Research Article, Res Rep Math Vol: 1 Issue: 1

# On Statistically (I) - Sequential Spaces

**Renukadeviz V ^{*} and Prakash B**

Center for Research and Post Graduate Department of Mathematics, Ayya Nadar Janaki Ammal College, Inida

***Corresponding Author :** **Renukadevi V**

Center for Research and Post Graduate Department of Mathematics, Ayya Nadar Janaki Ammal College(Autonomous), Sivakasi 626124, Tamil Nadu, India

**Tel:** 04562-254100

**E-mail: **renu_siva2003@yahoo.com

**Received:** July 21, 2017 **Accepted:** August 21, 2017 **Published:** August 28, 2017

**Citation:** Renukadeviz V, Prakash B (2017) On Statistically (I) - sequential spaces. Res Rep Math 1:1.

## Abstract

In this paper, we answer the question “Is the product of two statistically sequential spaces statistically sequential?” raised by Zhongbao and Fucai. By an example we see that products of two statistically sequential spaces need not be statistically sequential. Also, we give a necessary and sufficient condition for the product to be in most general case of I-sequential spaces. And we point out the error in proposition 2.2 of. Further, we prove that a subspace of a statistically sequential space need not be statistically sequential and we find a necessary and sufficient condition for a subspace to be in most general case of an I-sequential space. Finally, we develop the properties of the most general concept of I-sequential spaces.

### Keywords: Phrases statistical convergence; I-convergent sequence; Sequential space

## Introduction

The concept of convergence of a sequence of real numbers has been extended to statistical convergence independently by Fast [1,2] and Schoenberg [3]. If then K_{n} will denote the set {K: k≤n} and |Kn| stands for the cardinality of K: The natural density of K is defined by

if the limit exists [4,5]. A sequence {X_{n}} in a topological space X is said to converge statistically [6] (or shortly s-converge) to x∈X, if for every neighborhood U of , Any convergent sequence is statistically convergent but the converse is not true [7]. But in general, s-convergent sequences satisfy many of the properties of ordinary convergent sequences in metric spaces. It has been discussed and developed by many authors [8-16].

The concept of I-convergence of real sequences [17,18] is a generalization of statistical convergence which is based on the structure of the ideal I of subsets of the set of natural numbers. In the recent literature, several works on I-convergence including remarkable contributions by salat et al. have occurred [15-20]. The idea of I-Convergence has been extended from the real number space to a topological space [14] and to a normed linear space [21].

I-convergence coincides with the ordinary convergence if I is the ideal of all finite subsets of and with the statistical convergence if I is the ideal of subsets of of natural density zero.

Throughout this paper, (X, τ) will stand for a topological space and I for a nontrivial ideal of , the set of all positive integers. X_{n}→x denotes a sequence{X_{n}} converging to x Let X be a space and P⊂X A sequence {X_{n}} converging to x in X is eventually in P if for some it is frequently in P if is eventually in P for some subsequence of{Xn}. Let P be a family of subsets of X. Then ∪P and ∩P denote the union ∪{P|P∈P}, and the intersection ∩{P|PP}, respectively. We recall the following definition [22].

**Definition 1**

If X is a nonvoid set, then a family of sets is I ⊂ 2^{X} is an ideal if (i) A, B∈I implies A∪B∈I and (ii)A ∈I, B⊂A, implies B∈I.

The ideal is called nontrivial if I ≠ ∅ and X∉I. A nontrivial ideal I is called admissible if it contains all the singleton sets. Several examples of nontrivial admissible ideals may be seen in [17].

**Definition 2**

A sequence {X_{n}}. in X is said to be I-convergent to x_{0}∈X if for any nonvoid open set U containing x_{0}, We call x_{0} as the I-limit of the sequence {X_{n}}[14] .

**Definition 3**

Let X be a space. P ⊂ X is called a sequential neighborhood of x in X, if each sequence convergent to x∈X is eventually in P. A subset U of X is called sequentially open if U is a sequential neighborhood of each of its points. X is called a sequential space [24] if each sequentially open subset of X is open.

