Download example 4.9

SEMI – STAR GENERALIZED HOMEOMORPHISMS IN
TOPOLOGICAL SPACES
By
Dr.K.Chandrasekhara Rao, Department of Mathematics,Mohamed Sathak Engineering
College, Kilakarai, 623806, Tamil nadu.
and
K.Joseph, 14, Bazaar Street, Ramachandrapuram, 622505, Tamil nadu.
Abstract
In this paper, we introduce two classes of maps called semi – star generalized
homeomorphisms and s*gc-homeomorphisms which are generalizations of homeomorphisms,
semi - homeomorphisms due to Biswas [4], semi – homeomorphisms due to Crossely and
Hildebrand [6], g–homeomorphisms and gc-homeomorphisms of H.Maki et al [11], sghomeomorphisms, sgc- homeomorphisms, gs – homeomorphisms and gsc – homeomorphisms
of Devi et al [8].
Key words: s*g-homeomorphisms,s*gc-homeomorphisms.
1. INTRODUCTION
Let (X, ) denote a topological space unless explicitly stated. AX is said to be semistar generalized closed [5] (written as s*g-closed) if A  U whenever AU and U is semiopen in X.
The complement of a s*g-closed set is called s*g-open. In this paper, we introduce two
classes of maps called semi – star generalized homeomorphisms and s*gchomeomorphisms which are generalizations of homeomorphisms, semi homeomorphisms due to Biswas [4], semi – homeomorphisms due to Crossely and
Hildebrand [6], g–homeomorphisms and gc-homeomorphisms of H.Maki et al [11], sghomeomorphisms, sgc- homeomorphisms, gs – homeomorphisms and gsc –
homeomorphisms of Devi et al [8].
2. PRELIMINARIES
Throughout this paper, (X, ) (Y, ) and (Z, ) always denote the toplogical spaces
on which no separation axioms are assumed unless explicitly satted. A subset A is said to be
semi - open [9] if there exists an open set U of X such that UAU. The complement of a
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semi - open set is said to be semi - closed. We denote the closure and interior of A by A- and
A0 respectively. The intersection (union) of all semi - closed sets (semi-open sets) containing
A (contained in A) is denoted by A - (A0).S.O.( X, ), (S.C. (X, ) denotes the set of all semi
- open (semi – closed) sets of (X, ). Ac denotes complement of A in X unless explicitly
stated. S*.G.O. (X, ) (G.O. (X, ) denotes the set of all s*g – open (g-open) sets of (X, )
S.G.O.
(X, ) (G.S.O. (X, ) denotes the set of all sg-open (gs-open) sets of (X, )
DEFINITION 2.1.
A subset A is said to be generalized open [10] (Written shortly as g - open ) if FA0
whenever FA and F is closed in X. The complement of a g - open set is called a generalized
closed (written shortly as g – closed) set.
DEFINITION 2.2.
A subset A of a space X is said to be semi - generalized closed (Written shortly as sg closed [3]), if A U whenever AU and U is semi - open in X. The complement of a sg closed set is said to be semi - generalized open (Written shortly as sg - open).
DEFINITION 2.3.
A subset A of a space X is said to be generalized semi – open [1] (Written shortly as
gs - open ), if FA0 whenever FA and F is closed in X. The complement of a gs - open set
is called a generalized semi - closed (written shortly as gs - closed) set.
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DEFINITION 2.4.
A map f : (X, )  (Y, ) is called g-continuous [2] (sg - continuous [14], gs continuos map if f-1 (V) is g – closed (sg – closed, gs – closed ) in
(X, ) for every closed set
V of (Y, ).
DEFINITION 2.5.
A map f: (X, )  (Y, ) is called a g-closed [12] (sg-closed) [7], gs – closed [7] map
if f (V) is g – closed (sg – closed, gs – closed ) in (Y, ) for every closed set V of (X, )
DEFINITION 2.6.
A map f: (X, )  (Y, ) is called a g-open [12] (sg-open [7], gs – open [7] ) map if f
(V) is g – open (sg – open, gs – open ) in (Y, ) for every open set V of (X, )
DEFINITION 2.7.
