International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 1
ISSN 2229-5518
S -compact and
S -closed spaces
Alias B. Khalaf* and Nehmat K. Ahmed**
Abstract - The objective of this paper is to obtain the properties of s -compact and s –closed spaces by using nets, filterbase and s -complete accumulation points.
Index Terms— s -open sets, semi open sets, s -compact spaces, s –closed spaces, s -complete accumulation points.
—————————— ——————————
of X is said to be semi-open [14] (resp., pre-open [15], -
It is well- Known that the effects of the investigation of
open [16], -open [1] regular open [19] and regular -open
[22]) set if A cl int A , (resp., A int clA , A int cl int A ,
properties of closed bounded intervals of real numbers,
A cl int clA ,
A int clA
and
A int clA ). The
spaces of continuous functions and solution to differential
equations are the possible motivations for the formation of the
complement of S
-open (resp., semi-open, pre-open, -open
notion of, compactness. Compactness is now one of the most important, useful, and fundamental notions of not only general topology but also of other advanced branches of mathematics. Recently Khalaf A.B. et al. [13] introduced and
investigated the concepts of S -space and S -continuity. A
, -open, regular open, regular -open) set is said to be S -
closed (resp., semi-closed, pre- closed, -closed , -closed, regular closed, regular -closed ). The intersection of all S - closed (resp., semi-closed , pre- closed, -closed) sets of X
semi open subset A of a topological space (X , ) is said to be
containing a subset A is called the
S -closure (resp., semi-
S -open if for each x X
there exists a -closed set F such
closure, pre-closure -closure) of A and denoted by S clA
(resp., sclA, pclA , clA). The union of all S -open (resp.,
that x F A
. The aim of this paper to give some
semi-open ,pre-open, -open) set of X contained in A is
characterizations of S -compact spaces in terms of nets and
filter-bases. We also introduce the notion of S -complete accumulation points by which we give some characterizations of S -compact spaces. Throughout the present paper, (X , ) and (Y , ) or simply X and Y denote topological space . In [14] Levine initiated semi open sets and their properties. Mathematicians gives in several papers interesting and different new types of sets. In [1] Abd-El-Moonsef defined the class of -open set. In [18] Shareef introduced a new class of
semi-open sets called S P - open sets. We recall the following
called the S -interior (resp., semi-interior, pre-interior, - interior) of A and denoted by S intA (resp., sintA , pintA ,
intA). The family of all S -open (resp., semi-open, pre-
open, -open -open, regular -open, regular open, S - closed, semi-closed , pre- closed, -closed , -closed , regular -closed , and regular closed) subset of a topological space X is denoted by S O(X) (resp., SO(X), PO(X),
O(X), O(X), R O(X), RO(X), S C(X), SC(X), PC(X),
C(X), C(X) , R C(X) and RC(X)). A subset A of X is
definitions and characterizations. The closure (resp., interior)
called -open [21] if for each x A
, there exists an open set B
of a subset A of X is denoted by clA (resp., intA). A subset A
such that x B int clB A . A subset A of a space X is called
-semi-open [12] (resp., semi- -open [7] if for each
x A ,
————————————————
*Department of Mathematics, Faculty of Science, University of Duhok, Duhok City, Kurdistan- Region, Iraq.
there exists a semi-open set B such that x B clB A (resp.,
x B SclB A .
Definition 1.1 [15]. A topological space (X , ) is said to be :
1- Extremely disconnected if clV
for every V .
at the Department of Mathematics, College of Education, Salahaddin
University.
E-mail: nehmatbalin@yahoo.com
2- Locally indiscrete, if every open subset of X is closed.
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3- Hyper-connected if every non-empty open subset of X is dense.
F S clV
every F .
) for any S -open set V containing x and
It is clear from the definition above, that S
-converges
A SC(X) if and only if clA cl int A .
(resp., S
-accumulates) of a filter bases in a topological spaces
The following results can be found in [13].
