on semi complete bornological vector...

Republic of Iraq

Ministry of Higher Education & Scientific Research

AL-Qadisiyah University

College of Computer Science and Mathematics

Department of Mathematics

On Semi –Complete Bornological Vector Space.

A Research

Submitted by

Abbas Shakir Mansor

To the Council of the department of Mathematics ∕ College of Education,

University of AL-Qadisiyah as a Partial Fulfilment of the Requirements for the

Bachelor Degree in Mathematics

Supervised by

Fatma Kamel Majeed

A. D. 2018 A.H. 1439

كلمة شكر

البد لنا ونحن نخطو خطواتنا األخيرة في الحياة الجامعية من وقفة نعود إلى أعوام قضيناها في رحاب الجامعة مع أساتذتنا الكرام الذين قدموا لنا الكثير باذلين بذلك جهودا كبيرة في بناء جيل الغد لتبعث

األمة من جديد ...

واالمتنان والتقدير والمحبة وقبل أن نمضي تقدم أسمى آيات الشكر إلى الذين حملوا أقدس رسالة في الحياة ...

إلى الذين مهدوا لنا طريق العلم والمعرفة ...

لى جميع أساتذتنا األفاضل.......إ

"كن عالما .. فإن لم تستطع فكن متعلما ، فإن لم تستطع فأحب العلماء ،فإن لم تستطع فال تبغضهم"

والشكروأخص بالتقدير

الى

فاطمه كامل مجيد م االستاذه

The Contents

Subject Page

Chapter One

1.1 Bornological Space 1

1.2 Subspace of Bornological Space 4

Chapter Two

2.1 Semi- Bounded Sets 7

2.2 Semi- Bounded Linear Maps 8

Chapter three

3.1 Bornological semi-convergence 12

Abstract

We introduce the concept of semi- bounded set in Bornological

spaces, semi bounded linear map and study some of their basic properties.

We study the definition of semi-convergence of nets, some

theorems of these concept and some results.

introduction

For the first time in (1977), H. Hogbe–NIend [6] introduced the

Concept of Bornology on a set and study Bornological Construction. In

chapter one study Bornology on a set , Bornological subspace, convex

Bornological space and Bornological vector space . Semi bounded sets

were first introduced and investigated in [2], we introduce in chapter two

semi bounded sets have been used to define and study many new

bornological proprietiesi. The convergent net in convex bornological

vector space and its results studied [1]. A semi- convergent net in convex

bornological vector space and some properties have been studied in

chapter three.

Chapter One

1

1.1 Bornological Space

In this section we introduce the basic definitions, notions and the

theories of bornological vector spaces and construction of bornological

vector space.

Definition (1.1.1)[6]:

A bronology on a set 𝑋 is a family 𝛽 of subset 𝑋 satisfying the following

axioms :

(i) 𝛽 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑣𝑒𝑟𝑖𝑛𝑔 𝑋, 𝑖. 𝑒. 𝑋 = ⋃ 𝛽𝐵∈𝛽 ;

(ii) 𝛽 is hereditary under inclusion i.e. if 𝐴 ∈ 𝛽 and 𝐵 is a subset of 𝑋 contained

in 𝐴, then 𝐵 ∈ 𝛽

(iii) 𝛽 𝑖𝑠 𝑠𝑡𝑎𝑏𝑙𝑒 𝑢𝑛𝑑𝑒𝑟 𝑓𝑖𝑛𝑖𝑡𝑒 𝑢𝑛𝑖𝑜𝑛.

A pair (𝑋, 𝛽) consisting of a sat 𝑋 and a bornology 𝛽 on 𝑋 is called a

bornological space, and the elements of 𝛽 are called the bounded subset of 𝑋.

Example (1.1.2)[1]:

Let 𝑋 = {𝑎, 𝑏, 𝑐}, 𝛽 = {∅, {𝑎}, {𝑏}, {𝑐}, {𝑎, 𝑏}, {𝑎, 𝑐}, {𝑏, 𝑐}, 𝑋}

(i) since 𝑋 ∈ 𝛽, then 𝑋 is converging itself, we have 𝛽 is converging of 𝑋

(ii) since 𝛽 is the set of all subset of 𝑋 i.e.𝛽 = 𝑝(𝑋)and 𝛽 hereditary under

inclusion i.e. 𝐴 ∈ 𝛽, 𝐵 ⊂ 𝐴,then 𝐵 ∈ 𝛽

(iii) since 𝛽 = 𝑝(𝑋)then ⋃ 𝐴𝑖 ∈ 𝛽 𝑛𝑖=1 for all 𝐴𝑖 ∈ 𝛽 𝑖 = 1,2, … , 𝑛

Example (1.1.3)[6]:

Let 𝑅 be a field with absolute value. The collection

𝛽={A≤ R:A is a bounded subset of R in the usual sense for the absolute value}.

Then 𝛽 is a bornology on 𝑅 called the Canonical bornology of 𝑅.

Example (1.1.4)[6]:

Let 𝐸 be a vector space over 𝐾 (we shall assume that 𝐾 is either 𝑅 or 𝐶).