A map f: (X, )  (Y, ) is called a gc-irresolute [2] (sg-irresolute [14], gs –
irresolute) map if f-1 (V) is g – closed (sg – closed, gs – closed ) in (X, ) for every g - closed
(sg – closed, gs – closed ) set V of (Y, )
DEFINITION 2.8.
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A function f: (X, )  (Y, ) is pre - semi - open [6] (pre-semi-closed ) if f (V) is
semi - open - (semi - closed) in (Y, ) for every semi - open - (semi-closed) set V of (X, )
DEFINITION 2.9.
A function f: (X, )  (Y, ) is irresolute [6] if f-1(V) is semi - open in (X, ) for
every semi - open set V of (Y, )
DEFINITION 2.10.
A map f: (X, )  (Y, ) is said to be semi-homeomorphism(B)[4] (simply s.h.(B)) if
(i) f is continuous ii) f is semi – open (i.e. f (U) is semi – open in (Y, ) for every open set U
of (X, ) and iii) f is bijective.
DEFINITION 2.11.
A map f: (X, )  (Y, ) is said to be semi-homeomorphism (C.H.) [6] (simply s.h.
(C.H.)) if i) f is irresolute ii) f is pre-semi – open and iii) f is bijective. It is clear that every
homeomorphism is a semi – homeomorphism(C.H.).
DEFINITION 2.12.
A bijection f: (X, )  (Y, ) is said to be g – homeomorphism [11] (sg –
homeomorphism [8], gs – homeomorphism [8] ) if f is both g – continuous and g – open (sgcontinuous and sg – open, gs-continuous and gs-open).
DEFINITION 2.13.
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A bijection f: (X, )  (Y, ) is called a gc – homeomorphism [11] (sgc –
homeomorphism [8], gsc – homeomorphism [8] ) if f and its inverse f-1 are gc – irresolute (sgirresolute, gs-irresolute).
DEFINITION 2.14.
A space (X, ) is called a Tb space [7](semi-T½ [3], T½ [10]) if every gs-closed set is
closed. (every sg-closed set is semi-closed, every g-closed set is closed)
3.
HOMEOMORPHIC AND SEMI-HOMEOMORPHIC IMAGES OF
T½ AND SEMI – T ½ SPACES
PROPOSITION 3.1.
For each x  X, {x} is semi-closed or its complement {x}c is s*g-closed in (X, )
Proof
Suppose that {x}is not semi – closed in (X, ) Since {x}c is not semi – open, the space
X itself is the only semi – open set containing {x}c . Therefore, {x}c-  X holds and so {x}c is
s*g – closed.
THEOREM 3.2.
A space ((X, ) is T½ if and only if = S*G.O (X, ) = G.O (X, )
Proof. Necessity
Since every open set is s*g – open and hence g-open,   S*.G.O (X,)  G.O (X,).
But in a T½ space, every s*g – closed set is closed and hence G.O. (X, )  S*.G.O. (X, ) 
Consequently,  = S*.G.O. (X, ) = G.O. (X, )
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Sufficiency
Let  = S* G.O. (X, ) = G.O. (X, ) Now, every g – closed set is closed. This proves
the sufficiency.
REMARK 3.3
The following example shows that if  = S*.G.O. (X, ) then (X, ) is not T½..
EXAMPLE 3.4
Let X = {a, b, c} and  = { , {a}, X }. Now,  = S*.G.O. (X, ) but (X, ) is not T½.
THEOREM 3.5
If  = S.O. (X, ) = S*.G.O. (X, ) then (X, ) is semi- T½. but not conversely.
Proof
Since  = S.O. (X,) the set of all sg – closed sets of (X, ) and the set of all s*g –
closed sets of (X, ) are equal to each others. Hence every sg – closed set is semi – closed.
But the converse is not true from the following example.
EXAMPLE 3.6
Let X = {a,b,c} and  = { , {a}, {b}, {a,b}, X}. Now X is semi - T½. but  = S*.G.O.