**Definition 4**

O is I-sequentially open if and only if no sequence in X \O has an I-limit in O [1].

**Definition 5**

A subset A of a space X is said to be I-sequentially closed set if for every sequence {X_{n}}in A with {X_{n}}I-converges to x, then x∈A [1].

**Definition 6**

A topological space is I-sequential when any set O is open if and only if it is I-sequentially open.

**Definition 7**

A space X is called locally I-sequential space if every point of x∈X has a neighborhood which is an I-sequential space.

Even though we mainly deal with I-sequential spaces, we see the basic definition for s-sequential space since it will be useful for the examples which deal with s-sequential spaces. An I-sequential space X is statistically sequential if .

**Definition 8**

A space X is called statistically sequential(or shortly s-sequential) space [6] if for each non-closed subset A⊂X, there is a point x∈X \A and a sequence in Astatistically converging to x.

There is another way to define s-sequential space.

**Definition 9**

A subset A of a space X is said to be a statistically sequentially open set (s-sequentially open) [25] if for any sequence {X_{n}} statistically converge to x and x∈A then

A topological space is s-sequential when any set O is open if and only if it is s-sequentially open.

**Definition 10**

A subset K of the set is called statistically dense [6] if d(k)=1.

**Definition 11**

A subsequence S of the sequence L is called statistically dense in L [4] if the set of all indices of elements from S is statistically dense.

**Remark 12**

1. The limit of an I-convergent sequence is uniquely determined in Hausdorff spaces [14,6] .

2. If a sequence (x_{n})_{n∈N} converges to x in the usual sense, then it statistically converges to x . But the converse is not true in general.

3. A sequence (x_{n})_{n∈N} is statistically convergent if and only if each of its statistically dense subsequence is statistically convergent.

4. If a sequence{X_{n}} I-converges to x , then every subsequence is I-convergent for every I′∈I

**Lemma 13**

Let X be a topological space and A⊂X Then the following hold [1].

(a) A is I-sequentially open.

(b) X\A is I-sequentially closed.

**I-sequential spaces 14**

In this section, we answer the question 2 [25]: Is the product of two s-sequential spaces s-sequential?

By the Example 1 that product of two s-sequential spaces need not be s-sequential. Also, we give the necessary and sufficient condition for the product to be in most general case of I-sequential space. And we point out the error in proposition 2 of [1].

**Proposition 1**

The disjoint topological sum of any family of I-sequential spaces is I-sequential.

Proof. Let X be the disjoint sum of the family {X_{i}}_{i∈Λ} of I-sequential spaces. U∩ X_{i} is not I-sequentially open. Thus, there is a point X_{i}∈ U∩ X_{i} and a sequence I-converges to Xi and therefore in X. Hence U is not I-sequentially open. Therefore, X is an I-sequential space.

The following Example 1 shows that the products of two s-sequential spaces need not be s-sequential.

**Example 1**

For each let be a sequence statistically converges to x such that if m≠l. Take Let X′ be the disjoint topological sum of Sn where and X be the space obtained from X′ by identifying x_{n} to ∞. Now let f: X′ → X be a natural mapping. Since each Sn is a s-sequential space and hence X′ by Proposition 1. By Theorem 1 in [25], X is a s-sequential space.

The open set in X is as follows :

1. Each point x_{n,m} is isolated;

2. Each open neighborhood of the point ∞ is a set V of the form

where each M_{n} is a dense subsequence of Sn. Next we define Y: Let

Now yet

Topologize Y as follows:

Let each point of be open and be a countably local base at y, where . Then Y is a metric space.

Now let . Then Thus, A is not closed in X, Y, but statistically sequentially closed in X x Y since A has no non trivial s-convergent sequence. Suppose A has a s-convergent sequence statistically converges to (,y), Then statistically converges to ∞ which implies for some m, . But the corresponding sequence in statistically converges to y which is a contradiction to . Therefore, X Y is not a s-sequential space.