Let 𝑝 be a semi-normed 𝑝 if 𝑝(𝐴) is bounded subset of 𝑅 in the sense of

(Example (1.1.3). The subset of 𝐸 which are bounded for the semi-normed 𝑝

2

form a bornology on 𝐸 called the canonical bornology on the semi-normed

space (𝐸, 𝑃)

(i) since ∀𝑥 ∈ 𝐸, 𝑝(𝑥) ∈ 𝑅, then ∃𝑟 > 0 such that 𝑝(𝑋) ≤ 𝑟 implies ∀𝑥 ∈

𝐸, {𝑋}is a bounded subset for the semi-normed, i.e. {𝑥} ∈ 𝛽, then β is a covering

of 𝐸

(ii) 𝑖𝑓 𝐴 ∈ 𝛽 𝑎𝑛𝑑 𝐵 ⊆ 𝐴, then 𝑝(𝐴)is bounded subset of 𝑅 in the sense for the

absolute value since 𝐵 ⊆ 𝐴 then 𝑝(𝐵) ⊆ 𝑝(𝐴), then 𝑝(𝐵)is a bounded of

𝑅(every subset of a bounded set is bounded)

(iii) 𝑖𝑓 𝐴1, … , 𝐴𝑛 ∈ 𝛽, then 𝑝(𝐴1), … , 𝑝(𝐴𝑛)are bounded subsets of 𝑅.

⋃ 𝑝(𝐴𝑖) 𝑛𝑖 is a bounded subset of 𝑅 by (iv); proposition 1,2.3) i.e. 𝑝(⋃ 𝐴𝑖) 𝑛

𝑖 is

bounded subset of 𝑅. Then ⋃ 𝐴𝑖 𝑛𝑖 is a bounded subset of semi-normed 𝑝,

i.e. ⋃ 𝐴𝑖 𝑛𝑖 ∈ 𝛽, then 𝛽 is abonology on 𝐸.

Definition (1.1.5)[2]:

A base of abornology 𝛽 on 𝑋 is any subfamily 𝛽° of 𝛽 such that every

element of 𝛽 is contained in an element of 𝛽°

Example (1.1.6):

If 𝑋 = {1,2,3} , 𝛽 = {∅, {1}, {2}, {1,2}, {1,3}, {2,3}, 𝑋}

Let 𝛽° = {𝑋} be a base for 𝛽. It is clear that each element of 𝛽 is contained in an

element of 𝛽°

Proposition (1.1.7)[2]:

Let (𝑋, 𝛽) be a bornological space and 𝛽° is the base of 𝛽, then the

following hold :

(𝑖)𝑋 = ⋃ 𝛽°

𝛽°∈𝛽

(𝑖𝑖) A finite union of elements of 𝛽° is contained in an element of 𝛽°.

Conversely; if 𝑋 ≠ ∅ and 𝛽° is a family of subset of 𝑋 satisfying : (i) and (ii)

then ∃ abornology 𝛽 𝑜𝑛 𝑋 such that 𝛽° is base for 𝛽.

3

Proof :-Since 𝛽° forms a base of abornology 𝛽 an a set 𝑋, then every element of

𝛽 is contained in an element of 𝛽° 𝑖. 𝑒. ∀𝐵 ∈ 𝛽, ∃𝛽° ∈ 𝛽 such that 𝐵 since 𝛽 is

covering of 𝑋.

𝑖. 𝑒. 𝑋 = ⋃ 𝐵

𝐵∈𝛽

, 𝑡ℎ𝑒𝑛 𝑋 = ⋃ 𝛽°

𝛽°∈𝛽

𝑖. 𝑒. 𝛽° 𝑐𝑜𝑛𝑣𝑒𝑟𝑠 𝑋, 𝑤ℎ𝑒𝑛𝑐𝑒 (𝑖)

(𝑖𝑖)𝐿𝑒𝑡 𝐵𝑖 ∈ 𝛽° ∀𝑖 = 1,2, … , 𝑛

To prove that the union ⋃ 𝛽°𝛽°∈𝛽 is contained in as element of 𝛽°

Since 𝛽° ⊆ 𝛽 𝑡ℎ𝑒𝑛 𝐵𝑖 ∈ 𝛽 ∀ 𝑖 =

1,2, … , 𝑛 (𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑜𝑛 (1.1.1)) 𝑡ℎ𝑒𝑛 ⋃ 𝐵𝑖 ∈ 𝛽𝑛𝑖=1 , 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛(1.1.5)

We have ⋃ 𝐵𝑖𝑛𝑖 is contained in an element of 𝛽°

Conversely; let 𝛽 = {𝐵 ⊆ 𝑋: ∃𝐵° ∈ 𝛽°, 𝐵 ⊆ 𝛽°}

Then to show that 𝛽 is a bornology on 𝑋 we must satisfy the following

conditions :

(𝑖)∀𝐵° ∈ 𝛽° → 𝐵∘ ⊆ 𝑋 → 𝐵° ∈ 𝛽° → 𝐵° ∈ 𝛽 (𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 (1.1.1))

Then 𝐵° ∈ 𝛽

Since 𝛽° covers 𝑋, then 𝛽 is covering to 𝑋

(𝑖𝑖)𝐼𝑓 𝐵 ∈ 𝛽, 𝐴 ⊆ 𝑋, 𝐴 ⊆ 𝐵 , 𝑠𝑖𝑛𝑐𝑒 𝐴 ⊆ 𝐵 ⊆ 𝐵°

For some 𝐵° ∈ 𝛽°, 𝑡ℎ𝑒𝑛 𝐴 ⊆ 𝛽°, 𝑖. 𝑒. 𝐴 𝑖𝑠 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑑 𝑖𝑛 𝑎𝑛 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝛽°

(by definition (1.1.1), 𝐴 ∈ 𝛽.