Proposition 1.3: If a space X is T1 , then SO(X) S O(X) .
implies S
- - converges (resp., S
- - accumulates) but the
Proposition 1.4: If a topological space X is locally indiscrete, then every semi-open set is S open set.
converses are not true in general as shown in the following
examples
Example 2.3: Let X {a, b, c , d} and { , X ,{a , b},{a, b, c}} we get
S O(X) { , X ,{a, b},{a, b, c},{a, b, d}} and let
{X ,{c , d},{{b, c , d}} ,
S O(X) {X , } .
then S
- -converges to a point a, but does not S -
Corollary 1.6: For any space X, S O(X , ) S O(X , )
Proposition 1.7: If a topological space (X , ) is -regular,
converges to a because the set {a,b} S O(X) contains a, but
there exists no B such that B {a, b} . Also S - -
then S O(X)
accumulates to a point b, but does not S
-accumulates to
Proposition 1.8: 1-Every Sp -open set is S -open .
2- An S -open set is regular -open .
3- A regular closed set is S -open .
4- Every Regular open set is S -closed .
Definition 1.9: A subset A of a space X is said to be S -open
[18] (resp., SC [4]) if for each x A SO(X) there exists a pre-
b, because the set {a,b} S O(X) contains b, but there exist
{c, d} such that {a, b} {c, d} .
Theorem 2.4: Let (X , ) be a topological space and let be a filter base on X. Then the following statements are equivalent:
1- There exists a filter base finer than {Ux } , where {Ux }
is the family of S -open sets of X containing x.
closed (resp., closed) set F such that x F A .
converges to x.
1 finer than and S -
Definition 1.10: A filter base in a space X is s-converges
[20] (resp., rc-converges [11] ) to a point x if for every semi- open (resp., regular closed) subset U of X containing x there
Proof: Let 1 be a filter base which is finer than both and
{Ux } . Then S - converges to x since it contains {Ux } .
exist B such that B clU (resp., B U ).
Conversely, let
1 be the filter base which is finer than
Definition 1.11: A filter base in a space X is s-accumulates [20] (resp., rc- accumulates [11] )to a point x if for every semi- open (resp., regular closed) subset U of X containing x there exist B such that B clU (resp., B U ).
In this section, we introduce new classes of topological
and which converges to x. Then must contain {Ux } by definition.
Corollary 2.5: If is a maximal filter base in a topological space (X , ) , then S -converges (resp., S - converges) to a point x X if and only if S -accumulates (resp., S -
-accumulates) to a point x.
Proof: Let be a maximal filter base in X and S - accumulates to a point x X , and then by Theorem 2.4, there
space called S -compact and
S - closed spaces .
exists a filter base
1 finer than and S
-converges to x.
Definition 2.1: A filter base is S -convergent (resp., S -
But is maximal filter base. Thus it is S
-convergent to x.
convergent) to a point x X , if for any S -open set V
Theorem 2.6: Let be a filter base in a topological space
containing x, there exists F such that F V
(resp.,
(X , ) . If S
F S clV ) .
x X , then rc-converges (resp., rc-accumulates) at a point
Definition 2.2: A filter base is S -accumulates (resp., S -
x X .
accumulates) to a point x X
, if F V
(resp.,
Proof: Suppose that be a filter base S - converges to a
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International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 3
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point
x X . Let V be any regular closed set containing x, by
F V
(resp., F S clV ). Hence S -converges (resp., S -
Theorem 2.1.15, then V is S -open set containing x. since
S - converges (resp., S - accumulates) to a point x X ,
converges) to a point x X .
Theorem 2.11 : Let be a filter base in a topological space
There exists F such that F V
(resp., F V ). This
( X , ) and E is any -closed set containing x. If there exists
shows that rc-converges (resp., rc-accumulates ) to a point
x X .
The following example Show that the converse of Theorem
2.6 is not true in general.
Example 2.7: Let X {a, b, c , d} and { , X ,{a, c}} , then the family RC(X) { , X} and S O(X) { , X,{a, c},{a,b,c},{a,c ,d}} and
= {{b},{a, b, c}, X} . Then is rc-converges (resp., rc-
accumulates) to a point c X , but not S - converges (resp.,
S - accumulates) to a point c X .
Theorem 2.8: Let be a filter base in a topological space
(X , ) . If S - converges (resp., S - accumulates ) to a point
x X , then s- converges (resp., s- accumulates) at a point
x X .
