# Reconstruction of the core convex topology and its applications in vector optimization and convex analysis

## Abstract

In this paper, the core convex topology on a real vector space , which is constructed just by operators, is investigated. This topology, denoted by , is the strongest topology which makes into a locally convex space. It is shown that some algebraic notions existing in the literature come from this topology. In fact, it is proved that algebraic interior and vectorial closure notions, considered in the literature as replacements of topological interior and topological closure, respectively, in vector spaces not necessarily equipped with a topology, are actually nothing else than the interior and closure with the respect to the core convex topology. We reconstruct the core convex topology using an appropriate topological basis which enables us to characterize its open sets.

Furthermore, it is proved that is not metrizable when X is infinite-dimensional,
and also it enjoys the Hine-Borel property. Using these properties, -compact sets are
characterized and a characterization of finite-dimensionality is provided. Finally, it is shown that the properties of the core convex topology lead to directly extending various important results in convex analysis and vector optimization from topological vector spaces to real vector spaces.

Keywords: Core convex topology, Functional Analysis, Vector optimization, Convex Analysis.

2

###### Contents:

## 1 Introduction

Convex Analysis and Vector Optimization under real vector spaces, without any topology, have been studied by various scholars in recent years [1, 2, 3, 8, 16, 18, 19, 20, 29, 30]. Studying these problems opens new connections between Optimization, Functional Analysis, and Convex analysis. Since (relative) interior and closure notions play important roles in many convex analysis and optimization problems [4, 18, 21], due to the absence of topology, we have to use some algebraic concepts. To this end, the concepts of algebraic (relative) interior and vectorial closure have been investigated in the literature, and many results have been provided invoking these algebraic concepts; see e.g. [1, 2, 3, 11, 16, 18, 20, 23, 24, 29, 30] and the references therein. The main aim of this paper is to unifying vector optimization in real vector spaces with vector optimization in topological vector spaces.

In this paper, core convex topology (see [10, 19]) on an arbitrary real vector space, , is dealt with. Core convex topology, denoted by , is the strongest topology which makes a real vector space into a locally convex space (see [10, 19]). The topological dual of under coincides with its algebraic dual [10, 19]. It is quite well known that when a locally convex space is given by a family of seminorms, the locally convex topology is deduced in a standard way and vice versa. In this paper, is reconstructed by a topological basis. It is known that algebraic (relative) interior of a convex set is a topological notion which can be derived from core convex topology [10, 19]. We provide a formula for -interior of an arbitrary (nonconvex) set with respect to the algebraic interior of its convex components. Furthermore, we show that vectorial closure is also a topological notion coming from core convex topology (under mild assumptions). According to these facts, various important results, in convex analysis and vector optimization can be extended easily from topological vector spaces (TVSs) to real vector spaces. Some such results are addressed in this paper. After providing some basic results about open sets in , it is proved that, is not metrizable under topology if it is infinite-dimensional. Also, it is shown that enjoys the Hine-Borel property. A characterization of open sets in terms of there convex components is given. Moreover, -convergence as well as -compactness are characterized.

The rest of the paper unfolds as follows. Section 2 contains some preliminaries and Section 3 is devoted to the core convex topology. Section 4 concludes the paper by addressing some results existing in vector optimization and convex analysis literature which can be extended from TVSs to real vector spaces, utilizing the results given in the present paper.

## 2 Preliminaries

Throughout this paper, is a real vector space, is a subset of , and is a nontrivial nonempty ordering convex cone. is called pointed if . , , and denote the cone generated by , the convex hull of , and the affine hull of , respectively.

For two sets and a vector , we use the following notations:

is the set of all subsets of and for ,

The algebraic interior of , denoted by , and the relative algebraic interior of , denoted by , are defined as follows [16, 19]:

where is the linear hull of . When we say that is solid; and we say that is relatively solid if . The set is called algebraic open if . The set of all elements of which do not belong to and is called the algebraic boundary of . The set is called algebraically bounded, if for every and every there is a such that

If is convex, then there is a simple characterization of as follows: if and only if for each there exists such that for all

###### Lemma 2.1.

Let be a vector basis for , and be nonempty and convex. if and only if for each there exists scalar such that

###### Proof.

Assume that for each there exists scalar such that Let . There exist a finite set and positive scalars such that

Let and . Considering , we have

where stands for the line segment joining ,. Since is convex,

and then, due to the convexity of again,

This implies

which means Furthermore, the convexity of guarantees that for all Thus The converse is obvious. ∎

Some basic properties of the algebraic interior are summarized in the following lemmas. The proof of these lemmas can be found in the literature; see e.g. [1, 2, 10, 16, 19].

###### Lemma 2.2.

Let be a nonempty set in real vector space . Then
the following propositions hold true:

1. If is convex, then ,

2. ,

3. for each

4. for each

5. If , then is absorbing (i.e. ).

###### Lemma 2.3.

Let be a convex cone. Then
the following propositions hold true:

i. If , then is a convex cone,

ii. ,

iii. If are convex and relatively solid, then

iv. If is a convex (concave) function, then is continuous.

