CHAPTER 4
OPEN, CLOSED AND CONTINUOUS FUNCTIONS VIA
GENERALIZED PREOPEN SETS
4.1
INTRODUTION
In 1970, Levine [48] introduced and studied the notions of
generalized closed (g-closed) sets. In 1982, Malghan [58] introduced and
studied the concept of generalized closed (briefly, g-closed) functions. After
that several topologists like Sundaram [90], Mashhour et al [60] and Devi et
al [29] introduced and studied g-open functions pre-closed functions and agclosed functions respectively.
This chapter contains four sections.
Second section deals with
stronger form of preopen function namely (s, p) - open function and (s, p) closed function and obtain some of their properties.
In the third section, we introduce (p, gp) - closed functions , (p, gp)
- open functions, strongly gp-closed functions, strongly gp-open functions,
pre gp-closed functions, pre gp-open functions and (gp, gp)- open functions
and investigate some of their properties.
Moreover in the last section, we define Rarely gp-continuous
functions, I. gp-continuous functions and study some of their properties.
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4.2 SOME MORE PROPERTIES OF PREOPEN FUNCTIONS
In this section, we study some results on preopen functions and
also we introduce and investigate some properties of (s, p) - open, (s, p) closed functions in topological spaces.
DEFINITION 4.2.1 [60]: A function f: X —» Y is called preopen if the
image of each open set in X is preopen in Y.
Clearly, every open function is preopen but not the converse which is shown
by the following example
EXAMPLE 4.2.2: Let X = { p, q, r},
x = {
§
, {p}, {p, q}, X },
Y = {a, b, c}, a = { <)>, {a}, {a, b}, Y} then it is clear that x and a are
topologies on X and Y respectively. If f : X —» Y is a function defined by
f(p) = a, f(q) = c and f(r) = b. It is clear that f is a preopen function but not
open function. Since f _1({p, q}) = {a, c} which is not in o.
DEFINITION 4.2.3:[70] A function f: X —> Y is called semipreopen if the
image of each open set in X is a semipreopen set in Y.
DEFINTION 4.2.4 [12]: A function f : X —> Y is called semiopen if the
image of each open set in X is a semiopen set in Y.
Clearly, every preopen (resp. semiopen) function is semipreopen but
converses are not true in general.
EXAMPLE 4.2.5 : Let f : X —> Y be semiopen but not preopen. Let X =
{ P, q, r}, x= { <|>, {p}, {p, q},X }, Y= {a, b, c}, a ={<(>, {a}, {a, b}, Y}
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then it is clear that x and o are topologies on X and Y respectively.
If f: X —> Y is a function defined by f(p) = a, f(q) = c and f(r) = b. Then f is
semipreopen but not open. Since f *'({p, q}) = {a, c} which is not in a.
EXAMPLE 4.2.6: Let f: X —> Y be semipreopen but not semiopen
LetX= { p, q, r}, x = { <J>. {p}, {p, q}, X }, Y = {a, b, c},
cr = { ((), {a},
{b, c}, Y} then it is clear that x and a are topologies on X and Y
respectively. If f: X —> Y is a function defined by f(p) = a, f(q) = c and f(r)
= b.
Then f is semiopen but not semipreopen, since {a,c} is not
semipreopen.
We recall the following
DEFINITION 4.2.7 [33]: A function f : X —> Y is preclosed if for each
closed set F of X, f(F) is preclosed in Y.
We define a stronger form of preopen function in the following,
DEFINITION 4.2.8 : A function f : X —> Y is said to be (s,p) - open if
image of each semiopen set of X is preopen in Y.
Clearly, every (s, p) - open function is preopen but not converse.
EXAMPLE. 4.2.9: Let f: X —> Y is preopen but not (s,p) - open.
DEFINITION 4.2.10 : A function f: X —> Y is said to be (s,p) - closed if
image of each semiclosed set of X is preclosed in Y.
Clearly, every (s, p) - closed function is preclosed but converse is not true.
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EXAMPLE 4.2.11 : Let f : X —> Y is a (s, p) - closed function but not
preclosed function. X =Y = {a,b,c} and x = {X, <|>, {a}, {b,c}} and
o = {Y , c|), {a}}. Define a function f : X —> Y by f(a) =b, f(b) = a and
f(c) = c. Then f is preclosed but not (s, p) - closed function, since for the
preclosed set {c} in Y, f ({c}) = {c} is not a semi-closed in X.
