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Journal of Al-Nahrain University
Science
Vol.17 (1), March, 2014, pp.160-166
Minimal and Maximal Beta Open Sets
Qays Rashid Shakir
Operations Management Techniques Department-Technical College of Management Baghdad,
Baghdad-Iraq.
E-mail: qays.shakir@gmail.com.
Abstract
The present paper deals and discusses new types of sets all of these concepts completely
depended on the concept of Beta open set. The importance concepts which introduced in this paper
are minimal -open and maximal -open sets. Besides, new types of topological spaces introduced
which called Tminand Tmaxspaces. Also we present new two maps of continuity which called
minimal -continuous and maximal -continuous. Additionally we investigated some fundamental
properties of the concepts which presented in this paper.
Keywords: minimal -open set, maximal
-continuous
-open
set, minimal
-continuous
and maximal
contained in A is called the interior of A and
denoted by AO and the intersection of all
closed subset of X which contain A is called
the closure of A and denoted by A.
Introduction
Minimal and maximal sets play an
important role in the researches of generalized
topological spaces, Nakaoka and Oda
introduced these concepts in [1] and [2] and
they used them to investigate many topological
properties. In this paper we introduced
the notion of minimal -open and maximal
-open and their complements.
Definition (6) [5]:
A subset A of a space X is called a
o
-open
set if A A .The complement of a -open
set is defined to be a -closed set.
Definition (7) [5]: Let X and Y be topological
spaces and f:XY is a map then f is called a
Definition (1) [1]:
A proper nonempty open subset O of a
topological space X is said to be minimal open
set if any open set which is contained in O is
or O.
-continuous function if f1(A) is a -open
set in X for every open set A in Y.
Minimal and Maximal
Definition (2) [2]:
A proper nonempty open subset O of a
topological space X is said to be maximal open
set if any open set which is contains O is O or
X.
-open sets
Definition (8):
A proper -open subset B of a topological
space X is said to be a minimal -open set if
any -open set which is contained in B is
or B.
Definition (3) [3]:
A proper nonempty closed subset O of a
topological space X is said to be minimal
closed set if any closed set which is contained
in O is or O.
Definition (9):
A proper nonempty -open subset B of a
topological space X is said to be a maximal
-open set if any -open set which contains
B is X or B.
Definition (4) [3]:
A proper nonempty closed subset O of a
topological space X is said to be maximal
closed set if any closed set which is contains O
is O or X.
Definition (10):
A proper nonempty -closed subset F of a
topological space X is said to be a minimal
-closed set if any -closed set which is
contained in F is or F.
Definition (5) [4]:
Let A be a subset of a topological space X,
then the union of all open subset of X which
160
Qays Rashid Shakir
XDF and this contradict being F is
minimal -closed.
let F be a -closed subset of X, suppose
that there is a -closed K such that
KF thus XFXKbut X-K is proper
-open set. Contradiction to the assumption of
being X-F is maximal -open.■
Definition (11):
A proper nonempty -closed subset F of a
topological space X is said to be a maximal
-closed set if any -closed set which
contains F is X or F.
Remarks (12):
(1) The family of all minimal -open (resp.
minimal -closed) sets of a topological
space X is denoted by MiO(X) (resp.
MiC(X) ).
(2) The family of all maximal -open (resp.
maximal -closed) sets of a topological
space X is denoted by MaO(X)(resp.
Theorem (16):
Let U and V be maximal -open subsets
of a Topological space X, then UUV X or
U=V.
Proof:
If UUV X then the proof is complete.
If not, i.e. UUV X so we have to show that
U=V.
Since UUV X so U UUV and
V UUV .
But U is maximal -open set, so UUV X
or UUV U
Thus UUV U and so VU.
Now since V UUV and V is maximal
f-open set, so UUV X or UUV V ,but
UUV X so UUV V and hence UV
Therefore U=V.■
MaC(X)).
Remark (13):
The concept of minimal -open, maximal
-open, minimal -closed and maximal
f-closed are independent of each other as in the
following example. Example (14): let X = {a,
b, c} and
{a, b}, X} so
= { , {a},
O(X){,{a},{a,b},{a,c},X},
MiO(X) {{a}},
MiC(X) {{c},{b}},
MaO(X) {{a,b},{a,c}},
MaC(X) {{b,c}}
Theorem (17):
Let U be a maximal -open and V be a
-open subsets of a Topological space X then
UUV X or VU.
So the following table show that the new sets
are independent each other.
Table (1).
Proof:
If UUV X then the proof is complete.
If UUV X so U UUV and V UUV .