Although the (relative) algebraic interior is usually defined in vector spaces without topology, in some cases it might be useful under TVSs too. It is because the algebraic (relative) interior can be nonempty while (relative) interior is empty. The algebraic (relative) interior preserves most of the properties of (relative) interior.

Let be a real topological vector space (TVS) with topology . We denote this space by The interior of with respect to topology is denoted by . A vector is called a relative interior point of if there exists some open set such that The set of relative interior points of is denoted by

###### Lemma 2.4.

Let be a real topological vector space (TVS) and . Then . If furthermore is convex and , then .

The algebraic dual of is denoted by , and exhibits the duality pairing, i.e., for and we have . The nonnegative dual and the positive dual of are, respectively, defined by

If is a convex cone with nonempty algebraic interior, then

The vectorial closure of , which is considered instead of closure in the absence of topology, is defined by [1]

is called vectorially closed if .

## 3 Main results

This section is devoted to constructing core convex topology via a topological basis. Formerly, the core convex topology was constructed via a family of separating semi-norms on ; see [19]. In this section, we are going to construct core convex topology directly by characterizing its open sets. The first step in constructing a topology is defining its basis. The following definition and two next lemmas concern this matter.

###### Definition 3.1.

[22] Let be a subset of , where stands for the power set of . Then, is called a topological basis on if and moreover, the intersection of each two members of can be represented as union of some members of .

The following lemma shows how a topology is constructed from a topological basis.

###### Lemma 3.1.

If is a topological basis on , then the collection of all possible unions of members of is a topology on .

Lemma 3.2 provides the basis of the topology which we are looking for. The proof of this lemma is clear according to Lemma 2.2.

###### Lemma 3.2.

The collection

is a topological basis on .

Now, we denote the topology generated by

by ; more precisely

The following theorem shows that is the strongest topology which makes into a locally convex TVS. This theorem has been proved in [19] using a family of semi-norms defined on . Here, we provide a different proof.

###### Theorem 3.1.

i. is a locally convex TVS;

ii. is the strongest topology which makes into a locally
convex space.

###### Proof.

By Lemmas 3.1 and 3.2, is a topology on .

Proof of part i: To prove this part, we should show that
is a Hausdorff space, and two operators addition
and scalar multiplication
are -continuous.

Continuity of addition: Let and let be a -open set containing . We should find two -open sets and containing and , respectively, such that . Since is a basis for , there exists such that

Defining

by Lemma 2.2, we conclude that and Convexity of , implies that and are the desired -open sets, and hence the addition operator is -continuous.

Continuity of scalar multiplication: Let , , and be a -open set containing . without lose of generality, assume that . We must show that there exist and a -open set containing such that

Since , by considering in the definition of algebraic interior, there exists such that

Define

We get , , and by Lemma 2.2, . Furthermore, is balanced (i.e. for each ), because is convex and . Now, we claim that

(1) |

To prove (1), let with . Therefore

Thus,

Hence,

This proves (1). Setting and proves the continuity of the scalar multiplication operator.

Now, we show that is a Hausdorff space. To this end,
suppose . Consider such
that , and set and
. It is not difficult to see that
and while . This implies that is a Hausdorff space.

ii. Let be an arbitrary topology on which makes
into a locally convex space. Let be a locally convex basis of
topology For each we have
(by Lemma 2.4) and hence . Thus we have ,
which leads to and completes the
proof.∎

The interior of with respect to topology is denoted by The following theorem shows that the algebraic interior (i.e. ) for convex sets is a topological interior coming from .

###### Theorem 3.2.

[19, Proposition 6.3.1] Let be a convex set. Then

###### Proof.

The proof of the following result is similar to that of Theorem 3.2.

###### Theorem 3.3.

If is a convex subset of , then , where denotes the relative interior of with respect to the topology .

It is seen that the convexity assumption plays a vital role in Theorems 3.2 and 3.3. In the following two results, we are going to characterize the -interior of an arbitrary (nonconvex) nonempty set with respect to the of its convex components. Since and is a basis for the set could be written as union of some subsets of which are algebraic open.

###### Lemma 3.3.

Let be a nonempty subset of real vector space Then could be uniquely decomposed to the maximal convex subsets of , i.e. where are non-identical maximal convex subsets (not necessary disjoint) of (Here, sets are called convex components of ).

