chapter 3 -continuous functions in ideal topological...

Chapter 3

rpsI-continuous functions in ideal

topological spaces

One of the important and basic concepts in the theory of classical

point set topology is continuous functions [1], [21], [22], [23], [44], [19].

This concept has been extended to the setting of ideal topological

spaces. Abd EI-Monsef et al.[2] introduced I-open sets and I-continuous

functions and investigated their properties. After that Abd EI-Monsef

et al.[3], [4], Devi et al.[9], Ekici[18], Hatir et al.[28], [29], [30], [31], Ke-

skin et al.[36], Khan et al.[37], [40], Noiri et al.[49], Renukadevi et

al.[53] working in the field of general topology have shown more inter-

est in studying the concept of generalizations of continuous functions

via ideals.

In this chapter rpsI-continuous functions, rpsI-irresolute func-

tions, totally rpsI-continuous, rpsI-totally continuous, strongly rpsI-

46

CHAPTER 3. rpsI-continuous functions in ideal topological spaces 47

continuous functions are introduced and also their properties are dis-

cussed.

3.1 rpsI-continuous Functions

In this section, we establish that the composition of two rpsI-

continuous functions need not be rpsI continuous function. Further

some charaterizations of rpsI-continuous function under certain con-

ditions have been discussed.

Definition 3.1.1. A function f : (X, τ, I) → (Y, σ) is called rpsI-

continuous if f−1(V ) is rpsI closed in (X, τ, I) for each closed set V

in (Y, σ).

Definition 3.1.2. A function f : (X, τ, I) → (Y, σ) is called pgprI-

continuous if f−1(V ) is pgprI closed in (X, τ, I) for each closed set V

in (Y, σ).

Theorem 3.1.3. Every continuous function is rpsI-continuous.

Proof. Follows from the fact that every closed set is rpsI-closed.

Remark 3.1.4. The converse of the above theorem need not be true

as seen from the following example.

Example 3.1.5. Let X = Y = {a, b, c, d}, τ = {φ, X, {a}, {b, d},

{a, b, d}}, σ = {φ, Y, {b, c, d}}, I = {φ, {a}} and J = {φ}. Define


a function f : (X, τ, I) → (Y, σ) by f(a) = a, f(b) = b, f(c) = c,

f(d) = d. Then the inverse image of every closed set in Y is rpsI-

closed in X. Therefore f is rpsI-continuous. But f is not continuous

because f−1({a}) = {a} not closed in X and {a} is closed in Y .

Theorem 3.1.6. Every semi preI-continuous function is rpsI-contin

uous.

Proof. It follows from the fact that every semi pre I-closed set is rpsI-

closed.



Example 3.1.8. Let X = Y = {a, b, c, d}, τ = {φ, X, {a}, {b}, {a, b},

{b, c}, {a, b, c}}, σ = {φ, X, {a}, {b, d}, {a, b, d}}, I = {φ, {a}} and

J = {φ}. Define a function f : (X, τ, I) → (Y, σ, J) by f(a) = c,

f(b) = c, f(c) = d, f(d) = c. Then the inverse image of every closed

set in Y is rpsI-closed in X. Therefore f is rpsI-continuous. But f

is not semi pre I-continuous because the subset {c} is closed in Y ,

f−1({c}) = {a, b, d} is not semi pre I-closed in X.

Theorem 3.1.9. Every pgprI-continuous function is rpsI-continuous.

Proof. The proof follows from the fact that every pgprI-closed set is

rpsI-closed.






J = {φ}. Define a function f : (X, τ, I) → (Y, σ, J) by f(a) = a,

f(b) = b, f(c) = c, f(d) = d. Then the inverse image of every closed

set in Y is rpsI-closed in X. Therefore f is rpsI-continuous. But f

is not pgprI-continuous because the subset {a, c} is closed in Y and

f−1({a, c}) = {a, c} is not pgprI-closed in X.

Theorem 3.1.12. Every preI-continuous(resp.αI-continuous, resp.rI-

continuous) function is rpsI-continuous.

Proof. The proof follows from the fact that every preI-closed(resp.αI-

closed, resp.rI-closed) set is rpsI-closed.






f(b) = b, f(c) = c, f(d) = d. Then the inverse image of every

closed set in Y is rpsI-closed in X. Therefore f is rpsI-continuous.


