**Turkish Journal of Analysis and Number Theory**

## On Irresolute Topological Vector Spaces-II

**Muhammad Asad Iqbal**^{1}, **Muhammad Maroof Gohar**^{2}, **Moiz ud Din Khan**^{1,}

^{1}Mathematics COMSATS Institute of Information Technology, Park Road, Chak Shahzad, 45550 Islamabad, PAKISTAN

^{2}Mathematics, G.C. University, Lahore, Pakistan

### Abstract

In this paper, we continue the study of Irresolute topological vector spaces. Notions of convex, bounded and balanced set are introduced and studied for Irresolute topological vector spaces. Along with other results, it is proved that: 1. Irresolute topological vector spaces are semi-Hausdorff spaces. 2. Every Irresolute topological vector space is semi-regular space. 3. In Irresolute topological vector spaces, as well as is convex if is convex. 4. In Irresolute topological vector spaces, is bouned if is bounded. 5. In Irresolute topological vector spaces, is balanced if is balanced and 6. In Irresolute topological vector spaces, every semi compact set is bounded.

**Keywords:** topological vector space, irresolute topological vector space, irresolute mapping, semi open set

Received January 06, 2016; Revised April 21, 2016; Accepted April 29, 2016

**Copyright**© 2016 Science and Education Publishing. All Rights Reserved.

### Cite this article:

- Muhammad Asad Iqbal, Muhammad Maroof Gohar, Moiz ud Din Khan. On Irresolute Topological Vector Spaces-II.
*Turkish Journal of Analysis and Number Theory*. Vol. 4, No. 2, 2016, pp 35-38. http://pubs.sciepub.com/tjant/4/2/2

- Iqbal, Muhammad Asad, Muhammad Maroof Gohar, and Moiz ud Din Khan. "On Irresolute Topological Vector Spaces-II."
*Turkish Journal of Analysis and Number Theory*4.2 (2016): 35-38.

- Iqbal, M. A. , Gohar, M. M. , & Khan, M. U. D. (2016). On Irresolute Topological Vector Spaces-II.
*Turkish Journal of Analysis and Number Theory*,*4*(2), 35-38.

- Iqbal, Muhammad Asad, Muhammad Maroof Gohar, and Moiz ud Din Khan. "On Irresolute Topological Vector Spaces-II."
*Turkish Journal of Analysis and Number Theory*4, no. 2 (2016): 35-38.

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### 1. Introduction

This paper is devoted to the study of Irresolute topological vector spaces. The notion was defined by Moiz and Asad ^{[1]} in 2016. The notion is defined although analogously but is independent of linear topological space. The part of this paper deals with the boundedness, convexity and balancedness of sets in Irresolute topological vector spaces.

The motivation behind the study of this paper is to investigate such structures in which the topology is endowed upon a vector space which fails to satisfy the continuity condition for vector addition and scalar multiplication or either. We are interested to study such structures for irresolute mappings in the sense of Levine. It is proved in ^{[1]} that:

• Irresolute topological vector space is not topological vector space in general.

• Open hereditary property holds in irresolute topological vector spaces.

• Every semi-open set is translationally invariant in irresolute topological vector spaces

• A homomorphism between irresolute topological vector spaces is irresolute if it is irresolute at identity element.

In this paper, notions of convex, bounded and balanced set are introduced and studied for Irresolute topological vector spaces.

### 2. Preliminaries

Throughout in this paper, and are always representing topological spaces on which separation axioms are not considered untill and unless stated. We will represent field by and the set of all real numbers by and ∈ are assumed here negligible small but positive real numbers.

