On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
On Properties of Strongly Irresolute Topological Vector Spaces
Received: 22/11/2006
Accepted: 31/12/2007
Anwar Al-Nayef*
&
Omar Jaradat**
Ahmad Al-Omari***
‫ملخص‬
،‫في هذا البحث قمنا بدراسة خصائص الفضاءات التبولوجية المتجهة المترددة التامة‬
‫( واثبم ممات أن القم مميط‬Semi-Hausdorff) ‫ل فضم مماءات الشم ممبا ه م موا در‬
‫ى المحيط في الفضاءات التبولوجية المتجهمة المتمرددة‬
‫وتم ممط ء وم مماء وص م م‬
‫القصوى ل مجمو ات المحدبة تقع‬
.‫التامة‬
Abstract
In this paper we obtain several characteristics of semi-Hausdorff of
strongly irresolute topological spaces and show that the extreme point of
convex subset of strongly irresolute topological spaces X lies on the
boundary.
Keyword: topological vector spaces, irresolute topological vector spaces,
strongly irresolute topological vector spaces, semi- Hausdorff and extreme
points.
2000 AMS Subject Classification: 54C08, 54C10, 54C30.
1. Introduction:
In 1963, N. Levine introduced the concept of semi-open set by defining a
subset A of a topological space X to be semi-open if there exists an open set
U in X such that A contains U and A is contained in the closure of U in
X . He proved that a set A in a topological space X is semi-open if and only if
A is contained in the closure of the interior of A in X . SO (X ) will denote the
class of all semi-open sets in a topological space X . We note that every open set in
a topological space X is a semi-open set but clearly a semi-open set may not be an
*
**
***
Associate Professor, Department of Mathematics & Statistics, Mutah University.
Assistant Professor, Department of Mathematics & Statistics, Mutah University.
Full time Lectrer, Department of Mathematics & Statistics, Mutah University.
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On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
open set in X . A point x  X is said to be a semi-interior point of A if and
only if there exists U  SO (X ) such that x U  A . The set of all semiinterior point of A is said to be the semi-interior of A and is denoted be
sInt (A ) . The complement of an semi-open set is called semi-closed. The
intersection of all semi-closed sets containing A is called the semi-closure of A ,
denoted by sCl (A ) see[5]. Recently [1] and [2] defined the notion of strongly
irresolute topological vector space (ITVSs) and investigated its properties. The
purpose of the present paper is to continue to explore further properties,
characteristics of semi-Hausdorff. The paper show that the extreme points of
convex subset of irresolute topological vector space (ITVSs) lies on the boundary.
2. Preliminaries:
Throughout this paper, (X , ) and (Y ,  ) ( or simply, X and Y ) denote
topological spaces on which no separation axioms are assumed unless explicitly
stated. Let (Y , ) be a topological space. A map f : X  Y is irresolute if
1
( B)  SO( X ) whenever B  SO(Y ) [4]. A map f : X  Y is pre-semiopen if f ( A)  SO (Y ) whenever A SO( X ) [4]. Finally, a map f : X  Y is
f
s-continuous if f
1
( B)   for every B  SO(Y ) [3] .
Definition 2.1 [2]: Let  be a topology on a real vector space X such that
(a) The addition map S : X  X  X
(b) The scalar multiplication M : R  X  X ,
are both irresolute, then the pair ( X , SO ( x)) is called an irresolutetopological vector space (ITVS).
If the addition map S is s-continuous, instead of irresolute, then
( X , SO ( x)) is called strongly- irresolute topological vector space (SITVS).
Note that every SITVS is an ITVS, but the converse, in general, is not true see
[1, 2]. A subset A of an ITVS ( X , SO ( x)) is called semi-neighborhood ( resp;
neighborhood) of x  X , if there exists U  SO( X ) ( resp; U  such
that x  U  A . The set of all semi-neighborhoods of x  X is denoted by
N x ( X ) or simply N x . In particular, the set of all semi-neighborhoods of the zero
vector 0 of X is denoted by N 0 ( X ) or N 0 .
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On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
Theorem 2.1 [1] Let ( X , SO ( x)) be an ITVS .(or ISTVS) For x  X , the
following hold.
(a) If U  N x then x U
(b) If U  N x and V a neighborhood of x then U  V  N x .
(c) If U  N x , then there exists V  N x such that U  N y , for all y  V .
(d) If U  N x and U  V , then V  N x .
(e) If U  N 0 , then  U  N 0 for all   R,   0 .
(f) U  N 0 if and only if x  U  N x .
A subset A of a vector space X is called balanced if A  A for   1
and absorbing if for every x  X there exists   0 such that x  A for
   . It is called absolutely convex if it is both convex and balanced.