But f is not preI-continuous because the subset {a, c} is closed in Y ,

f−1({a, c}) = {a, c} is not preI-closed in X.






But f is not αI-continuous because the subset {a, c} is closed in Y ,

f−1({a, c}) = {a, c} is not αI-closed in X.






But f is not rI-continuous because the subset {c} is closed in Y ,

f−1({c}) = {c} is not rI-closed in X.

The following theorem gives a characterization of rpsI-continuous

functions.

Theorem 3.1.17. Suppose f : (X, τ, I) → (Y, σ, J) is a function then

the followings are equivalent.

(i) f is rpsI-continuous.


(ii)The inverse image of each open set in (Y, σ) is rpsI-open in (X, τ, I).

(iii)The inverse image of each closed set in (Y, σ) is rpsI-closed in

(X, τ, I).

Proof. Suppose (i) holds. Let V be open in Y . Then V c is closed

in Y . By (i) f−1(V c) is rpsI-closed in X. But f−1(V c) = (f−1(V ))c

which is rpsI-closed in X. Therefore f−1(V ) is rpsI-open in X. This

proves (i)⇒ (ii). The implications(ii) ⇒ (iii) and (iii) ⇒ (i) follows

easily.

Remark 3.1.18. Composition of two rpsI-continuous functions need

not be rpsI-continuous in general as seen from the following example.

Example 3.1.19. Let X = Y = Z = {a, b, c}, τ = {φ, X, {a}, {b},

{a, b}}, I = {φ, {a}}, σ = {φ, Y, {a, b}}, J = {φ, {a}} and η =

{φ, Z, {a}, {a, b}}, K = {φ, {a}} . Define a functions f : (X, τ, I) →

(Y, σ, J) by f(a) = a, f(b) = c, f(c) = b and g : (Y, σ, J) → (Z, η, K)

by g(a) = c, g(b) = a, g(c) = b. Then both f and g are rpsI-

continuous. Let A = {b, c} is closed in Z. Then (g ◦ f)−1(A) =

f−1(g−1({A})) = f−1(g−1({b, c})) = f−1({a, c}) = {a, b} which is not

rpsI-closed in X.


3.2 Diagram

As a consequence of the Theorems [3.1.6, 3.1.9 and 3.1.12], Re-

marks [3.1.7, 3.1.10 and 3.1.13] and Examples [3.1.8, 3.1.11, 3.1.14,

3.1.15 and 3.1.16] the following implication diagram holds.

rpsI-continuous

semi preI-continuous

pgprI-continuouspreI-continuous

αI-continuous rI-continuous

❄

✛

♦✒

✲

In this diagram, A → B means A implies B but B does not imply A.


3.3 rpsI-totally continuous functions

In 1960, Levine [42] introduced stong continuous maps. In 1984,

Noiri [48] introduced and studied perfectly continuous maps. In this

section, the notion of totally rpsI-continuous function, rpsI-totally

continuous function and strongly rpsI-continuous function are intro-

duced and their properties are discussed.

Definition 3.3.1. A function f : (X, τ, I) → (Y, σ, J) is called rpsI-

totally continuous function if the inverse image of every rpsI-open

subset of Y is clopen in X.

Theorem 3.3.2. A function f : (X, τ, I) → (Y, σ, J) is rpsI-totally

continuous if and only if the inverse image of every rpsI-closed subset

of Y is clopen in X.

Proof. Let F be any rpsI-closed set in Y . Then F c is rpsI-open in

Y . By definition, f−1(F c) is clopen in X. But f−1(F c) = (f−1(F ))c

which is clopen in X. This implies f−1(F ) is clopen in X.

Conversely suppose V is rpsI-open Y , then V c is rpsI-closed in Y . By

hypothesis f−1(V c) is clopen in X. But f−1(V c) = (f−1(V ))c which

is clopen in X, which implies f−1(V ) is clopen in X.

Hence the inverse image of every rpsI-open set in Y is clopen in X.

Thus f is rpsI-totally continuous.


Theorem 3.3.3. Every rpsI-totally continuous function is perfectly

continuous.

Proof. Suppose f : (X, τ, I) → (Y, σ, J) is rpsI-totally continuous.