(X, )  S.O. (X, )
THEOREM 3.7
A space (X, ) is semi -T½ if and only if for each x  X, {x} is semi-open or semi –
closed.
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Proof. Necessity
Suppose that {x}is not semi – closed for some x  X. Since X is the only semi – open
set containing {x}c, {x}c-  X holds and so {x}c is s*g – closed and hence semi – closed in
the semi - T½ space (X, ) Therefore {x} is semi – open.
Sufficiency
It follows from Theorem 4.8 of [14].
THEOREM 3.8
If A is s*g – closed, then A – A contains no non-empty semi-closed set.
Proof:
Let F  A – A where F is semi-closed in X. Now A  Fc where Fc is semi-open.
Since A is s*g – closed, A  Fc. Consequently, A c  F. This implies that   F and hence
F = .
THEOREM 3.9
Let f: (X, )  (Y,) be a mapping. Suppose that f is continuous and presemi-closed. Then f-1(A) is s*g – closed in (X, ) for every s*g – closed set A of (Y ,).
Proof
Let A be s*g – closed in Y. Let U be semi – open in X such that f-1(A)U.
Since f is continuous, we have f ((f-1(A)  Uc))  f (f-1(A))  f (Uc)  f (f-1(A))  f (f-1(Ac))
 AAc = A -A.
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Since f is pre – semi – closed, f(f-1(A)  Uc ) is semi – closed. Since A - A
contains no non – empty semi – closed set if A is s*g – closed by Theorem 3.8. Hence
f-1(A)Uc = . Consequently, f-1(A)  U. Therefore, f-1(A) is s*g – closed in (X, ).
THEOREM 3.10
Let (Y, ) be a topological space such that S.O.(Y, ) = . If f: (X, )  (Y, ) is
onto, continuous, pre-semi – closed and (X, ) is semi - T½, then Y is semi - T½ .
Proof
Let A be a s*g – closed set of (Y, ). By using Theorem 3.9 f-1 (A) is s*g – closed in
(X, ) Since X is semi - T½,, f-1 (A) is semi – closed and hence A is semi – closed. Since S.O.
(Y,) = , every s g – closed set in (Y,) is s*g – closed by the fact found in Theorem 3.5.
and hence Y is semi - T½ .
COROLLARY 3.11
The homeomorphic image f(X) = Y is semi-T1/2 if X is semi-T1/2 and S.O.(Y, ) = .
The proof follows from the known fact that every homeomorphism is a semihomeomorphism (C.H).
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PROPOSITION 3.12
Let H be an open and s*g – closed set of (X, ) and A be a subset of H. Then A is
semi-closed in (X, ) if and only if A is semi-closed in (H, /H).
Proof:
Let A be semi-closed in (X, ).
Then A A0  (AH)x0 = A0H  H0 =A0H
Thus, A is semi-closed in (H, /H).
Conversely, let A be semi-closed in
(H, /H).
Consequently, A  (A0)H
= (A  H)0x =A0x. Since H is open and s*g – closed it is closed. Hence A  (A0)x and A
is semi-closed in X.
THEOREM 3.13
Let H be an open set of (X, ), If (X, ) is a semi-T ½ space, then the subspace
(X, /H) is semi-T ½.
Proof
By using Theorem 3.7, it is enough to show that any singleton of (H, /H), say {x} is
semi–open or semi-closed in (H, /H). Since (X, ) is semi-T½ , {x} is semi–open or semiclosed in (X, ). If {x} is semi – open in (X, ) then {x}  H = {x}  S.O.(H) [13], Theorem
1). If {x} is semi-closed in (X, ), then it is semi-closed in (H, /H) by Proposition 3.12.
Hence (H, /H) is semi-T½.
If H is preopen, then we obtain Theorem 5.5 of [14].
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THEOREM 3.14
Let ( XxY,  x ) be the product space of (X, ) and (Y, ). Suppose that there exists
an open singleton {y} of (Y, ). Then, if {X x Y,  x  } is semi-T1/2, the factor space (X, )
is semi-T1/2.