Semi open sets in topological spaces were firstly appeared in 1963 in the paper of N. Levine ^{[3]}. With the invent of semi open sets and semi continuity, many interesting concepts in topology were further generalized and investigated by number of mathematicians. A subset of a topological space is said to be semi-open if, and only if, there exists an open set *0* in such that or equivalently if * * denotes the collection of all semi-open sets in the topological space The complement of a semi-open set is said to be semi closed; the semi closure of denoted by is the intersection of all semi-closed subsets of containing ^{[4]}. It is known that if, and only if, for any semi-open set containing , is non-empty. Every open set is semi-open and every closed set is semi-closed. It is known that union of any collection of semi-open sets is semi-open set, while the intersection of two semi-open sets need not be semi-open. The intersection of an open set and a semi-open set is semi-open set. A subset of a topological space is said to be semi compact if for every cover of * * by semi open sets of , there exists a finite subcover.

Remember that, a set is a semi-open neighbourhood of a point if there exists such that A set is semi open in if, and only if, is semi open neighbourhood of each of its points. If a semi open neighbourhood of a point is a semi open set, we say that is a semi open neighbourhood of . If and then where and are topological spaces and is a product space. It is worth mentioning that a set semi-open in the product space cannot be expressed as product of semi-open sets in the components spaces. Basic properties of semi-open sets are given in ^{[3]} and of semi closed sets in ^{[4]} and ^{[5]}, and references there in.

If is a vector space then denotes its identity element, and *for *a fixed and denote the left and the right translation by , respectively. The addition mapping is defined by and the scalar multiplication mapping is defined by

**Definition 2.1**: Let be single valued function between topological spaces (continuity not assumed). Then:

1. is termed as semi-continuous ^{[3]}, if and only if, for each open in , there exists

2. is termed as irresolute ^{[4]}, if, and only if, for each there exists . Note that the function is irresolute at if for each semi open set in containing , there exists a semi open set in containing such that

Recall that a topological vector space is a vector space over a topological field (most often the real or complex numbers with their standard topologies) that is endowed with a topology such that:

1. Addition mapping defined by is continuous function.

2. Multiplication mapping defined by is continuous function (where the domains of these functions are endowed with product topologies).

Equivalently, we have a topological vector space over a topological field (most often the real or complex numbers with their standard topologies) that is endowed with a topology such that:

1. for each and for each open neighbourhood of in , there exist neighbourhoods and of and respectively in , such that

2. for each and for each open neighbourhood in containing , there exist neighbourhoods of in and of in such that

Equivalently, we have: topological Vector Space over the field ( *or** ** *) with a topology on such that is a topological group and is a continuous mapping.

### 3. Irresolute Topological Vector Spaces

**Definition 3.1**: ^{[2]} A mapping form a space onto a space is said to be semi-quotient provided a subset of is open in if and only if is semi-open in . Equivalently, every semi-quotient mapping is semi-continuous and every quotient mapping is semi-quotient.

**Remark 3.2**: ^{[2]} Every surjective continuous mapping which is either s-open or s-closed is a semi-quotient mapping. Let be a topological space and a set be a mapping. Define, It is easy to see that the family is a generalized topology on .(*i.e.* and union of any collection of sets in, is again in generated by ): we call it the semi-quotient generalized topology. But, need not be a topology in . It happens if is an extremely disconnected space, because in this case the intersection of two semi-open sets in is semi-open. It is trivial fact that in the later case, is the finest topology on such that is semi-continuous.

For more detail we refer the reader to ^{[2]}.

**Lemma 3.3**: Let be a canonical projection mapping then for any subset of , . We know that, if is a group and is a subgroup of then the collection of all left cosets of in that is is not a group until H is normal in .

**Definition 3.4**: A mapping form a topological space to itself is called irresolute-homeomorphism if it is bijective, irresolute and pre-semi-open.

**Definition 3.5**: A mapping is said to be semi-open if the image of a semi-open set is open.

**Lemma 3.6**: ^{[1]} Let and be subsets of an irresolute topological vector space. Then

**Lemma 3.7**: ^{[1]} Let be an irresolute topological vector space. For given and nonzero each translation mapping and multiplication mapping where is irresolute homeomorphisim onto itself.

**Lemma 3.8**: ^{[1]} If * * is an irresolute homeomorphism, then for all

**Lemma 3.9**: ^{[1]} Let be an irresolute topological vector space:

i. If then for every

ii. If then for every

iii. For every semi-open neighbourhood of , there exists a semi-open neighbourhood of such that

**Theorem 3.10**: For a closed subspace of an irresolute topological vector space the semi-quotient mapping is s-open mapping, that caries semi-open sets to open.