Next we see the necessary and sufficient condition for the product of I-sequential spaces to be I-sequential.

**Proposition 2**

Every I-sequential space is a quotient of a topological sum of I-convergent sequences.

Proof. Let X be an I-sequential space. For each x∈X and for each sequence {S_{n}} in XI-converging to x, let I(S,x)={s_{n}/n=1,2,3,…}{x} be a topological space, where each sn is a discrete point and neighborhood U of x is such that where be the set of all I-convergent sequences. Now we consider a mapping

1. f is onto.

For each point x∈X there is a constant sequence S in X such that I(S,x)={s_{n}/n=1,2,3,…}∪{x} that is, there exists I(S,x)X{s}⊂ X* and f((x,S))=x Therefore, f is onto.

2. f is continuous.

Let U be an open set in X and (x′,S)∈f-1(U). Then there is a sequence S in X such that x′∈I(S,x) )={s_{n}/n=1,2,3,…}∪{x} that is, (x′,S)∈ I(S,x)x{s}⊂ X* and f((x′,S))= x′. If (x′,S). is an isolated point, then there is nothing to prove. If (x′,S)=( x,S),then there exists I′∈I such that S_{n}∈U for and hence which is open in I (S, X) and hence open in X*. Therefore, f^{-1}(U) is open in X*. Therefore, f is continuous.

3. f is quotient.

Suppose U ⊂ X and f^{-1}(U) is open in X*. If x_{0}∈U and {S_{n}} is I-convergent to x_{0} in X, then which is an open neighborhood of in Then as a subset of x{S}, there exists I′∈I such that for and so for . Hence U is I-sequentially open and thus open. Therefore, f is a quotient mapping.

Here I(S_{n},x) is not second countable, when the set of all finite subsets of is a proper subset to I and not a Hausdorff space, when the set of all finite subsets of properly contains I. Hence for such I, I(S_{n},x) is not a metrizable space and hence its topological sum.

In Proposition 2 in [1], is mentioned as a metric space. But it is true only when I is the set of all finite subset of .

Also, the following Proposition 3 is not true in general.

**Proposition 3**

Every I-sequential space X is a quotient of some metric space.

Suppose Proposition 3 is true. Then by Proposition 2 [1], X is an I-sequential space if it is a quotient image of a metric space. And hence I-sequential space and sequential space coincide, by Corollary 1.14 in [23] which is a contradiction to Example 2 in [25].

**Lemma 2.5**

Let X be an I-sequential space. If f: X→Y preserves I-convergence sequence, then f is continuous.

Proof. Let U be an open subset in Y. Since X is an I-sequential space, it is enough if we prove that f^{-1}(U) is an I-sequentially open subset of X.

Let {X_{n}}be a sequence in X which I-converges to x∈f^{-1}(U) By our assumption, {f(x_{n})}I-converges to f(x)∈U Since U is open in Y, for some for . This implies for . and hence f^{-1}(U) is an I-sequentially open subset of X. Therefore, f is continuous.

**Theorem 1**

Let X* and Y* be the spaces obtained from X, Y as in Proposition 2. Then the following hold

1. X* is an I-sequential space

2. (XxY)* is homeomorphic to X*xY*

3. X*xY* is an I-sequential space.

**Proof 1**

Let U be an I-sequential open subset of X* and (x′,S)∈U where S is an I-convergent sequence {x_{n}}in X with its limit x and x∈S This implies (x_{n}, S) I-converges to (x, S) in X*. If (x′,S)= (x_{n},S) for some n, then there is nothing to prove. If then since U is an I-sequentially open subset of X*. This implies is open in X* which is contained in U. Therefore, U is open in X*. Therefore, X* is an I-sequential space.

