Download 88 CHAPTER 5 KURATOWSKI CLOSURE OPERATORS IN GTS

CHAPTER 5
KURATOWSKI CLOSURE OPERATORS IN GTS
VIA NETS AND FILTERS
In this chapter we introduce two new operators in a generalized topological
space and we prove that the new operators satisfy kuratowski closure axioms and
hence induce topologies.
A set (D,≤) is called a directed set if ≤ is reflexive and transitive and for
λ1,λ2∈D ∃ λ3∈D such that λ1≤λ3 and λ2≤λ3.
If X is a non empty set, then any function from D to X is called a net where D
is a directed set .
A net f:D→X where X is a topological space is said to converge to x∈X if for
each open set U containing x, ∃ λ0∈D such that f(λ) = xλ∈U ∀λ≥λ0. We write f→x
or (xλ)→x.
5.1 NET CLOSED SET
Definition 5.1.1
Let X be a generalized topological space and (xλ)λ∈D be a net in X. (xλ) is said to
converge to x∈X if for every open set U containing x, ∃λ0∈D such that xλ∈U
∀λ≥λ0.
Example 5.1.2:
Let X = {a,b,c} and D = P(X) – {φ}
( D,⊂ ) is a directed set.
Consider G = {φ,{a,c},{a,b},X} and G is a generalized topology.
Define f : D→X as f{a} = a, f{b} = b, f{c} =c,
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f{a,b} =b, f{b,c} = b, f{a,c} = b, f{a,b,c} = b
fքa,f→b and fքc.
Definition 5.1.3:
Let (X,G) be a generalized topological space. A subset A of X is said to be net closed
if no net in A can converge to a point outside A.
Example 5.1.4:
X = {a,b,c}
G = {φ.{a,c},{a,b},X}
Take A = {b}. The only net in A is constant net where xλ = b ∀λ.
Let us find the limits of this constant net.
Since {a,c} is a neighborhood of a, not containing b, (xλ)քa. Also (xλ)քc,(xλ)→b.
The net in A can not converge to any point outside A and hence A is net closed.
Take B = {a,c}. Consider the constant net xλ = a ∀λ.
{a,b} is a neighborhood of b. xλ∈{a,b} ∀λ.
X is a neighborhood of b. xλ∈X ∀λ. xλ = a ∀λ is a net in B and xλ→b∉B.
Hence B is not net closed.
Theorem5.1.5:
In a generalized topological space, every closed set is net closed.
Proof:
Let A be a closed set in X, a generalized topological space. Let (xλ) be a net in A.
Suppose (xλ) does not converge then
nothing to say. Suppose (xλ)→x,
we claim that x∈A. If not, then AC is an open set containing x,
hence ∃λ0 ∋ xλ∈AC ∀λ≥λ0. In particular xλ0∈ AC which contradicts the fact that
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(xλ) is a net in A. Hence if (xλ)in A converges to x then x∈A which implies A is net
closed.
Result 5.1.6:
The converse of the theorem is not true.
Example 5.1.7:
X = {a,b,c}
G = {φ,{a,b},{a,c},X},
Let A = {b,c}.
Let (xλ)→a. Since {a,b} is an open set containing a, ∃ λ1 ∋ xλ ∈{a,b} ∀λ≥λ1. Since
{a,c} is an open set containing a, ∃ λ2 ∋ xλ∈{a,c} ∀λ≥λ2. Now consider λ3 such that
λ3 ≥λ1 and λ3 ≥λ2. Then xλ3∈{a,b}, xλ3∈{a,c} and xλ3∈{b,c}.
This is a
contradiction. Hence any net in A can not converge to a point outside A which implies
A is net closed. Now A is not closed. Hence a net closed set need not be closed.
Result 5.1.8:
Converse is true if X is a topological space.
Proof:
Let X be a topological space and A be net closed in X. In a topological space (xλ) in A
converges to x iff x∈Ā implies ∃ a net (xλ) in A which converges to x. Since A is net
closed, x∈A. Hence Ā ⊂ A which gives A is closed.
Theorem 5.1.9:
In a generalized topological space if A is singleton and net closed then A is closed.
Proof:
X is a generalized topological space.