Proof
The product space (XxY,  x  ) contains a subspace (X x {y},  x  / X x {y})
which is homeomorphic to (X, ). Since X x {y} is open in (X x Y,  x  ), it follows from
Theorem 3.13. that the subspace ( X x {y},  x  / X x {y}) is semi-T ½. Since every
homeomorphism is a semi-homeomorphism (C.H) by definition 2.11, (X, ) is semi-T1/2. [14]
Corollary 5.3).
If {y} is preopen, we obtain Theorem 5.6 of [14].
PREPOSITION 3.15
A topological space (X, ) is a Tb space if and only if  =S*G.O(,T) = S.G.O.(X, )
= G.S.O.(X, ) = G.O.(X, ).
Proof : Necessity
Since every gs-closed set is closed in a Tb space, we have G.S.O.(X, )  S.G.O.
(X, )  S*.G.O.(X, )  . But always   S*.G.O.(X, )  S.G.O.(X, )  G.S.O(X, ).
Consequently,  = S*.G.O(X, ) = S.G.O.(X., ) = G.S.O.(X, ) = G.O.(X, ).
Sufficiency
 = G.S.O.(X, ) implies every gs-closed set is closed and hence (X, ) is a Tb space.
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4 SEMI – STAR GENERALIZED HOMEOMORPHISMS
DEFINITION 4.1
A map f: (X, )  (Y, ) is called a semi-star generalized continuous (written shortly
as s*g-continuous, s*g-irresolute) map if f-1(V) is s*g –closed in (X,) for every closed (s*gclosed) set V of (Y, ).
DEFINITION 4.2
A map f: (X, )  (Y, ) is called a s*g –closed (s*g-open) map if f(V) is s*g closed
(s*g –open) in (Y, ) for every closed (open) set V of (X, ).
DEFINTION 4.3
A bijection f: (X, )  (Y, ) is called a semi-star generalized homeomorphism
(written shortly as s*g-homeomorphism) if f is s*g-continuous and s*g-open.
PROPOSITION 4.4
The composition of two s*g-homeomorphisms is not always a s*g-homeomorphism.
Proof
Let X = Y = Z = {a,b,c},  =  = { , {a,}, {b,c}, X } and  = {  , {a}, {c}, {a,c},
{b,c}, Z}. Let f: (X,  )  (Y, ) be defined as f(a) = b, f(b) = a and f(c) = c. Let g: (Y, ) 
(Z, ) be the identity mapping. Now f and g are s*g-homeomorphisms. But gof is not a s*ghomeomorphim, since (gof)({a}) = {b} is not s*g-open in Z where {a} is open in X.
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PROPOSITION 4.5
Let f: (X,  )  (Y, ) be a map. Then the following are true.
a)
Every homeomorphism is a s*g-homeomorphism.
b)
Every s*g-homeomorphism is a g-homeomorphism.
c)
Every s*g-homeomorphism is a sg-homeomorphism.
d)
Every s*g-homeomorphism is a gs-homeomorphism.
Proof
Obvious from definitions.
But the converses of (a), (b), (c) and (d) are not true in general as can be seen from the
following examples.
EXAMPLE 4.6
Let X = Y = {a,b,c} and  = {  , {a}, {b,c}, X} = . Let f: (X,  )  (Y,). be
defined as f(a) = b, f(b) = a and f(c) = c. Then f is a s*g-homeomorphism but not a
homeomorphism. However, evey s*g-homeomorphism from a T
1/2
space onto itself is a
homeomorphism.
EXAMPLE 4.7
Let X = Y = { a, b, c },  = {  , {a}, {a,b}, X} and  = { , {a}, Y}.
Let f : (X,  )  (Y, ) be defined as f(a) = a, f(b) = c, f(c) = b. Then f is a ghomeomorphism but not a s*g-homeomorphim. However, every g-homeomorphism from a
T 1/2 space onto itself is a s*g-homeomorphism by Theorem 3.2.
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EXAMPLE 4.8
Let X = Y = {a, b, c},  = {  , {a}, X} and  = { , {b}, {a,b}, Y}.
