Download Chapter 2 h -Closed sets in Topological Spaces

Chapter 2
h∗-Closed sets in Topological
Spaces
2.1
Introduction
The concept of generalised closed sets play a significant role in topology. There are many research papers which deals with different types
of generalised closed sets. Many concepts of general topology have
been generalized, by considering the concept of the semi-open sets due
to Levine[44]. The study of generalised closed sets in the topological
spaces was initiated by Levine[45] and the concept of T 21 spaces was introduced. In 1987, Bhattacharyya and Lahiri [9] introduced the class
of semi generalised closed sets. In 1990, Arya and Nour [6] defined the
generalised semi closed sets. The modified forms of generalised closed
sets was studied by Balachandran, Devi, Maki and Sundaram [7, 15].
Veera Kumar [79] introduced a weak form of closed sets namely g ∗
closed sets between closed sets and generalised closed sets and investigated some of their properties. ω-closed sets and η̂ ∗ -closed sets were
introduced by Sheik John [71] and J. Antony Rex Rodrigo [3] respectively.
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In this chapter by using ω-open sets, a new class of sets in topological spaces namely h∗ -closed sets which is properly placed between the
class of semi-open set and the class of gs-closed sets have been introduced. Also, the relationship among related generalized closed sets
were investigated. Moreover as application, four new spaces namely
Th∗ -space, αTh∗ -space, sTh∗ -space and h∗ Tgs -space were introduced and
some new characterizations have been obtained.
2.2
h∗-closed sets and their basic properties
A new class of sets called h∗ -closed sets is introduced in topological
spaces and some basic properties are investigated.
Definition 2.2.1. A subset A of a space (X, τ ) is said to be h∗ -closed
in (X, τ ) if scl(A) ⊆ int(U ) whenever A ⊆ U and U is ω-open in
(X, τ ). The class of h∗ -closed sets in (X, τ ) is denoted by h∗ C(X, τ ).
That is h∗ C(X, τ ) = {A ⊂ X : A is h∗ -closed in X}.
First it is proved that the class of h∗ -closed set properly lies between
the class of semi-closed sets and the class of gs-closed sets.
Lemma 2.2.2. [71] A set A is ω-open if and only if F ⊆ int(A)
whenever F is semi-closed and F ⊆ A.
Proposition 2.2.3. Every semi-closed subset of (X, τ ) is h∗ -closed
but not conversely.
Proof. Let A be any semi-closed subset of (X, τ ). Let A ⊆ U and
U be ω-open in X. Then by Lemma 2.2.2, A = scl(A) ⊆ int(U ) and
hence A is h∗ -closed.
✷
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Example 2.2.4. Let X = {a, b, c, d} and τ = {∅, {a}, {a, b}, X}. The
set {a, b, c} is h∗ -closed but not semi-closed in (X, τ ).
Proposition 2.2.5. Every closed (resp. α-closed) set is h∗ -closed but
not conversely.
Proof. Let A be any closed (resp. α-closed) set, then it is a semiclosed set and hence A is a h∗ -closed set by Proposition 2.2.3.
✷
Example 2.2.6. Let X = {a, b, c} and τ = {∅, {a, b}, X}. Then the
set {b, c} is h∗ -closed but neither α-closed nor closed in (X, τ ).
Proposition 2.2.7. Let (X, τ ) be a topological space and A ⊆ X. If
A is a h∗ -closed set then the following are true.
a. A is a gs-closed set.
b. A is a gsp-closed set.
c. A is an η̂ ∗ -closed set.
d. A is a πsg-closed set.
Proof. The proof of (a),(b) and (c) follows from the fact that every
open set is ω-open and also from the result that spcl(A) ⊆ scl(A). (d)
Since every π-open set is open and so ω-open the result follows.
✷
Remark 2.2.8. The converse of the Proposition 2.2.7 is not true.
Example 2.2.9. Let X = {a, b, c, d} and τ = {∅, {a, b}, {c, d}, X}.
The set {a, b, c} is η̂ ∗ -closed, gs-closed, gsp-closed, πsg-closed but not
h∗ -closed.
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Remark 2.2.10. h∗ -closedness is independent of pre-closedness,
g-closedness ,
αg-closedness ,
gα-closedness,
sg-closedness,
gp-closedness and gpr-closedness. It follows from the following
examples.
Example 2.2.11. Let X = {a, b, c} and τ = {∅, {a}, {b, c}, X}. Let
A = {a, b}. The set A is g-closed and hence αg-closed, gp-closed.
Moreover A is gα-closed and sg-closed. Also A is pre-closed and gprclosed. But A is not a h∗ -closed set.
Example 2.2.12. Let X = {a, b, c} and τ = {∅, {a}, {b}, {a, b}, X}.
Let A = {a}. The set A is h∗ -closed. But A is not g-closed (resp.
not αg-closed, not gα-closed, not gp-closed, not gpr-closed and not
pre-closed).
Example 2.2.13. Let X = {a, b, c} and τ = {∅, {a}, X}. The set
{a, b} is h∗ -closed but not sg-closed.
Proposition 2.2.14. If a set A is g # -closed (resp. g # s-closed) then
A is h∗ -closed.
Proof. Let A be a g # -closed (resp. g # s-closed) set in the space
(X, τ ). Let A ⊆ U and U be ω-open. Every ω-open set is αg-open.
Therefore cl(A) ⊆ U (resp. scl(A) ⊆ U ). Hence by Lemma 2.2.2,
scl(A) ⊆ int(U ).
✷
Example 2.2.15. Let X = {a, b, c} and τ = {∅, {a}, {a, b}, X}. Then
the set {a, c} is h∗ -closed but neither g # -closed nor g # s-closed.
Remark 2.2.16. The following examples support that h∗ -closed sets
are independent of ω-closed sets, η̂-closed sets and β-closed sets.
