Download ON SOMEWHAT β-CONTINUITY, SOMEWHAT β

International Journal of General Topology
Vol. 4, Nos. 1-2, January-December 2011, pp. 1 – 7; ISSN: 0973-6751
ON SOMEWHAT β-CONTINUITY, SOMEWHAT
β-OPENNESS AND HARDLY β-OPENNESS
M. Caldas1, S. Jafari2 and G. Navalagi3
ABSTRACT
A new class of functions, called somewhat  -continuous functions, has been defined and studied by making
use of  -open sets. Characterizations and properties of somewhat -continuous functions are presented.
2000 Mathematics Subject Classification: 54A40.
Key words and phrases: Topological spaces, -open sets, -continuity, somewhat  -continuity.
1. INTRODUCTION AND PRELIMINARIES
Recent progress in the study of characterizations and generalizations of continuity has
been done by means of several generalized closed sets. The first step of generalizing
closed set was done by Levine in 1970 [12]. The notion of generalized closed sets has
been studied extensively in recent years by many topologist because generalized closed
sets are the only natural generalization of closed sets. More importantly, they also suggest
several new properties of topological spaces. As a generalization of closed sets, β-closed
sets were introduced and studied by Abd El Monsef et al. ([2], [3], [4]) and Andrijevi ć
[1]. In this paper somewhat β-continuous, somewhat β-open and hardly β-open functions
are introduced and their properties are studied.
Throughout this paper, (X, τ) and (Y, σ) (or simply, X and Y) denote topological
spaces on which no separation axioms are assumed unless explicitly stated. For a subset
A of a space (X, τ), cl(A), int(A) and X\A denote the closure of A, the interior of A and
the complement of A in X, respectively. Abd El Monsef et al. [2] and Andrijevi ć [1]
introduced the notion of β-open set, which Andrijevi ć called semipreopen, completely
independent of each other. They characterized the most important properties of β-open
sets. Throughout this paper, we adopt the word β-open for the sake of clarity. A subset
A of a topological space (X, τ) is called β-open if A ⊆ cl(int(cl(A))). The complement of a
β-open set is called β-closed. The intersection (resp. union) of all β-closed (resp. β-open)
sets containing (resp. contained in) A in X is called the β-closure (resp. β-interior) of A
and is denoted by clβ(A) (resp. intβ(A)) [4]. By βO(X) or βO(X, τ), we denote the collection
of all β-open sets of X. For some more information concerning β-open sets and their
applications we refer the interested reader to [6], [7], [10] and [11].
2. SOMEWHAT -CONTINUOUS FUNCTIONS
Definition 1: A function f : (X; τ) → (Y, σ) is said to be somewhat β-continuous provided
that if for V ∈ σ and f–1(V) ≠ 0/ there is U ∈ βO(X, τ) of X such that U ≠ 0/ and U ⊂ f–1(V).
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International Journal of General Topology
Definition 2: A function f : (X, τ) → (Y, σ) is called somewhat continuous [8], if for
V ∈ σ and f–1(V) ≠ 0/ , there exists and open set U of X such that U ≠ 0/ and U ⊂ f–1(V).
It is clear that every continuous function is somewhat continuous and every
somewhat continuous is somewhat β-continuous. But the converses are not true as it is
shown by the following examples.
Example 2.1: ([8], Example 1). Let X = {a, b, c} with topologies τ = { 0/ , {b}, {a, c}, X}
and σ = { 0/ , {a, b}, X}. Then the identity function f : (X, τ) → (Y, σ) is somewhat continuous.
But f is not continuous.
Example 2.2: Let X = {a, b, c} with topologies τ = { 0/ , {a}, {b, c}, X} and σ = { 0/ , {a}, X}.
Define a function f : (X, τ) → (X, σ) by f(a) = c, f(b) = a and f(c) = b. Then f is somewhat
β-continuous. But f is not somewhat continuous.
Recall that, a subset E of a topological space (X. τ) is said to be β-dense in X if clβ(E)
= X, equivalently if there is no proper β-closed set C in X such that E ⊂ C ⊂ X.
Theorem 2.3: For a surjective function f : (X, τ) → (Y, σ), the following statements
are equivalent:
(i) f is somewhat β-continuous;
(ii) If C is a closed subset of Y such that f–1(C) ≠ X, then there is a proper β-closed subset
F of X such that f–1(C) ⊂ F;
(iii) If E is a β-dense subset of X, then f(E) is a dense subset of Y.