Let f: (X,  )  (Y, ) be defined as f(a) = b, f(b) = a and f(c) = c. Now, f is a sghomeomorphism but not a s*g - homeomorphism. However, every sg - homeomorphism
from a Tb space onto itself is a s*g - homeomorphism following the result in proposition 3.15.
EXAMPLE 4.9
From example 4.7, f is a gs - homeomorphism but not a s*g - homeomorphism.
However, every gs - homeomorphism from a Tb space onto itself is a s*g - homeomorphism.
PROPOSITION 4.10
Let f: (X, )  (Y, ) be a map. Then the following are true.
a)
s*g - homeomorphisms and semi - homeomorphisms(B) are
independent of each other.
b)
s*g - homeomorphisms and semi - homeomorphisms (C.H.) are
independent of each other.
c)
s*g - homeomorphisms and gc - homeomorphisms are independent
of each other.
d)
s*g - homeomorphisms and sgc - homeomorphisms are
independent of each other.
e)
s*g - homeomorphisms and gsc - homeomorphisms are
independent of each other.
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Proof
a) From example 4.6, f is a s*g - homeomorphism but not a s.h. (B).
From example 4.7, f is a s.h.(B) but not a s*g - homeomorphism. However, every s*g
- homeomorphism from a T1/2 space onto a semi - T1/2 space is a s.h. (B).
b)
i)
Let X and Y be the topological spaces of proposition 4.4.
Let f: (X, )  (Y, ) be defined as f(a) = b, f(b) = a and f(c) = c. Now f is a s*g homeomorphism but not a s.h.(C.H).
ii)
Let X = Y = { a,b,c},  = { , {a}, X} and  = { , {b}, {a,b}, Y}. Let
f: (X, )  (Y, ) be defined as f (a) = b, f(b) = a and f(c) = c. Then f is a s.h. (C.H.) but not
a s*g - homeomorphism. Also f is a sgc - homeomorphism.
c)
Let X = Y = {a,b,c},  = { , {a}, {a,b}, X} and  = { , {a}, {b}, {a,b}, Y}. Let
(X, )  (Y, ) be defined as f {a} = b, f(b} = a and f{c} = c.
f:
Now, f is a gc -
homeomorphism but not a s*g - homeomorphism.
Let X = Y = { a,b,c},  = { , {a}, {b,c}, X} and  = { , {a}, {b}, {a,b},{b,c}, Y}.
Let f : (X, )  (Y, ) be the identity mapping. Now, f is a s*g – homeomorphism but not a
gc-homeomorphism. However, if X and Y are T1/2 spaces, then f is a gc - homeomorphism if
and only if f is a s*g - homeomorphism following the fact found in Theorem 3.2.
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d)
i) By (b) ii),f is a sgc- homeomorphism but not a s*g - homeomorphism.
ii) Let X and Y = {a,b,c},  = { , {a}, {b,c}, X} and
 = { , {a}, {c}, {a,c}, {b,c}, Y}. Let f: (X, )  (Y, ) be the identity
mapping. Now f is a s*g - homeomorphism but not a sgc - homeomorphism. However, if X
and Y are Tb spaces, then f is a s*g-homeomorphism if and only if f is a sgc-homeomorphism
following the result in proposition 3.15.
e)
i) Let X = Y = { a, b, c},  = { , {a}, {b}, {a,b}, X} and  = {  , {a, b},
Y}. Let f : (X,  )  (Y, ) be the identity mapping. Now, f is a s*g-homeomorphism but
not a gsc-homeomorphism.
ii)
Let X = Y = {a, b, c},  = { , {a}, {a,b}, X} and  = {  , {a}, {b}, {a, b},
{b,c}, Y}. Let f : (X,  )  (Y, ) be defined as f(a) = b, f(b) = a and f(c) = c. Now, f is a
gsc-homeomorphism but not a s*g-homeomorphism. However, if X and Y are Tb spaces,
then f is a s*g-homeomorphism if and only if f is a gsc-homeomorphism.
PROPOSITION 4.11
For any bijection f : (X,  )  (Y, ), the following statements are equivalent:
a)
f-1 : (Y, )  (X,  ) is s*g-continuous.
b)
f is s*g-open.
c)
f is s*g-closed.