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Example 2.2.17. Let X and τ be as in Example 2.2.15. The set
{a, c} is h∗ -closed but neither ω-closed nor β-closed.
Example 2.2.18. Let X = {a, b, c} and τ = {∅, {a}, {b}, {a, b}, X}.
Let A = {a}. The set A is h∗ -closed but A is not η̂-closed.
Example 2.2.19. Let X and τ be as in Example 2.2.9. The set A =
{a, b, c} is ω-closed, η̂-closed sets and β-closed. But A is not h∗ -closed.
Proposition 2.2.20. If a set A is g ∗ -closed then A is h∗ -closed but
not conversely.
Proof. Let A be a g ∗ -closed set. Let A ⊆ U and U be ω-open,
then U is g-open. Therefore cl(A) ⊆ U . That is scl(A) ⊆ U . By
Lemma 2.2.2, scl(A) ⊆ int(U ) implies A is h∗ -closed.
✷
Example 2.2.21. Let X and τ be as in example 2.2.13. The set {a, b}
is h∗ -closed but not g ∗ -closed.
Remark 2.2.22. From the above discussions and known results, the
following implications are obtained.
closed
g # -closed
g ∗ -closed
g-closed
η̂ ∗ -closed
α-closed
semi-closed
αg-closed
gα-closed
g # s-closed
η̂-closed
h∗-closed
gs-closed
ω-closed
sg-closed
β-closed
πsg-closed
pre-closed
gpr-closed
gp-closed
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gsp-closed
Remark 2.2.23. The union (intersection) of any two h∗ -closed sets
is not h∗ -closed as shown in the following examples.
Example 2.2.24. Let X = {a, b, c} and τ = {∅, {a}, {b}, {a, b}, X}.
The sets {a} and {b} are h∗ -closed but {a} ∪ {b} = {a, b} is not a
h∗ -closed set.
Example 2.2.25. Let X = {a, b, c, d} and τ = {∅, {a}, {a, b}, X}.
The sets {a, b, c} and {a, b, d} are h∗ -closed but {a, b, c} ∩ {a, b, d} =
{a, b} is not a h∗ -closed set.
Definition 2.2.26. [12] Let (X, τ ) be a topological space and A ⊆ X.
The set of all semi-limit points of A is said to be a semi-derived set of
A and it is denoted by Ds [A].
Theorem 2.2.27. If D[A] ⊆ Ds [A] for each subset A of a space (X, τ )
then the union of two h∗ -closed sets is a h∗ -closed set.
Proof. Let A and B be h∗ -closed subsets of X and U be an ω-open
set such that A ∪ B ⊆ U . Then scl(A) ⊆ int(U ) and scl(B) ⊆ int(U ).
Since for each subset A of X, Ds [A] ⊆ D[A], then cl(A) = scl(A) and
cl(B) = scl(B). Therefore scl(A ∪ B) ⊆ cl(A ∪ B) = cl(A) ∪ cl(B) =
scl(A) ∪ scl(B) ⊆ int(U ). Hence A ∪ B is a h∗ -closed set.
✷
Theorem 2.2.28. If a subset A of (X, τ ) is h∗ -closed then scl(A) − A
contains no non-empty closed set but not conversely.
Proof. Let F ⊆ scl(A) − A be a non empty closed set. Then A ⊆
X − F . X − F is open and hence ω-open. Since A is h∗ -closed
scl(A) ⊆ int(X − F ) = X − cl(F ). That is, F ⊆ cl(F ) ⊆ X − scl(A).
Therefore F ⊆ scl(A) ∩ (X − scl(A)) = ∅. Hence scl(A) − F contains
no non-empty closed set.
✷
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Example 2.2.29. Let X
=
{a, b, c, d} be with the topology
τ = {∅, {a, b}, {c, d}, X}. Let A = {b, c}. scl(A) − A = X − {b, c} =
{a, b} contains no non-empty closed set. But A is not h∗ -closed.
Theorem 2.2.30. If a set A is h∗ -closed then scl(A) − A contains no
non-empty ω-closed set but not conversely.
Proof. Let F ⊆ scl(A) − A be a non empty ω-closed set. Then A ⊆
X − F , since A is h∗ -closed, scl(A) ⊆ int(X − F ) = X − cl(F ). That
is, F ⊆ cl(F ) ⊆ X − scl(A). Therefore F ⊆ scl(A) ∩ (X − scl(A)) = ∅.
Hence scl(A) − F contains no non-empty ω-closed set.
✷
Example 2.2.31. Let X and τ be as in example 2.2.29. The set
A = {b, c} is such that scl(A) − A contains no non-empty ω-closed
set. But A is not h∗ -closed.
Lemma 2.2.32. [12] Let A be a subset of a topological space X.
a. Then scl(A) = scl(scl(A)).
b. Let A ⊆ Y ⊆ X, where Y is a subspace. Let A ∈ SO(X). Then
A ∈ SO(Y ).
Proposition 2.2.33. Let A and B be any two subsets of a space
(X, τ ). If A is h∗ -closed such that A ⊆ B ⊆ scl(A), then B is h∗ closed.
Proof. Let U be an ω-open set of (X, τ ) such that B ⊆ U . Since A ⊆
U and A is h∗ -closed, scl(A) ⊆ int(U ). Now scl(B) ⊆ scl(scl(A)) =
scl(A) ⊆ int(U ) and hence B is h∗ -closed.
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✷
Remark 2.2.34. The converse of the Proposition 2.2.33 need not be
true. Consider the topological space X = {a, b, c} with the topology
τ = {∅, {a, b}, X}. Let A = {c}, B = {b, c}. Then A and B are
h∗ -closed sets in (X, τ ) but A ⊆ B 6⊆ scl(A).