**Proof 2**

Let be defined by where S is an I-convergent sequence with its limit in XxY and (x,y)∈S. Since projection mappingsπ_{1},π_{2} are continuous and by Proposition 1 in [1], π_{1}(s) and π_{2}(s) are I-convergent sequences in X and Y. respectively. Therefore, f is well defined.

(a) f is one-one and onto

Since product of I-convergent sequence is again I-convergent sequence, f is onto. Also, we easily check that f is one-one.

(b) f is continuous

Clearly, f preserve I-convergence and by Theorem, f is continuous.

(c) f ^{-1} is continuous

Let U be an open set in (XxY)* and If y′ and y′′ are the limit point of S′ and S′′, respectively, then S′× S′′ I-converges to ( y′, y′′). This implies that there exists i∈I such that for all Now let is open in X* and is open in Y*. This implies For are not a limit point, we can easily find an open set in. Therefore is open in X*xY*.

Therefore, (XxY)*is homeomorphic to X*xY*.

3. By (a), (XxY)*is an I-sequential space. By(b) and Proposition 2.1 in [1], X*xY* is an I-sequential space.

By Lemma 3 and Definition 6, we have the following Theorem 2.

**Theorem 2**

Let X be a topological space. Then the following are equivalent:

(a) X is an I-sequential space.

(b) Every I-sequentially open subset of X is open.

(c) Every I-sequentially closed subset of X is closed.

**Theorem 3**

Each I-sequentially open (closed) subspace of an I-sequential space is I-sequential.

Proof. Let X be an I-sequential space. Suppose that Y is an I-sequentially open subset of X. By Theorem 2, Y is open in X. Let U be an arbitrary I-sequentially open subset in Y and let {xn} be a sequence in X I-converges to x∈U Then x∈Y and since Y is an I-sequentially open subset of X. there exists I′∈I such that

Then by Remark 1.12, is an I-convergent subsequence in Y. Since U is an I-sequentially open subset in Y , there exists I′′∈I such that is in U . That is, is in U. Therefore, U is I-sequentially open in X . and hence open in X . This implies U . is open in Y, since Y is open in X . Therefore, Y is an I-sequential space.

If Y is an I-sequentially closed subset of X , then by Theorem 2, Y is closed in X . Let F be an I-sequentially closed subset in Y and let {x_{n}} be a sequence in F I-convergent to x Since Y is closed, x∈Y and hence x∈F. Therefore, F is an I-sequentially closed set in X and hence it is closed in X . Since Y is closed in X, F is closed in Y.

**Proposition 4**

Every locally I-sequential space X is I-sequential.

Proof. Let U be an I-sequentially open set in X . Since X is a locally I-sequential space, there exists an I-sequential neighborhood V of x With out loss of generality, assume that V is open and an I-sequential space. Since U is I-sequentially open, U ∩V is an I-sequentially open subset of V . Since U is an I-sequential space, x∈U∩V is open in V and hence in X which is contained in U. Therefore, U is open and hence X is an I-sequential space.

In the rest of the section, the space X* and the mapping φ= f: X* X are as in Proposition 2.3.

**Theorem 4**

The product of two I-sequential spaces X and Y is I-sequential iff φ_{X} xφ_{Y} is a quotient mapping.

Proof. Suppose φ_{X} xφ_{Y} is a quotient mapping. By (c) in Theorem 1 , X*xY*is I-sequential. By Proposition 2 in [1] and by assumption, XxY is I-sequential.

Conversely, suppose that XxY is I-sequential. Then is a quotient mapping (in proof of Proposition 3 ) and f-1: X*xY*→(XxY)*is also a quotient mapping (defined in proof of Theorem 1 (b) ). Since composition of two quotient mapping is again quotient, is a quotient mapping.

Since a s-sequential space is not closed under Cartesian product like a sequential space, naturally, one can arise a question that “ Is subspace of a s-sequential space, s-sequential?”. The answer is not as shown by the following Example 2.

