IK -SEQUENTIAL TOPOLOGY Текст научной статьи по специальности «Математика»

URAL MATHEMATICAL JOURNAL, Vol. 9, No. 2, 2023, pp. 46-59

DOI: 10.15826/umj.2023.2.004

-SEQUENTIAL TOPOLOGY

H. S. Behmanush, M. Kücükaslan^

Mersin University, Science Faculty, Department of Mathematics, 33100 Yenisehir, Mersin, Turkey

[email protected] [email protected]

Abstract: In the literature, I-convergence (or convergence in I) was first introduced in [11]. Later related notions of I-sequential topological space and I *-sequential topological space were introduced and studied. From the definitions it is clear that I*-sequential topological space is larger(finer) than I-sequential topological space. This rises a question: is there any topology (different from discrete topology) on the topological space X which is finer than I*-topological space? In this paper, we tried to find the answer to the question. We define IK-sequential topology for any ideals I, K and study main properties of it. First of all, some fundamental results about IK-convergence of a sequence in a topological space (X, T) are derived. After that, IK-continuity and the subspace of the IK-sequential topological space are investigated.

Keywords: Ideal convergence, IK-convergence, Sequential topology, IK-sequential topology.

1. Introduction

The notion of convergence of real or complex valued sequences was generalized using asymptotic density and was called statistical convergence by Fast [7] and Steinhause [20] in the same year 1951, independently. After some years P. Kostyrko, T. Salat, W. Wilczynki [11] gave a generalization of statistical convergence and called it as ideal convergence (or converges in ideal). Various fundamental properties (convergence in I and I*) were investigated. Later B.K. Lahiri and P. Das in [12] discussed convergence in I and in I * and investigate some additional results related to mentioned concepts [4, 8-10, 15-17].

The concept of I*-convergence of functions was extended to IK-convergence by M. Macaj and M. Sleziak in [13] in 2011. The authors of [2, 3, 5, 6, 14] gave further properties and results about IK-convergence.

In first part of this paper we introduce IK-sequential topological (seq.-top.) space, which is a natural generalization of I*-seq.-top. space. Later we discuss the IK-continuity of the function and in last two section we write about IK-subspace and IK-connectedness. We will use further the abbreviation T.S. for a topological space.

2. Definition and preliminaries

In this part, we give some known definitions and necessary results.

Definition 1 [7, 20]. Let Ac N, and for m € N let the .set

Am := {x € A : x < m} and |Am| stand for the cardinality of Am. Natural density of A is defined by

p{A) := iim

m^<Xi m

whenever the limit exists. A real sequence x = (xj) is said to statistically converges to x0 if for any

e> 0,

P({n : |xi - Xo| > e}) =0

holds.

Definition 2 [11]. Let I be any subfamily of P(N), with P(N) being the family of all subsets of N. Then, I is called an ideal on N if the following requirements hold:

(i) finite union of sets in I is again in I;

(ii) any subset of a set in I is in I.

I is admissible if all singleton subsets of N belong to I. The ideal I is non-trivial if I = 0 and I = P (N). A non-trivial ideal I is called proper if N is not in I.

The family of finite subsets of the N is an admissible non-trivial ideal denoted by Fin and the family of the subsets of N with natural density zero is also an admissible non-trivial ideal denoted by Ig. The set of all non-trivial admissible ideals will be denoted as NA throughout the study.

Example 1. [11] Consider the decomposition of N as N = UPj where all Pj are infinite subsets of N and are mutually disjoint. Take the family

I = {N C N : N intersect only finite number of Pjs}.

Then, I belongs to NA.

Definition 3 [19]. Assume F C P(N). The collection F is a filter on N if

(i) a finite intersection of elements of F is in F and

(ii) if CeFACCD, then DeF.

If empty set is not in F then F is proper. If I e NA then the collection

F = {N C N : NC e I}

is a filter on N. It is known as the I-associated filter.

Definition 4 [21]. In a T.S. (X, T) a sequence X = (xj) C X is called to converging in I to a, point x e X if

{i e N : xj e u} e F(I)

holds for each neighborhood u of x. The point x is referred to as the ideal limit of the sequence x = (xj) and it is represented by xj — x (or I — lim xj = x).

Remark 1.

(i) Statistical and Ig — convergence are coincide.