Proposition 2.2.35. If a subset A of (X, τ ) is ω-open and h∗ -closed
then A is semi-closed in (X, τ ).
Proof. Since A ⊆ A and A is ω-open and h∗ -closed, scl(A) ⊆ int(A) ⊆
A. Thus A is semi-closed.
✷
Proposition 2.2.36. In a topological space X, for each x ∈ X, {x}
is ω-closed or its complement X − {x} is h∗ -closed in (X, τ ).
Proof. Suppose that {x} is not ω-closed in (X, τ ) then X −{x} is not
ω-open and the only ω-open set containing X − {x} is X. Therefore
scl(X − {x}) ⊆ X = int(X). Hence X − {x} is h∗ -closed in (x, τ ). ✷
Theorem 2.2.37. If A is a subset of a topological space (X, τ ), then
the following are equivalent.
(1) A is regular open.
(2) A is open and h∗ closed.
Proof. (1) ⇒ (2): Let U be an ω-open set in X containing A. Since
every regular open set set is open, A ∪ int(cl(A)) ⊆ A = int(A) ⊆
int(U ). Hence scl(A) ⊆ int(U ) and therefore A is h∗ -closed.
(2) ⇒ (1): Since A is open and h∗ -closed, then scl(A) ⊆ int(A) = A
and so A ∪ int(cl(A)) ⊆ A. Therefore int(cl(A)) ⊆ A. Since every
open set is pre-open, then A ⊆ int(cl(A)). Therefore A = int(cl(A))
hence A is regular open.
✷
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Theorem 2.2.38. If a subset A of X is h∗ -closed, then
ωcl({x}) ∩ A 6= ∅ for each x ∈ scl(A).
Proof. Suppose that x ∈ scl(A) and ωcl({x}) ∩ A = ∅, then A ⊆
(ωcl({x}))c and (ωcl({x}))c is ω-open. By assumption, scl(A) ⊆
int((ωcl({x}))c ), which is a contradiction to x ∈ scl(A).
✷
Theorem 2.2.39. [71] Let A ⊂ Y ⊂ X and suppose that Y is closed
in X and A is ω-open in X. Then A is ω-open in Y .
Theorem 2.2.40. [71] If A and B are ω-open, then A ∩ B is ω-open.
Theorem 2.2.41. [71] If A ⊂ B ⊂ X where A is ω-open relative to
B and B is open in X, then A is ω-open in X.
Theorem 2.2.42. Let F ⊂ A ⊂ X and A be open in X.
(a). If F is h∗ -closed in A and A is closed in X, then F is h∗ -closed
in X.
(b). If F is h∗ -closed in X, then F is h∗ closed in A.
Proof. (a). Let U be an ω-open set in X such that F ⊂ U . Then
clearly F ⊂ U ∩ A and U ∩ A is ω-open in X, by Theorem 2.2.40.
Since U ∩ A ⊂ A ⊂ X and U ∩ A is ω-open in X and A is
closed in X, U ∩ A is ω-open in A by Theorem 2.2.39. Given F
is h∗ -closed relative to A. Hence sclA (F ) ⊆ int(U ∩ A). That is
scl(F )∩A ⊆ int(U ∩A). Also A is closed, hence semi-closed. Thus
scl(A) = A and since scl(F ) ⊆ scl(A) = A, scl(F ) ∩ A = scl(F ).
Hence, scl(F ) ⊆ int(U ∩ A) ⊆ int(U ) implies that F is h∗ -closed
in X.
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(b). Let F ⊆ V where V is ω-open in A. Since A is open in X by
Theorem 2.2.41, V is ω-open in X. Hence scl(F ) ⊆ int(V ) by
hypothesis. Thus scl(F ) ∩ A ⊆ int(V ) ∩ A ⊂ int(V ) implies
sclA (F ) ⊆ int(V ). Ultimately, F is h∗ closed relative to A.
✷
Definition 2.2.43. For every set E ⊂ X, the h∗ -closure of E is defined
to be the intersection of all h∗ -closed sets containing E. In symbols,
h∗ cl(E) = ∩{A : E ⊆ A, A ∈ h∗ C(X, τ )}.
Lemma 2.2.44. For any E ⊆ X, E ⊆ h∗ cl(E) ⊆ cl(E).
Proof. Follows from Proposition 2.2.5
✷
Lemma 2.2.45. If A ⊂ B, then h∗ cl(A) ⊂ h∗ cl(B).
Proof. Follows from Definition 2.2.43
✷
Lemma 2.2.46. If a subset A of (X, τ ) is h∗ -closed then h∗ cl(A) = A
but not conversely.
Proof. From Definition 2.2.43, the proof follows.
Example 2.2.47. Let X
=
{a, b, c} and τ
✷
=
{∅, {a}, X}.
Let A = {a}, here h∗ cl(A) = A but A is not h∗ -closed.
2.3
h∗-open sets
In this section, h∗ -open sets h∗ -interior and h∗ -neighbourhood in topological spaces are defined and certain characterization of these sets are
obtained.
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Definition 2.3.1. A subset A of a space (X, τ ) is said to be h∗ -open
in (X, τ ) if its complement Ac is h∗ -closed in (X, τ ). The family of all
h∗ -open sets in (X, τ ) is denoted by h∗ O(X, τ ).
The following three propositions are the analogue of Propositions
2.2.7,2.2.5,2.2.3, and 2.2.14.
Proposition 2.3.2. For any topological space (X, τ )
1. Every h∗ -open set is a gs open set.
2. Every h∗ -open set is η̂ ∗ -open.
3. Every h∗ -open set is πsg-open.
4. Every h∗ -open set is gsp-open.
Proposition 2.3.3. Every open (resp. α-open, semi-open) set is
h∗ -open.
Proposition 2.3.4. Every g # -open (resp. g # s-open) is h∗ -open.