Since U is maximal -open and U UUV so
by definition of Maximal -open we have that
UUV X or UUV Ubut UUV X so
UUV U and hence VU.■
Theorem (15):
Let F be a subset of a topological space X,
then F is a minimal -closed if and only if XF is maximal -open set.
Theorem (18):
Let U be a maximal -open subset of a
Topological space X with xX/ Uthen
X/ UV for any -open subset of X with
xV.
Proof:
Let xX/ Uand xV, so VU,
thus
by
(17)
we
have
that
Proof:
let F be a minimal -closed, so X-F is
-open. We have to show that X-F is maximal
-open suppose not, so there is a -open
subset D of X such that XFD hence
UUV X X \UI X \V
X\ UV. ■
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Journal of Al-Nahrain University
Science
Vol.17 (1), March, 2014, pp.160-166
V I W V I X I W
Set Theory
V I WI UUV by (2.9 )
V I WI UUWI V Set Theory
V I WI UUV I WI VSet Theory
UI VUV I W sinceUI V W
Theorem (19):
Let F be a minimal -closed and K be a
-closed subsets of a topological space X then
FI K or FK.
Proof:
If FI K then the proof is complete.
If FI K then we have to show that
FK.
Since FI K then FI K F and
FI K K.
But F is minimal -closed, so we have
FI K F or FI K .
Thus FI K F
So FK.■
UUWI V Set Theory
X I V sinceU UW X Thus VI WV
V
implies VW but V is maximal -open
therefore
V=W
or
VUW X but
VUW X so V=W. ■
Theorem (22):
U, V and W be maximal -open subsets of
a Topological space X which are different
from each other, then UI V UI W
Proof:
Let UI V UI W
Theorem (20):
Let F and K be minimal -closed subsets
of a topological space X then FI K or
FK.
UI VUWI V UI WUWI V
UI WUV UI VUW
X UV X UW
V W
But V is maximal -open and W is a proper
Proof:
If FI K then the proof is complete.
If FI K then we have to show that
FK.
Since
FI K so FI K F or
FI K K.
Since F is minimal -closed so we have
FI K F or FI K . But FI K
hence FI K F which means FK.
Now since K is minimal -closed so we have
FI K K or FI K . But FI K
hence FI K K which means KF.
Therefore F=K. ■
subset of X so V=U, this result contradicts the
fact that U, V and W are different from each
other. Hence UVUW ■
Theorem (23):
Let F be a minimal -closed subset of a
Topological space X , if xFthen FK for
any -closed subset K of X containing x.
Proof:
Suppose xKand FK so FI KF
and FI K since xFI K
But F is minimal -closed so FI K F. or
FI K .
hence FI K F which contract the relation
FI K F. Therefore FK.■
Theorem (21):
Let U, V and W be maximal -open
subsets of a Topological space X such that
U V, if UI V W, then either U=W or
V=W.
Proof:
Suppose that UI V W, if U=W then the
proof is complete.
If U Wwe have to show that V=W
Theorem:
Let F
-closed
and
sets if
F () be minimal
F UF then there exists
A
OAsuch that FFO .
162
Qays Rashid Shakir
Let X is Tmaxspace. Suppose that X is
not Tmin, so there is a proper -open subset
Proof:
First we have to show that
FI FO ,
FO X\ F and
FI FO then
F U F X \ F which
suppose that
so
A
is
K of X which is not minimal, this mean there
exist an -open subset of X with HK.
Thus we get that H is not maximal which is
contradict of being X is Tmax.■
a
FI FO . and hence
FI FO F and FI FO FO
since FI FO F and F is minimal -closed
then FI FO F or F I FO
thus FI FO F and hence FO F. Now
since FI FO FO and FO is minimal
-closed then FI FO FO or FI FO .
Thus FI FO FO and hence F FO .
Therefore F FO .■
contradiction. So
Theorem (29):
A topological space X is
Tmin space if
and only if every nonempty proper -closed
subset of X is maximal -closed set in X.
Proof:
let F be a proper -closed subset of X and
suppose F is not maximal.
So there exists an -closed subset K of X with
KX such that FK.
Thus XKXF. Hence X-F is a proper
-open which is not minimal and this
contradicts of being X is Tminspace.
Tminand Tmaxspace
Definition (24):
A topological space X is said to be
Tminspace if every nonempty proper
Suppose U is a proper -open subset of X.
thus X-U is a proper -closed subset of X, so
X-U is maximal -closed subset of X. and by
(15) U is minimal -open. thus X is Tmin
-open subset of X is minimal -open set.