(ii) Classical convergence and Fin—convergence are coincide.

Lemma 1 [1]. Assume that I,Ii and I2 be ideals on the set N and consider a T.S. (X, T), then

1. If I e NA, then every convergent sequence is I-convergent sequence which converges to same point.

2. If I1 C I2 and (xj) CX is a sequence which xj —1 x, then xj ^ x.

3. If X the Hausdorff space, then the limit of every convergent sequence is unique.

3. Z -convergence of sequence

In this part we will investigate some results related to ZK-convergence of sequences which is a generalized form of Z*-convergence of sequences. If we consider Fin instead of K, then we will have Z*-convergence.

Definition 5 [6]. In a T.S. (X, T) a sequence x = (x) C X is called to be Z*-converging to x0 € X if 3M € F(Z) s.t. the sequence

= i Xi, i € M, yi :=\ x, i/M

is Fin convergent to x.

That is, for each neighborhood u of x,

{i € N : yi eu}e F (Fin),

or So,

This implies that Therefore,

{i € M : yi / u} U {i € MC : yi / u} € Fin. {i € M : xi / u} U {i € MC : x / u} € Fin. {i € M : yi / u} € Fin. {i € M : yi € u} € F (Fin).

It is clear that this definition is the same as the definition given in [6]. In the definition of Z*-convergence of sequence if we consider an arbitrary ideal K instead of the ideal Fin then it yields the definition of ZK-convergence of a sequence. That is, ZK -convergence is the generalized form of Z*-convergence.

Definition 6 [13]. Let Z and K stand for the ideals of N and consider a T.S. (X, T). The sequence x = (xi) C X is ZK-convergent to a point x € X if 3M € F(Z) s.t. the sequence

yi

xiy i € M, x, i/M,

IK

K-converges to x. We represent it as ZK — lim(xi) = x or xi ^ x .

Definition 7. Let Z and K stand for the ideals of N and (X, T) represent a T.S. Consider the sequences x = (xi) C X and y = (yi) C X. Define a relation as

x y ^ {i : xi = yi} € I.

The relation is an equivalence relation. That is,

1. V x = (xi) CX, {i : xi = xi} = 0 € I ^ x x.

2. Let x y. Since {i : yi = xi} = {i : xi = yi} € I, then y x.

3. Let x y and y z. Then, A := {i : xi = yi} € F(I) and B := {i : yi = zi} € F(I). So, {i : xi = zi} = A n B € F (I). Hence, x i holds.

Lemma 2. Let I and K stand for the ideals of N and consider the T.S. (X, T) and the sequences

IK

x = (xj) C X. Assume xj — x for any x € X and i = (ij) C X is a sequence s.t. x i. Then, I k

the sequence tj — x.

IK

Proof. Let xj — x, then 3M € F(I) s.t. the following sequence

Jxj, i € M, yj = \ x, i € M

is K-convergent to x. Since (xj) (tj). So Vi € M, xj = tj. Therefore, the following sequence

'tj, i € M,

□

yj \ x, i € M

IK

is K-convergent to x which shows that tj — x holds.

The Definition 7 gives the possibility that the definition of IK-convergence of a sequence can be rewritten as follows:

Definition 8. Let I and K stand for the ideals of N and consider the T.S. (X, T). A sequence x = (xj) C X is IK-convergent to the point x € X if there exist a sequence t = (tj) C X s.t. x i

and U K x holds.

In the following lemma we demonstrate that Definition 6 and Definition 8 are equivalent for any ideals I and K and for any T.S. (X, T).

Lemma 3. Let I and K stand for the ideals of N and consider the T.S. (X, T) and

IK ~ ~ K

x = (xj) C X be a sequence. Then, xj — x iff 3t = (tj) C X s.t. x t and tj — x hold.

IK

Proof. Let xj — x holds. Then, 3M € F(I) s.t. the following sequence

Jxj, i € M, yj = \ x, i € M

is K-convergent to x. Let us chose (tj) = (yj) Vi € N. Then, the proof will complete if we show that x y.

Consider the fact {i € N : xj = yj} = {i € M : xj = yj} € F(I). Hence, x i. Conversely, let x = (xj) and t = (tj) be sequences s.t. x i and tj — x hold. Since x i, then

M = {i € N : xj = tj} € F(I)

holds. Define a sequence

Jxj, i € M, yj = \ x, i € M.