Proof: (i) ⇒ (ii): Let C be a closed subset of Y such that f–1(C) ≠ X. Then Y\C is an
open set in Y such that f–1(Y\C) = X\f–1(C) ≠ 0/ . By (i), there exists a β-open set U in X
such that U ≠ 0/ and U ⊂ f–1(Y\C) = X\f–1(C). This means that f–1(C) ⊂ X\U and X\U =
F is a proper β-closed set in X.
(ii) ⇒ (i): Let V ∈ σ and f–1(V) ≠ 0/ . Then Y\V is closed and f–1(Y\V ) = X\f–1(V) ≠ X.
By (ii), there exists a proper β-closed set F of X such that f–1(Y\V) ⊂ F. This implies that
X\F ⊂ f–1(V) and X\F ∈ βO(X) with X\F ≠ 0/ .
(ii) ⇒ (iii): Let E be a β-dense set in X. Suppose that f(E) is not dense in Y. Then there
exists a proper closed set C in Y such that f(E) ⊂ C ⊂ Y. Clearly f–1(C) ≠ X.
By (ii), there exists a proper β-closed subset F such that E ⊂ f–1(C) ⊂ F ⊂ X. This is a
contradiction to the fact that E is β-dense in X.
(iii) ⇒ (ii): Suppose (ii) is not true. This means that there exists a closed set C in Y
such that f–1(C) ≠ X but there is not proper β-closed set F in X such that f–1(C) ⊂ F. This
means that f–1(C) is β-dense in X. But by (iii), f(f–1(C)) = C must be dense in Y, which is a
contradiction to the choice of C.
Definition 3: If X is a set and τ and τ∗ are topologies on X, then τ is said to be
β-equivalent (resp. equivalent [8]) to τ* provided if U ∈ τ and U ≠ 0/ then there is a
β-open (resp. open) set V in (X, τ) such that V ≠ 0/ and V ⊂ U and if U ∈ τ∗ and U ≠ 0/
then there is a β-open (resp. open) set V in (X, τ) such that V ≠ 0/ and V ⊂ U.
On Somewhat  -continuity, Somewhat  -openness and Hardly  -openness
3
Now consider the identity function f : (X, τ) → (X, τ∗) and assume that τ and τ∗ are
β-equivalent. Then f : (X, τ) → (X, τ∗) and f–1 : (X, τ∗) → (X, τ) are somewhat β-continuous.
Conversely, if the identity function f : (X, τ) → (X, τ∗) is somewhat β-continuous in both
directions, then τ and τ* are β-equivalent.
Theorem 2.4: If f : (X, τ) → (Y, σ) is a surjection somewhat β-continuous and τ* is a
topology for X, which is β-equivalent to τ, then f : (X, τ∗) → (Y, σ) is somewhat
β-continuous.
Proof: Let V be an open subset of Y such that f–1(V) ≠ 0/ . Since f : (X, τ) → (Y, σ) is
somewhat β-continuous, there exists a nonempty β-open set U in (X, τ) such that U ⊂
f–1(V). But by hypothesis τ* is β-equivalent to τ. Therefore, there exists a β-open set U* in
(X, τ*) such that U* ⊂ U. But U ⊂ f–1(V). Then U* ⊂ f–1(V); hence f : (X, τ*) → (Y, σ) is
somewhat β-continuous.
Theorem 2.5: Let f : (X, τ) → (Y, σ) be a somewhat β-continuous surjection and σ∗ be
a topology for Y, which is equivalent to σ. Then f : (X, τ) → (Y, σ∗) is somewhat
β-continuous.
Proof: Let V* be an open subset of (Y, σ∗) such that f–1(V*) ≠ 0/ . Since σ∗ is equivalent
to σ, there exists a nonempty open set V in (Y, σ) such that V ⊂ V*. Now 0/ ≠ f–1(V) ⊂
f–1(V*). Since f : (X, τ) ⊂ (Y, σ) is somewhat β-continuous, there exists a nonempty β-open
set U in (X, τ) such that U ⊂ f–1(V). Then U ⊂ f–1(V*); hence f : (X, τ) → (Y, σ∗) is somewhat
β-continuous.
Theorem 2.6: Let f : (X, τ) → (X, σ) be a function and X = A ∪ B, where A, B ∈ τ. If the
restriction function f A : (X, τ) → (X, σ) and f B : (X, τB) → (X, σ) are somewhat
β-continuous, then f is somewhat β-continuous.