Definition (25):
A topological space X is said to be Tmax
space if every nonempty proper -open subset
of X is maximal -open set.
space. ■
Theorem (30):
A topological space X is
Tmax space if
and only if every nonempty proper -closed
subset of X is minimal -closed set in X.
Example (26):
Let X={a, b, c} and {,{a,b},{c},X}
thus O(X)= , it is clear that {a, b} and {c}
are maximal and minimal -open sets thus the
space X is both Tminand Tmax.
Proof:
let F be a proper -closed subset of X,
suppose F is not minimal -closed in X, so
there is a proper -closed subset of X such
that KF
Thus XFXKbut X-K is proper -open
in X so X-F is not maximal in X.
Contradiction to the fact X-F is maximal
-open.
let U be a proper -open subset of X, then
X-U is a proper -closed subset of X and so it
is minimal -closed set. By (15) we get that U
is maximal -open. ■
Remark (27):
Tminand Tmaxspaces are identical.
Theorem (28):
A space X is
Tmin if
and only if it is
Tmax.
Proof:
Let X is
Tmin space. Suppose that X is
not Tmax, so there is a proper -open subset
K of X which is not maximal, this mean there
exist a -open subset of X with KH .
Thus we get that H is not minimal which is
contradict of being X is Tmax.
Theorem (31):
Every pair of different minimal
sets of Tminare disjoint.
163
-open
Journal of Al-Nahrain University
Science
Vol.17 (1), March, 2014, pp.160-166
Proof: Let U and V be minimal f-open subsets
of Tminspace X such that UV to show
Definition (35):
Let X and Y be topological spaces, a map
f : XYisn called maximal -continuous if
that UI V suppose not i.e. UI V .
So UI V U andUI V V .
Since
UI V Uand U is minimal -open then
UI V U or UI V thus UI V U.
Now since UI V V and V is minimal
then
UI V V orUI V
-open
thus UI V V .
Hence we get that U=V this result contradicts
the fact that U and V are different.
Therefore UI V .■
f 1(U) is maximal -open in X for any open
subset U of Y.
Example (36):
Let X =
Y = {a, b, c} and
f :(X,) (Y,)is the identity map, where
={ , {a}, {b},{a, b}, X} and ={ , {a, c},
Y} then f is maximal -continuous since the
only proper open subset of Y is {a, c}
and f 1({a, c}){a, c}is maximal
X.
Theorem (32):
Union of every pair of different maximal
-open sets in Tmaxspace X is X.
Theorem (37):
Every minimal
-continuous.
Proof: Let U and V be maximal -open
subsets of Tmaxspace X such that UV to
that UUV X suppose not i.e.
UUV X .
So U UUV andV UUV .
Since U UUV and U is maximal -open
then UUV U orUUV X .
Thus UUV U… (1).
Now since V UUV and V is maximal
-open then UUV V or UI V X
Thus UUV V … (2)
Hence from (1) and (2) we get that U=V this
result contradicts the fact that U and V are
different. Therefore UI V X .■
Remark 38:
The converse is not true in general as in
the following example.
Example (39):
Let X =
Y = {a, b, c} and
is
f :(X,) (Y,) the identity map, where
={ , {a}, {c}, {a, c}, X} and ={ , {a, c},
Y} then f is -continuous but f is not minimal
-open
-continuous since f1({a,c}){a,c} is not
minimal -open since {a}O(X) and
{a}{a,c} .
f 1(U) is minimal -open in X for any open
Theorem (40):
Let X and Y be topological spaces, if
f : XYis an -continuous onto map and
X is Tmin space then f is minimal
subset U of Y.
Y
=
{a,
b,
c}
-continuous.
and
f :(X,) (Y,)is the identity map, where
={ , {a}, {a, b}, X} and ={ , {a}, Y}
then f is minimal -continuous since the only
proper
open
subset
of
and f 1({a}){a}is minimal
Y
is
map is
f : XY be a minimal -continuous
map and U be open subset of Y. then f 1(U)
is minimal -open in X and so f 1(U) is
-open subset of X.■
Definition (33):
Let X and Y be topological spaces, a map
f : XYis called minimal -continuous if
Example (34):
Let X =
-continuous
in
Proof:
Let
show
Continuity with Minimal and Maximal
Sets
-open
Proof:
It is clear that the inverse image of and Y
are f-open subsets of X. So let U be a proper
open subset of Y. Since f is f-continuous so
{a}
-open in X.
164
Qays Rashid Shakir
f 1(U) is proper f-open subset of X, but X is
Tminso f 1(U) minimal f-open.■
maximal -continuous since f 1({a}){a} is
not maximal -open since {a,c} {a}.