Let U be any open subset of Y . Since every open set is rpsI-open,

U is rpsI-open in Y and also f is rpsI-totally continuous, f−1(U) is

clopen in X. This proves the theorem.



Example 3.3.5. Let X = Y = Z = {a, b, c}, τ = {φ, X, {a}, {b, c}},

I = {φ, {a}}, σ = {φ, Y, {a}}, J = {φ, {a}}. Define a function

f : (X, τ, I) → (Y, σ, J) by f(a) = a, f(b) = b, f(c) = c. Then

the inverse image of every open set in Y is clopen in X. Thus f is

perfectly continuous. But f is not rpsI-totally continuous since the

rpsI-open set {a, b} of Y , f−1({a, b}) = {a, b} is not clopen in X.

The following theorem gives a characterization of rpsI-totally contin-

uous function.

Theorem 3.3.6. Let f : X → Y be a function, where X and Y are

ideal topological spaces. Then the following are equivalent.

(i) f is rpsI-totally continuous


(ii) for each x ∈ X and each rpsI-open set V in Y with f(x) ∈ V ,

there is a clopen set U in X such that x ∈ U and f(U) ⊆ V .

Proof. (i) ⇒ (ii). Suppose f is rpsI-totally continuous and V be any

rpsI-open set in Y containing f(x) so that x ∈ f−1(V ). Since f is

rpsI-totally continuous, f−1(V ) is clopen in X. Let U = f−1(V ), then

U is clopen in X and x ∈ U . Hence f(U) = f(f−1(V )) ⊆ V .

(ii) ⇒ (i). Let V be rpsI-open set in Y . Let x ∈ f−1(V ) be any

arbitrary point. This implies f(x) ∈ V . Then there is a clopen

set f(G) ⊆ X containing x such that f(G) ⊆ V , which implies

G ⊆ f−1(V ) and x ∈ G ⊆ f−1(V ). This implies f−1(V ) is clopen

neighbourhood of each of its points. Hence it is clopen set in X. Thus

f is rpsI-totally continuous.

Theorem 3.3.7. If a function f : (X, τ, I) → (Y, σ, J) is rpsI-totally

continuous then it is continuous.

Proof. Let V be an open set in Y . Then V is rpsI-open in Y . Since

f is rpsI-totally continuous, f−1(V ) is both open and closed in X.

Thus f is continuous.




{b, c}, {a, b, c}} , I = {φ, {a}}, σ = {φ, Y, {a, c}, {d}, {a, c, d}} and


J = {φ}. Define a function f : (X, τ, I) → (Y, σ, J) by f(a) = d,

f(b) = a, f(c) = c, f(d) = b. Then the inverse image of every closed

set in Y is closed in X. Thus f is continuous. But f is not rpsI-totally

continuous since for the subset {d} is rpsI-open in Y , f−1({d}) = {a}

is open but not closed in X.

Theorem 3.3.10. The composition of two rpsI-totally continuous

functions is rpsI-totally continuous.

Proof. Let f : (X, τ, I) → (Y, σ, J) and g : (Y, σ, J) → (Z, η, K) be

any two rpsI-totally continuous functions. Let V be rpsI-open set

in Z. Since g is rpsI-totally continuous, g−1(V ) is clopen and hence

open in Y . Since every open set is rpsI-open, g−1(V ) is rpsI-open in

Y . Also f is rpsI-totally continuous, f−1(g−1(V )) = (g ◦ f)−1(V ) is

clopen in X. Hence g ◦ f is rpsI-totally continuous function.

Definition 3.3.11. A function f : (X, τ, I) → (Y, σ, J) is called

strongly rpsI-continuous if the inverse image of every rpsI-open set

of Y is open in X.

The next theorem follows easily as a direct consequence of

Definitions.


continuous then f is strongly rpsI-continuous.




Example 3.3.14. Let X = Y = {a, b, c}, τ = {φ, X, {a}, {a, b}},

I = {φ, {b}}, σ = {φ, Y, {b}, {a, b}}, J = {φ, {a}}. Define a function

f : (X, τ, I) → (Y, σ, J) by f(a) = b, f(b) = f(c) = a. Then the

inverse image of every rpsI-open set in Y is open in X. Thus f is

strongly rpsI-continuous. But f is not rpsI-totally continuous since

the subset {b} is rpsI-open in Y , f−1({b}) = {a} is open but not

closed in X.