LEMMA 4.2.12 [31, Lemma 6.3] : If f : X —» Y is a preclosed function,
then for each subset S c Y and each open set U c X containing f _I(S) there
exist V € PO(Y) such that S c V and f ‘(V) c U.
PROOF: (i) Let f is preclosed function. Let S c Y and U is open set of X
containing f '5(S). Since f _1(S) c U. Put V = Y - f(X -U). Then V is
preopen in Y, S c: V and f'(V) cz U.
(ii) Let F be any closed set of X. put S = Y - f (F), then we have f '(S) cz X
- F and X - F is open set of X. There exists a preopen set V of Y such that S
= Y - f (F) cz V and f 4(V) c X - F. Thus, we obtain f (F) = Y - V, which is
preclosed. Hence f is preclosed.
THEOREM 4.2.13: A f : X —> Y is (s,p) - open iff every subset AcX,
f(sInt(A)) c: Int (Cl(f(A)).
PROOF. Let f be a (s,p)-open function we have f(sInt(A)) cz f( A) for each A
c X and by definition, f(sInt(A)) is a preopen set in Y and by definition,
f(sInt(A)) c Int(Cl(A)). Conversely, Let the given condition holds ture and
G any semiopen set in X. Then f(G) = f(sInt(G)) cz Int(Cl(f(G)) which
implies that f(G) <z Int(Cl(f(G)). Thus f(G) is a preopen set in Y and hence f
is a (s,p)-open function. Similarly we prove the following
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THEOREM 4.2.14 : A f : X —» Y is (s,p) -closed iff every subset FcX,
Cl(Int(f(F)) c f(sCl(F)).
PROOF. Let f be a (s,p)-closed function. We have f(F) c f(sCl(F)) for
each subset F of X and by hypothesis f(sCl(f(F)) is a preclosed set in Y and
by definition Cl(Int(f(F)) c f(sCl(F)). Conversely, let the given condition
holds true and H is any semi-closed set in X then f(H) = f(sCl(H)) =>
Cl(Int(f(H)) which implies that Cl(Int(f(H)) <z f(H).
Thus f (H) is a
preclosed set in Y by definition and hence f is a (s, p) - closed function.
THEOREM 4.2.15 :
Let X,Y.Z be three topological spaces and let
f: X —> Y and g:Y —> Z be two mapping with gof : X —> Z is a semiopen
function then
(i)
If f is continuous and surjective, then g is semiopen.
(ii)
if g is irresolute and injective, then f is semiopen.
PROOF, (i) Let U be an arbitrary open set in Y. Since gof is semiopen and
f is continuous and surjective then g(U) = gof{ f ’(U)} is a semiopen set in
Z. This shows that g is a semiopen function.
(ii) Since g is injective, we remark that f (A) = g '(g(f(A)) for every subset
A of X. Let U be an arbitrary open set in X, then by hypothesis gof (U) is a
semiopen set in Z. Again f (U) = g ’'(gof(U)) is semiopen in Y. Since g is
irresolute and injective. This shows that f(U) is semiopen in Y. Hence, f is
a semiopen function.
THEOREM 4.2.16:
Let X, Y, Z be three topological spaces and let
f: X —» Y and g:Y —» Z be two mapping with gof : X —» Z is a preopen
function then
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(i)
If f is continuous and surjective, then g is preopen.
(ii)
if g is preirresolute and injective, then f is preopen.
PROOF, (i) Let U be an arbitrary open set in Y. Since gof is preopen and
f is continuous and surjective then g (U) = gof {f'(U)} is a preopen set in Z.
This shows that g is a preopen function.
(ii) Since g is injective, we remark that f (A) = g '(g (f (A)) for every subset
A of X. Let U be an arbitrary open set in X, then by hypothesis gof (U) is a
preopen set in Z. Again f (U) = g ‘‘(gof (U» is preopen in Y. Since g is
irresolute and injective. This shows that f(U) is preopen in Y. Hence, f is a
preopen function.
THEOREM 4.2.17:
Let X, Y, Z be three topological spaces and let
f: X -» Y and g:Y —> Z be two mapping with gof: X -» Z is a (s, p) - open
function then
(i)
If f is irresolute and surjective, then g is (s,p) - open.
(ii)
if g is preirresolute and injective, the f is (s,p) - open.
PROOF, (i) Let U be an arbitrary semiopen set in Y. Since gof is (s,p) open and f is irresolute and surjective then g(U) = gof {f *‘(U)} is a preopen
set in Z. This shows that g is a (s,p) - open function.