(𝑖𝑖𝑖)𝑖𝑓 𝐵1, 𝐵2 ∈ 𝛽 𝑡ℎ𝑒𝑛 ∃𝐵°, 𝐵° ∈ 𝛽° such that

𝐵1 ⊆ 𝐵°(by condition (i)) and 𝐵2 ⊆ 𝐵°

Then 𝐵1 ∪ 𝐵2 ⊆ 𝐵° ∪ 𝐵° for some 𝐵∘

” ∈ 𝛽° (by condition (ii))

then (by the definition (1.1.1))𝐵1 ∪ 𝐵2 ∈ 𝛽

4

Then 𝛽 is a bornology on 𝑋, and (by definition (1.1.5)) 𝛽° is a base for 𝛽, we say

𝛽 is a bornology generated by a base

Definition (1.1.8)[6]:-

A bornology an a set 𝑋 is said a bornology with countable base if and only

if 𝛽 possesses a base consisting of an increasing sequence of bounded sets (an

element of bornology).

1.2 Subspaces Of Bornological Space


Let (𝑋, 𝐵) be a bornological space and let 𝑌 ⊆ 𝑋. The bornology for 𝑌 is

the collection

𝛽𝛾 = {𝐺 ∩ 𝑌: 𝐺 ∈ 𝛽}. The bornological space (𝑌, 𝛽𝛾) is called a subspace of

(𝑋, 𝛽),and 𝛽𝛾 the relative bornology on 𝑌.

Example (1.2.2):

Let 𝑋 = {1,2,3}, 𝛽 = {{∅}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}}, and the base

for the bornology is

𝛽° = {{1,2}, {1,3}, 𝑋}, and let 𝑌 = {1,2}. Then we have

∅ ∩ 𝑌 = ∅

{1,2} ∩ 𝑌 = 𝑌

{1} ∩ 𝑌 = {1}

{1,3} ∩ 𝑌 = {1}

{2} ∩ 𝑌 = {2}

{2,3} ∩ 𝑌 = {2}

{3} ∩ 𝑌 = ∅

5

𝑋 ∩ 𝑌 = 𝑌

Then the subspace 𝛽𝛾 = {∅, {1}, {2}, 𝑌}, and a base of the bornological subspace

is 𝛽𝛾 = {{1}, 𝑌}.


Let 𝐸 be a vector space over the field 𝐾 (the real or complex field). A

bornology 𝛽 on 𝐸 is said to be a bornology compatible with a vector space of 𝐸

or to be a vector bornology on 𝐸, if 𝛽 is stable under vector addition homothetic

transformations and the formation of circled hulls, in other words, if the sets

𝐴 + 𝐵, 𝜆𝐴, ⋃ 𝛼𝐴|𝛼|≤1 belongs to 𝛽 whenever 𝐴 and 𝐵 belong to 𝛽 and 𝜆 ∈ 𝐾.


A convex bornological space is bornological vector space for which the

disked hull of every bounded set is bounded i.e. it is stable under the formation of

disked hull

A separated bornogical vector space (𝐸, 𝛽) is one where {0} is the only bounded

vector subspace of 𝐸.

Example (1.2.5)[6]:-

The Von-Neumann bornology of topological vector space, let 𝐸 be a

topological vector space. The collection

𝛽 = {𝐴 ⊆ 𝐸: 𝐴 𝑖𝑠 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑎 𝑡𝑜𝑝𝑜𝑙𝑜𝑔𝑖𝑐𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑠𝑝𝑎𝑐𝑒 𝐸} forms a

vector bornology on 𝐸

Called the Von-Neumann bornology of 𝐸. Let us verify that 𝛽 is indeed a vector

bornology on 𝐸, if 𝛽° is a base of circled neighborhoods of zero in 𝐸, it is clear

that a subset 𝐴 of 𝐸 is bounded if and only if for every 𝐵 ∈ 𝛽° there exists 𝜆 > 0

such that 𝐴 ⊆ 𝜆𝐵. Since every neighborhood of zero is absorbent, 𝛽 is a

6

converting of 𝐸. 𝛽 is obviously hereditary and we shall see that it is also stable

under vector addition.

Let𝐴1, 𝐴2 ∈ 𝛽 and𝛽° , there exists 𝐵° such that 𝐵° + 𝐵° ⊆ 𝐵°(proposition (1.2.3)).

Since 𝐴1 and 𝐴2 are bounded in 𝐸, there exists positive scalars 𝜆 𝑎𝑛𝑑 𝜇 such that

𝐴1 ⊆ 𝜆𝐵° and 𝐴2 ⊆ 𝜇𝐵°

. With 𝛼 = max(𝜆, 𝜇) we have

𝐴1 + 𝐴2 ⊆ 𝜆𝐵° + 𝜇𝐵°

⊂ 𝛼𝐵° + 𝛼𝐵°

⊂ 𝛼(𝐵° + 𝐵°

) ⊂ 𝐵°

Finally since 𝛽° is stable under the formation of circled hulls (resp. under

homothetic transformation). Then so is 𝛽, we conclude that 𝛽 is a vector

bornology on 𝐸. If 𝐸 is locally convex, then clearly 𝛽 is a convex bornology.

Moreover, since every topological vector space has a base of closed

neighborhood of 0, the closure of each bounded subset of 𝐸 is again bounded.


Let (𝑋𝑖 , 𝛽𝑖)𝑖∈𝐼 be a family of bornological vector space indexed by a non-

empty set 𝐼 and let 𝑋 = ∏ 𝑋𝑖𝑖∈𝐼 be the product of the sets 𝑋𝑖. For every 𝑖 ∈

𝐼 , 𝑙𝑒𝑡 𝑝𝑖: 𝑋 → 𝑋𝑖 be the canonical projection the product bornological 𝑋 is the

initial bornology on 𝑋 for the maps 𝑝𝑖. The set endowed with the product

bornology is called bornological product of the space (𝑋𝑖 , 𝛽𝑖).