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Proof
Obvious.
PROPOSITION 4.12
Let f : (X,  )  (Y, ) be a bijective and s*g-continuous map. Then the following
statements are equivalent :
a)
f is a s*g-open map.
b)
f is a s*g-homeomorphism.
c)
f is a s*g-closed map.
Proof
Obvious.
5. s*gc-HOMEOMORPHISMS
DEFINITION 5.1
A bijection f : (X,  )  (Y, ) is said to be a s*gc-homeomorphism if f and f-1 are
s*g-irresolute.
PROPOSITION 5.2
Let f : (X,  )  (Y, ) be a map. Then the following are true.
a)
Every homeomorphism is a s*gc-homeomorphism.
b)
Every s*gc-homeomorphism is a g-homeomorphism.
c)
Every s*gc-homeomorphism is a sg-homeomorphism.
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d)
Every s*gc-homeomorphism is a gs-homeomorphism
e)
Every s*gc-homeomorphism is a s*g-homeomorphism
Proof
a)
Obvious by Theorem 1.8 of S.G. Crossley and Hildebrand [6] and Theorem 3.10.
b)
It follows from the result that every open set is s*g-open and every s*g-open set is gopen.
c)
It follows from the result that every s*g-open set is sg-open.
d)
It follows from the result that every s*g-open set is gs-open.
e)
It follows from the result that every open set is s*g-open.
But the converses of (a), (b), (c), (d) and (e) are not true in general as can be seen from
the following examples.
EXAMPLE 5.3
From example 4.6, f is a s*gc-homeomorphism but not a homeomorphism. However,
every s*gc-homeomorphism from a T1/2 space onto itself is a homeomorphism.
EXAMPLE 5.4
From example 4.7, f is a g-homeomorphism, sg-homeomorphism and gshomeomorphism but not a s*gc-homeomorphism. However, every g-homeomorphism from a
T1/2 space onto itself is a s*gc-homeomorphism.
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Every sg-homeomorphim from a Tb space onto itself is a s*gc-homeomorphim.
Every gs-homeomorphism from a Tb space onto itself is a s*gc-homeomorphism.
EXAMPLE 5.5
By example 4.10 (d) (ii), f is a s*g-homeomorphism but not a s*gc-homeomorphism.
However, every s*g-homeomorphism from a T1/2 space onto itself is a s*gc-homeomorphism.
PROPOSITION 5.6
The composition of two s*gc-homeomorphisms is always a s*gc-homeomorphism.
Proof
Obvious.
PROPOSITION 5.7
Let f : (X,  )  (Y, ) be a map. Then the following are true.
a)
s*gc-homeomorphisms and s.h (B) are independent of each other.
b)
s*gc-homeomorphisms and s.h (C.H) are independent of each other.
c)
s*gc-homeomorphisms and gc-homeomorphisms are independent of each other.
d)
s*gc-homeomorphisms and sgc-homeomorphisms are independent of each other.
e)
s*gc-homeomorphisms and gsc-homeomorphisms are independent of each other.
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Proof
a)
From example 4.6, f is a s*gc-homeomorphism but not a s.h(B).
From example 4.7, f is a s.h(B) but not a s*gc-homeomorphism. But, however, every
s*gc-homeomorphism from a T1/2 space onto a semi-T1/2 space is a s.h.(B).
b)
From example 4.6, f is a s*gc-homeomorphism but not a s.h.(C.H).
From example 4.10 (b) (ii), f is a s.h.(C.H) but not a s*gc-homeomorphism.
c)
Let X = Y = {a, b, c},  = {  , {b}, {a, b}, X} and  = {, {a,b}, Y}
Let f : (X,  )  (Y, ) be defined as f(a) = b, f(b) = a and f(c) = c.
Now f is a gc-homeomorphism but not a s*gc-homeomorphism. Similarly, we can
construct s*gc-homeomorphism which is not gc-homeomorphism.
However, if X and Y are T1/2 spaces, then f is a gc-homeomorphism if and only if f is
a s*gc-homeomorphism.
d)
From example 4.10 (b) (ii), f is a sgc-homeomorphism but not a s*gchomeomorphism.