Theorem 3.3.15. Let X be a discrete topological space and Y be any

ideal space and f : (X, τ, I) → (Y, σ, J) be a function where I = φ. If

f is strongly rpsI-continuous then f is rpsI-totally continuous.

Proof. Let V be rpsI-open in Y . Since f is strongly rpsI-continuous,

f−1(V ) is open in X. Also since X is a discrete space, f−1(V ) is both

open and closed in X and so f is rpsI-totally continuous.

Definition 3.3.16. A function f : (X, τ, I) → (Y, σ, J) is called to-

tally rpsI-continuous if f−1(V ) is rpsI-clopen in (X, τ, I) for each

open set V in (Y, σ).

Theorem 3.3.17. Every totally rpsI-continuous function is rpsI-

continuous.


Proof. proof follows from Definition.



Example 3.3.19. Let X = Y = {a, b, c}, τ = {φ, X, {a}, {a, b}},

I = {φ, {b}}, σ = {φ, Y, {b}, {a, b}} and J = φ. Define a function

f : (X, τ, I) → (Y, σ) by f(a) = b,f(b) = a, f(c) = c. Then the

inverse image of every closed set in Y is rpsI-closed in X. Thus f is

rpsI-continuous. But it is not totally rpsI-continuous since the set

{b} is open in Y , f−1({b}) = {a} is rpsI-open and not rpsI-closed in

X.

Theorem 3.3.20. Every rpsI-totally continuous function is totally

rpsI-continuous.

Proof. Suppose f : (X, τ, I) → (Y, σ, J) is rpsI-totally continuous. Let

U be any open subset of Y . Then U is rpsI-open in Y and f−1(U) is

clopen in X. This proves the theorem.



Example 3.3.22. Let X = Y = Z = {a, b, c}, τ = {φ, X, {a}, {b, c}},

I = J = {φ, {a}}, σ = {φ, Y, {a}}. Define a function f : (X, τ, I) →

(Y, σ, J) by f(a) = a, f(b) = b, f(c) = c. Then the inverse image of


every open set in Y is rpsI-clopen in X. Therefore f is totally rpsI-

continuous. Also f is not rpsI-totally continuous because the subset

{a, b} rpsI-open in Y , f−1({a, b}) = {a, b} is not clopen in X.

3.4 Diagram

As a consequence of the Theorems [3.3.3, 3.3.7, 3.3.12 and 3.3.20],

Remarks [3.3.4, 3.3.8, 3.3.13 and 3.3.21] and Examples [3.3.5, 3.3.9,

3.3.14 and 3.3.22] the following implication diagram holds.

rpsI-totally continuous

■✒

❘

✴

continuous strongly rpsI-continuous

perfectly continuous totally rpsI-continuous

In this diagram, A → B means A implies B but B does not imply A.


3.5 rpsI-irresolute Functions

Continuous functions and irresolute functions give new path to-

wards research. In 1972, Crossely and Hildebrand [8] introduced the

notion of irresoluteness. Many different forms of irresolute functions

have been introduced over the years. In this section, we introduce

rpsI-irresolute functions and study about their properties.

Definition 3.5.1. A function f : (X, τ, I) → (Y, σ, J) is called rpsI-

irresolute if f−1(V ) is rpsI closed in (X, τ, I) for each rpsI-closed set

V in (Y, σ).

Theorem 3.5.2. Every rpsI-irresolute function is rpsI-continuous.

Proof. Suppose f : (X, τ, I) → (Y, σ, J) is rpsI-irresolute function.

Let A be any closed subset of Y . Then A is semi pre I-closed in

Y . Thus A is rpsI-closed in Y. Since f is rpsI-irresolute, f−1(A) is

rpsI-closed in X. This proves the theorem.

Theorem 3.5.3. If f : (X, τ, I) → (Y, σ, J) and g : (Y, σ, J) →

(Z, η, K) are two functions. Let h = g ◦ f . Then

(i) h is rpsI-continuous if f is rpsI-irresolute and g is rpsI-continuous.

(ii) h is rpsI-irresolute if both f and g are rpsI-irresolute and

(iii) h is rpsI-continuous if f is rpsI-continuous and g is continuous.