Proof: Suppose that is S - converges to a point x X . Let
V be any semi-open set containing x, then by Theorem 1.2
F such that F E (resp., F SclE ), then
S - accumulates (resp., S -accumulates) at a point
x X .
Definition 2.12: A topological space ( X , ) is said to be S - compact (resp., S -closed) if for every cover {V : } of X, by S -open sets, there exists a finite subset 0 of such that X {V : 0 } (resp., X {S clV : 0 } )
It is clear that S -compact is S -closed, but not conversely as shown in the following examples
countable topology. Since X is T1 , then by Proposition 1.3 the family of open, semi-open set and S - open sets are identical.
Hence X is an S - closed because every open set in X is dense
clV cl int V , so clV is regular closed , by Theorem
, not S
- compact [19] p-194.
Proposion 1.8 clV is S -open set containing x. Since is S - converges (resp., S - accumulates ) to a point x X , then there exists B such that B clV (resp., B clV ). This implies that is s- converges (resp., s- accumulates) at a
Theorem 2.14: If every -closed cover of a space has a finite subcover, then X is S -compact.
Proof: Let {V : } be any S - open cover of X, and
x X , then for each x V ( x ) and , there exists a -
point
x X .
closed sets x X such that x F ( x) V ( x ) , so the family
The converse of Theorem 2.8 is not true in general as shown
in the following example:
SO(X) { , X ,{a, b},{a, b, c},{a, b, d}} S O(X) , considering the family = {{c},{a, b, c}, X} . Then is filter base s-converges (resp., s-accumulates) to a point b X , but not S - converges (resp., S - accumulates) to a point b X .
Theorem 2.10: Let be a filter base in a topological space
(X , ) and E is any -closed set containing x. If there exists F such that F E (resp., F S clE ). Then S - converges (resp., S - converges) to a point x X .
Proof: Let V be any S -open set containing x. then V is semi-
{F ( x ) : } is a -closed cover of X, then by hypothesis, this family has a finite subcover such that
X {F ( xi ) ; i 1,2,...n} {V ( xi ) ; i 1, 2,...n} . Therefore
X {V ( xi ) : i 1, 2,...n} . Hence X is S -compact.
The following theorem shows the relation between S - compact and some other compactness
Theorem 2.15: Every semi-compact space, is S -compact
spaces.
Proof: Let {V : } be any S - open cover of X. Then
{V : } is a semi--open cover of X. Since X is semi- compact, there exists a finite subset 0 of such that
open and for each
x V , there exist a -closed set E such
X {V : 0 } . Hence X is S -compact.
that x E V . By hypothesis, there exists F such that
F E V (resp., F S clE S clV ). Which implies that
The following example shows that the converse of
Theorem 2.15 is not true in general.
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International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 4
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Example 2.16: Let X= R with the topology {X , ,{0}} . Then ( X , ) is not semi-compact, since the space X is
hyperconnected, then by Proposion 1.5
regular T1 space there exist Gx SO(X) and
x Gx SclGx V ( x ) . Then by Proposition 1.3, the family
{Gx : x X} is an S -open cover of X. Since X is S -closed
S O( X ) = {, X }. Then ( X , ) is S -compact.
space, then there exists a subfamily {Gxi
: i 1,2,...n}such
Theorem 2.17: If a topological space ( X , ) is T1 and S -
that
n n
X S clGxi V ( xi )
. Thus X is compact.
compact space , then it is semi-compact.
i 1
i 1
Proof: Suppose that X is T1 and S -compact space. Let
{V : } be any semi-open cover of X. Then for
every x X , there exist ( x) such that x V ( x ) . Since X is T1 by proposition 1.3, the family {V : } is S -open cover of X. Since X is S -compact, there exists a finite subset
0 of in X such that X {V : 0 } . Hence X is semi- compact.
Theorem 2.18: If a topological space (X , ) is locally indiscrete. Then S -compact space X is semi-compact . Proof: Follows from Theorem 1.4.
The following example shows that the condition of s-
regularity in Theorem 2.22 can not be dropped
Example 2.23: In Example 2.20 (X , ) is neither s-regular nor compact but it is S -closed because the only non-empty S - open set in (X , ) is X itself.