From example 4.10 (e)(i), f is a s*gc – homeomorphism but not a sgchomeomorphism. However, if X and Y are Tb spaces, then, f is a s*gc –
homeomorphism if and only if. f is a sgc – homeomorphism.
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e)
From example 4.10 (e) (i), f is a s*gc-homeomorphism but not a gschomeomorphism.
From example 4.10 (e)(ii), f is a gsc-homeomorphim but not a s*gc –
homeomorphism. However, if X and Y are Tb spaces, then f is a s*gc – homeomorphism if
and only if f is a gsc – homeomorphism.
In order to state the algebraic structure of the set of all s*g-homeomorphisms and
s*gc-homeomorphisms, we introduce the following notations.
5.8.
h(X, ) = { f : f : (X, )  (X, ) is a homeomorphism}.
s*gch(X, ) = { f : f : (X, )  (X, ) is a s*gc-homeomorphism}.
s*gh(X, ) = { f : f : (X, )  (X, ) is a s*g-homeomorphism}.
gsh(X, ) = { f : f : (X, )  (X, ) is a gs-homeomorphism}.
sgh(X, ) = { f : f : (X, )  (X, ) is a sg-homeomorphism}.
gh(X, ) = { f : f : (X, )  (X, ) is a g-homeomorphism}.
THEOREM 5.9
a)
h(X, )  s*gch(X, )  s*gh(X, )  gsh(X, ).
b)
s*gh(X, )  sgh(X, ).
c)
s*gh(X, )  gh(X, ).
d)
The set s*gch(X, ) is a group which contains h(X, ) as a subgroup.
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Proof (a), (b), and (c)
They follow from propositions 4.5 and 5.2.
d) A binary operation  : s*gch(X, ) x s*gch(X, ) s*gch(X, ) is defined by
 (f, h) = h o f (the composition of f and h) for f and h  s*gch(X, ). It follows from the
definitions that s*gch (X, ) is a group with  . Since h(X, ) is a subset of s*gch(X, ) and
it is a group with the binary operation  of s*gch(X, ), h(X, ) is a subgroup of s*gch(X, ).
THEOREM 5.10
Let f : (X, )  (Y,) be a s*gc-homeomorphism. Then, it induces an isomorphism
from the group s*gch(X, ) onto s*gch (Y, ).
Proof
The homomorphism f* : s*gch(X, )  s*gch(Y, ) is induced from f by f*(h)
= fohof-1 for every h  s*gch(X, ). The f* is a bijection and also f* is an isomorphism by
usual argument.
But the converse is not always true as can be seen from the following example.
Example 5.11
Let X = Y = {a, b, c},  = {  , {a}, X} and  = {, {a,b}, Y}.
Let f : (X,  )  (Y, ) be defined as f(a) = c, f(b) = b and f(c) = a.
Let hc : (X,  )  (X,  ) and ha : (Y, )  (Y, ) denote maps defined by hc(a) =a,
hc(b) = c, hc(c) = b and ha(a) = b, ha(b) = a, ha(c) = c respectively. Then, we have that
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s*gch(X,  ) = {hc, Ix} and s*gch(Y, ) = {ha, Iy} hold where Ix :(X,  )  (X, ) and Iy :
(Y, )  (Y, ) are the identity maps. Since f*(hc) = f o hc of-1 = ha and f*(Ix) = Iy, the
induced homomorphism f* : s*gch(X,  )  s*gch(Y, ) is an isomorphism. However, f is
not a s*gc-homeomorphism.
REMARK 5.12
From remark 4.21 of Devi et.al [8], definitions, Theorems and examples, we have the
following diagram of implications :
gc-homeomorphism
semi - homeomorphism(B)
homeomorphism
s*gc - homeomorphism
semi - homeomorphism (C.H)
Where A
gsc-homeomorphism
B (resp. A
sgc - homeomorphism
g - homeomorphism
gs - homeomorphism
s*g - homeomorphism
sg - homeomorphism
B) represents A implies B (resp. A does not always imply B).
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