Remark (41):
The converse is not true in general as in
the following example.
Remark (49):
Minimal -continuous and maximal
-continuous maps are independent of each
other and the following examples show that.
Example (42):
In (34) f is minimal f-continuous but X is
not Tmin.
Example (50):
In (36) f is maximal
f1({a,c}){a,c}is -open
minimal -continuous.
Theorem (43):
Let X and Y be topological spaces, if
f : XYis a -continuous onto map and X
Tminspace then f is maximal -continuous.
so
subset of X but
maximal
-open in
Theorem (52):
Let f : XYbe a map and X and Y be
topological spaces, then f is maximal (resp.
minimal) -continuous if and only if f 1(F)
is minimal (resp. maximal) -closed subset of
X for each closed subset F of Y.
Proof: let F be a closed set in Y. thus Y-F is
-open.■
Remark (44):
The converse is not true in general as in
the following example.
Example (45): In (36) f is maximal
-continuous but X is not Tminspace.
-continuous
but f is not
since f1({b}){b} is not maximal
X.
f 1(U) is a proper -open
X is Tmin so f 1(U) is
Theorem (46):
Every maximal
-continuous.
since
Example (51):
In (34) f is minimal -continuous but it is
not maximal -continuous
Proof:
It is clear that the inverse image of and Y
are -open subsets of X. So let U
be a proper open subset of Y. Since f is
-continuous
-continuous
f1(YF) is maximal (resp.
1
1
maximal) -open. but f (YF) Xf (F)
1
so f (F) is minimal (resp. maximal) open and so
closed.
map is
Theorem (53):
Let X,Y and Z be topological spaces, if
f : XYis a minimal ( respect. maximal)
-continuous map and g:YZ is a
continuous map then gf : XZ is a
minimal (resp. maximal ) -continuous map.
Proof:
Let
f : XYbe a maximal -continuous
map and U be open subset of Y. then f 1(U)
is maximal -open in X and so f 1(U) is
-open subset of X.■
Proof: Let U be an open subset of Z, since g is
continuous so g1(U) is an open subset of Y.
But f is minimal (respect. maximal)
Remark (47):
The Converse is not true in general as in
the following example.
-continuous thus f1(g1(U)) gf 1 is a
minimal (respect. maximal) -open subset of
Example (48):
Let X =
X.■
Y = {a, b, c} and
f :(X,) (Y,)is the identity map, then
where ={ , {a}, {a, c}, X} and ={ , {a},
Y} then f is -continuous but f is not
Conclusion
In this paper we get some theorems
presented to reveal many various properties of
165
Journal of Al-Nahrain University
Science
Vol.17 (1), March, 2014, pp.160-166
the minimal -open and maximal -open and
their complements and we defined two types
of topological spaces and finally we defined
continuity over the new sets which produced
here.
References
[1] Nakaoka F., and Oda N., “Some
applications of minimal open sets”, Int. J.
Math. Math. Sci. 27-8, 471-476, 2001.
[2] Nakaoka F., and Oda N., “ Some properties
of maximal open sets”, Int. J.Math. Math.
Sci. 21, 1331-1340, 2003.
[3] Nakaoka F., and Oda N., “On minimal
closed sets”, Proceeding of Topological
Spaces and its Applications, 5, 19-21,
2003.
[4] Munkers J.R., “Topology”, Pearson
Education (India), 2001.
[5] Abd El-Monsef M. E., El-Deeb S. N.,
Mahmoud R. A., “ –open sets and
-continuous mappings”, Bulletin of the
Faculty of Science, Assiut University, 12,
pp.77-90, 1983.
الخالصة
يناقش البحث الحالي أنواعا جديدة من المجموعات
وكل هذه المفاهيم تعتمد على مفهوم.ويتعامل معها
المجموعة المفتوحة من النمط بيتا وأهم هذه المفاهيم التي
قدمت في هذا البحث هي المجموعة المفتوحة بيتا االصغرية
كما تم تقديم نوعين.و المجموعة المفتوحة بيتا االكبرية
جديدين من الفضاءات التبولوجيا سميت الفضاء من النمط
كذلك قمنا بتعريف. Tmax و الفضاء من النمط
Tmin
نمطين جديدين من الدوال المستمرة هما الدالة المستمرة
االصغرية من النمط بيتا و الدالة المستمرة االكبرية من النمط
باالضافة الى ذلك تم دراسة بعض الخواص االساسية،بيتا
.للمفاهيم المطروحة في هذا البحث
166