Remark 2.3.5. The union(intersection) of any two h∗ -open set is not
h∗ -open.
Theorem 2.3.6. A subset A of a topological space (X, τ ) is h∗ -open
if and only if cl(K) ⊆ sint(A) whenever K ⊆ A and K is ω-closed.
Proof. Let A be a h∗ -open set in (X, τ ) and suppose K ⊆ A, where
K is ω-closed. Then X − A is h∗ -closed and it is contained in the
ω-open set X − K. Therefore scl(X − A) ⊆ int(X − K) that is
X −int(X −K) ⊆ X −scl(X −A). Thus cl(K) ⊆ sint(A). Conversely,
let K be an ω-closed set such that cl(K) ⊆ sint(A) whenever K ⊆ A,
it follows that X − A ⊆ X − K and X − sint(A) ⊆ X − cl(K). That
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is scl(X − A) ⊆ int(X − K) with X − A ⊆ X − K and X − K is
ω-open. Hence X − A is h∗ -closed and A becomes a h∗ -open set.
✷
In this view, the following corollary is obtained.
Corollary 2.3.7. A subset A of a topological space X is said to be
h∗ -open if and only if F ⊆ sint(A) whenever F ⊆ A and F is ω-closed
in (X, τ ).
Lemma 2.3.8. [71] Suppose that B ⊆ A ⊆ X, B is an ω-closed set
relative to A and A is open and ω-closed in (X, τ ). Then B is ω-closed
in (X, τ ).
Lemma 2.3.9. [71] Let A ⊆ Y ⊆ X and suppose that A is ω-closed
in (X, τ ). Then A is ω-closed relative to Y .
Theorem 2.3.10. Let B and Y be subsets of (X, τ ) such that B ⊆ Y
and Y is both open and closed. If B is a h∗ -open set relative to Y ,
then B is h∗ -open in X.
Proof. Let B be a h∗ -open set relative to Y such that F ⊆ B,
where F is ω-closed in X. From Lemma 2.3.9, F is ω-closed in Y .
Therefore clY (F ) ⊆ sintY (B). Since sintY (B) is semi-open in Y and
Y is open, then sintY (B) = sint(sintY (B)) ⊆ sint(B). Thus, Y ∩
cl(F ) ⊆ Y ∩ sint(B). Therefore Y ⊆ sint(B) ∪ (X − cl(F )). Hence
cl(F ) ⊆ cl(Y ) = Y ⊆ sint(B) ∪ (X − cl(F )) ⊆ sint(B). Thus B is
h∗ -open relative to X.
✷
Theorem 2.3.11. Let X0 be an open and semi-closed subset of (X, τ ).
If V is h∗ -open relative to X, then X0 ∩ V is h∗ -open relative to X0 .
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Proof. Suppose F ⊆ X0 ∩ V , where F is ω-closed in X0 then by
Lemma 2.3.8, F is ω-closed in X and hence cl(F ) ⊆ sint(V ). Thus
cl(F )∩X0 ⊆ sint(V )∩X0 . The set sint(V )∩X0 is semi-open in X. By
Lemma 2.2.32, sint(V ) ∩ X0 is semi-open in X0 . So, sint(V ) ∩ X0 =
sintX0 (sint(V ) ∩ X0 ) ⊆ sintX0 (V ∩ X0 ). Therefore, cl(F ) ∩ X0 =
clX0 (F ) ⊆ sintX0 (V ∩ X0 ) and so X0 ∩ V is h∗ -open relative to X0 . ✷
Lemma 2.3.12. Let B and Y be subsets of (X, τ ) such that B ⊆ Y .
If B is h∗ -open in (X, τ ) and Y is open and closed then B is h∗ -open
relative to Y .
Proof. From the Theorem 2.3.11, B = B ∩ Y is h∗ -open relative to
Y.
✷
Proposition 2.3.13. If sint(A) ⊂ B ⊂ A and if A is h∗ -open then B
is h∗ -open.
Proof. Suppose sint(A) ⊂ B ⊂ A and A is h∗ -open. Then Ac ⊂
B c ⊂ scl(Ac ) and since Ac is h∗ -closed, by Proposition 2.2.33 B c is
h∗ -closed. Hence B is h∗ -open.
✷
Definition 2.3.14. A space X is said to be a sω-space if the intersection of a semi-closed set with a ω-closed set is ω-closed.
Example 2.3.15. The set X = {a, b, c} with the topology τ1 = {∅, {a},
{b, c}, X} is a sω-space. Consider the set X = {a, b, c, d} with the
topology τ2 = {∅, {a}, {a, d}, X}. Here {c, d} is semi-closed and {b, c, d}
is ω-closed but {c, d} ∩ {b, c, d} = {c, d} is not ω-closed. Therefore
(X, τ2 ) is not a sω-space.
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Lemma 2.3.16. Let the space (X, τ ) be a sω-space. Let Aj (1 ≤ j ≤ n)
be h∗ -open sets of (X, τ ) and semi-separated, that is scl(Aj ) ∩ Ai = ∅
for i 6= j. Then ∪nj=1 Aj is h∗ -open.
Proof. Assume that F ⊆ ∪nj=1 Aj , where F is an ω-closed set in
(X, τ ). Then for each i, F ∩ scl(Ai ) ⊆ (∪nj=1 Aj ) ∩ scl(Ai ) = ∪nj=1 (Aj ∩
scl(Ai ))
(A1 ∩ scl(Ai )) ∪ . . . ∪ (Ai ∩ scl(Ai )) ∪ . . .
=
∪ (An ∩ scl(Ai )) = Ai . Since the space (X, τ ) is a sω-space, F ∩ scl(Ai )
is ω-closed and Ai is h∗ -open, then cl(F ∩scl(Ai )) ⊆ sint(Ai ) for each i.