Since xj = tj hold Vi € M, then we can write

. = ixj, i € M, j \ x, i € M.

Because i = (tj) is K-convergent to x, the sequence y = (yj) is also K-convergent to x. Hence, the sequence x = (xj) is IK-convergent to the point x and this completes the proof. □

4. Zr-seq.-top. space

In this section, we are going to define a new topology on the X using the ideal Z and K and investigate some properties of the new T.S. This topology will be an extended version of the Z*-seq.-top. space which was discussed in [18]. If we take Z = Fin, then Zr-seq.-top. space is coincide with Z*-T.S.

Definition 9. Let Z and K stand for the ideals of N and consider the T.S. (X, T). Then r

1. A set F C X is Zr-closed, if for each (xi) C F with xi ^ x, then x € F.

2. A set V C X is Zr-open, if its complement VC is Zr-closed.

Remark 2. Consider the T.S. (X, T). An O C X is Zr-open iff each sequence in X — O has Zr-limit in X — O.

Proof. The proof is evident from Definition 9. Therefore, it is omitted here. □

Definition 10. Let I and K stand for the ideals of N and consider the T.S. (X, T). For any

_xK-

subset A С X define a set A (it is called -closure of A) by

_iK jk

A := (x е X : 3(xj) С A, xi ^ x}.

_T/C -Tk'- —TK:

It is clear that 0 = 0, X = X, and А С A holds \M С X.

r- —X>z

Remark 3. A subset С of the T.S. X is XK closed set iff С = С.

Proof. Proof is obvious from the Definition 10. So, it is omitted here. □

Lemma 4. Let Z and K stand for the ideals of N and let (X, T) represent a T.S. For any subset A C X, Zr-closure of A is Zr-closed.

Proof. We must show that

It is clear that

-IK

_-tk. _TK

(A ) =A

A с (A1 )

_j/c - _2>c jK, _2>c

Let. x £ (A ) . Then, there exist a sequence (xi) C A s.t. xi y x holds. Since (x^ C A , then there exist sequences (x™) C A s.t. xf ^ xi. Therefore there exist the sets Mn € F(Z) s.t.

{i € Mn : xf/ un} € K

for each neighborhood un of xi. Choose mi the i where xl is belonging to neighborhood u1 of xi, similarly m2 the i where x2 is belonging to neighborhood u2 of x2. If we continue this process and take mp the i where xpp is belonging to neighborhood un of xp. The obtained sequence (xmp)

IK jK

belongs to A. The theorem will be proved if we show that xm ^ x. Since xi ^ x, so 3M € F(Z)

s.t. the sequence

= xi, i е M, к Vi = i x, i/M, yi ^ x.

к

So,

{i € M : xj / u} € K

for each neighborhood u of x. Now,

{i € M : un C u} C {i € M : xj / u} € K.

Therefore,

{i € M : un C u} € K

and

{i € M : xmp / u} C {i € M : un C U} € K

j/C _j/C

hold. So, —>• x and x £ A . □

Definition 11. Let I and K stand for the ideals of N and (X, T) represent a T.S. Then, for A C X, IK-interior of A is defined as

TIC _Tic.

\o-

:=A-(X-A ).

Proposition 1. Let V be a subset of T.S. X, then V is IK-open iff VoI = V. Proof. Let V be an IK-open set. Then, X — V is IK-closed set and

clI k (X —V) = X —V

holds. So, we have

VoI = V — (X — V) = V.

Conversely assume that

IK

VoI = V

holds. From the definition of IK-interior of V we have

IK

V = V-(X-V ).

Hence,

V n x — v = 0.

Consequently

X — VIK C X — V.

Thus,

X — V = X — V

is satisfied. Therefore, X — V is IK-closed and V is IK-open. □

Definition 12 [21]. A sequence (xj) in a T.S. X is I-eventually in a subset A of X if

{i € N : xj € A} € F(I).

K

K

Definition 13. Let I and K stand for the ideals of N and consider the T.S. (X, T). A sequence x = (xj) C X is IK-eventually in a subset V of X. If there exist a sequence y = (yj) C X s.t. y x and y is K-eventually in V.

In the next theorem, we will provide a sequence characterization of IK— open set.