(ii) Since g is injective, we remark that f (A) = g ‘(g (f (A)) for every subset
A of X. Let U be an arbitrary semiopen set in X, then by hypothesis gof (U)
is a preopen set in Z. Again f (U) = g ‘(gof (U)) is preopen in Y. Since g is
irresolute and injective. This shows that f (U) is preopen in Y. Hence, f is a
(s, p) - open function.
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THEOREM 4.2.18 : Let f: X —> Y be presemiopen and g : Y —> Z be (s,p) open then gof: X —> Z is (s, p) - open.
PROOF. Let U be an arbitrary semiopen set in X. Since f is presemi-open,
f (U) is semiopen in Y, again since g is (s, p) - open, and f(U) is semiopen,
g(f(U) = gof (U) is preopen in Z.
THEOREM 4.2.19 : Let f: X —> Y be open and g:Y —> Z be preopen then
gof: X —> Z is a preopen.
PROOF: Obvious.
THEOREM 4.2.20 : Let f: X —> Y be closed and g:Y —> Z be preclosed
then gof: X —» Z is preclosed.
PROOF: Obvious.
THEOREM 4.2.21 : Let f: X —» Y be presemiclosed and g:Y —» Z be
(s, p) -closed then gof: X —> Z is (s, p) - closed.
PROOF : Let F be any arbitrary subset of X. Since f is presemiclosed,
f (F) is semiclosed in Y. Again, since g is (s, p)-closed and f(F) is semiclosed, g(f(F)) = gof(F) is preclosed in Z. This shows that gof is (s, p) closed.
THEOREM 4.2.22 : Let f: X —» Y be preclosed and g:Y —> Z be
M - preclosed then gof: X —» Z is preclosed.
PROOF: Let F be any closed subset of X. Since f is preclosed, f(F) is
preclosed in Y. Again, since g is M - preclosed and f (F) is preclosed, g
(f(F)) = gof(F) is preclosed in Z. This shows that gof is preclosed.
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THEOREM 4.2.23 : Let f : X —> Y be preopen and g :Y —» Z be
M - preopen then gof: X —» Z is preopen.
PROOF: Similar to proof of the Theorem 4.2.22.
THEOREM 4.2.24 : Let f: X —> Y be (s, p) - open and g :Y —> Z be
M -preopen then gof: X —> Z is (s, p) - open.
PROOF : Let U be anyu semiopen set in X. Since f is (s, p) - open, f(U) is
preopen in Y. Again, since g is M - preopen and f (U) is preopen, g(f(U)) =
gof (U) is preopen in Z. This shows gof is (s, p) - open.
THEOREM 4.2.25 : Let f : X
Y be (s, p) - closed and g :Y -> Z
be M -preclosed then gof: X —> Z is (s, p) - closed.
PROOF: Similar to proof of the Theorem 4.2.24.
THEOREM 4.2.26. Let X, Y and Z be the topological spaces and Let
f : X —» Y and g : Y
—>
Z be two mappings and gof :X-)Za weakly
preopen, then if f is continuous and surjective then g is weakly preopen.
PROOF. Let U be an arbitrary open set in Y then f '(U) is open in X since
f is continuous. Also gof is weakly preopen and f is surjective, g(U) =
(gof)(f -'(U) c pint (gof(Cl(f -’(U)) <= pint(gof(f'1 (C1(U))) = pint(g(Cl(U))).
This shows that g is weakly preopen.
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4.3
GENERALIZED PREOPEN FUNCTIONS
In this section, we introduce (p, gp) - closed functions, (p,gp) -
open functions, strongly gp-closed functions, strongly gp-open functions, pre
gp-closed functions, pre gp-open functions and (gp,gp)- open functions and
investigate some of their properties.
DEFINITION 4.3.1: A f: X —> Y is called (p, gp) - closed if image of each
gp - closed set of X is preclosed in Y.
THEOREM 4.3.2: Every (p, gp) - closed function is preclosed but not
conversely.
PROOF : Let f: X —> Y is called (p, gp) - closed function and F be a closed
set in X it is also gp-closed in X. Then f (F) is preclosed set in Y. Hence f is
preclosed function.
EXAMPLE 4.3.3: Let X = {a, b, c} =Y, x = {X, (|), {a}} and a = { Y, <|>,
{a,b}}. Let f: X —> Y be the identity function. Then the function f is
preclosed but not (p, gp)-closed function, since for the gp-closed set {a, b}
in X, f ({a, b}) = {a, b} is not a preclosed set in Y.