###### Proof.

For every , let be the set of all maximal convex subsets of containing (the nonemptiness of such is derived from Zorn lemma). Set the index set (this type of defining enables us to avoid repetition). For every define Hence, where are non-identical maximal convex subsets of To prove the uniqueness of s, suppose such that are non-identical maximal convex subsets of Let and ; then , and hence there exists such that This means , and the proof is completed.∎

###### Theorem 3.4.

Let be a nonempty subset of real vector space . Then

where are convex components of

###### Proof.

Let There exists such that Set Clearly and hence . Furthermore it is easy to verify that each chain (totaly ordered subset) in has an upper bound within . Therefore, using Zorn lemma, has a maximal element. Let be maximal element of . Obviously, is a convex component of . It leads to the existence of an such that . Therefore, To prove the other side, suppose for some Since is convex, by Theorem 3.2, ∎

###### Example 3.1.

Consider

as a subset of It can be seen that while However, where are convex components of as follows

∎

The following theorem shows that the topological dual of is the algebraic dual of . In the proof of this theorem, we use the topology which induces on . This topology, denoted by , is as follows:

where

###### Theorem 3.5.

[10]

###### Proof.

The following result provides a characterization of finite-dimensional spaces utilizing and the topology induced by on .

###### Theorem 3.6.

is finite-dimensional if and only if

###### Proof.

Assume that is finite-dimensional. Since there is only one topology on which makes this space a TVS, we have

To prove the converse, by indirect proof assume that . Let be an ordered basis of ; and denote the vector of coordinates of with respect to the basis . It is easy to show that, for each , where

Define by Since is a norm on , thus is also a norm on We denote this norm by .

Since is the strongest locally convex topology on , we have , where stands for the unit ball with On the other hand, every -open set containing origin, contains an infinite-dimensional subspace of (see [26]). Hence, This implies , and the proof is completed. ∎

Theorem 3.7 demonstrates that the vectorial closure () for relatively solid convex sets, in vector space , is a topological closure coming from . The closure of with respect to is denoted by

###### Theorem 3.7.

Let be a convex and relatively solid set. Then

###### Proof.

Since is a TVS, it is easy to verify that To prove the converse, let and . Without loss of generality, we assume , and then we have and , where stands for the algebraic interior of with respect to the subspace . We claim that To show this, assume that ; then there exists a functional such that and for each . Therefore the set

is a open neighborhood of with which contradicts Now, we restrict our attention to subspace . By Theorem 3.2, , and thus, by [18, Lemma 1.32], , which means . Therefore , and the proof is completed. ∎

###### Corollary 3.1.

Let be a nonempty subset of with finite many convex components. Then

where are convex components of .

###### Proof.

The equality may not be true in general; even if each convex component, , is relatively solid. For example, consider as the set of rational numbers in . The convex components of are singleton, which are relatively solid. Therefore , while

A TVS is called metrizable if there exists a metric such that -open sets in coincide with -open sets in . Now, we are going to show that is not metrizable when is infinite-dimensional. Lemma 3.4 helps us to prove it. This lemma shows that every algebraic basis of the real vector space is far from the origin with respect to topology.

###### Lemma 3.4.

Let be an algebraic basis of . Then

###### Proof.

Define

We claim that the following propositions hold true;

a. ,

b. ,

c. .

To prove (a), assume that and Then there exist positive scalars and finite sets such that;

and .

Hence,

where

Furthermore,

Therefore , and hence is a convex set.

To prove (b), first notice that for each ; then due to the convexity of , from (a) we get , because of Lemma 2.1.

We prove assertion (c) by indirect proof. If , then

for some , and some finite set . Also, and values are positive and . Since is a linear independent set, we have , and also . Hence which is a contradiction. This completes the proof of assertion (c).

Now, we prove the lemma invoking . Since is convex, by theorem 3.2 and claim (b), is a open neighborhood of . On the other hand, (c) implies Thus, ∎

###### Theorem 3.8.

If is infinite-dimensional, then is not metrizable.

###### Proof.

Suppose that is an infinite-dimensional metrizable TVS with metric Then for every choose such that and is linear independent. This process generates a linear independent sequence such that, (i.e. is convergent to zero). Furthermore, this sequence can be extended to a basis of , say . This makes a contradiction, according to Lemma 3.4, because ∎

In every topological vector space, compact sets are closed and bounded, while the converse is not necessarily true. A topological vector space in which every closed and bounded set is compact, is called a Hiene-Borel space. Although it is almost rare that an infinite-dimensional locally convex space be a Hine-Borel space (for example, normed spaces), the following result proves that is a Hine-Borel space.

###### Theorem 3.9.

enjoys the Hine-Borel property. Moreover, every compact set in lies in some finite-dimensional subspace of

###### Proof.

Let be a -closed and -bounded subset of . First we claim that the linear space , i.e. the smallest linear subspace containing , is finite-dimensional. To see this, by indirect proof, assume that is an infinite-dimensional subspace of ; then there exists a linear independent sequence in . Thus, the sequence is a linear independent set in and also is -bounded. Hence, . Furthermore, has a basis , containing This makes a contradiction, according to Lemma 3.4, because Hence, is a closed and bounded set contained in a finite-dimensional space . Therefore, by traditional Hine-Borel theorem in finite-dimensional spaces, is a compact set in , and hence is a -compact set in ∎

One of the important methods to realize the behavior of a topology defined on a nonempty set is to perceive (characterize) the convergent sequences. Assume that a sequence is -convergent to some (i.e, ). Thus, is a -compact set. Therefore, by Theorem 3.9, lies in a finite-dimensional subspace of