Proof: Let U be any open subset of Y such that f–1(U) ≠ 0/ . then (fA)–1(U) ≠ or (fB)–1(U)
≠ 0/ or both (fA)–1(U) 0/ or (fB)–1(U) ≠ 0/ . Suppose (fA)–1(U) ≠ 0/ . Since fA : (X, τA) → (X, σ)
is somewhat β-continuous, there exists a β-open set V in A such that V ≠ 0/ and V ⊂
(fA)–1(U) ⊂ f–1(U). Since V is β-open in A and A is open in X, V is β-open in X. Therefore
f is somewhat β-continuous.
The proof of the other cases are similar.
Theorem 2.7: If f : (X, τ) → (X, σ) is somewhat β-continuous and g : (X, σ) → (X, η
)
is continuous, then fog : (X, τ) → (Z, η
) is somewhat β-continuous.
3. TWO FORM OF WEAK OPENNESS
In this section, we define and characterize two new weak forms of openness, i.e.
somewhat β-open and hardly β-open functions.
Definition 4: A function f : (X, τ) → (Y, σ) is said to be somewhat β-open provided
that if U ∈ τ and U ≠ 0/ , then there exists a β-open set V in Y such that V ≠ 0/ and V ⊂
f (U).
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International Journal of General Topology
We have the following obvious characterization of somewhat β-openness.
Theorem 3.1: A function f : (X, τ) → (Y, σ) is somewhat β-open if and only if for any
A ⊂ X, int(A) ≠ 0/ implies that int(f(A)) ≠ 0/ .
Theorem 3.2: For a function f : (X, τ) → (Y, σ), the following statements are
equivalent:
(i) f is somewhat β-open;
(ii) If D is a β-dense subset of Y, then f–1(D) is a dense subset of X.
Proof: (i) ⇒ (ii): Suppose D is a β-dense set in Y. We want to show that f–1(D) is a
dense subset of X. Suppose that f–1(D) is not dense in X. Then there exists a closed set B
in X such that f–1(D) ⊂ B ⊂ X. By (i) and since X\B is open, there exists a nonempty
β-closed subset E in Y such that E ⊂ f(X\B). Therefore E ⊂ f(X\B) ⊂ f(f–1(Y\D)) ⊂ Y\D.
That is, D ⊂ Y\E ⊂ Y. Now, Y\E is a β-closed set and D ⊂ Y\E ⊂ Y. This implies that D
is not a β-dense set in Y, which is a contradiction to the fact that D is β-dense in Y.
Therefore, f–1(D) is a dense subset of X.
(ii) ⇒ (i): Suppose D is a nonempty open subset of X. We want to show that intβ(f(D))
≠ 0/ . Suppose intβ(f(D)) = 0/ . Then clβ(Y\f(D)) = Y. Therefore, by (ii) f–1(Y\f(D)) is dense in
X. But f–1(Y\f(D)) ⊂ X\D. Now X\D is closed. Therefore f–1(Y\f(D)) ⊂ X\D gives X =
cl(f–1((Y\f(D)) ⊂ X\D. This implies that D = 0/ which is contrary to D ≠ 0/ . Therefore
intβ(f(D)) ≠ 0/ . This proves that f is somewhat β-open.
Theorem 3.3: For a bijective function f : (X, τ) → (Y, σ), the following statements are
equivalent:
(i) f is somewhat β-open;
(ii) If C is a closed subset of X such that f(C) ≠ Y, then there is a β-closed subset F of Y
such that F ≠ Y and f(C) ⊂ F.
Proof: (i) ⊂ (ii): Let C be any closed subset of X such that f(C) ⊂ Y. Then X\C is an
open set in X and X\C ≠ 0/ . Since f is somewhat β-open there exists a β-open set V in Y
such that V ≠ 0/ and V ⊂ f(X\C). Put F = Y\V. Clearly F is β-closed in Y and we claim F
≠ Y. If F = Y, then V = 0/ which is a contradiction. Since V ⊂ f(X\C), f(C) = (Y\f(X\C)) ⊂
Y\V = F.
(ii) ⇒ (i): Let U be any nonempty open subset of X. Then C = X\U is closed set in X
and f(X\U) = f(C) = Y\f(U) implies f(C) ≠ Y. Therefore, by (ii), there is a β-closed set F of
Y such that F ≠ Y and f(C) ⊂ F. Clearly V = Y\F ∈ βO(Y, σ) and V ≠ 0/ . Also V = Y\F ⊂
Y\f(C) = Y\f(X\U) = f(U).