THEOREM 4.3.4 : A surjective function f: X —» Y is (p,gp) - closed iff
for each subset B of Y and each gp-open set U of X containing f '(B), there
exists a preopen set V of Y such that BcV and f '(V) <z U.
PROOF: Necessity. Suppose that f is (p, gp) - closed. Let B be any subset
of Y and U be gp - open set of X containing f '(B). Put V = Y Then V is preopen in Y, B <z V and f ‘(V) a U.
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f(X - U).
Sufficiency. Let F be any gp - closed set of X. Put B = Y - f (F), then
we have f ‘(B) cz X - F and X - F is gp -open. There exists a preopen set V
of Y such that B = Y - f (F) c V and f '(V) c X - F. Therefore we obtain
f (F) = Y - V and hence f(F) is preclosed in Y. This shows that f is (p, gp) closed.
DEFINITION 4.3.5: A f: X —» Y is called (p, gp) - open if image of each
gp - open set of X is preopen in Y.
Clearly (p, gp) - open is preopen but not conversely.
EXAMPLE 4.3.6: In example 4.3.3, The function f is preopen but not a
(p,gp) - open function , since for the gp-open set {c} in X, f({c}) = {c} is not
preopen in Y.
THEOREM 4.3.7 : If f : X —> Y is (p,gp) - open iff for each subset B of Y
and each gp -closed set U of X containing f _1(B), there exists a preclosed
set V of Y such that B c V and f '(V) c U.
PROOF: Proof is similar to 4.3.4.
We define the following
DEFINITION 4.3.8: A function f: X —> Y is called strongly gp -closed if
the image of gp - closed set of X is closed in Y.
Clearly, every strongly gp-closed function is closed but not converse.
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EXAMPLE 4.3.9: Let X = {a, b, c} =Y, x = {X, <|>, {a}} and
a = { Y, <(), {a},{b},{a,b}}. Define a function f: X —» Y by f(a) = b, fib) = a
and fic) = c.
Then the function f is closed but not strongly gp-closed
function, since for the gp-closed set {b} in X, fi{b}) = {a} is not a closed set
in Y.
THEOREM 4.3.10 : A surjective function f: X —» Y is strongly gp-closed
iff for each subset B of Y and each gp-open set U of X containing f _1(B),
there exists a open set V of Y such that B c V and f '(V) c: U.
PROOF: Necessity. Suppose that f is strongly gp - closed. Let B be any
subset of Y and U be gp - open set of X containing f '(B). Put V = Y f (X - U). Then V is open in Y, B c V and f ‘‘(V) c U.
Sufficiency. Let F be any gp - closed set of X. Put B = Y - f (F), then
we have f '(B) c X - F and X - F is gp -open in X. There exists a preopen
set V of Y such that B = Y - f (F) c V and f _1(V) c X - F. Therefore we
obtain f (F) = Y - V and hence f (F) is closed in Y. This shows that f is
strongly gp - closed.
DEFINITION 4.3.11: A function f: X —> Y is called strongly gp -open if
the image of gp-open set of X is open in Y.
Clearly, every strongly gp - open function is open but converse is not true in
general.
EXAMPLE 4.3.12: In Example 4.3.9, the function f is an open function but
not strongly gp-open function, beause for the gp-open set {a,c} in X, f({a,c)}
= f b,c} is not an open set in Y.
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DEFINITION 4.3.13: A function f: X —» Y is called (gp, gp) - closed if the
image of gp - closed set of X is gp - closed in Y.
Clearly, every (gp, gp) - closed function is pre gp-closed. But converse is not
true.
EXAMPLE 4.3.14: Let X = {a, b, c} =Y, x ={ X, <|>, {a}} and a = { Y, <|>,
{a}, {a,c}}. Then the identity function f : X
—>
Y is pre gp-closed but not
(gp,gp) - closed function because for the gp-closed set {a,c} in X, f({a,c}) =
{a,c} is not a gp-closed set in Y.
DEFINITION 4.3.15: [57]: A function f: X —> Y is called pre gp -closed if
the image of preclosed set of X is gp-closed in Y.
DEFINITION 4.3.16: A function f : X —> Y is called pre generalized
preopen (briefly pre gp -open) if the image of preopen set of X is gp-open
in Y.
EXAMPLE 4.3.17: In the example 4.3.14, the function f is pre gp-open
function
THEOREM 4.3.18 : A surjective function f: X —> Y is pre gp-open iff for
each subset B of Y and each preclosed set U of X containing f '(B), there
exists a gp-closed set V of Y such that BcV and f _1(V) <z U.