Theorem 1. Let I and K stand for the ideals of N and consider the T.S. (X, T). A subset u of X is

IK

-open iff each IK-convergent sequence to x0 € u is IK-eventually in u.

r r -

Proof. Let u is I -open. Then, X — u is IK-closed and X — u = X — u holds. Let

IK

x = (xj) C X be a sequence s.t. xj — x and x € u. Then, 3M € F(I) s.t. the sequence

, = ixj, i € M, j \ x, i / M

is K-convergent to x. Since u is a neighborhood of x, then we have

H = {i € N : xj / u} € K.

If we choose yj = tj, then

{i € N : yj = xj} = {i € N : tj = xj} = M € F(I)

holds. So, (yj) (xj) holds and (yj) is eventually in u. Conversely, let x = (xj) C X is a sequence which is

IK

-convergent sequence to a point x u and

r r -

it is IK-eventually in u. Assume that u is not I -open subset of X. So there exists xo € X — u

which x0 / X — u. This means that there exists a sequence (xj) C X — u which is IK-convergence

to x0 € u. So, (xj) is IK-eventually in u.

Therefore, 3y = (yj) C X which x y and y is K-eventually in u. This implies that y is

K-eventually in u which is not in case. □

Theorem 2. Let I and K stand for the ideals of N and consider the T.S. (X, T). A subset C C X is IK-closed iff

C = n{A : A is IK— closed and C C A}.

P r o o f. Let

C = n{A : A is IK— closed and C C A}.

K IK

Let x be any element of IK-closure of C. Then there exists (xj) C C s.t. xj y x. Let x / C so

x / n{A : A is IK— closed and C C A}.

This implies that 3IK-closed subset F of X s.t. x / A, but C is IK-closed and it is a subset of A, which is a contradiction.

The converse is obvious. □

Theorem 3. Let I and K be ideals of N and (X, T) be a T.S. A function clIK : P(X) — P(X) —

defined as clIK (A) = A is satisfying Kuratowski closure axioms (K1) clIK (0) = 0 and clIK (X)= X,

(K2) A C clXK (A) VA CX,

(K3) cljk (A) = cljK (clXK (A) VA C X,

(K4) clxK (A U B) = clxK (A) U clxK (B) VA, B C X.

Proof. (K1) and (K2) are clear from the definition of Zr-closure function. By Lemma 4,

cljK(A) is closed. So, cljK(cljK(A)) = cljK(A). Therefore, (K3) holds.

To prove (K4), let x € clXK(A) U clXK(B). Then, x € clXK(A) or x € clXK(B). Without lost

IK

of generality assume that x € clIk(A). So, 3(xi) C A s.t. xi y x. Therefore, 3(xi) C A U B s.t. IK

xi ^ x. So, x € cljK (A) U cljK (B).

I K

Conversely, let x € clIK (A U B). Then, there exist a sequence (xi) C (A U B) s.t. xi y x. Assume that x / cIik (A) and x / cIik (B). So, neither set A nor set B contains a sequence s.t. Zr-converges to the point x. Consequently, there is not any sequence in the A U B which is convergent to x. But x € clIK (A U B) which is a contradiction. Hence,

clIk (A U B) = clIk (A) U clIk (B)

holds. □

Corollary 1. A subset A of X is Zr-closed iff cIik (A) = A and a subset O C X is

Zr

-open

iff X — O is Zr-closed.

Theorem 4. Let Z and K stand for the ideals of N and consider the T.S. (X, T). Then,

7Ik := {A C X : clIK (X — A) = X — A}

is a topology over the set X.

Proof. By (K1), it is clear that X € 7Jk and 0 € 7Jk hold. Let A,B € 7Jk be arbitrary sets. To prove A U B € 7Jk we must to prove that

X — A U B = clIK (X — A U B)

holds. By (K2), we have

X — A U B C clIk(X — A U B).

Now, let x € cIik(X — A U B) be an arbitrarily element. Then, 3(xi) C X — (A U B) s.t. it is Zr-convergent to x. This implies that (xi) is not subset of A U B. So, (xi) is neither subset of A nor subset of B. Therefore, (xi) C X — A or (xi) C X — B which Zr-converges to point x. So, x € cljK(X — A) or x € cljK(X — B). Since X — A and X — B are closed sets, then

x € (X — A) U (X — B) = X — A U B

holds.