Chapter Two

7

2-1 Semi- Bounded Sets

In this section we give the concept of semi- bounded sets and prove

some results on semi bounded sets in Bornological space.


A subset 𝐴 of a Bornological space 𝑋 is said to be a semi-bounded

(written s- bounded) if and only if there exists a bounded subset Β of 𝑋

such that Β ⊆ 𝐴 ⊆ Β where Β is the set of all upper and lower bounds of

Β which continues Β. 𝑆Β(X) will denoted the class of all s- bounded

subsets of 𝑋.

Remark (2.1.2)[2]:-

In any Bornological space, every bounded set is s- bounded. But the

converse is not true in general.

Example(2.1.3)[2]:-

Let 𝑅 the real line with the canonical Bornology, and let Β = [a, b]

is bounded set in 𝑅, let 𝐴 = [𝑎, ∞),then 𝐴 is s- bounded in 𝑅, but is not

bounded in 𝑅.

Remark(2.1.4)[2]:-

In any Bornological space 𝑋

(i) If {𝐴𝛼}𝛼𝜖Λ is a collection of s- bounded subset of 𝑋, then ∩𝛼𝜖Λ 𝐴𝛼

need not be s- bounded in 𝑋.

(ii) The union of two s-bounded sets in 𝑋 is s- bounded in general.

Proof:- (ii) Let 𝐴1, 𝐴2 is s-bounded, then there exists a bounded set

Β1, Β2 in 𝑋 such that Β1 ⊆ A1 ⊆ Β1 and Β2 ⊆ A2 ⊆ Β2

, where Β1 , Β2

is

the sets of all upper and lower bounds which contains Β1, Β2 in 𝑋. Thus

Β1 ∪ Β2 ⊆ A1 ∪ A2 ⊆ Β1 ∪ Β2 , where Β1 ∪ Β2 is a bounded set in 𝑋 and

Β1 ∪ Β2 is the set of all lower bounds of Β1 ∪ Β2, then A1 ∪ A2 is s-

bounded set in 𝑋.

Example (2.1.5)[2]:-

Let 𝑅 be the real line with canonical Bornology, then 𝐴 =

[𝑎, 𝑏), Β = [b, 𝔞) are s-bounded sets in 𝑅 and [𝑎, 𝑏) ∪ [𝑏, 𝑐) = [𝑎, 𝑐) is s-

bounded in 𝑅.

Theorem (2.1.6)[2]:-

Let 𝐴 an s-bounded set in the Bornological space 𝑋 and 𝐴 ⊆ 𝐻 ⊆ ��,

where A is the set of upper and lower bounded of 𝐴. Then 𝐻 is s-

bounded, where 𝐻 ⊆ 𝑋.

8

Proof:- Since 𝐴 is s-bounded in 𝑋 then there exists an bounded set Β such

that Β ⊆ 𝐴 ⊆ B, n Β ⊆ 𝐻. But A ⊆ Β (since ⊆ A ⊆ Β ). Hence Β ⊆ 𝐻 ⊆

B and 𝐻 is s-bounded.

Remark (2.1.7)[2]:-

If (𝑋, Β) is a Bornological space, and 𝑌 is a subspace of 𝑋, Β ⊆ 𝑌,

then Β𝑌 = A ∩ 𝑌 is the set of all upper and lower bounds of 𝐴.

Theorem (2.1.8)[2]:-

Let 𝐴 ⊆ 𝑌 ⊆ 𝑋 where 𝑋 is a Bornological space and 𝑌 is a

subspace, then 𝐴𝜖𝑆Β(𝑋) if and only if 𝐴𝜖𝑆Β(Y).

Proof:- Let 𝐴𝜖𝑆Β(𝑋) , then there exists a bounded set Β in 𝑋 such that

Β ⊆ 𝐴 ⊆ B, where Β𝑋 is the set of all upper andlower bounds which

contains Β in 𝑋. Now Β ⊆ 𝑌 (since Β ⊆ 𝐴 ⊆ 𝑌), and thus

Β ⊆ Β ∩ 𝑌 ⊆ 𝐴 ∩ 𝑌 ⊆ 𝑌 ∩ Β𝑋 𝑜𝑟 Β ⊆ 𝐴 ⊆ Β𝑌

(by remark (2.1.7)) Β𝑌 =

𝒀 ∩ Β𝑋 is the set of all upper and lower bounds of Β ∩ Y in 𝑌, and Β =

Β ∩ Y is bounded in 𝑌, then 𝐴𝜖𝑆Β(𝑌).

Conversely,

Let 𝐴𝜖𝑆Β(𝑌), then there exists a bounded set Β𝑌 in 𝑌 such that Β𝑌 ⊆ 𝐴 ⊆

Β𝑌 where Β𝑌

is the set of all upper and lower bounds of Β𝑌 in 𝑌. Β𝑌 in 𝑋,

then 𝐴𝜖𝑆Β(𝑋).

Example(2.1.9)[2]:-

Let 𝑅 be the real line with canonical Bornology and𝑌 = {0} and

𝐴 = {0}. Then 𝐴 is bounded in 𝑌 hence 𝐴𝜖𝑆Β(𝑌) and (by theorem 2.1.8)

𝑆Β(𝑋) .