Therefore,
cl(F )
⊆
cl(∪ni=1 (F
∩
scl(Ai )))
⊆
∪ni=1 cl(F ∩ scl(Ai )) ⊆ ∪ni=1 sint(Ai ) ⊆ sint(∪ni=1 Ai ). Hence ∪nj=1 Aj
is h∗ -open.
✷
Proposition 2.3.17. If a set A is h∗ -closed, then scl(A) − A is
h∗ -open.
Proof. Suppose A is h∗ -closed. Let F ⊆ scl(A) − A where F is
ω-closed.
By Theorem 2.2.30 F
= ∅.
Therefore cl(F ) ⊆
sint(scl(A) − A) and by Theorem 2.3.6, scl(A) − A is h∗ -open.
✷
The converse of the Proposition 2.3.17 is not true as seen from the
following example.
Example 2.3.18. Let X = {a, b, c, d} and τ1 = {∅, {c}, {d}, {a, c},
{c, d}, {a, c, d}, X}. Let A = {c}. Then scl(A) − A = {a, c} − {c} =
{a} is h∗ -open but A is not h∗ -closed.
Proposition 2.3.19. Let A be a subset of a topological space X. For
any x ∈ X, x ∈ h∗ cl(A) if and only U ∩ A 6= ∅ for every h∗ -open set
U containing x.
27
Proof. Necessity: Suppose that x ∈ h∗ cl(A). Let U be a h∗ -open set
containing x such that U ∩ A = ∅ and so A ⊆ U c . But U c is h∗ -closed
/ U c, x ∈
/ h∗ cl(A) which is contrary
and hence h∗ cl(A) ⊆ U c . Since x ∈
to the hypothesis.
Sufficiency: Suppose that every h∗ -open set of X containing x meets
A. If x ∈
/ h∗ cl(A), then there exists a h∗ -closed set F of X such
that A ⊂ F and x ∈
/ F . Therefore, x ∈ F c and F c is a h∗ -open set
containing x. But F c ∩ A = ∅. This is contrary to the hypothesis. ✷
Definition 2.3.20. For any A ⊆ X, h∗ interior of A is defined as the
union of all h∗ -open set contained in A. That is h∗ int(A) = ∪{U :
U ⊆ A and U ∈ h∗ O(X, τ )}.
Proposition 2.3.21. For any A ⊆ X, int(A) ⊆ h∗ int(A).
Proof. Since every open set is h∗ -open the proof follows.
✷
Proposition 2.3.22. For any two subsets A1 and A2 of X.
i) If A1 ⊂ A2 then h∗ int(A1 ) ⊆ h∗ int(A2 ).
ii) h∗ int(A1 ∪ A2 ) ⊃ h∗ int(A1 ) ∪ h∗ int(A2 ).
Proposition 2.3.23. If A is h∗ -open, then A = h∗ int(A).
Proof. Clearly the proof follows from the Definition 2.3.20
✷
Remark 2.3.24. Converse of the proposition 2.3.23 is not true as it
can be seen by the following example.
Example 2.3.25. Let X = {a, b, c} and τ = {∅, {a}, X}. Consider
the set A = {b, c}. Then h∗ int(A) = {b} ∪ {c} = {b, c} = A. But A is
not h∗ -open.
28
Proposition 2.3.26. Let A be a subset of a space X, then the following are true.
i) (h∗ int(A))c = h∗ cl(Ac ).
ii) h∗ int(A) = (h∗ cl(Ac ))c .
iii) h∗ cl(A) = (h∗ int(Ac ))c .
Proof.
(i): Let x ∈ (h∗ int(A))c . Then x ∈
/ h∗ int(A). That is,
every h∗ -open set U containing x is such that U 6⊆ A. Thus every
h∗ -open set U containing x is such that U ∩ Ac 6= ∅. By Proposition 2.3.19, x ∈ h∗ cl(Ac ) and therefore, (h∗ int(A))c ⊆ h∗ cl(Ac ).
Conversely, let x ∈ h∗ cl(Ac ). Then by proposition 2.3.19, every h∗ -open set U containing x is such that U ∩ Ac 6= ∅. By
Definition 2.3.20, x ∈
/ h∗ int(A), hence x ∈ (h∗ int(A))c and so
h∗ cl(Ac ) ⊂ (h∗ int(A))c . Thus (h∗ int(A))c = h∗ cl(Ac ).
(ii): It follows by taking complements in (i).
(iii): It follows by replacing A by Ac in (i).
✷
Definition 2.3.27. For any A ⊆ X, h∗ kernal of A is defined as the
intersection of all h∗ -open sets containing A. In symbol, h∗ ker(A) =
∩{U : A ⊂ U and U ∈ h∗ O(X, τ )}.
Example 2.3.28. Let X = {a, b, c} and τ = {∅, {a}, X}. If A =
{b, c}, then h∗ ker(A) = X. If B = {a}, then h∗ ker(B) = {a}.
Lemma 2.3.29. For subsets A, B and Aα (α ∈ ∧) of a topological
space X, the following are true.
1. A ⊂ h∗ ker(A).
29
2. If A ⊂ B, then h∗ ker(A) ⊂ h∗ ker(B).
3. h∗ ker(h∗ ker(A)) = h∗ ker(A).
4. If A is h∗ -open, then A = h∗ ker(A).
5. h∗ ker(∪{Aα |α ∈ ∧}) ⊃ ∪{h∗ ker(Aα )|α ∈ ∧}.
6. h∗ ker(∩{Aα |α ∈ ∧}) ⊂ ∩{h∗ ker(Aα )|α ∈ ∧}.
Proof.