Let {Ai} be a collection of Zr-open subsets of X. Then, clIK (X — Ai) = X — Ai Vi € N. By considering (K2), we have

Hien(X — Ai) C cljK ( HneN (X — Ai)).

Let x € cljk nneN (X — Ai) be an arbitrary element. Then, 3(xi) C nneN(X — Ai) which is Zr-convergent to x. Then, (xi) C (X — Ai) Vi € N. Since X — Ai are closed sets, then x € X — Ai Vi N. Therefore,

x € Hi€N(X — Ai).

Hence, the set 7Jk is a topology and (X, 7Jk) is a T.S. □

Definition 14. The T.S. (X, 7jk) is called as IK-sequential T.S. For abbreviation we will show it by IK-seq.-top. An IK -seq.-top. (X, 7jk) is said to be IK-discrete space if 7jk = P(X).

Theorem 5. Let I, K, I\, Ki, I2 and K2 stand for ideals of N and (X, T) represents a T.S. Let Ii C I2 and K C K2. Then,

1. Txk2 x 71% ,

2. 71 K x 71K .

Proof. Let u be

any IK2 -open subset of X. Then, X—u is IK2-closed and clXK2 (X—u) = X—u hold. To prove u is IKl-open subset ofX, we will show that

cl1K1 (X — u) C X — u.

!K1

Let x € cljk1 (X — u) be any point. Then, there exists (xj) C X — u s.t. xj ^ x. Since

Ki C K2, then by Proposition 3.6 in [13], Xj ^ x. So, x € clXK2(X — u). Therefore, x € X — u. Hence X — u is IK2-closed set and u is IK2 -open subset of X.

The second one can be proved by using the fact that if Ii C I2, then,

x j r x

implies x j ^ x,

it easily can be proved. □

Theorem 6. Let I and K stand for the ideals of N and (X, T) represent a T.S. Then, every I*-open set is IK-open set.

Proof. If we take K = Fin then I*-open set will be IK-open set. □

Theorem 7. Let I and K stand for the ideals of N and (X, T) represent a T.S. Then, every IK-open set is K-open set.

Proof. Let u be an arbitrary IK-open subset of X. Then, X — u is IK-closed and

cljK (X — u) = X — u.

To prove u is K open, it is sufficient to show that X — u is K-closed, i.e,

X -v = X -vK.

-K T , ^r;-K

It is clear that X — v C X — v . Let x € X — v be an arbitrary element s.t. 3(xj) C X — v satisfying xj x.

Then, by Lemma 3.5 in [13] we have xj ^ x . So, x € cljK (X — u) = X — u. Hence, the theorem proved. □

Proposition 2. Let I and K stand for the ideals of N and (X, T) represent a T.S. Then, the following statements are true:

1. If Kc I, then, each I-open set is -open set.

2. If the space X is a first countable space and the ideal I has additive property with respect to K (see Definition 3.10 in ¡13]), then, each 1K-open set is I-open set.

3. If I c K, then every K-open set is 1K-open.

Proof. The proof is obvious from Proposition 3.7 and Theorem 3.11 of [13]. □

5. Ir-continuity of functions

In this section we will define Ir —continuous and sequential Ir-continuous functions. We will prove that in any Ir-sequential T.S. these two concepts coincide. Also, we will state some theorems that give the definition of Ir-continuous function in different words and ways. At the end of this section we will see that the combination of Ir-continuous functions is Ir-continuous.

Definition 15. Let I and K stand for the ideals of N and (X, 7Jk) (Y, 7^K) represent Ir-seq.-top. spaces. A function f, from X to Y is said to be

(i) Ir-continuous which provides that inverse image of any Ir-open subset of Y is Ir -open in X.

r IK jK

(ii) Sequentially Ir-continuous which provides that f (xi) — f (x) V(xj) C X with xi — x.

Theorem 8. Let I and K stand for the ideals of N and (X, 7Jk) (Y, 7^K) represent Ir-seq.-top. spaces; and f, from X to Y be a function. Then, f is Ir-continuous iff it is sequentially Ir-continuous.

Proof. Let f be an Ir-continuous function. Then, inverse image of any Ir -open subset of r jK

Y is Ir-open subset in X. Let (xi) C X be a sequence with xi — x. Then, there exists M € F(I)

s.t. the following sequence

'xi, i € M,

U : \x, i/M is K-convergent to x. That is, for each neighborhood u of x we have

{i € N : ti € u} € F(K).