Proof. (i) Let A be any closed set in Z. Since g is rpsI-continuous,

g−1(A) is rpsI-closed in Y . Also since f is rpsI-irresolute, f−1(g−1(A))

is rpsI-closed in X. Thus h is rpsI-continuous.

(ii) Let A be any closed set in Z, Then A is rpsI-closed in Z. Since g is

rpsI-irresolute, g−1(A) is rpsI-closed in Y . Also f is rpsI-irresolute,

f−1(g−1(A)) is rpsI-closed in X. Thus h is rpsI-irresolute.

(iii) Let B be closed set in Z. since g is continuous, g−1(B) is closed in

Y . Also f is rpsI-continuous, f−1(g−1(B)) is rpsI-closed in X. Thus

h is rpsI-continuous.

Theorem 3.5.4. If f : (X, τ, I) → (Y, σ, J) is rpsI-totally continuous

and g : (Y, σ, J) → (Z, η, K) is rpsI irresolute, then g ◦ f : (X, τ, I) →

(Z, η, K) is rpsI-totally continuous.

Proof. Let V be rpsI-open set in Z. Since g is rpsI-irresolute, g−1(V )

is rpsI-open in Y and hence f−1(g−1(V )) = (g ◦ f)−1(V ) is clopen in

X. Thus g ◦ f is rpsI-totally continuous function.

Theorem 3.5.5. If f : (X, τ, I) → (Y, σ, J) is rpsI-totally continuous

and g : (Y, σ, J) → (Z, η, K) is rpsI-continuous, then g◦f : (X, τ, I) →

(Z, η, K) is perfectly continuous.

Proof. Let V be open set Z. Then g−1(V ) is rpsI-open in Y and

f−1(g−1(V )) = (g ◦ f)−1(V ) is clopen in X. Hence g ◦ f is totally

continuous function.


3.6 rpsI-connectedness and rpsI-compactness

The notions of compactness and connectedness are fundamental

notions of not only in general topology but also of other advanced

branches of Mathematics. Many researchers have investigated the

properties of compactness and connectedness. The productivity and

fruitfulness of these notions of compactness and connectedness moti-

vated mathematicians to generalize these notions.

In topology and related branches of Mathematics a connected space is

a topological space that cannot be represented as the union of two dis-

joint non-empty open sets. Connectedness is one of the principal topo-

logical properties that are used to distinguish topological spaces. In

2008, Ekici [15] introduced connectedness in ideal topological spaces.

In 1990, Hamlett [26] introduced compactness in ideal topological

spaces. In this section, the concept of rpsI-connectedness in ideal

topological spaces is introduced and their properties are discussed.

Definition 3.6.1. An ideal topological space (X, τ, I) is said to be

rpsI-connected if X cannot be written as the disjoint union of two

non-empty rpsI-open sets in X. If X is not rpsI-connected it is said

to be rpsI-disconnected.

A subset of X is rpsI-connected if it is rpsI-connected as a subspace

of X.


Theorem 3.6.2. For an ideal topological space (X, τ, I), the following

are equivalent

(i) X is rpsI-connected.

(ii) X and φ are the only subsets of X which are both rpsI-open and

rpsI-closed.

Proof. Suppose (i) holds. Let V be a rpsI-open and rpsI-closed subset

of X. Then V c is both rpsI-open and rpsI-closed. Therefore X = V ∪

V c, a disjoint union of two non-empty rpsI-open sets which contradicts

(i). Thus V = φ or X. This proves (i) ⇒ (ii).

Suppose (ii) holds. Suppose X and φ are the only subsets of X which

are both rpsI-open and rpsI-closed. If X = A∪B, where A and B are

disjoint non-empty rpsI-open subsets of X. Then A is both rpsI-open

and rpsI-closed. Also, A 6= φ and A 6= X, therefore A is a proper

non-empty subset of X which is both rpsI-open and rpsI-closed which

is a contradiction to our assumption. This proves (ii) ⇒ (i).

Theorem 3.6.3. Every rpsI-connected space is connected.

Proof. Let X be a rpsI-connected space. Suppose X is not connected,

then X = A ∪ B where A and B are disjoint non-empty open sets in

X. Since every open set is rpsI-open, X is not rpsI-connected and so

X is connected.