PROOF: Necessity. Suppose that f is pre gp-open. Let B be any subset of
Y and U be an preclosed set of X containing f ~'(B). Put V = Y - f(X - U).
Then V is gp-closed in Y, B cz V and f _I(V) c U.
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Sufficiency. Let F be any preopen set of X. Put B = Y - f (F), then
we have f ’(B) c X - F and X - F is preclosed. There exists a preopen set V
of Y such that B = Y - f (F) cz V and f !(V) cz X - F. Therefore we obtain
f (F) = Y - V and hence f (F) is gp-open in Y. This shows that f is pre gpopen.
DEFINITION 4.3.19: A function f: X -> Y is called (gp, gp) - open if the
image of gp - open set U of X f (U) is gp - open in Y.
Clearly, every (gp, gp) - open function is pre gp-open but converse is
not true.
EXAMPLE 4.3.20: In the Example 4.3.14, the function f is pre gp-open but
not (gp,gp) - open function, since for the gp-open set {b} in X, f({b}) = {b}
is not a gp-open set in Y.
THEOREM 4.3.21 : A surjective function f : X —> Y is (gp,gp)-open iff
for each subset B of Y and each gp-closed set U of X containing f _1(B),
there exists a gp-closed set V of Y such that BcV and f _1(V) cz U.
PROOF: Necessity. Suppose that f is (gp, gp) - open. Let B be any subset
of Y and U be an gp-closed set of X containing f '(B). Put V = Y - f(X - U).
Then V is gp-closed in Y, B cz V and f'(V) c U.
Sufficiency. Let F be any gp-open set of X. Put B = Y - f (F), then
we have f '(B) cX-F and X - F is gp-closed. There exists a gp-closed set
V of Y such that B = Y- f(F)cV and f '(V) cX-F. Therefore we obtain
f (F) = Y - V and hence f (F) is gp-open in Y. This shows that f is (gp, gp) open.
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THEOREM 4.3.22: If the function f : X —> Y is strongly gp - open (resp.
strongly gp - closed) and the function g : Y —> Z is gp-open (resp. gp closed) then the composition gof : X —> Z is (gp,gp) - open (resp.(gp,gp) closed).
PROOF: Let U be an arbitrary gp-open (resp. gp-closed) set in X then f (U)
is open (resp. closed) in Y. Again f (U) is open (resp. closed) in Y and g is
gp-open [gp-closed], g (f (U)) = gof (U) is gp - open (resp.gp - closed) in Z.
This shows that function gof is (gp, gp) - open (resp. (gp, gp) - closed).
THEOREM 4.3.23: If the function f: X —> Y is pre gp - open (resp. pre gp
- closed) and the function g : Y —>Z is (gp, gp)-open (resp.(gp, gp) - closed)
then the composition gof: X —> Z is pre gp - open (resp. pre gp - closed).
PROOF: Let U be an arbitrary preopen (preclosed) set in X then f (U) is gpopen (resp. gp - closed) in Y. Again f (U) is gp-open (resp. gp-closed) in Y
and g is (gp, gp) - open (resp. (gp, gp)-closed), g (f (U)) = gof (U) is gp-open
(resp. gp-closed) in Z. This shows that function gof is pre gp - open (resp.
pre gp - closed).
THEOREM 4.3.24: If the function f : X —> Y is (gp, gp) - open (resp.
(gp, gp) - closed) and the function g : Y -> Z is (gp,gp) - open (resp. (gp,gp)
- closed) then the composition gof: X -> Z is (gp,gp) - open (resp. (gp,gp) closed).
PROOF: Let U be an arbitrary gp-open (resp. gp-closed) set in X then
f (U) is gp-open (resp. gp-closed) in Y. Again f (U) is gp-open (resp. gpclosed) in Y and g is gp-open (resp gp-closed), g (f (U)) = gof (U) is gpopen (resp. gp-closed) in Z. This shows that function gof is (gp, gp) - open
(resp. (gp, gp) - closed).
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THEOREM 4.3.25: If the function f: X —» Y is (gp, gp) - open (resp.
(gp, gp) - closed) and the function g: Y —>Z is strongly gp-open (resp.
strongly gp-closed) then the composition gof: X —> Z is strongly gp - open
(resp. strongly gp - closed)
PROOF: Let U be an arbitrary gp-open (resp. gp-closed) set in X then f (U)
is gp-open (resp. gp-closed) in Y.