If (𝑋, 𝐵) is a Bornological space and 𝐴 ⊆ 𝑋. Then 𝐴 is semi

unbounded if and only if 𝑋 − 𝐴 is semi bounded and (written s-

unbounded)

Remark (2.1.11)[2]:-

In any Bornological space, every Bornological unbounded set is s-

unbounded and the converse is true if 𝑋 is finite (since the Bornological

space is hereditary under inclusion)

Remark (2.1.12)[2]:-

In any Bornological space

(i) The union of any family of s-unbounded sets need not be s-unbounded

(ii) The intersection of two s-unbounded sets is s-unbounded.

(iii) If 𝑋 is a finite set, then the only Bornology on 𝑋 is 𝑝(𝑋) and so 𝛣 =

SΒ(𝑋) also the collection of s-unbounded set in 𝑋 is 𝛣 itself.

9

2.2 Semi- Bounded Linear Maps

In this section we defined semi- bounded linear maps and prove

some results on semi bounded linear map in Bornological space


A map 𝑓 from a Bornological space 𝑋 into a Bornological space 𝑌 is

said to be semi- bounded map if the image 𝑓 of every bounded subset of

𝑋 is semi bounded in 𝑌.

Remark (2.2.2)[2]:-

(i) Obviously the identity map of any Bornological space is semi-

bounded by (2.1.2)

(ii) Let Β1 and Β2 be two Bornologies on 𝑋 such that Β1 is finer than Β2

(i. e Β1 ⊂ Β2) (or Β2 is coarser than Β1) if the inclusion map (𝑋, Β1) →

(𝑋, Β2) is semi- bounded.


Let 𝐸 and 𝐹 be two Bornological vector spaces. A semi- bounded

linear map of 𝐸 into 𝐹 is that map which is bouth linear and semi

bounded at the same time.


A semi bounded linear functional on a Bornological vector space 𝐸

is a semi bounded linear map of 𝐸 into the scalar field 𝐾 (𝑅 or 𝐶).

Remark (2.2.5)[2]:-

Every bounded map is s- bounded by (2.1.2), and the converse is not

true.

Example (2.2.6)[2]:-

Let 𝑅 be the real line with the canonical Bornology, then the map

𝑓: 𝑅 \{0} → 𝑅 such that 𝑓(𝑥) =1

𝑥 ∀ 𝑥𝜖𝑅\{0} is s- bounded map but it is

not bounded map since, let 𝐹 = (0,1]is bounded set in 𝑥𝜖𝑅\{0}, then the

image of this bounded set under 𝑓 is [1, ∞) is s- bounded in 𝑅 but it is not

bounded in 𝑅.


Let 𝑋 and 𝑌 be Bornological spaces. A map 𝑓: 𝑋 → 𝑌 is said to be a

𝑠∗- bounded map if and only if image of every s- bounded subset of 𝑋

under 𝑓 is s- bounded in 𝑌.


10

Let 𝑋 and 𝑌 be Bornological spaces. A map 𝑓: 𝑋 → 𝑌 is said to be a

𝑠∗∗- bounded map if and only if image of every s- bounded subset of 𝑋

under 𝑓 is s- bounded in 𝑌.

Proposition (2.2.9)[2]:-

Let 𝑋, 𝑌 and 𝑍 be Bornological spaces. A map 𝑓: 𝑋 → 𝑌 bounded (s-

bounded) map and 𝑔: 𝑌 → 𝑍 be s- bounded (𝑠∗∗- bounded) map then 𝑔 ∘

𝑓: 𝑋 → 𝑍 is s- bounded (𝑠∗∗-bounded) map.

Proof:- Let 𝐹 be bounded in 𝑋, since 𝑓 is bounded (s- bounded) map,

then 𝑓(𝐹) is bounded (s- bounded) in 𝑌. But 𝑔 is s- bounded (𝑠∗∗-

bounded) map. Then 𝑔(𝑓(𝐹)) is s- bounded in 𝑍. But 𝑔(𝑓(𝐹)) = (𝑔 ∘

𝑓)(𝐹), then 𝑔 ∘ 𝑓 is s- bounded (𝑠∗∗- bounded)

Chapter

Three

11

3.Bornological semi-convergence net

In this section, we define semi-convergence of net in every convex

bornological vector space, and the main propositions and theorems about

this concept.


Let (𝑥𝛾)𝛾∈Γ

be a net in a convex bornological vector space E. We

say that (𝑥𝛾) bornologically semi-converges to 0 ( (𝑥𝛾)𝑏𝑠→ 0), if there

is an absolutely convex set and semi-bounded 𝑆 of E and a net (𝜆𝛾) in 𝐾

converging to 0, such that 𝑥𝛾 ∈ 𝜆𝛾𝑆, for every 𝛾 ∈ 𝛤. Then A net (𝑥𝛾)

bornologically semi-converges to a point 𝑥 ∈ 𝐸 and ( 𝑥𝛾

𝑏𝑠→ 𝑥) when

((𝑥𝛾 − 𝑥)𝑏𝑠→ 0).

Remark (3.1.2)[1]:-

Let 𝐸 be a convex bornological vector space. A net (𝑥𝛾)𝛾∈Γ

in 𝐸 is

bornologically semi- convergent to a point 𝑥 ∈ 𝐸, if there is a decreasing

net (𝜆𝛾) of positive real numbers tending to zero such that the net (𝑥𝛾−𝑥

𝜆𝛾)

is a semi-bounded.

Theorem (3.1.3)[1]:-

Every bornologically semi-convergent net is semi-bounded.