The following example shows that the converse of the above the-

orem need not true.

Example 3.6.4. Consider the ideal topological space X = {a, b, c, d},

τ = {φ, X, {a}, {b, d}, {a, b, d}}, I = {φ, {a}}. In this ideal space, X is

connected but not rpsI-connected. Because the set X = {a}∪{b, c, d}

which are non-empty disjoint rpsI-open sets.

Theorem 3.6.5. Let (X, τ, I) be an ideal topological space. If X is

rpsI-connected, then X cannot be written as the union of two disjoint

non-empty rpsI-closed sets.

Proof. Suppose not, X = A ∪ B where A and B are rpsI-closed sets,

A 6= φ, B 6= φ and A ∩ B = φ. Then A = Bc and B = Ac. Since A

and B are rpsI-closed, A and B are rpsI-open sets. Therefore X is

not rpsI-connected. Hence the proof.

Theorem 3.6.6. Let f : (X, τ, I) → (Y, σ, J) be a function. If X is

rpsI-connected and f is rpsI-irresolute, surjective, then Y is rpsI-

connected.

Proof. Suppose that Y is not rpsI-connected. Let Y = A ∪ B, where

A and B are disjoint non-empty rpsI-open sets in Y . Since f is rpsI-

irresolute and onto, X = f−1(A) ∪ f−1(B) and f−1(A) ∩ f−1(B) =

f−1(A ∩ B) = f−1(φ) = φ, where f−1(A) and f−1(B) are disjoint


non-empty rpsI-open sets in X. This contradicts to the fact that X

is rpsI-connected. Hence Y is rpsI-connected.

Theorem 3.6.7. Let f : (X, τ, I) → (Y, σ, J) be a function. If X is

rpsI-connected and f is rpsI-continuous, surjective, then Y is con-

nected.

Proof. Let X be a rpsI-connected space and f be rpsI-continuous

surjective. Suppose Y is disconnected, then Y = A ∪ B, where A

and B are disjoint non-empty open subsets of Y . Since f is rpsI-

continuous surjective, X = f−1(A) ∪ f−1(B) also f−1(A) ∩ f−1(B) =

f−1(A ∩ B) = f−1(φ) = φ, where f−1(A) and f−1(B) are disjoint

non-empty rpsI-open subsets of X. This contradicts the fact that X

is rpsI-connected. Hence Y is connected.

Theorem 3.6.8. If a surjective map f : (X, τ, I) → (Y, σ, J) is

strongly rpsI-continuous and X is a connected space, then Y is rpsI-

connected.

Proof. It is similar to that of Theorem 3.6.6.

Definition 3.6.9. An ideal topological space (X, τ, I) is said to be

pgprI-connected if X cannot be written as the union of two disjoint

non-empty pgprI-open sets.


Theorem 3.6.10. For an ideal topological space (X, τ, I), the follow-

ings are equivalent.

(i) X is pgprI-connected

(ii) The only subsets of X which are both pgprI-open and pgprI-

closed are the empty set and X.

Proof. The proof is similar to that of Theorem 3.6.2.

Theorem 3.6.11. If f : (X, τ, I) → (Y, σ) is rpsI-totally continuous

surjection and X is connected then Y is rpsI-connected.

Proof. Suppose Y is not rpsI-connected. Then Y = A ∪ B where

A and B are disjoint rpsI-open sets in Y . Also X = f−1(Y ) =

f−1(A ∪ B) = f−1(A) ∪ f−1(B) and f−1(A ∩ B) = φ where f−1(A)

and f−1(B) are non-empty clopen sets in X, because f is rpsI-totally

continuous function. This implies X is not connected. Hence Y is

rpsI-connected.

Theorem 3.6.12. If f : (X, τ, I) → (Y, σ) is totally rpsI-continuous

function from an rpsI-connected space X onto any space Y , then Y

is indiscrete.

Proof. Suppose Y is not indiscrete. Let A be a proper non-empty

subset of Y . Since f is totally rpsI-continuous, f−1(A) is a proper


non-empty rpsI-clopen subsets of X, which is contrary to the fact

that X is rpsI-connected.