Again f (U) is gp - open (resp.
gp - closed) in Y and g is strongly gp-open (resp. strongly gp - closed), g (f
(U)) = gof (U) is open (resp. closed) in Z. This shows that function
gof is
strongly gp - open (resp. strongly gp-closed).
THEOREM 4.3.26: If the function f : X —> Y is M preopen (resp.
M -preclosed) and the function g : Y —»Z is pre gp-open (resp. pre
gp-closed) then the composition gof : X —> Z is pre gp - open(resp. pre
gp - closed)
PROOF: Let U be an arbitrary preopen [resp. preclosed] set in X then
f(U) is preopen (resp. preclosed) in Y. Again f (U) is preopen (resp.
preclosed) in Y and g is pre gp-open (resp. pre gp-closed), g (f (U)) =
gof (U) is gp - open (resp.gp - closed) in Z. This shows that function gof is
pre gp - open (resp. pre gp-closed).
THEOREM 4.3.27: If the function f : X —> Y is open (resp. closed) and
the function g : Y —>Z is gp-open (resp. gp - closed) then the composition
gof: X —> Z is gp - open (resp. gp - closed)
PROOF: Let U be an arbitrary open (resp.closed) set in X then f (U) is open
(resp. closed) in Y. Again f (U) is open (resp.closed) in Y and g is gp-open
(resp.gp-closed), g (f (U)) = gof (U) is gp-open (resp.gp-closed) in Z. This
shows that function gof is gp - open (resp.gp-closed).
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THEOREM 4.3.28: If the function f : X -> Y is gp - open (resp.
gp - closed) and the function g : Y -»Z is strongly gp-open (resp. strongly
gp-closed) then the composition gof: X —> Z is open (resp. closed).
PROOF: Let U be an arbitrary open (resp. closed) set in X then f(U) is
gp-open (resp. gp-closed) in Y. Again f (U) is gp-open (resp. gp-closed) in
Y and g is gp-open (resp. gp-closed), g (f (U)) = gof (U) is open (resp.
closed) in Z. This shows that function gof is open (resp. closed).
THEOREM 4.3.29: Let f and g be two functions. Then
i)
If f is gp - closed and g is strongly gp - closed, then the
composition gof is closed.
ii)
If f is closed and g is gp - closed, then the composition gof is
gp - closed.
PROOF: (i)
Let F be closed subset of X then f (F) is gp - closed in Y.
Since f(F) is gp -closed in Y and g is strongly gp -closed, g(f(F)) = gof (F) is
closed set in Z. This shows that gof is closed function.
(ii)
Let F be closed subset of X. Since f is closed then f (F) is closed in Y.
Again since g is gp-closed, g(f(F)) is gp - closed set in Z. But g(f(F)) =
gof(F) Therefore gof(F) is gp - closed set in Z. This shows that gof is
gp - closed function.
THEOREM 4.3.30 : A surjective function f: X —» Y is (gp,gp) - closed iff
for each subset B of Y and each gp-open set U of X containing f "'(B), there
exists a gp-open set V of Y such that B c V and f ‘(V) <z U.
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PROOF: Necessity. Suppose that f is (gp, gp) - closed . Let B be any
subset of Y and U be gp - open set of X containing f '(B). Put V = Y f(X - U). Then V is gp-open in Y, B c V and f *(V) c U.
Sufficiency. Let F be any gp - closed set of X. Put B = Y - f (F), then
we have f '(B) cX-F and X - F is gp - open. There exists a gp-open set
Vof Y such that B = Y - f (F) c= V and f '(V) c X - F. Therefore we obtain
f(F) = Y - V and hence f (F) is gp - closed in Y. This shows that f is (gp, gp)
- closed.
DEFINITION 4.3.31: A function f: X -> Y is called perfectly gp-open if
the image of every gp-open set of X is clopen in Y.
THEOREM 4.3.32: If a function f: X —> Y is perfectly gp-open then f is
strongly gp - open.
PROOF : Let U be any gp-open set of X and f: X —> Y is perfectly gp-open
function then f(U) is clopen in Y. But every clopen is open and hence f (U)
is open in Y. This shows that f is strongly gp-open.
Converse of above theorem is not true in general.
DEFINITION 4.3.33 : A function f: X -» Y is called perfectly open if the
image of every open set of X is clopen in Y.
THEOREM 4.3.34: If f : X
—>
Y is perfectly gp-open, the f is perfectly
open.
PROOF : Let U be an arbitrary subset of X and f : X —» Y is perfectly
gp-open. Then f(U) is clopen in Y. But every clopen set is open and hence
f (U) is open in Y. This shows that f is perfectly open.
