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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
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 48
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.
Remark 3.1.7. The converse of the above theorem need not be true
as seen from the following example.
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 49
Remark 3.1.10. The converse of the above theorem need not be true
as seen from the following example.
Example 3.1.11. 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) = 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.
Remark 3.1.13. The converse of the above theorem need not be true
as seen from the following example.
Example 3.1.14. 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) = 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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 50
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.
Example 3.1.15. 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) = 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 αI-continuous because the subset {a, c} is closed in Y ,
f−1({a, c}) = {a, c} is not αI-closed in X.
Example 3.1.16. 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) = 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 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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 51
(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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 52
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 53
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 54
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.
Remark 3.3.4. The converse of the above theorem need not be true
as seen from the following example.
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
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 55
(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.
Remark 3.3.8. The converse of the above theorem need not be true
as seen from the following example.
Example 3.3.9. Let X = Y = {a, b, c, d}, τ = {φ, X, {a}, {b}, {a, b},
{b, c}, {a, b, c}} , I = {φ, {a}}, σ = {φ, Y, {a, c}, {d}, {a, c, d}} and
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 56
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.
Theorem 3.3.12. If a function f : (X, τ, I) → (Y, σ, J) is rpsI-totally
continuous then f is strongly rpsI-continuous.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 57
Remark 3.3.13. The converse of the above theorem need not be true
as seen from the following example.
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 58
Proof. proof follows from Definition.
Remark 3.3.18. The converse of the above theorem need not be true
as seen from the following example.
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.
Remark 3.3.21. The converse of the above theorem need not be true
as seen from the following example.
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
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 59
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 60
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 61
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 62
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 63
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 64
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
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 65
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.
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 66
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
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 67
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
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 68
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 :
CHAPTER 3. rpsI-continuous functions in ideal topological spaces 69
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.
Theorem 3.6.19. If a function f : (X, τ, I) → (Y, σ, J) is rpsI-totally
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.