Definition 5: A function f : (X, τ) → (Y, σ) is said to be hardly β-open provided that
for each β-dense subset A of Y that is contained in a proper open set, f–1(A) is β-dense
in X.
We have the following characterizations of hardly β-openness.
On Somewhat  -continuity, Somewhat  -openness and Hardly  -openness
5
Theorem 3.4: A function f : (X, τ) → (Y, σ) is hardly β-open if and only if intβ(f–1(A))
= 0/ for each set A ⊂ Y having the property that intβ(A) = 0/ and A containing a nonempty
closed set.
Proof: Assume f is hardly β-open. Let A ⊂ Y such that intβ(A) = 0/ and let F be a
nonempty closed set contained in A. Since intβ(A) = 0/ , Y\A is β-dense in Y. Because F ⊂
A, Y\A ⊂ Y\F ≠ Y. Therefore f–1(Y\A) is β-dense in X. Thus X = clβ(f–1(Y\A)) = clβ(X\
f–1(A)) = X\intβ(f–1(A)) which implies that intβ(f–1(A)) = 0/ .
For the converse implication assume that intβ(f–1(A)) = 0/ for every A ⊂ Y having the
property that int β (A) = 0/ and A contains a nonempty closed set. Let A be a
β-dense subset of Y, that is contained in the proper open set U. Then intβ(Y\A) = 0/ and
0/ ≠ Y\U ⊂ Y\A. Thus Y\A contains a nonempty closed set and hence intβ(f–1(Y\A)) =
0/ . Then intβ(f–1(Y\A)) = intβ(X\f–1(A)) = X\clβ(f–1(A)) and hence f–1(A) is β-dense in X.
Theorem 3.5: Let f : (X, τ) → (Y, σ) be a function. If intβ(f(A)) ≠ 0/ for every A ⊂ X
having the property that intβ(A) ≠ 0/ and there exists a nonempty closed set F for which
f–1(F) ⊂ A then f is hardly β-open.
Proof: Let B ⊂ U ⊂ Y where B is β-dense and U is a proper open set. Let A = f–1(Y\B)
and F = Y\U, obviously f–1(F) = f–1(Y\U) ⊂ f–1(Y\B) = A. Also intβ(f(A)) = intβ(f(f–1(Y\B)))
⊂ intβ(Y\B) = 0/ . Therefore we must have that 0/ = intβ(A) = intβ(f–1(Y\B)) = intβ(X\f–1(B))
which implies that f–1(B) is β-dense. It follows that f is hardly β-open.
Theorem 3.6: If f : (X, τ) → (Y, σ) is hardly β-open, then intβ(f(A)) ≠ 0/ for every A ⊂
X having the property that intβ(A) ≠ 0/ and f(A) contains a nonempty closed set.
Proof: Let A ⊂ X such that intβ(A) ≠ 0/ and let F be a nonempty closed set for which
F ⊂ f(A). Suppose intβ(f(A)) = 0/ . Then Y\f(A) is β-dense in Y and Y\f(A) ⊂ Y\F which is
a proper open set. Since f is hardly β-open, f–1(Y\f(A)) is β-dense in X. But f–1(Y\f(A)) =
X\f –1 (f(A)) and hence int β (f –1 (f(A))) = 0/ . It follows that int β (A) = 0/ which is
a contradiction.
Theorems 3.5 and 3.6 are reversible provided that f is surjective. Thus we have the
following characterization for surjective hardly β-open functions.
Theorem 3.7: If f : (X, τ) → (Y, σ) is surjective, then the following conditions
are equivalent:
(i) f is hardly β-open.
(ii) intβ(f(A)) ≠ 0/ for every A ⊂ X having the property that intβ(A) ≠ 0/ and there exists
a nonempty closed F ⊂ Y such that F ⊂ f(A).
(iii) intβ(f(A)) ≠ 0/ for every A ⊂ X having the property that intβ(A) ≠ 0/ and there exists
a nonempty closed set F ⊂ Y such that f–1(F) ⊂ A.
Proof: (i) ⇒ (ii): Theorem 3.6
(ii) ⇒ (iii): Since f is surjective f–1(F) ⊂ A implies that F ⊂ f(A).
(iii) ⇒ (i): Theorem 3.5.
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International Journal of General Topology
4. -RESOLVABLE SPACES AND -IRRESOLVABLE SPACES
Definition 6: A topological space (X, τ) is said to be β-resolvable (resp. resolvable
[9]) if there exists a β-dense (resp. dense) set A in (X, τ) such that X\A is also β-dense
(resp. dense) in (X, τ). A space (X, τ) is called β-irresolvable (resp. irresolvable) if it is not
β-resolvable (resp. resolvable).