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NOTE 4.3.35: From above results we have following implication diagram
Str. gp-open
(p, gp) - open
(Str. gp-closed)
((p, gp) - closed)
Here ( A------- ► B) and (A ◄-------► B) represents A implies B but not
conversely (A and B are independent of each other).
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4.4
RARELY gp - CONTINUOUS FUNCTIONS
In this section we define Rarely gp-continuous functions, I.
gp-continuous functions and study some of their properties.
We recall that a rare set is a set R such that Int(R) = <j) [84].
A nowhere dense set, is a set R in which Int(Cl(R)) = 4> if Cl(R)is codense.
DEFINITION 4.4.1: A function f: X —» Y is called Weakly gp-continuous
if for each x
g
X and each open set G containing f(x), there exists
U e 0(X, x) (resp. U e GPO(X)) such that f (U) c Cl (G).
DEFINITION 4.4.2: A function f: X —> Y is called rarely gp-continuous if
for each x g X and each G g 0(Y, f(x», there exist a rare set RG with G n Cl
(Rg) = 0 andUG GPO(X, x) such that f(U) c G u R0.
THEOREM 4.4.3: The following statements are equivalent for a function
f: X -» Y :
(i)
f is rarely gp-continuous at x
(ii)
For each set G
g
g
X.
0(Y, f(x)), there exists U
g
GPO(X, x) such
g
GPO(X, x) such
that Int [f (U) n (Y \ G)] = 0.
(iii)
For each set G
g
0(Y, f(x)), there exists U
that Int [f(U)]ciCl (G)].
PROOF: (i) -> (ii): Let G
fact that Int (C1(G))
g
g
0(Y, f(x». By f(x)
g
G c Int (Cl (G» and the
0(Y, f(x», there exist a rare set Rc with Int(Cl(G)) \
C1(Rg) = 0, and a gp-open set Uc X containing x such that f(U) c
Int(Cl(G)) uR0. We have Int [ f(U) n (Y - G)] = Int [f(U>] n Int(Y - G) c
Int [C1(G) u Rg] n (Y - C1(G)) c (C1(G) u Int(R0» n (Y - C1(G)) = 0.
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(ii) -» (iii): It is straightforward.
(iii) -» (i): Let G e 0(Y, f(x)). Then by (iii), there exists U <e GPO(X, x)
such that Int[f(U)] c C1(G). We have f(U) = [f(U) - Int(f(U))] u Int(f(U)) c
[f(U> - Int(f(U)>] u C1(G) = [f(U) - Int(f(U)>] uGu (C1(G) - G) = [f(U) Int(f(U))] n (Y - G) u G u (C1(G) - G). Set R* = [f (U) - Int (f (U))] n (Y G) and R** = (Cl (G) - G). Then R* and R** are rare sets. Moreover RG =
R* u R** is a rare set such that Cl (R0) n G = 0 and f (U) cGu RG. This
shows that f is rarely gp -continuous.
We define the following notion which is a new generalization of
gp-continuity.
DEFINITION 4.4.4: A function f: X —> Y is I. gp-continuous at x e X if
for each set G e 0(Y, f(x)), there exists U e GPO(X, x) such that Int (f(U))
c: G.
If f has this property at each point x e X, then we say that f is I.gpcontinuous on X.
EXAMPLE 4.4.5: Let X = Y = {a, b, c} and t = a = {X,0, {a}}. Then a
function f: X —> Y defined by f (a) = f (b) = a and f(c) = c is I.
gp-continuous.
REMARK 4.4.6: Since, if f: X —> Y is gp-continuous, then for each point
x e X and each open set V containing f(x), there exists U e GPO(X, x) such
that f (U) <= V. Then, it should be noted that I.gp-continuity is weaker than
gp-continuity and stronger than rare gp-continuity.
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THEOREM 4.4.7: Let Y be a regular space. Then the function f: X
Y is
I.gp-continuous on X if and only if f is rarely gp-continuous on X.
PROOF: We prove only the sufficient condition since the necessity
condition is evident Remark 4.4.6. Let f be rarely gp-continuous on X and x
g
X. Suppose that f(x)
g
G, where G is an open set in Y. By the regularity
of Y, there exists an open set Gj
g
0(Y, f(x)) such that Cl (Gi) c G. Since
f is rarely gp-continuous, then there exists U
g
GPO(X, x) such that Int [f
(U)] <= Cl (GO Theorem 4.4.3. This implies that Int [f (U)] c= G and
therefore f is I.gp-continuous on X.