Proof:- Let 𝐸 be a convex bornological vector space and a net (𝑥𝛾)

bornologically semi-converges to a point 𝑥 ∈ 𝐸. i.e. (𝑥𝛾 − 𝑥)𝑏𝑠→ 0, there

is an absolutely convex subset and semi-bounded 𝑆 of 𝐸 and a net (𝜆𝛾)

of scalars tend to 0, such that 𝑥𝛾 ∈ 𝑆 and (𝑥𝛾 − 𝑥) ∈ 𝜆𝛾𝑆 for every 𝛾 ∈

Γ.

Since (𝜆𝛾) is a net of scalars tend to 0, and 𝐸 is a convex bornological

vector space then 𝜆𝛾 𝑆 is a semi-bounded subset of 𝐸 ,

12

i.e. {(𝑥𝛾 − 𝑥)𝛾∈𝛤

} ⊆ {(𝑥𝛾 − 𝑥)𝛾∈𝛤

⊆ 𝜆𝛾𝑆} ( every subset of semi-

bounded is semi-bounded [3])

implies (𝑥𝛾 − 𝑥) is a semi-bounded subset of 𝐸, then (𝑥𝛾) is semi-

bounded.

Theorem (3.1.4)[1]:-

Let 𝐸 and 𝐹 be convex bornological vector space. Then the image of

a bornologically semi-convergent net under a semi-bounded linear map of

E into F is a bornologically semi-convergent net.

Proof:- Let (𝑥𝛾) be a net semi- converges bornologically to a point 𝑥 in

𝐸, and let 𝑢: 𝐸 ⟶ 𝐹 be a semi-bounded linear map. Since (𝑥𝛾 − 𝑥)𝑏𝑠→ 0

in 𝐸, then there is an absolutely convex set and semi-bounded 𝑆 of E

and a net (𝜆𝛾) of scalars tending to 0, such that 𝑥𝛾 ∈ 𝑆 and 𝑥𝛾 − 𝑥 ∈ 𝜆𝛾

𝑆 for every 𝛾 ∈ 𝛤.

Then 𝑢(𝑥𝛾 − 𝑥) ∈ 𝑢(𝜆𝛾𝑆) for every 𝛾 ∈ 𝛤. Since 𝑢 is a linear map and

(𝜆𝛾) is a net of scalars, then 𝑢(𝑥𝛾) − 𝑢(𝑥) ∈ 𝜆𝛾𝑢(𝑆), since 𝑢 is a semi-

bounded map, we have 𝑢(𝑆) is semi-bounded and disk (if absolutely

convex set then it is disked [6]) when 𝑆 is semi-bounded and disked by

(definition 2.1.1) we have 𝑢(𝑥𝛾) − 𝑢(𝑥)𝑏𝑠→ 0 then 𝑢(𝑥𝛾) bornologically

semi-converges to a point 𝑢(𝑥) in 𝐹.

Theorem (3.1.5)[1]:-

Suppose that (𝑥𝛾)𝛾∈Γ

and (𝑦𝛾)𝛾∈Γ

are bornologically semi-

convergent nets in a convex bornological vector space 𝐸, (𝜆𝛾) is a

convergent net in 𝐾 such that 𝑥𝛾

𝑏𝑠→ 𝑥, 𝑦𝛾

𝑏𝑠→ 𝑦 and 𝜆𝛾 ⟶ 𝜆 then

(i) 𝑥𝛾 + 𝑦𝛾

𝑏𝑠→ 𝑥 + 𝑦;

(ii) 𝑐𝑥𝛾

𝑏𝑠→ 𝑐𝑥, for any number c ∈ K;

(iii) 𝜆𝑦𝑥𝛾

𝑏𝑠→ 𝜆𝑥 .

Proof:-

13

(i) Since 𝑥𝛾

𝑏𝑠→ 𝑥, 𝑦𝛾

𝑏𝑠→ 𝑦 in 𝐸 i.e. 𝑥𝛾 − 𝑥

𝑏𝑠→ 0, 𝑦𝛾 − 𝑦

𝑏𝑠→ 0 in 𝐸, then there

exist an absolutely convex and semi-bounded sets 𝑆1, 𝑆2 of 𝐸 and nets

(𝛼𝛾), (𝛽𝛾) of scalars tending to 0, such that :

(𝑥𝛾 − 𝑥) ∈ 𝛼𝛾𝑆1 and (𝑦𝛾 − 𝑦) ∈ 𝛽𝛾𝑆2 for every 𝛾 ∈ Γ . Then 𝑥𝛾 − 𝑥 +

𝑦𝛾 − 𝑦 ∈ 𝛼𝛾𝑆1 + 𝛽𝛾𝑆2 = (𝛼𝛾 + 𝛽𝛾)(𝑆1 + 𝑆2) − 𝛼𝛾𝑆2 − 𝛽𝛾𝑆1.

Since 𝛼𝛾 ⟶ 0 and 𝛽𝛾 ⟶ 0.

Then ((𝑥𝛾 + 𝑦𝛾) − (𝑥 + 𝑦)) ∈ 𝛼𝛾𝑆1 + 𝛽𝛾𝑆2 ⊆ (𝛼𝛾 + 𝛽𝛾)(𝑆1 + 𝑆2)

Now, if 𝛼𝛾 ⟶ 0 and 𝛽𝛾 ⟶ 0 in 𝐾 then 𝛼𝛾 + 𝛽𝛾 ⟶ 0 in 𝐾 and 𝑆1+𝑆2

is semi-bounded and absolutely convex set when 𝑆1 and 𝑆2 are semi-

bounded and absolutely convex sets and since 𝐸 is a bornological vector

space

implies ((𝑥𝛾 + 𝑦𝛾) − (𝑥 + 𝑦)) 𝑏𝑠→ 0 then (𝑥𝛾 + 𝑦𝛾)

𝑏𝑠→ (𝑥 + 𝑦).