1. Clearly follows from Definition 2.3.27.
2. Suppose x ∈
/ h∗ ker(B). Then there exists a subset U ∈ h∗ O(X, τ )
such that B ⊂ U with x ∈
/ U . Since A ⊂ B, x ∈
/ h∗ ker(A). Thus
h∗ ker(A) ⊂ h∗ ker(B).
3. Follows from (1) and Definition 2.3.27.
4. By Definition 2.3.27 and A ∈ h∗ O(X, τ ), it follows that h∗ ker(A) ⊂
A. By (1), A ⊂ h∗ ker(A). Therefore A = h∗ ker(A).
5. For each α ∈ ∧, h∗ ker(Aα ) ⊂ h∗ ker( ∪ Aα ). Therefore, h∗ ker(∪{Aα |α ∈
α∈∧
∗
∧}) ⊃ ∪{h ker(Aα )|α ∈ ∧}.
6. Suppose that x ∈
/ ∩{h∗ ker(Aα )|α ∈ ∧}, then there exists an α0 ∈
∧ such that x ∈
/ h∗ (ker(Aα0 )) and there exists a h∗ -open set U
such that x ∈
/ U and Aα0 ⊂ U . Hence ∩ Aα ⊂ Aα0 ⊂ U and x ∈
/
α∈∧
U . Therefore x ∈
/ h∗ ker(∩{Aα |α ∈ ∧}). Thus h∗ ker(∩{Aα |α ∈
∧}) ⊂ ∩{h∗ ker(Aα )|α ∈ ∧}.
✷
Remark 2.3.30. In (5) and (6) of Lemma 2.3.29, the equality does
not necessarily hold as shown by the following example.
30
Example 2.3.31. Let X = {a, b, c} and τ = {∅, {a}, X}. Let A =
{a}, B = {b}. Now h∗ ker(A) ∪ h∗ ker(B) = {a, b} and h∗ ker(A ∪ B) =
X. Also let P = {a, b} and Q = {b, c}. Here h∗ ker(P ∩ Q) = {b} and
h∗ ker(P ) ∩ h∗ ker(Q) = {a, b} ∩ X = {a, b}.
Definition 2.3.32. Let A be a subset of a topological space (X, τ ).
A point x ∈ X is said to be a h∗ -limit point of A if for each h∗ -open
set U containing x, U ∩ (A − {x}) 6= ∅. The set of all h∗ -limit points
of A is called a h∗ -derived set of A and is denoted by Dh∗ (A).
Theorem 2.3.33. For subsets A, B of a space X, the following statements hold true:
i) Dh∗ (A) ⊂ D(A), where D(A) is the derived set of A.
ii) If A ⊂ B, then Dh∗ (A) ⊂ Dh∗ (B).
iii) Dh∗ (A) ∪ Dh∗ (B)
⊂
Dh∗ (A ∪ B) and Dh∗ (A ∩ B)
⊂
Dh∗ (A) ∩ Dh∗ (B).
iv) Dh∗ (Dh∗ (A)) − A ⊂ Dh∗ (A).
v) Dh∗ (A ∪ Dh∗ (A)) ⊂ A ∪ Dh∗ (A).
Proof.
i) Since every open set is h∗ -open, the proof follows.
ii) Follows by the Definition 2.3.32.
iii) Follows by (ii).
iv) If x ∈ Dh∗ (Dh∗ (A)) − A and U is a h∗ -open set containing x,
then, U ∩ (Dh∗ (A) − {x}) 6= ∅. Let y ∈ U ∩ (Dh∗ (A) − {x}). Since
y ∈ Dh∗ (A) and y ∈ U , U ∩ (A − {y}) 6= ∅. Let z ∈ U ∩ (A − {y}).
Then z 6= x for z ∈ A and x ∈
/ A. Hence U ∩ (A − {x}) 6= ∅.
Therefore, x ∈ Dh∗ (A).
31
v) Let x ∈ Dh∗ (A ∪ Dh∗ (A)). If x ∈ A, the result is obvious. So
let x ∈ Dh∗ (A ∪ Dh∗ (A)) − A, then there exists a h∗ -open set
U containing x such that U ∩ (A ∪ Dh∗ (A) − {x}) 6= ∅. Thus
U ∩ (A − {x}) 6= ∅ or U ∩ (Dh∗ (A) − {x}) 6= ∅. Now, it follows
similarly from (iv) that U ∩ (A − {x}) 6= ∅. Hence x ∈ Dh∗ (A).
Therefore in any case Dh∗ (A ∪ Dh∗ (A)) ⊂ A ∪ Dh∗ (A).
✷
Theorem 2.3.34. For any subset A of a space X, h∗ cl(A) =
A ∪ Dh∗ (A).
Proof. Since Dh∗ (A) ⊂ h∗ cl(A), A∪Dh∗ (A) ⊂ h∗ cl(A). On the other
hand, let x ∈ h∗ cl(A). If x ∈ A, then the proof is complete. If x ∈
/ A,
each h∗ -open set U contains x intersects A at a point distinct from
x, so x ∈ Dh∗ (A). Thus h∗ cl(A) ⊂ Dh∗ (A) ∪ A which completes the
proof.
✷
Definition 2.3.35. Let X be a topological space. Let x ∈ X. A
subset N of X is said to be a h∗ -neighbourhood of x if and only if
there exists a h∗ -open set U such that x ∈ U ⊆ N .
Theorem 2.3.36. Let (X, τ ) be a topological space. Let x ∈ X. Then
every neighbourhood N of x is a h∗ -neighbourhood of X, but not conversely.
Proof. Let N be a neighbourhood of a point x ∈ X. Then there
exists an open set U such that x ∈ U ⊆ N . Since every open set is a
h∗ -open set, U is a h∗ -open set such that x ∈ U ⊆ N . This implies N
is a h∗ -neighbourhood of x.