Theorem 2.24: Let X be an almost- regular space if X is S -
compact, then it is nearly compact.
Proof: Let {V : } be any regular open cover of X, Since X
is almost- regular space then for each x X and each regular open set V ( x ) , there exist an open set Gx such that
x Gx clGx V ( x ) . But clGx is regular closed for each
In general S -compact spaces and compact spaces are not
comparable as shown in the following examples.
x X . Therefore X
xX
clGx
( x )
V ( x )
This implies that the
family {clGx : x X }is an S -open cover of X. Since X is
S -compact, then there exists a subfamily
not
S -compact, but it is compact.
n n
{clGxi : i 1,2,...n} such that X clGxi V ( xi ) . Thus X
{X , ,G (0,1 1 ), n 2, 3,...} then (X , ) is not compact [19],
is nearly compact.
i 1
i 1
p. 76, but it is S -compact since the only S -open subset of
(X , ) are X and .
S -compact.
Theorem 2.21: Let ( X , ) be a topological space. If X is -
: } be any S
- open cover of X, Then V is
regular and S -compact , then X is compact.
Proof: Let {V : } be any open cover of X. Since X is - regular By Proposition 1.7, {V : } forms an S -open cover of X. Since X is S -compact, there exists a finite subset
0 of such that S int F 0 S cl(X \ F 0 ) , hence X is
semi-open for each , since X is semi-regular for each
x X and V ( x ) , there exists a semi-open set Gx such that
x Gx SclGx V ( x ) . Then the family {Gx : x X} is semi-open cover of X. Since X is s-closed space, then there exists a
n n
subfamily {Gxi : i 1,2,...n} such that X SclGxi V ( xi ) .
compact.
Thus X is S
-compact.
i 1
i 1
Theorem 2.22: If X is an s- regular S -closed T1 space, then
X is compact.
Theorem 2.26: For any topological space (X , ) . The following statements are equivalent:
Proof: Let {V : } be any open cover of an s-regular and
1- (X , ) is S
-compact spaces (resp., S
-closed)
S -closed T1 space X, then for each and for each
2- For any S
-open cover {V
: } of X, there exists a finite
x X there exists ( x) such that x V ( x ) . Since X is s-
subset of such that X {V
: } (resp.,
0 0
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X {S clV : 0 } )
3- Every maximal filter base in X S - converges (resp. S -
a filter base on X. By hypothesis, S - accumulates (resp.,
S - -accumulates) to some point x X . This implies that for
- converges ) to some point x X .
every S -open set V containing x, F V
(resp.,
4- Every filter base in X S - accumulates (resp., S - -
F S clV ), for every F and every .Since
accumulates) to some point
x X .
x F
,there exist such that x F
. Hence X \ F is
0 0 0
5- For every family {F : } of S -closed subsets of X such
S -open set containing x and F 0
X \ F 0
(resp.,
that {F : }
there exists finite subset 0 of such
S int F 0 S cl(X \ F 0 )
). Which contracting the fact that
that {F : 0 }
(resp., {S int F : 0 } )
S - accumulates to x (resp., S
- -accumulates). So the
proof: 1 2. Straightforward.
assertion in (5) is true.
2 3 Suppose that for every S -open cover {V : } of X ,
5 1. Let {V
: } be S
-open cover of X.
there exist a finite subset 0 of such that
Then {X \V
: } is a family of S
-closed subsets of X such
X {V : 0 } (resp., X {S clV : 0 } and let
that {X \V
: } . By hypothesis, there exists a finite
{F : } be a maximal filter base. Suppose that does
subset of such that {X \ V
: }
(resp.,
0 0
not S -converges (resp., S - - converges ) to any point of X.
Since is maximal, by Corollary 5.1.4 dose not S - accumulates (resp., S - -accumulates) to any point of X.
{S int(X \V ) : 0 } . Hence X {V : 0 } (resp.,
X {S clV : 0 } . This shows that X is S -compact (resp.,
S -closed) .
This implies that for every x X there exists S -open set
Theorem 2.27: If a topological space (X , ) is S
-closed and
Vx and F ( x ) such that F ( x ) Vx
(resp.,
T1 -space then it is nearly compact.