Theorem 4.1: For a topological space (X, τ), the following statements are equivalent:
(i) (X, τ) is β-resolvable;
(ii) (X, τ) has a pair of β-dense sets A and B such that A ⊂ (X\B).
Proof: (i) ⇒ (ii): Suppose that (X, τ) is β-resolvable. There exists a β-dense set A in (X,
τ) such that X\A is β-dense in (X, τ). Set B = X\A, then we have A = X\B.
(ii) ⇒ (i): Suppose that the statement (ii) holds. Let (X, τ) be β-irresolvable. Then X\B
is not β-dense and clβ(A) ⊂ clβ(X\B) ≠ X. Hence A is not β-dense. This contradicts the
assumption.
Theorem 4.2: For a topological space (X, τ), the following statements are equivalent:
(i) (X, τ) is β-irresolvable (resp. irresolvable);
(ii) For any β-dense (resp. dense) set A in X, intβ(A) ≠ 0/ (resp int(A) ≠ 0/ ).
Proof: We prove the first statement since the proof of the second is similar.
(i) ⇒ (ii): Let A be any β-dense set of X. Then we have clβ(X\A) ≠ X, hence intβ(A)
≠ 0/ .
(ii) ⇒ (i): Suppose that (X, τ) is a β-resolvable space. There exists a β-dense set A in
(X, τ) such that X\A is also β-dense in (X, τ). It follows that intβ(A) = 0/ , which is a
contradiction; hence (X, τ) is β-irresolvable.
Definition 7: A topological space (X, τ) is said to be strongly β-irresolvable if for
a nonempty set A in X intβ(A) = 0/ implies intβ(clβ(A)) = 0/ .
Theorem 4.3: If (X, τ) is a strongly e-irresolvable space and clβ(A) = X for a nonempty
subset A of X, then clβ(intβ(A)) = X.
Proof: The proof is clear.
Theorem 4.4: If (X, τ) is a strongly β-irresolvable space and intβ(Clβ(A)) ≠ 0/ for a
nonempty subset A of X, then intβ(A) ≠ 0/ .
Proof: The proof is clear.
Theorem 4.5: Every strongly β-irresolvable is β-irresolvable.
Proof: This follows from Theorems 3.6 and 4.4.
Theorem 4.6: If f : (X, τ) → (Y, σ) is a somewhat β-open function and intβ(B) = 0/ for
a nonempty subset B of Y, then intβ(f–1(B)) = 0/ .
Proof: Let B be a nonempty set in Y such that intβ(B) = 0/ . Then clβ(Y\B) = Y. Since f
is somewhat β-open and Y\B is β-dense in Y, by Theorem 3.2 f–1(Y\B) is β-dense in X.
Then clβ(X\f–1(B)) = X. Hence intβ(f–1(B)) = 0/ .
On Somewhat  -continuity, Somewhat  -openness and Hardly  -openness
7
Theorem 4.7: If f : (X, τ) → (Y, σ) be a somewhat β-open function. If X is irresolvable,
then Y is β-irresolvable.
Proof: Let B be a nonempty set in Y such that clβ(B) = Y. We show that intβ(B) ≠ 0/ .
Suppose not, then clβ(Y\B) = Y. Since f is somewhat β-open and Y\B is β-dense in Y, we
have by Theorem 3.2 f–1(Y\B) is dense in X. Then int(f–1(B)) = 0/ . Now, since B is β-dense
in Y and using again Theorem 3.2 f–1(B) is dense in X. Therefore we have that for the
dense set f–1(B) that int(f–1(B)) = 0/ , which is a contradiction to Theorem 4.2. Hence we
must have intβ(B) ≠ 0/ for all β-dense sets B in Y. Hence by Theorem 4.2, Y is β-irresolvable.
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1
Miguel Caldas
2
Saeid Jafari
Departamento de Matemáatica Aplicada,
College of Vestsjaelland South Herrestraede 11
Universidade Federal Fluminense,
4200 Slagelse, DENMARK. E-mail: [email protected]
Rua Máario Santos Braga, s/n
3
24020-140, Niterói, RJ-BRASIL
Department of Mathematics, KLE Societys G.H.College,
E-mail: [email protected]
G. Navalagi
Haveri-581110, Karnataka, INDIA.
E-mail: [email protected]