✷
Example 2.3.37. Let X = {a, b, c} and τ = {∅, {a}, X}. The set
{a, b} is a h∗ -neighbourhood of the point b, since the h∗ -open set {b} is
32
such that b ∈ {b} ⊆ {a, b}. However the set {a, b} is not a neighbourhood of the point b, since no open set U exists such that b ∈ U ⊆ {a, b}.
Theorem 2.3.38. Every h∗ -open set is a h∗ -neighbourhood of each of
its points, but not conversely.
Proof. Suppose N is h∗ -open. Let x ∈ N . Now N is a h∗ -open set
such that x ∈ N ⊆ N . Since x is an arbitrary point of N , it follows
that N is a h∗ -neighbourhood of each of its points.
✷
Example 2.3.39. Let X = {a, b, c, d} and τ = {∅, {a}, {a, d}, X}.
The set {b, c} is a h∗ -neighbourhood of each of its points, but the set
{b, c} is not a h∗ -open set in X.
Theorem 2.3.40. If F is a h∗ -closed subset of X and x ∈ F c then
there exists a h∗ -neighbourhood N of x such that N ∩ F = ∅.
Proof. Let F be a h∗ -closed subset of X and x ∈ F c . Then F c is a h∗ open set of X. So by Theorem 2.3.38, F c contains a h∗ -neighbourhood
of each of its points. Hence there exists a h∗ -neighbourhood N of x
such that N ⊆ F c . Hence N ∩ F = ∅.
2.4
✷
Applications
As application of h∗ -closed sets, four new topological spaces namely
Th∗ -spaces, α Th∗ -spaces, s Th∗ -spaces and
h∗ Tgs -spaces
are introduced.
Using these spaces, certain relation between the spaces T 21 and α Tb are
obtained.
Definition 2.4.1. A space (X, τ ) is said to be a Th∗ -space if every
h∗ -closed set is closed.
33
Example 2.4.2. Let X = {a, b, c} and τ = {∅, {a}, {a, b}, {a, c}, X}.
The space (X, τ ) is a Th∗ -space.
Example 2.4.3. Let X = {a, b, c} and τ = {∅, {a, b}, X}. The set
{b, c} is h∗ -closed but not a closed set. Hence the space (X, τ ) is not
a Th∗ -space.
Proposition 2.4.4. Every Tb -space is a Th∗ -space but not conversely.
Proof. Let (X, τ ) be a Tb -space and A be a h∗ -closed set. By Proposition 2.2.7, A is a gs-closed set of (X, τ ). Since (X, τ ) is a Tb -space,
then A is closed.
✷
Example 2.4.5. Let X = {a, b, c} and τ = {∅, {a}, {b, c}, X}. The
space (X, τ ) is a Th∗ -space but it is not a Tb -space,since the set {b} is
gs-closed but not closed.
Theorem 2.4.6. If (X, τ ) is a Th∗ -space, then every singleton of X
is either ω-closed or open. The converse is not true.
Proof. Suppose that {x} is not ω-closed in X. Then {x}c is not
ω-open and the only ω-open set containing {x}c is the space X itself. Therefore scl({x}c ) ⊂ X = int(X), so {x}c is h∗ -closed. By
hypothesis {x}c is closed implies {x} is open.
✷
Example 2.4.7. Let X = {a, b, c} and τ = {∅, {a}, {b}, {a, b}, X}.
Here the set {c} is ω-closed where as the sets {a} and {b} are open.
But (X, τ ) is not a Th∗ -space.
Remark 2.4.8. Th∗ -ness is independent of α Tb -ness and T 12 -ness.
Example 2.4.9. The space (X, τ ) as in Example 2.4.7 is an α Tb -space
and hence a T 12 -space but not a Th∗ -space.
34
Example 2.4.10. The space (X, τ ) as in Example 2.4.5 is a Th∗ -space.
But (X, τ ) is not even a T 21 -space.
Definition 2.4.11. A space (X, τ ) is said to be an α Th∗ -space if every
h∗ -closed set is α-closed.
Example 2.4.12. The space (X, τ ) as in Example 2.4.2 is an
α Th∗ -space.
Example 2.4.13. Let X = {a, b, c} and τ = {∅, {a}, X}. The set
{a, b} is h∗ -closed set but it is not α-closed. Therefore (X, τ ) is not
an α Th∗ -space.
Theorem 2.4.14. If (X, τ ) is an α Th∗ -space, then every singleton of
X is either ω-closed or α-open, but not conversely.
Proof. The proof is similar to Proposition 2.4.6
✷
Example 2.4.15. Let X and τ be as in Example 2.4.7. Here the set
{c} is ω-closed where as the sets {a} and {b} are open. But (X, τ ) is
not an α Th∗ -space.
Remark 2.4.16. α Th∗ -ness is independent of α Tb -ness and T 21 -ness.
Example 2.4.17. The space (X, τ ) as in Example 2.4.7 is a α Tb -space
and T 12 -space but not an α Th∗ -space.
Example 2.4.18. The space (X, τ ) as in Example 2.4.5 is a
α Th∗ -space
but neither an α Tb -space nor T 12 -space.
Definition 2.4.19. A space (X, τ ) is said to be a s Th∗ -space if every
h∗ -closed set is semi-closed.
Example 2.4.20. The space (X, τ ) as in Example 2.4.5 is a
s Th∗ -space.
35
Example 2.4.21. The space (X, τ ) as in Example 2.4.13 is not a
s Th∗ -space.
Lemma 2.4.22. [14] For a topological space (X, τ ), the following conditions are equivalent:
a) Every gs-closed set of (X, τ ) is semi-closed.
b) For each x ∈ X, {x} is closed or semi-open in (X, τ ).
c) For each x ∈ X, {x} is closed or open in (X, τ ).
d) (X, τ ) is a T 21 space.
Proposition 2.4.23. Every T 12 -space is a s Th∗ -space, but not conversely.