F ( x ) S clVx ). The family {Vx : x X} is an S -pen cover of
: } be any regular open cover of X. Then
X and by hypothesis, there exists a finite number of points
{V
: } is a S
-open cover of X. Since X is S
-closed , there
x1 , x2 ,..., xn of X such that X {V( xi ) : i 1, 2,...n} (resp.,
exists a finite subset of such that X {V
: } .
X {S clV( xi ) : i 1, 2,...n} ). Since is a filter base on X, there exists a F0 such that F0 {F (xi ) : i 1,2,..., n} . Hence
0 0
Hence X is nearly compact.
F0 V ( xi )
(resp., F0 S clV ( xi ) ) for i 1,2,...n which
S compact spaces
implies that F0 ({V( xi ) : i 1, 2,...n}) (resp.,
F0 ({S clV(xi ) :i 1,2,...n) F0 X Therefore, we
obtain F0 . Which is contradict the fact that , thus
is S - converges to some point x X
Definition 3.1: A point x in X is said to be S -complete accumulation point of a subset A of X if
Card ( A U ) Card(A) for each U S (X , x) . Where Card(A)
denotes the cardinality of A .
3 4. Let be any filter base on X. Then, there exists a
an S
-adherent point of a filter on X if it lies in the S -
maximal filter base 0 such that 0 . By hypothesis 0 is
S -converges (resp., S - - converges) to some point x X .
closure of all sets of .
-compact spaces if and only if
For every F and every S -open set V containing x, there
each infinite subset of X has S
-complete accumulation point.
exists F0 0
such that F0 V
(resp., F0 S clV
), hence
-compact and S an infinite subset
F0 F V F (resp., S clV F . This shows that S -
accumulates at x (resp., S - -accumulates).
of X. Let K be the set of points x in X which are not S - complete accumulation points of S. Now it is obvious that for
4 5. Let {F
: } be a family of S
-closed subsets of X
each point x in K , we are able to find U( x) S O(X , x) such
such that {F : } . If possible suppose that every finite
that Card(S U( x )
) Card(S) . If K is the Whole space , then
subfamily {F i
: i 1, 2,...,n} . Therefore A Y X form
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{U( x ) : x X} is S -cover of X. By hypothesis X is S - compact, so there exists a finite subcover {U( xi ) ; i 1, 2,..., n} such that S {U( xi ) S ; i 1,2,...,n} , then
Card(S) max{Card(U(xi ) S); i 1, 2,...n} , which does not agree with what we assumed. This implies that S has an S - complete accumulation. Now assume that X is not S - compact and that every infinite subset S of X has an S - complete accumulation point in X. it follows that there exists an cover with no finite subcover. Set
min {Card(): , wher is anS cov er of X} . Fix ,
( x) such that V {x : (x)} . This implies that
{x : (x)} is a subset of X V( x) . Then the collection
{V( x ) :x X} is S - cover of X. By hypothesis of theorem, X
is S -compact and so has a finite subfamily
{V( xi ) : i 1,2,...n} such that X {V( xi ) : i 1, 2,...n} .Suppose that the corresponding elements of be { (xi)} where i=1,2,…,n, since is well-ordered and { ( xi)} where i=1,2,…,n is finite. The largest elements of { ( xi)} exists. Suppose it is
{ (xi)} . Then for { ( xi)} . We have
{x : } (X V( xi ) ) X V( xi ) . Which is impossible.
for which Card( )
and {U : U } X . Let N denote the
i 1
i 1
set of natural numbers, then by hypothesis Card(N) [ By well-ordering of ] . By some minimal well-ordering “~” , suppose that U is any member of . By minimal well- ordering “~”, we have
Card({V : V , V ~ U}) Card({V : V }) . Since can not
have any subcover with cardinality less that , then for each
U we have X {V : V , V ~ U ] . For each U choose a point x(U) X {V {x(V )} : V , V ~ U } . We are always
able to do this, if not, one can choose a cover of smaller
This shows that has at least one S -adherent point in X .
(ii) (i): Now it is enough to prove that each infinite subset has an S -complete accumulation point by utilizing above theorem. Suppose that S X is an infinite subset of X.