Proof: Let be semi-open set in_{ }. Then, because is an irresolute topological vector space. Therefore, is open, this proves that is s-open mapping.

**Definition 3.11**: Let be a linear subspace of which means and for all

**Theorem 3.12**: Let be an irresolute topological vector space and if is linear subspace of , so is

Proof: Let be a linear subspace of , which means that, and By Lemma 3.6, Since, scalar multiplication is an irresolute homeomorphism, therefore by Lemma 3.8, it maps the semi-closure of a set into the semi-closure of its image, namely for every Therefore, is linear subspace.

**Definition 3.13**: A subset of an irresolute topological vector space is said to be balanced if

**Theorem 3.14**: Let be an irresolute topological vector space, For every if is balanced, so is

Proof: By lemma 3.8, semi-closures of a set maps onto the semi-closures of its image, thus for every If is balanced, then for Hence, is balanced.

**Lemma 3.15**: ^{[2]} Let be an irresolute topological group, and the collection of all semi-open neighbourhoods of . Then:

1. For every there is a such that

2. For every and every there is such that

**Theorem 3.16**: Let be an irresolute topological vector space then for any

Proof: Firstly we have to show that, For this, assume that this implies that there exists semi-open set containing and hence gives: Or Or This shows that, Conversely, Let this implies that there exists, such that: and Let and Thus for some This gives This proves that, Hence, Therefore, we have

**Theorem 3.17**: Let be an irresolute topological vector space. Let be the collection of all semi-open neighbourhoods of then for every the collection is a semi-open neighbourhood system for

Proof: Since for the irresolute topological vector spaces the translation mappings are irresolute homeomorphism. Therefore, this result is obvious. i.e. Thus by algebra we can deduce that This gives that is semi-open neighbourhood system for The is semi-open by Lemma 3.9(i).

**Theorem 3.18**: Let be an irresolute topological vector space. is collection of all semi-open neighbourhoods of then the topology is semi- space if and only if

Proof: Let so because Assume is semi-Hausdorff, the set is semi-open neighbourhood of Now, there exists with Therefore, This implies, Conversely, let and we have to show that is semi-. For that let and with and let us indicate how to construct two disjoint semi-open neighbourhoods one for and one for Using translations we can assume Since, there exist some such that Using Lemma 3.9(iii), there is some such that: We still have and is semi-open. Since is semi-open neighbourhood of therefore, is also semi-open neighbourhood of by Lemma 3.9(ii). would be the semi-open neighbourhood for Therefore, is semi-Hausdorff. If then we can construct a semi-open neighbourhood for we can clearly see that and also and This implies, is semi- space.

**Theorem 3.19**: Every irresolute topological vector space is semi-Hausdroff space.

Proof: We only need to separate and with Let is semi-open neighbourhood of , then by Lemma 3.9(iii), there exists semi-open neighbourhood of such that: Since is semi-open neighbourhood of , then is semi-open neighbourhood of by algebra. We claim that, If then and thus where and are some members of Therefore, we have which is a contradiction. This proves that every irresolute topological vector space is semi-Hausdroff space.

**Theorem 3.20**: Every irresolute topological vector space is semi-regular space if the space is extremally disconnected.

Proof: Let be semi-closed subset of an irresolute extremally disconnected space and let Now is semi-open set and is a semi-open neighbourhood of By Lemma 3.9(iii), there exists semi-open neighbourhood of such that: Now, is also semi-open neighbourhood of , and let Then, moreover Let and Then, and and clearly, and are semi-open sets. Further, if then we must have, and where and are some members of and But then we would have, A contradiction to the fact that Hence, Therefore, every extremally disconnected irresolute topological vector space is semi-regular space.