A = {a} is net closed. Take any b≠a
Consider the constant net xλ = a ∀λ.
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(xλ) is a net in A and A is net closed, hence (xλ)քb. There exists a neighborhood Ub
of b such that a∉Ub. Let U = ∪{ Ub / b≠a}.
U is open and UC = {a} = A and hence A is closed.
Result 5.1.10:
By definition φ is net closed.
By definition X is net closed.
Theorem 5.1.11:
In a generalized topological space, arbitrary intersection of net closed sets is net
closed.
Proof:
Let X be a generalized topological space. Let {Aα/α∈I} be a collection of net closed
sets. Let A = ∩Aα where α∈I. Let (xλ) be a net in A. Suppose (xλ) does not converge
then nothing to prove. If (xλ)→x, since (xλ) is a net in A, (xλ) is a net in Aα for each
α∈I. Since for each α, Aα is net closed, x∈Aα for each α. Hence x∈A. Therefore
A is net closed.
Theorem 5.1.12:
In a generalized topological space, the union of two net closed sets is net closed.
Proof:
Let A and B be two net closed sets in X, a generalized topological space.
Let (xλ) be a net in A∪B and let (xλ)→x.
Case 1:
Suppose ∃ λ0 such that (xλ)∈A ∀λ≥λ0. Then (xλ) where λ≥λ0 is a sub net of (xλ). This
sub net is a net in A and this sub net converges to x. Since A is net closed, x∈A and
hence x∈A∪B
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Case 2:
Suppose such a λ0 does not exists then for each λ0 we have a λ≥λ0 and xλ∈B. Now
we get a subnet in B. This subnet converges to x and B is net closed and hence x∈B
which implies x∈A∪B. Hence A∪B is net closed.
Theorem 5.1.13:
Let X be a generalized topological space. Then finite union of net closed sets is net
closed.
Proof:
Follows from previous Theorem.
Theorem 5.1.14:
Let X be a generalized topological space. {A/AC is net closed} is a topology.
Proof:
X is a generalized topological space. T = {A/AC is net closed}. By definition φ and X
belong to T. By theorem 5.1.11. T is closed under arbitrary union. By theorem 5.1.3,
T is closed under finite intersection. Hence T is a topology.
Definition 5.1.15:
Let X be a generalized topological space T = {A/AC is net closed}. T is a topology
called net topology on X.
Example 5.1.16:
X = {a,b,c},
G = {φ,{a,b},{a,c},X}
The closed sets with respect to G are X,{c},{b},φ. These sets are net closed. Let us
find out other net closed sets.
Consider A = {a}. The only net in A is constant net xλ = a ∀λ. Neighborhoods of b
are {a,b} and X. Hence (xλ)→b. Hence A is not net closed.
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Consider A = {a,b}. Neighborhoods of c are {a,c} and X. The constant net xλ = a ∀λ
is a net in A which converges to c. Hence A is not net closed.
Considser A = {a,c}. The constant net xλ = a ∀λ is a net in A which converges to b.
Hence A is not net closed.
Consider A = {b,c}. In example 5.1.7, we have seen that A is net closed.
Hence net closed sets are
φ,{b},{c},{b,c},X.
T = {A/AC is net closed} = {φ,a,{a,c},{a,b},X}. T is the net topology.
Theorem 5.1.17:
Let X be a generalized topology. Then the net topology is the smallest topology
containing the generalized topology.
Proof:
Let X≠φ and G be the generalized topology. Let T be the net topology.
If A is open with respect to G, then AC is closed which implies
AC is net closed and
hence A∈T. Hence T contains G. By theorem5.1.14. T is a topology. Now let T’ be
any topology containing G. If A∈T then AC is net closed with respect to G. Since
T’⊃G, AC is net closed with respect to T’.
Since T’ is a topology, AC is closed with respect to T’ and this implies A is open
with respect to T’. Hence T⊂T’.
Theorem 5.1.18:
Let (X,G) be a generalized topological space. Then the net topology is the topology
generated by having G∪{x} as sub basis.
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Proof:
X≠φ. G is generalized topology. S = G∪{x}. The topology generated by having S as
sub basis is the smallest topology containing S. By theorem5.1.17, the net topology is
the smallest topology containing S. Hence these two topologies are the same.