Let V be any I -open neighborhood of f (x). Then, f-1(V) is IK -open subset of X which contains the point x. So, it is a neighborhood of x. Therefore,

{i € N : ti € f-1(V)} € F(K),

implies that {i € N : f (ti,) € V} € F(K). Hence, the sequence

f (t ):= if (xi), i € M,

f (ti ):=\ f (x), i/M

jK r

is K-convergent to f (x). So, f (xi) ^ f (x). Hence, f is sequentially Ir-continuous function.

Conversely, let the function f be sequentially Ir-continuous and u is any IK -open subset of Y. Assume that f-1(u) is not Ir-open subset of X. Then, X — f-1(u) is not Ir-closed subset of X. So,

-1 jK -1

3(xi) CX — f 1(u) s.t. xi ^ x and x/ X — f 1(u), I k

i.e. xi / f -1(u) Vn and xi x which means x € f-1(u). Since f is Ir-sequentially continuous

IK

function then f (xi) — f (x). So, f (x) € u and f (xi) / u Vn. This is a contradiction. □

Lemma 5. Let I and K stand for the ideals of N and (X, 7Jk) (Y, 7^K) represent Ir-seq.-

top. spaces and f, from X to Y be an Ir-continuous function. If (yi) C Y be a sequence s.t.

IK -1 IK -1

yi — y, then f 1 (yi) — f 1(y).

ir 1

Proof. Let f be an I -continuous function. Let yj — y then 3M € F(I) s.t. the sequence

'Vi, i € M,

Sn = \ y, i/M is K-convergent to y. So, for each neighborhood u of Y,

{i € N : yj € u} € F(K).

Since f is IK-continuous function, then inverse image of any IK — open set in Y is IK-open in X, f-i(u) is open neighborhood of x in X. Then

{i € N : f-i(yj) € f-i(u)} € F(K).

Therefore,

f-i(s«) = i '"->>• I€M

f-i(y), i/M,

1K

is K-convergent to f-i(y) and hence f-i(yj) — f-i(y). □

Theorem 9. Let I and K stand for the ideals of N and (X, 71k ) (Y, 7^K) represent IK-seq.-top. spaces. Then the function f, from X to Y is IK-continuous iff

clx K (f-i(B) = f "i(clIK (B)

holds VB C Y.

Proof. Assume that function f, from X to Y is IK-continuous function. Let

x € clxk(f-i(B)).

-1 i k K

Then, 3(xj) C f i(B) s.t. xj - x. Since f is I -continuous so,

1K

f (xj) — f (x).

In another hand (xj) C B, so f (x) € cl1K(B) and x € f-i(cl1K(B)).

1K

Now, let x € f-i(cl1 k(B)), i.e. f(x) € cl1K(B). Therefore, 3(yj) C B s.t. xj — x. Then,

1K

by Lemma 5 there exists (xj) = (f-i(yj) C f-i(B) s.t. xj - x, where x = f i(y) holds. So, x € cl1 k(f-i(B)). Hence,

cl1 k (f-i(B ) = f-i(cl1 K (B).

Conversely, let

cl1K(f-i(B) = f-i(cl1K(B), VB € P(Y). Let u be IK -open subset of Y then

cl1 K (Y — B) = Y — B.

Let B = Y — u, then

cl1 K (f-i(Y — u)) = f-i(cl1 K (Y — u)) = f-i(Y — u). This shows that f-i(Y — u) is IK-closed. Hence, the following equality

f-i(Y — u) = X — f-i(u) implies that X — f-i(u) is IK-closed. Therefore f-i(u) is IK-open set. □

Corollary 2. Let I and K stand for the ideals of N and (X, 7Jk ) (Y, 7^K) represent Ir-seq.-top. spaces. A function f, from X to Y is Ir-continuous iff

intjK (f-1(B) = f-1(intjk (B) VB CY.

Definition 16. Let I and K stand for the ideals of N and (X, 7Jk ) (Y, 7^K) represent Ir -seq.-top. spaces and f, from X to Y be a function. The function f is Ir-continuous at a point x € X if inverse image of any neighborhood of f (x) is a neighborhood of x in X.