Theorem 2.2 [1] Let ( X , SO ( x)) be an ITVS. Then
(a) Every U  N 0 is absorbing.
(b) For every U  N 0 there exists a balanced V  N 0 such that V  U .
( X , SOX ( X )) be an ITVS. If A  X , then
sCl ( A)   ( A  U ) . In particular sCl ( A)  A  U for all U  N 0 .
Theorem 2.3 [1] Let
U N 0
Theorem 2.4 [1 ] Let ( X , SOX ( X )) be an SITVS. Then
(a) for every U  N 0 , there exists V  N 0 such that V  V  U .
(b) for every U  N 0 , there exists a semi-closed balanced V  N 0 such that
V U .
Definition 2.2 [2] An ITVS ( X , SOX ( X )) is called Locally convex if for all
x  X , every V  N x contains a convex U  N x .
Theorem 2.5 [2] An ITVS ( X , SOX ( X )) is Locally convex if and only if every,
V  N 0 contains a convex U  N 0 .
f : X  Y is
irresolute at x  X if for each V  N f ( x ) (Y ) there exists U  N x (X ) such that
f (U )  V , see [2].
Let ( X , SOX ( X )) and (Y , SOX (Y )) be ITVS A map
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On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
Main Results:
3. Semi-Hausdorff SITVS:
A topological space X is called semi-Hausdorff [9], if for each two distinct
points x and y in X, there exist disjoint sets U , V  SO ( X ) such that xU and
y V .
The following result provides a characterization for semi-Hausdorff of SITVS.
Theorem 3.1 Let X be SITVS. The following are equivalent.
(a) X is semi-Hausdorff.
(b) If x  X , x  0, then there exists U  N 0 such that xU .
(c) If x, y  X , x  y, then there exists V  N x such that y  V .
Proof: By continuity of translation, it is sufficient to prove the equivalence
between (a) and (b) only.
Let x  X , x  0. by assumption, there exist U , V  SO ( X ) such that
0 U , x V
and U  V   . Thus, U  N 0 , V  N x , and
xU . Conversely, let
x, y  X be such
that x  y  0 . There exists U  N 0 such that x  y  U . But by part (a) of
Theorem 2.4, there exists w N 0 such that w  w  U . By part (b) of Theorem
2.4, w can be assumed to be balanced. Let V1  x  w and V2  y  w and note
V1  V2   , since if z V1  V2 then
 ( z  x)  w , as it is balanced and z  y  w . It follows that
x  y  ( z  y )  (( z  x))  w  w  U which is a contradiction. So, we
must have V1  V2   . Finally, by the definition of
that
V1  N x , V2  N y
semi-neighborhood,
and
there
exist
V 1, V 2  SO (X )
such
that
x V1  V1 , y V2  V2 , and V1  V2   . This shows that X is semiHausdorff. This completes the proof.
The following result follows from Theorem 3.1.
Corollary 3.1 Let X be SITVS. The following are equivalent
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On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
(a) X is semi-Hausdorff.
(b)  {U : U  N 0 }  {0} .
(c)  {U : U  N x }  {x}
Theorem 3.2 A SITVS X is semi-Hausdorff if and only if every one-point set in
X is semi-closed.
Proof: Let x  X and y  X  {x} . Then y  x  0, and by assumption, there
exists U  N 0 such that y  x  U . By part (b) of Theorem 2.4, there exists
semi-closed and balanced V  N 0 such that V  U . It follows that y  x  V
that is y  x  X  V . Thus y  ( X  V )  {x} . But ( x  V )  {x}  SO( X ) ,
since V is semi-closed, and ( X  V )  {x}  X  {x} . This shows that
X  {x}  SO( X ) , hence {x} is semi-closed. For the converse, let x  X
and
assume
that
{x}
is
semi-closed.
Then by Theorem 2.3
{x}  sCl{x}  {U  {x} : U  N 0 }  {V : V  N x } where V  U  {x}  N x .
By Corollary 3.1, X is semi-Hausdorff. This completes the proof .
Since translation is semi-homeomorphism, and as a consequence of Theorem
3.2, we have the following
Corollary 3.2 A SITVS X is semi-Hausdorff if and only if {0} is semi-closed.
A space X is said to be semi compact if every cover of X by semi open sets
has a finite cover.
Theorem 3.3 Let C, K be disjoint sets in a SITVS X with C semi-closed, K semicompact. Then there exists U  N 0 ( X ) with ( K  U )  (C  U )   .
Proof: If K   , there is nothing to prove. Otherwise, let x  K by the
invariance with translation, we can assume x  0 . Then X  C is an semi-open
set of 0 . Since addition is irresolute and s-continuous, by 0  0  0  0 , there is
a neigborhood U and hence U  N 0 ( X ) such that 3 U  U  U  U  X  C .