We say that a function f: X —> Y is r.gp-open if the image of a
gp-open set is open.
THEOREM 4.4.8: If f: X -» Y be an r. gp-open rarely gp-continuous
function, then f is weakly gp - continuous.
PROOF: Suppose that x
g
X and G
g
0(Y, f(x)). Since f is rarely
gp-continuous, there exist a rare set Rc with Cl (Rg) n U = 0. where U
g
GPO(X, x) such that f(U) c= G u Rg- This means that (f (U) n (Y \ Cl (G))
c Rg. Since the function f is r.gp-open then f (U) n (Y \ Cl (G)) is open. But
the rare set RG has no interior points. Then f (U) n (Y \ Cl (G)) = 0.This
implies that f (U) c Cl (G) and thus f is weakly gp - continuous.
THEOREM 4.4.9: Let GPO(X,x) closed under finite intersections.
If f : X —> Y is rarely gp - continuous function, then the graph function
g: X —> X x Y, defined by g(x) = (x, f(x)) for every x in X is rarely
gp - continuous.
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PROOF: Suppose that x
g
X and W is any open set containing g(x).
It follows that there exist open sets U and V in X and Y, respectively, such
that (x; f(x))
G
g
g
U x V c W. Since f is rarely gp - continuous, there exists
GPO(X, x) such that Int [f (G)] c Cl (V). Let E = U n G. It follows that
E e GPO(X, x) and we have Int(g(E)) c: Int(U x f(G)) <z U x C1(V ) c
Cl (W). Therefore, g is rarely gp - continuous.
DEFINITION 4.4.10: Let A = {Gj} be a class of subsets of X. By rarely
union sets [42] of A we mean {Gj u RGi }, where each RGi is a rare set such
that each of {G* n Cl (RGj) is empty.
A topological space X is said to be rarely almost compact if the set X
is rarely almost compact relative to X.
THEOREM 4.4.11: Let f: X —» Y be rarely gp-continuous and K an GPOcompact set relative to X. Then f (K) is rarely almost compact subset
relative to Y.
PROOF: Suppose that Q is a open cover of f (K). Let B be the set of all V
in O such that V n f(K) * 0. Then B is an open cover of f (K). Hence for
each k
g
K, there is some Vk
g
B such that f(k)
g
Vk. Since f is rarely gp-
continuous there exist a rare set RVk with Vk \ Cl (RVk) = 0 and a gp-open
set Uk containing k such that f(Uk) c Vk u Rvk- Hence there is a finite
subfamily |Uk}k e
a
which covers K, where A is a finite subset of K. The
subfamily {Vk u RVk} ke a also covers f (K).
COROLLORY 4.4.12: Let f: X —» Y is rarely continuous and K is a GPOcompact set in X. Then f (K) is rarely almost compact subset of Y.
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LEMMA 4.4.13: (Long and Herrington [45]). If g: Y —> Z is continuous and
one-to-one, then g preserves rare sets.
THEOREM 4.4.14: If f: X —> Y is rarely gp-continuous and g : Y —> Z is
continuous and one-to-one, then g o f: X —> Z is rarely gp - continuous.
PROOF: Suppose that x e X and (g o f) (x) e V, where V is an open set in
Z. By hypothesis, g is continuous, therefore there exists an open set G <z Y
containing f(x) such that g (G) c V. Since f is rarely gp-continuous, there
exists a rare set RG with G n Cl (RG) = 0 and a gp-open set U containing x
such that f (U) cGu RG. It follows from Lemma 4.4.13 that g (RG) is a rare
set in Z. Since RG is a subset of Y \ G and g is injective, we have Cl (g (RG))
n V = 0. This implies that (g o f) (U) cVug (RG). Hence the result.
THEOREM 4.4.15: Let f: X -> Y be pre-gp-open and g : Y —> Z a function
such that g o f : X —> Z is rarely gp-continuous. Then g is rarely gpcontinuous.
PROOF: Let y e Y and x e X such that f(x) = y. Let G e O (Z, (g o f)
(x)). Since g o f is rarely gp-continuous, there exists a rare set RG with G n
Cl (Rg) = 0 and U e GPO(X, x) such that (g o f) (U) <z G uRG. But f (U)
(say V) is a gp-open set containing f(x). Therefore, there exists a rare set RG
with G n C1(Rg) = 0 and V e GPO(Y, y) such that g(V )cGuRg, i.e. g
is rarely gp-continuous.
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