**Example 2**

Let S_{n} be the s-convergent sequence with its limit an for eachnthat is and be a s-convergent sequence with its limit x. Let X be the topological sum of the sequence S_{n}, n= 1,2,3,… and. Since each S_{n}, n= 1,2,3,… and is s-sequential, X is s-sequential, by Proposition 1.

Now let Y be a space obtained from X by identifying a_{n} to x_{n}.

Let f: X→Y be the natural mapping. By Theorem 4 in [25], Y is a s-sequential space but not a s-Frechet Urysohn space.

Let W be a weak base consists of the following three types of collection of weak neighbourhood of x′

If x′≠x then

If x′=x, U∈T_{x} then implies that

is a non − thin subsequence S when .

Then W is a weak base of Y. But for all erefore, Y is not a s-Frechet Urysohn space by Theorem 4 in [25]. Now let

Then {x}is s-sequentially open in Y′. Suppose{x}is not a s-sequentially open. Then there is a sequence in Y′ s-converges to x. Then we get a contradiction to Y is not a s-Frechet Urysohn space.

Let A be a subset of Y with

If x_{n}∈A, for all n∈N′ where N′ is a non-thin subsequence of{X_{n}} then s-converges to x, by Remark 1.2 (3) in [25].

If x_{n}∈A, for all n∈N′ where N′ is a thin subsequence of Then A∩S_{n} is a non-thin subsequence of Sn for all n∈N′′ where N′′ is a non-thin subsequence of

Suppose A∩S is a non-thin subsequence of S, then A∩S itself is a s-convergent sequence.

Suppose A∩S is a thin subsequence of S, then S\A is again s-convergent to x. Now we form a one-to-one function f between and N′′ by nth element in goes to n^{th} element in N′ Do the same between S\A and goes to that is, m^{th} element of Then is a sequence in A s-converges to x. Suppose not is a non-thin subsequence of . Now we form an open set U′ from U as follows:

If is non-dense in S_{n} where then eliminate the elements of S_{n} from U to get U′ .

If is dense in S_{n} where and is a non-thin subsequence of S_{n} , then eliminate the element x_{n,m} from U to get U′ where. Then is dense in Sn, since both and are non-thin subsequences of S_{n} and S_{f(n)} respectively.

Now which is a non-thin subsequence of . Therefore, S\A is not a s-convergent sequence which is a contradiction.

Therefore, is a sequence in A s-convergent to x and hence Y is s-Frechet Urysohn space which is a contradiction. Therefore, {x} is s-sequentially open in Y′ but not open in Y′. Therefore, Y′ is not s-sequential.

Next we see the necessary and sufficient condition for which a subspace of an most general case I-sequential space is I-sequential.

**Theorem 5**

A subspace Y of an I-sequential space X is I-sequential iff is a quotient.

Proof. Let and and let be a mapping as in the proof of Proposition 3. Suppose φ′ is a quotient mapping and let be open in Y*where U⊂Y.

Since φ′ is quotient, it is enough if we prove that is open in Y′.

Let where Suppose ^{1}(U), then there is nothing to prove.

Suppose and

Let Since I(S,x), is open in X* and is a subspace of is open in

Now let

Then {Y_{n}}in Y I-converges to x∈U which implies (Y_{n},S′) I-converges to That is

Now let Then is open in. Therefore, is open in and Therfore is open in Y′ and hence U is open in Y. Therefore, φ_{y} is a quotient mapping. By Theorem 2 and Proposition 1 in [1], Y is an I-sequential space.

Conversely, let Y be an I-sequential space. By Proposition 3, φ_{y} is a quotient mapping. Now let is open in Y′ where U⊂Y Since is open in Y*, U is open in Y. Therefore, φ′ is a quotient mapping [26,27].

Finally, we give some easy propositions whose proofs are omitted.

**Proposition 5**

The continuous open or closed image of a I-sequential space is I-sequential.

**Proposition 6**

If the product space is I-sequential, so is each of its factors.

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