**Definition 3.21**: A set is said to be convex if for

**Theorem 3.22**: Let be an irresolute topological vector space. If is convex then so is

Proof: Convexity is a purely algebraic property, but (closure) semi-closures and (interior) semi-interior are topological concepts. The convexity of C implies, Let then and By Lemma 3.6: Thus, is convex.

**Theorem 3.23**: Let be an irresolute topological vector space. If is convex then is convex.

Proof: Suppose that is convex. Let This means there exist semi-open neighbourhoods and of such that: and Since, is convex. Therefore, Which proves that Namely is convex.

**Definition 3.24**: Let be an irresolute topological vector space. A subset is said to be bounded if for all semi-open neighbourhoods of , there exists such that for all That is, every semi-open neighborhood of zero contains after being blown up sufficiently.

**Theorem 3.25**: Let be an irresolute topological vector space. If is bounded, then is bounded.

Proof: Let is neighbourhood of then by theorem, there exist such that Since is bounded, for sufficiently large It follows that for large enough * ** *Thus, is bounded.

**Theorem 3.26**: Let be an irresolute topological vector space, and let be a collection of all semi-open neighbourhoods containing the identity. Then, for each there exists such that

Proof: Let Then by Lemma 3.9(iii), there exists such that Let Since is semi-open neighbourhood of , so Choose, then where Thus, Therefore,

**Theorem 3.27**: (Balanced) Let be an irresolute topological vector space, and be a collection of all semi-open neighbourhoods, then for each , there exists a balanced such that

Proof: Let then by definition, there exist semi-open neighbourhood of in and of in such that Clearly, is balanced. Furthur, is semi-open neighbourhood of (Since ) and is semi-open neighbourhood of by Lemma 3.9(ii). Thus we have a balanced and

**Lemma 3.28**: Let be an irresolute topological vector space. Then, where and

Proof: Let such that then there exists a semi-open neighbourhood such that Now, As is semi-open by Theorem 3.9 (2). So, Therefore, Conversely, Let where define for some then there exists semi-open neighbourhood and such that be semi-open in by the fact that So, or or Therefore, Hence,

**Theorem 3.29**: (Semi-interior) Let be an irresolute topological vector space. For every if is balanced and then is balanced.

Proof: Let be balanced subset of . By Lemma 3.8, for every Since, is balanced therefore Also, Since for we must require for the latter to be balanced.

**Theorem 3.30**: (Convergent sequence) Let be a semi-open neighbourhood of 0 in Irresolute topological vector space. Then, for every sequence

Proof: Let and consider the sequence This sequence converges to by the irresoluteness of the scalar multiplication

Thus, for sufficiently large

**Theorem 3.31**: Let be an irresolute topological vector space. Every semi-compact set is bounded.

Proof: Let be a semi-compact. We need to prove that it is bounded, namely, that for every semi-open neighbourhood of , for sufficiently large Let be semi-open neighbourhood of , then by theorem 3.27, there exists a balanced semi-open neighbourhood of such that By theorem 3.30, Since, is semi-compact therefore:

Thus, for all Which proves that K is bounded.

**Theorem 3.32**: A Cauchy sequence in an irresolute topological vector space is bounded.

Proof: Let be a Cauchy sequence. Let be the semi-open neighbourhood of , then by Lemma 4(iii), there exists a semi-open neighbourhood of such that: By definition of a cauchy sequence, there exists such that for all and in particular for all Set such that then for all Since for balanced sets for and since every semi-open neighbourhood of zero contains a balanced neighbourhood, this proves that the sequence in indeed bounded.

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[2] | Muhammad Saddique Bosan, s-Topological groups, 2015, (Ph.D Thesis). | ||

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[3] | N. Levine, Semi-Open Sets and Semi-Continuity in Topological Spaces, Amer. math. month., 70(1) (1963), 37-41. | ||

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[4] | Crossley, S.G. and Hildebrand, S.K. Semi-Topological Properties. Fundamental Mathematicae, 74(1972), 233-254. | ||

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[5] | Crossley, S.G. and Hildebrand, S.K. Semi-Closure, Texas J. Sci., 22(1971), 99-112. | ||

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