(ii) If 𝑥𝛾

𝑏𝑠→ 𝑥 then (𝑥𝛾 − 𝑥)

𝑏𝑠→ 0 and so there is an absolutely convex set

and semi-bounded 𝑆 of 𝐸 and a net (𝛼𝛾) of scalars tends to 0, such that

𝑥𝛾 ∈ 𝑆 and (𝑥𝛾 − 𝑥) ∈ 𝛼𝛾𝑆 for every 𝛾 ∈ 𝛤 𝑐(𝑥𝛾 − 𝑥) ∈ 𝑐𝛼𝛾𝑆 then

𝑐𝑥𝛾 − 𝑐𝑥 ∈ 𝛼𝛾(𝑐𝑆). Since 𝑐 ∈ 𝐾 and 𝐸 is a convex bornological vector

space then 𝑐𝑆 is an absolutely convex set and semi-bounded of 𝐸 when 𝑆

is an absolutely convex set and semi-bounded of 𝐸 by(definition 2.1)

𝑐𝑥𝛾 − 𝑐𝑥𝑏𝑠→ 0 , then 𝑐𝑥𝛾

𝑏𝑠→ 𝑐𝑥 .

(iii) If (𝑥𝛾) bornologically semi-converges to 𝑥 in 𝐸, then there is an

absolutely convex and semi-bounded set 𝑆 of 𝐸 and a net (𝛼𝛾) of scalars

tends to 0, such that

𝑥𝛾 ∈ 𝑆 and (𝑥𝛾 − 𝑥) ∈ 𝛼𝛾𝑆 for every 𝛾 ∈ Γ. 𝜆𝛾𝑥𝛾 − 𝜆𝑥 =

(𝑥𝛾 − 𝑥)(𝜆𝛾 − 𝜆) + 𝑥(𝜆𝛾 − 𝜆) + 𝜆(𝑥𝛾 − 𝑥). Now 𝜆𝛾 ⟶ 𝜆, then 𝜆𝛾 −

𝜆 ⟶ 0 and 𝑥𝛾

𝑏𝑠→ 𝑥. If (𝑥𝛾 − 𝑥)

𝑏𝑠→ 0 then 𝜆𝛾𝑥𝛾 − 𝜆𝑥 = 𝜆𝛾𝑥𝛾 − 𝜆𝑥 =

(𝑥𝛾 − 𝑥)(𝜆𝛾 − 𝜆) ∈ 𝛼𝛾 ((𝜆𝛾 − 𝜆)𝑆) i.e. 𝜆𝛾𝑥𝛾 − 𝜆𝑥 ∈ 𝛼𝛾 ((𝜆𝛾 − 𝜆)𝑆).

Since 𝐸 is a convex bornological vector space, then (𝜆𝛾 − 𝜆)𝑆 is an

14

absolutely convex and semi-bounded set when 𝑆 is an absolutely convex

and semi-bounded set of 𝐸 by (definition 2.1) 𝜆𝛾𝑥𝛾 − 𝜆𝑥𝑏𝑠→ 0 then 𝜆𝛾𝑥𝛾

𝑏𝑠→ 𝜆𝑥.

Theorem (3.1.6)[1]:-

A convex bornological vector space 𝐸 is separated if and only if

every bornologically semi-convergent net in E has a unique limit.

Proof:- Necessity: Let 𝐸 be a separated bornological vector space. If a

net (𝑥𝛾) in 𝐸 bornologically semi-converges to 𝑥 and 𝑦 . Then the net

𝑧𝛾 = 𝑥𝛾 − 𝑥𝛾 = 0 semi-converges to 𝑧 = 𝑥 − 𝑦.

Thus it suffices to show that the limit 𝑧 of the net (𝑧𝛾 = 0) must be the

element 0 .

Let (𝜆𝛾) be a net of real numbers tends to 0 and let 𝑆 be a semi-bounded

subset of 𝐸 such that 𝑧 − 𝑧𝛾 = 𝑧 ∈ 𝜆𝛾𝑆 for every 𝛾 ≥ 1 .

If 𝑧 ≠ 0, then the line spanned by 𝑧 (i.e. the subspace (𝐾𝑧) is contained

in 𝑆,that contradicts the hypothesis that 𝐸 is separated.

Sufficiency: Assume the uniqueness of limits, and suppose that there is

an element 𝑧 ≠ 0 such that the line spanned by 𝑧 is semi-bounded. Then

we can find a semi-bounded set 𝑆 ⊂ 𝐸 such that 𝑧 ∈ (1

𝛾) 𝑆 for every 𝛾 ≥

1 and hence the net (𝑧𝛾 = 𝑧) semi-converges to 0.

But clearly this net also semi-converges to 𝑧 , whence, by uniqueness of

limits, 𝑧 = 0 we have reached a contradiction.

Theorem( 3.1.7)[1]:-

Let 𝐸 be a convex bornological vector space and let (𝑥𝛾) be a net in

𝐸 then the following are equivalent:

(i) The net (𝑥𝛾) bornologically semi-converges to 0;

(ii) there is an absolutely convex set and semi-bounded 𝑆 ⊂ 𝐸 and a

decreasing net (𝛼𝛾) of positive real numbers, tends to 0, such that 𝑥𝛾 ∈

𝛼𝛾𝑆 for every 𝛾 ∈ 𝛤;

15

(iii) there is an absolutely convex and semi-bounded set 𝑆 ⊂ 𝐸 such that,

given any 𝜀 > 0 , we can find an integer 𝛤(𝜀) for which 𝑥𝛾 ∈ 𝜀𝑆

whenever 𝛾 ≥ 𝛤(𝜀).