Definition 3.6.13. A collection {Aα : α ∈ ∇} of rpsI open sets in a

topological space X is called a rpsI-open cover of a subset B of X if

B ⊂⋃{Aα : α ∈ ∇} holds.

Definition 3.6.14. A topological space X is rpsI-compact if every

rpsI open cover of X has a finite subcover.

Definition 3.6.15. A subset B of a topological space X is said to be

rpsI-compact relative to X if for every collection {Aα : α ∈ ∇} of

rspI open subsets of X such that B ⊂⋃{Aα : α ∈ ∇}, there exists a

finite subset ∇0 of ∇ such that B ⊂ {Aα : α ∈ ∇0}.

Theorem 3.6.16. (i) A rpsI-continuous image of a rpsI-compact

space is compact.

(ii) If a function f : (X, τ, I) → (Y, σ, J) is rpsI irresolute and a

subset B of X, then f(B) is rpsI-compact relative to Y .

Proof. (i) Let f : (X, τ, I) → (Y, σ, J) be a rpsI-continuous map from

a rpsI-compact space X onto a topological space Y . Let {vα : α ∈ ∇}

be an open cover of Y . Then {f−1(vα) : α ∈ ∇} is a rpsI-open

cover of X. Since X is rpsI-compact, it has a finite subcover say

{f−1(v1), f−1(v2), · · · , f

−1(vn)}. Since f is onto {v1, v2, · · · , vn} is a


cover of Y and so Y is compact.

(ii) Let {vα : α ∈ ∇} be any collection of rpsI open subsets of Y

such that f(B) ⊆⋃{vα : α ∈ ∇}. Then B ⊂

⋃{f−1(vα) : α ∈ ∇}

holds. By hypothesis, there exists a finite subset ∇0 of ∇ such that

B ⊂⋃{f−1(vα) : α ∈ ∇0}. Therefore we have f(B) ⊂

⋃{vα : α ∈

∇0} which shows that f(B) is rpsI compact relative to Y .

Theorem 3.6.17. If a function f : (X, τ, I) → (Y, σ, J) is strongly

rpsI-continuous, onto and X is a compact space, then Y is also rpsI-

compact.

Proof. Let f : (X, τ, I) → (Y, σ, J) be a strongly rpsI-continuous map

from a compact space X onto a topological space Y . Let {vα : α ∈

∇} be an rpsI-open cover of Y . Then {f−1(vα) : α ∈ ∇} is an

open cover of X. Since X is compact, it has a finite subcover say

{f−1(v1), f−1(v2), · · · , f

−1(vn)}. Since f is onto, {v1, v2, · · · , vn} is a

cover of Y and so Y is rpsI-compact.

Theorem 3.6.18. If a function f : (X, τ, I) → (Y, σ, J) is totally

rpsI-continuous, onto and X is a rpsI-compact space, then Y is also

compact.

Proof. Let f : (X, τ, I) → (Y, σ) be a totally rpsI-continuous map

from a rpsI-compact space X onto a topological space Y . Let {vα :


α ∈ ∇} be an open cover of Y . Then {f−1(vα) : α ∈ ∇} is an rpsI-

clopen cover of X. Since X is rpsI-compact, it has a finite subcover

say {f−1(v1), f−1(v2), · · · , f

−1(vn)}. Since f is onto, {v1, v2, · · · , vn}

is a cover of Y and so Y is compact.


continuous, onto and X is a compact space, then Y is also rpsI-

compact.

Proof. Let f : (X, τ, I) → (Y, σ, J) be a rpsI-totally continuous map

from a compact space X onto a topological space Y . Let {vα :

α ∈ ∇} be an rpsI-open cover of Y . Then {f−1(vα) : α ∈ ∇} is

clopen cover of X. Since X is compact, it has a finite subcover say

{f−1(v1), f−1(v2), · · · , f

−1(vn)}. Since f is onto, {v1, v2, · · · , vn} is a

cover of Y and so Y is rpsI-compact.

Theorem 3.6.20. Every semiI-compact is rpsI-compact.

Proof. Follows from the fact that every open set is rpsI-open.

CONCLUSION

In this chapter rpsI-continuous, rpsI-irresolute functions, rpsI-

connectedness and rpsI-compactness are defined and their properties

are investigated.

chapter 3 -continuous functions in ideal topological...

Documents