Corollary 3. Let I and K stand for the ideals of N and (X, 7Jk ) (Y, 7^K) represent Ir-seq.-top. spaces. Then, the function f, from X to Y is Ir-continuous iff it is Ir-continuous at every point x X.

Definition 17. Let I and K stand for the ideals of N and (X, 7Jk) (Y, 7^K) represent Ir-seq.-top. spaces and f, from X to Y be a function, f is said to be Ir-closure preserving if

f (cljK (A)) = cljk (f (A) VA CX. Theorem 10. The function f, from X to Y is Ir-continuous iff it is Ir-closure preserving. P r o o f. Let f : X — Y be an

Ir

-continuous function. Then, for any subset B of Y

clj k (f-1(B ) = f-1(clj k (B)

holds. Consider a set A C X s.t. f (A) is subset of Y. So,

clj k (f-1(f (A)) = f-1(clj k (f (A))

holds and it implies that f (cljK (A)) = cljK(f(A)) VA C X holds. Conversely, let f be Ir-closure preserving function, then

f (cljK (A)) = clj k (f (A)) VA CX.

Let u be any subset of Y, then f-1(u) is subset of X and

f (cljK (f-1(u))) = cljK (f (f-1(u) = cljK (u)

holds. So

clj K (f-1(u) = f-1(clj K (u) and by Theorem 9 the function f is Ir-continuous. □

Theorem 11. Let X, Y and Z be Ir -seq.-top. spaces. Let f, from X to Y and g, from Y to Z be Ir-continuous functions. Then g o f : X — Z is Ir-continuous functions.

P r o o f. Let u be any Ir-open subset of Z. Since g is Ir -continuous function then g 1(u) is Ir-open subset of Y and because f is Ir-continuous function therefore f-1(g-1(u)) is Ir-open subset of X hence (g o f )-1(u) is Ir-open subset of X. □

6. Subspace of IK-seq.-top. space

In this section subspaces of the IK-seq.-top. space and its properties under an IK-continuous function will be discussed.

Definition 18. Let (X, 71k) be an IK-seq.-top. space and Y C X. Then

CY : P (Y) - P (Y), CY (A) = Y n cl1K (A) is a Kuratowsky operator. Define a T.S. as (Y, ty), where

TY = {U nY ,Y€ 71K } CP (Y). This T.S. is called IK-subspace of X.

Lemma 6. Let Y be an IK-subspace of IK-seq.-top. space X. If set A is IK-open subset of Y and Y is an IK-subset of X. Then A is IK-open subset of X.

Proof. Let A be IK-open subset of Y. Then 3U € 71k s.t. A = Y n U. Since Y is an IK-open subset of X. Then A € 71k . □

Proposition 3. Let (X, 71k) and (Y, 7^K) be IK-sequential spaces, f : X — Y be IK-continuous function and A C X is IK -subspace of X. Then f/A : A — Y, the restriction f over A is IK-continuous function.

Proof. Let U be an IK-open subset of Y. Since f is IK-continuous function then f-i(U) is IK-open subset of X. That is f-i(U) € 71k.

In other hand f-4i(U) = A n f-i(U). So f-4i(U) is I -open subset of subspace A. Hence f/A is IK-continuous function. □

Lemma 7. If A is -subspace of -sequential T.S. X. Then the inclusion map j : A ^ X is -continuous.

Proof. If U is I -open in X then j (U) = U n A is IK -open in subspace Y hence j is

IK -continuous. □

Proposition 4. Let (X, 71k) and (Y, 7^K) be IK-sequential spaces, B C Y be subspace of Y and f : X — B be IK-continuous function. Then, h : X — Y obtained by expanding the range of f is IK-continuous.

Proof. To show h : X ^ Y is IK -continuous function, if B as subspace of Y then note that h is the composition of the map f : X ^ B and j : B ^ Y. □

7. Conclusion

In this article we defined the notion of Ir-closed (resp. Ir-open) set in a T.S. (X, T) and established some important results concerning this notion. Furthermore, we defined the Ir-seq.-top., which is a generalized form of the I*-sequential space. We also talked about Ir-continuity of functions and saw that in

Ir

-seq.-top. space the notion of continuity and sequential continuity are the same. And in the last section of the paper, subspace of Ir-sequential space have been studied and some important results established.

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