(iv) there is a semi-bounded disk 𝑆 ⊂ 𝐸 such that (𝑥𝛾) belongs to the

semi-normed . 𝐸𝑆 and semi-converges to 0 in 𝐸𝑆.

Proof:- (i) ⟹(ii):

For any integer 𝑝 ∈ 𝛤 there is 𝛤𝑝 ∈ 𝛤 such that if 𝛾 ≥ 𝛤p then λγ ≤1

p ;

hence 𝜆𝛾𝑆 ⊂ (1

𝑝) 𝑆, since 𝑆 is disk. We may assume that the net 𝛤𝑝 is

strictly increasing, and, for 𝛤𝑝 ≤ 𝐾 ≤ 𝛤𝑝+1 ,

Let 𝛼𝐾 =1

𝑝 . Then the net {𝛼𝐾} satisfies the conditions of assertion (ii).

Clearly (ii) ⟹ (iii).

To show that (iii) ⟹ (i), let , for every 𝛾 ∈ 𝛤, 𝜀 = 𝑖𝑛𝑓 {𝜀 > 0; 𝑥𝛾 ∈ 𝜀𝑆},

and 𝜆𝛾 = 𝜀𝛾 +1

𝛾 ,

then the net (𝜆𝛾) converges to 0 and 𝑥𝛾 ∈ 𝜆𝛾𝑆 for every 𝛾 ∈ 𝛤.

Thus the assertion (i, ii, iii) are equivalent. Suppose that the bornology of

𝐸 is convex. Clearly (iv) implies (i) with 𝜆𝛾 = 𝑃𝑠(𝑥𝛾) and 𝑃𝑆 the gauge of

𝑆, while (ii) implies that 𝑥𝛾 ∈ 𝐸𝑆 and 𝑃𝑆(𝑥𝛾) ≤ 𝛼𝛾 → 0.

Remark (3.1.8)[1]:-

It is clear that (xγ) for every 𝛾 ∈ 𝛤 bornologically semi-converges to

0 if and only if every subnet of (𝑥𝛾) bornologically semi- converges to 0.

Theorem (3.1.9)[1]:-

A net {𝑥𝛾} in a product convex bornological vector space ∏ 𝐸𝑖𝑖∈𝐼

bornologically semi-converges to 𝑦 if and only if the net {𝑥𝛾𝑖 }

𝑖∈𝐼

bornologically semi-converges to 𝑦𝑖 in convex bornological vector space

𝐸𝑖.

Proof:- Let 𝑥𝛾

𝑏𝑠→ 𝑦 = (𝑦1, 𝑦2, … , 𝑦𝑛, … ) in ∏ 𝐸𝑖𝑖∈𝐼 , since 𝑝𝑖 is semi-

bounded linear map, then by (Theorem 3.1.4) 𝑝𝑖(𝑥𝛾)𝑏𝑠→ 𝑝𝑖(𝑦), for each

16

𝑖 ∈ 𝐼, then 𝑥𝛾𝑖

𝑏𝑠→ 𝑦𝑖 in 𝐸𝑖. Suppose that 𝑥𝛾

𝑖𝑏𝑠→ 𝑦𝑖 for each 𝑖 ∈ 𝐼, then there

is an absolutely convex and semi-bounded set 𝑆𝑖 of 𝐸𝑖 , 𝑖 ∈ 𝐼 and a net

(𝜆𝛾𝑖 ) of scalars tending to 0, such that 𝑥𝛾

𝑖 ∈ 𝑆𝑖 and for each 𝑖 ∈ 𝐼𝑥𝛾𝑖 − 𝑦𝑖 ∈

𝜆𝛾𝑖 − 𝑆𝑖 , 𝛾 ∈ 𝛤 since a net 𝜆𝛾

𝑖 → 0, 𝑖 ∈ 𝐼, for each 𝛾 ∈ 𝛤 then there is a

diagonal net semi-converging to 0 such that when 𝑖 = 𝛾 then 𝜆𝛾𝑖 → 0.

Since 𝑥𝛾𝑖 − 𝑦𝑖 ∈ 𝜆𝛾

𝑛𝐵𝑖 whenever 𝑖 = 𝑛 and = 1,2, … , 𝑛, … .

If ≠ 𝑖 , since a disk 𝑆𝑖 and |𝜆𝛾𝑛| → 0 and if 𝑎 ∈ 𝑆𝑖 then 𝜆𝛾

𝑛𝑎 ∈ 𝑆𝑖 then

𝑥𝛾𝑖 − 𝑦𝑖 ∈ 𝜆𝛾

𝑛𝑆𝑖 then 𝑥𝛾𝑖 − 𝑦𝑖 ∈ 𝜆𝛾

𝑛𝑆𝑖 . Therefore 𝑥𝛾 = (𝑥𝛾1, 𝑥𝛾

2, … ) − 𝑦 ∈

𝜆𝛾𝑛 ∏ 𝑆𝑖 𝑖. 𝑒 𝑥𝛾

𝑠𝑏→ 𝑖∈𝐼 𝑦 .

23

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[3] Barreira, L. and Almeida, j. "Hausdorff Dimension in Convex Bornological

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on semi complete bornological vector...

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