According to Zorns Lemma, the infinite set S can be well- ordered. This means that we can assume S to be a net with a domain which is a well ordered index set. It follows that S has S - adherent point z. Therefore is an S -complete
accumulation point of S this shows that X is S -compact.
cardinality from . If H {x(U ):U } , then to finish the
-compact if and only if each
proof we will show that H has no S -complete accumulation
family of S
-closed subsets of X with the finite intersection
point in X. Suppose that z is a point of the space X. Since is
S -cover of X, then z is a point of some set W in . By the fact that U~V we have x(U ) W . It follows that
T {U : U and x(U ) W } {V : V , V ~ W } . But
Card(T ) . Therefore Card(H W ) . But
Card(H ) Card(N) , since for two distinct points U and W in
, we have x(U ) x(W ) , this means that H has no S - complete accumulation point in X which contradicts our assumptions. Therefore X is S -compact.
property has a non-empty intersection .
Proof: Given a collection of subsets of X. let
{X w : w } be the collection of their complements . Then the following statements hold.
i- is the collection of S -open sets if and only if is a
collection of S -closed sets.
ii- the collection covers of X if and only if the intersection
of all the elements of is non empty
v
iii- The finite sub collection {w ,...w } of covers X if and
only if the intersection of the corresponding elements
equivalent:
vi X wi
of is empty.
i- X is S -compact.
ii- Every net in X with well-ordered directed set as its domain
The statement (i) is trivial . While the statement (ii) and (iii)
follows from De-Morgan Law X v = (X v ) . The
accumulates to some point of X..
j
j
Proof:(i) (ii): Suppose that X is S -compact and
proof of theorem now proceeds in two steps. Taking the
contra positive of the theorem and the complement .
{x : } a net with a well-ordered set as domain. .
The statement X is S
-compact is equivalent to: Given any
Assume that has no S - adherent point in X. Then for each
point x in X there exists V( x ) S O(X , x) and an
collection of S
-open subsets of X, if covers X, then
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International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 7
ISSN 2229-5518
some finite sub collection of covers X. This statement is
is a filter base on X with no S -adherent point. By hypothesis
equivalent to its contra positive, Which is the following.
Given any collection of S -open sets, if no finite sub
S
-converges to some point z in X. Suppose F is an
collection of covers X, then does not cover X . Letting
be as earlier, the collection {X w : w } , and applying
arbitrary element of . Then for each V S O(X , z) , there
exists an F such that F V . Since is a filter base
(i) to (iii), we see that this statement is in turn equivalent to
there exists a such that F F F F V
where F is a
the following.
nonempty. This means that F V
is nonempty for every
Given any collection of S -closed sets , if every finite
intersection of elements of is non empty. This is just the
V S O(X , z) and correspondingly for each , z is a point of
condition of our theorem.
S clF . It follows that z S clF
Therefore, z is
Theorem3.6: A space X is S -compact if and only if each filter base in X has at least one S -adherent point.
Proof: Suppose that X is S -compact and {F : } is a
S adherent point of . Which is contradiction . This shows that X is S compact.
filter base in it. Since all finite intersections of F
,s are
[1]. M.E.Abd El-Monsef , S.N. El-Deeb and R.A Mahmoud.,
nonempty . It follows that all finite intersections of S cl(F ), s
are also nonempty. Now it follows from Theorem [F.I.P] that
S clF is nonempty. This means that has at least one S -
adherent point. Now suppose that is any family of S - closed sets. Let each finite intersection be nonempty the set F with their finite intersection establish the filter base .
Therefore S -accumulates to some point z in X. It follows
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Theorem3.7: A space X is S -compact if and only if each filter
base on X , with at most one S -adherent point, is S - convergent.
Proof: Suppose that X is S -compact, x is a point of X, and
is a filter base on X. The S -adherent of is a subset of {x} . Then the S -adherent of is equal to{x} , by Theorem 3.7 . Assume that there exists a V S O(X , x) such that for all
F , F (X V ) is nonempty . Then {F V : F } is a filter base on X . It follows that the S -adherence of is nonempty. However
S cl(F V ) ( S clF) (X V ) {x} (X V ) . But this
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F with F V . This shows s that S -converges to x.
To prove the converse , It suffices to show that each filter base in X has at least one S -accumulation point. Assume that
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