~
By defining U  U  (U )  U
~
we have that U is open, and hence
~ ~ ~ ~
3U  U  U  U  X  C . This means that
~
~
~
~
~
  {3x, x  U }  C  {2 x, x  U }  { y  x, y  C , x  U }  U  (C  U ) .
semi-open symmetric and
This concludes the proof for a single point.
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On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
Since K is semi-compact, then repeating the above argument for all x  K ,
we obtain symmetric semi-open sets V x such that ( x  2Vx )  (C  Vx )   . The
sets {V x : x  K } are a semi-open ( and open ) covering of K , but K is semicompact hence there is a finite number of points xi  K , i  1,..., n, such that
n
n
i 1
i 1
K   ( xi  V xi ) . Define the semi-neighborhood V of 0 by V   V x . Then
n
n
i 1
i 1

i

( K  V )  (C  V )   ( xi  V xi  V )  (C  V )   ( xi  2 V xi )  (C  V xi )  
This completes the proof.
Lemma 3.1 [6] let {U  :   } be a collection of semi-open sets in a
topological space X. Then
U 
is semi-open.
 
Lemma 3. 2 If U is semi-open and U  A   , then U  sCl ( A)   .
Proof: Suppose that there exists an x  U  sCl ( A) . Thus x  sCl ( A) and U is
a semi-neighbourhood of x and X  U is semi-closed set containing A , hence
sCl ( A)  X  U and
x  sCl ( A)
which
is
contradiction,
hence
U  sCl ( A)   .
Corollary 3.3. Let C, K be disjoint sets in a SITVS X with C semi-closed, K semicompact. Then there exists U  N 0 ( X ) with sCl ( K  U )  (C  U )   .
Proof: Since C  U is a union of semi-open sets y  U for y  C the proof
follows directly from Theorem 2.3, and Lemma 2.2.
Corollary 3.4 A SITVS X is semi-Hausdorff space.
Proof: Take K  {x} and C  {y} in Theorem 3.3.
Let ( X , || . || ) be normed space over K . We denote by X * the vector space of
all linear maps from X of X . And X * is called the algebraic duel of X . Note
that for any f  X * , and x  X we write  x, f   f ( x) , see [6].
Theorem 3.4 Let ( X , SO( X )) be ITVS and 0  f  X * , then f (G ) is semiopen in k whenever G is semi-open in X .
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On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
Proof: Let G be non-empty semi-open set . Then one can assume that
0  x0  X such that f ( x0 )  1 . For any a  G , it is required to show
f (a )  sInt ( f (G )) . Since G  N a (X ) by Theorem 2.1 we have
that
G  a  N 0 ( X ) also by Theorem 2 .2 G  a is absorbing that is absorbs x 0 ,
namely there exists an  0 , such that  x0  G  a whenever   
with
 . Now for any
 
with
  f (a)  we have
(   f (a)) x0  G  a , hence f ((   f (a) x0 )  f (G  a)
(  f (a) f ( x0 )  f (G  a)
(   f (a) (1)  f (G  a)  f (G )  f (a)
which
implies
that
  f (G ) and f (a)  [    ,   ] ,
thus
f (a)  Int ( f (G ))  sInt ( f (G )) hence f (G )  sInt ( fG)) .
Lemma 3.3 [6] Let X be a vector space and   K  X . For a  K
The following statements are equivalent
1) a is an extreme point of K
1
( x  y ) , then a  x  y
2
3) let x, y  k , be such that x  y , let   (0 , 1) , and a  x  (1   ) y .
Then we have either   0 , or   1 .
2) if x, y  k are such that a 
Theorem 3.5: Let ( X , SO( X )) be ITVS and K a convex subset of X . Then
( sInt ( K ))  (K )  
Proof: If sInt (K )   , the result is trivial. Suppose now that sInt (K )   and
let x  sInt (K ) . Then there exists V  N 0 ( X ) ( semi-neighbourhood ) such that
x  V  K . As the map  :   X where  (  )  x is continuous at   1 ,
for this the semi-neighbourhood x  V , there is an r  0 such that
x  x  V whenever  1  r . In particular, we have (1  r ) x  x  V  K and
(1  r ) x  x  V  K . Now consider x   (1  r ) x  (1   )(1  r ) x and take
  12 . Consequently, we have x  12 (1  r ) x  (1  12 )(1  r ) x , which implies that
x is not an extreme point of K .
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On Properties ………….…Anwar Al-Nayef & Omar Jaradat & Ahmad Al-Omari
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