(xn)⁢→𝑠⁢f⁢(x). Since B is bs-cl set then f⁢(x)∈B. We have x∈f−1⁢(B) then f−1⁢(B) is bs-cl set in E.

Let E and F are (bvs)s, if f:E→F is seq bs-con map then:

f ⁢ ( bs-clr ⁢ A ) ⊆ bs-clr ⁢ f ⁢ ( A ) for every A ⊆ E

(i) Let f be seq bs-cont. Since bs-clr ⁢f⁢(A) is bs-cl in f−1⁢(bs-clr ⁢f⁢(A)) is bs-cl set in E (Theorem 5) bs-clr ⁢f−1⁢(f⁢(A))=f−1⁢(bs-clr ⁢f⁢(A)). Since f⁢(A)⊆bs-clr ⁢f⁢(A)⇒A⊆f−1⁢(f⁢(A))⊆f−1⁢(bs-clr ⁢f⁢(A)), then bs-clr ⁢A⊆f−1⁢(bs-clr ⁢f⁢(A)) [since f−1⁢(f⁢(A)) is bs-cl set in E and by Remark 3]. We have f⁢(bs-clr ⁢A)⊆f−1⁢(bs-clr ⁢f⁢(A))

(ii) Let f be seq bs-con map. Since bs-clr ⁢B is bs-cl set in F, then f−1⁢(bs-clr ⁢B) is bs-cl set in E (Theorem 5). f−1⁢(bs-clr ⁢B)=f−1⁢(B). (1). Since B⊂bs-clr ⁢B then f−1⁢(bs-clr ⁢B)⊆f−1⁢(B). We have bs-clr ⁢f−1⁢(B)=bs-clr ⁢f−1⁢(bs-clr ⁢B)=f−1⁢(bs-clr ⁢B) [by (1)].

Let E, F and H be (bvs)s and let f:E→F be a seq bs-cont map at a point x, g:F→H be seq bs-con at f⁢(x) then g∘f:E→H be a seq bs-con map at a point x.

Let {xn} be a sequence in E such that xn⁢→𝑠⁢x since f is a seq bs-con map at x then f⁢(xn)⁢→𝑠⁢f⁢(x). Since {f⁢(xn)} a sequence in F. Since g is a seq bs-con map at f⁢(x) then g⁢(f⁢(xn))⁢→𝑠⁢g⁢(f⁢(x)). Then (g∘f)⁢(xn)⁢→𝑠⁢(g∘f)⁢(x). Hence g∘f is a seq bs-con map at x.