Proof. Let (X, τ ) be a T 21 -space. Let A be a h∗ -closed set, then it is
a gs-closed set. Since (X, τ ) is a T 12 -space, A is semi-closed in (X, τ )
by Lemma 2.4.22. Therefore (X, τ ) is a s Th∗ -space.
✷
Example 2.4.24. Let X = {a, b, c} and τ = {∅, {a}, {b, c}, X}. Then
(X, τ ) is a s Th∗ -space but not a T 21 -space.
Proposition 2.4.25. Every Th∗ -space is a α Th∗ -space (resp.
s Th ∗
-
space) but not conversely.
Proof. Let A be a h∗ -closed set in (X, τ ). Then by assumption it is
a closed set and therefore α-closed (resp. semi-closed) set in (X, τ ).
Hence (X, τ ) is an α Th∗ -space (resp. s Th∗ -space).
✷
Example 2.4.26. Let X = {a, b, c} and τ = {∅, {a}, {b}, {a, b}, X}.
(X, τ ) is a s Th∗ -space but not an α Th∗ -space and hence not a Th∗ -space.
36
Theorem 2.4.27. In a topological space (X, τ ) the following conditions are equivalent.
(i). (X, τ ) is a s Th∗ -space.
(ii). Every singleton of X is either ω-closed or semi-open.
Proof. (i) ⇒ (ii) Let x ∈ X and suppose that {x} is not ω-closed.
Then {x}c is not ω-open in X. Since X is the only ω-open set containing {x}c , scl({x}c ) ⊂ int(X) = X, implies {x}c is h∗ -closed. By
(i) {x}c is semi-closed. Thus {x} is semi-open.
(ii) ⇒ (i) Let A be a h∗ -closed set in X. Clearly A ⊂ scl(A). Let
x ∈ scl(A), by (ii) {x} is either ω-closed or semi-open.
Case 1: Suppose that {x} is ω-closed .If x ∈
/ A, then scl(A) − A contains the ω-closed set {x} and A is h∗ -closed, which is a contradiction.
Thus x ∈ A.
Case 2: Suppose that {x} is semi-open. Since x ∈ scl(A), {x}∩A 6= ∅.
So x ∈ A. In both cases scl(A) ⊂ A. Thus A = scl(A), implies A is
semi-closed. Hence (X, τ ) is a s Th∗ -space.
✷
Proposition 2.4.28. Every Tb -space is a s Th∗ -space (resp.α Th∗ -space)
but not conversely.
Proof. Follows from Proposition 2.2.7 and the fact that every closed
set is semi-closed and α-closed.
✷
Example 2.4.29. Let X = {a, b, c} and τ = {∅, {a}, {b}, {a, b}, X}.
(X, τ ) is a s Th∗ -space but it is not a Tb -space .
Definition 2.4.30. A space (X, τ ) is said to be a h∗ Tgs -space if every
gs-closed set is h∗ -closed.
37
Example 2.4.31. Let X = {a, b, c} and τ = {∅, {a}, X}. Then (X, τ )
is a
h∗ Tgs -space.
Example 2.4.32. Let X = {a, b, c} and τ = {∅, {a}, {b, c}, X}. The
set {a, b} is gs-closed but not h∗ -closed. Therefore (X, τ ) is not an
h∗ Tgs -space.
Proposition 2.4.33. Every T 21 -space is a
h∗ Tgs -space,
but not
conversely.
Proof. Let (X, τ ) be a T 12 -space. Let A be a gs-closed set in the
T 21 -space (X, τ ). Then A is semi-closed in (X, τ ) by Lemma 2.4.22.
Therefore (X, τ ) is a h∗ Tgs -space, since every semi-closed is a h∗ -closed
set.
✷
Example 2.4.34. The space (X, τ ) given in Example 2.4.31 is a h∗ Tgs space. The set {a, b} is g-closed but it is not closed. Therefore (X, τ )
is not a T 12 -space.
Proposition 2.4.35. Every Tb -space is a
h∗ Tgs -space
, but not con-
versely.
Proof. Let (X, τ ) be a Tb -space. Let A be a gs-closed set in (X, τ ).
Then by assumption it is closed ,hence it is h∗ -closed.
✷
Example 2.4.36. The space (X, τ ) given in Example 2.4.31 is a h∗ Tgs space. The set {a, b} is gs-closed but it is not closed. Therefore (X, τ )
is not a Tb -space.
Theorem 2.4.37. A space (X, τ ) is a Tb -space if and only if (X, τ )
is a
h∗ Tgs -space
and Th∗ -space.
38
Proof. Necessity: Follows from Proposition 2.2.7 and Proposition
2.4.35.
Sufficiency: Suppose (X, τ ) is both a Tb -space and a
A be a gs-closed set of (X, τ ). Since (X, τ ) is a
h∗ Tgs -space
h∗ Tgs -space,
Let
then A is
a h∗ -closed of (X, τ ). Since (X, τ ) is a Th∗ -space, then A is a closed
set of (X, τ ). Therefore (X, τ ) is a Tb -space..
✷
Theorem 2.4.38. A space (X, τ ) is a T 21 -space if and only if (X, τ )
is a
h∗ Tgs -space
and s Th∗ -space.
Proof. Necessity: Follows from Proposition 2.4.23 and Proposition
2.4.33
Sufficiency: Suppose (X, τ ) is both a
h∗ Tgs -space
and a s Th∗ -space.
Let A be a gs-closed set of (X, τ ). Since (X, τ ) is a
h∗ Tgs -space,
then
A is a h∗ -closed of (X, τ ). Since (X, τ ) is also a s Th∗ -space, then
A is a semi-closed set of (X, τ ). Therefore (X, τ ) is a T 21 -space by
Lemma 2.4.22.
✷
39