Theorem 5.1.19:
Let X≠φ. Let G1 and G2 be two generalized topologies in X where G1⊂G2. Then A is
net closed in G1 implies A is net closed in G2.
Proof:
Let (xλ) be a net in A.
Suppose (xλ)→x∈AC with respect to G2 then every set in G2 contains a tail of (xλ).
Since G1⊂G2, every set in G1 contains a tail of (xλ) which implies (xλ)→x with
respect to G1. Hence A is not net closed with respect to G2 implies A is not net closed
with respect to G1. Hence A is net closed in G1 implies A is net closed in G2.
Definition 5.1.20:
Let X be a generalized topological space. Let A⊂X. A is called net open set if AC is
net closed.
Example 5.1.21:
X = {a,b,c}
G = {φ,{a,c},{a,b},X}
Take A = {a,c} is net open since {b} is net closed.
Theorem 5.1.22:
In a generalized topological space, every open set is net open.
Proof:
Let A be an open set in X, a generalized topological space. Since A is open, AC is
closed which implies AC is net closed. Hence A is net open.
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Result 5.1.23:
Converse of the theorem is not true.
Example 5.1.24:
X = {a,b,c}
G = {φ,{a,b},{a,c},X}
A ={a} . AC is net closed as proved in example 5.1.7.
A is net open but not open.
Result 5.1.25:
Converse of the theorem 5.1.22 is true if X is a topological space.
Proof:
Follows from Result 5.1.8
Theorem 5.1.26:
In a generalized topological space the complement of a singleton set is net open iff it
is open.
Proof:
Follows from Theorem 5.1.9.
Theorem 5.1.27:
Let X be a generalized topological space A⊂X is net open iff whenever (xλ)→x∈A,
(xλ) is eventually in A.
Proof:
Suppose A is net open. Let f:D→X be a net. Let f(λ) = xλ and let (xλ)→x∈A. Suppose
(xλ) is a not eventually in A, then for each λ0 we have λ such that λ≥λ0 and xλ∈AC.
Let D’ = {λ/xλ∈AC}. D’ is a directed set and the direction in D’ is the direction in D.
Now (xλ)λ∈D’ is a net in AC and (xλ)λ∈D’→x. Since AC is net closed, we have x∈AC
which is a contradiction to the fact that x∈A. Hence (xλ) is eventually in A.
Conversely,
Suppose whenever (xλ)→x∈A, (xλ) is eventually in A, we claim that A is net open.
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Suppose not, then AC is not net closed. So there exists a net (xλ) in AC and
(xλ)→x∈A. Then by assumption (xλ) is eventually in A which contradicts the fact that
(xλ) is a net in AC. Hence A is net open.
5.2. CHARACTERIZATION:
Let G be the generalized topology in X. Let T be the net topology. We have seen that
net topology is the smallest topology containing the G. Here we give some equivalent
characterization using net closed and net open sets.
Theorem 5.2.1:
Let (X,G) be a generalized topological space. Let T be a topology containing G. Then
the following are equivalent.
1. T is the net topology of G.
2. A⊂X is net closed with respect to GT iff A is net closed with respect to T.
3. A⊂X is net open with respect to GT iff A is net open with respect to T.
Proof:
First we prove 1⇒2
Suppose T is the net topology of G. A is net closed with respect to G implies A is
closed with respect to T.Since T is a topology, A is closed in T implies A is net closed
with respect to T. Also A is net closed with respect to T implies A is closed with
respect to T and since T is the net topology of T, A is net closed with respect to G.
Hence 1⇒2 is proved.
Let us prove 2⇒1, It is given that T is a topology containing G and satisfies
statement2. We want to prove T is the net topology. Let A be a closed set in T. A is
closed with respect to T implies A is net closed with respect to T since T is a
topology. Now A is net closed with respect to T implies A is net closed with respect
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to G and hence A is closed with respect to net topology. Hence every closed set with
respect to T is a closed set with respect to net topology. Hence T⊂Net topology.
Now G⊂T⊂Net topology. But net topology is the smallest topology containing G.
Hence T≡ Net topology. Hence 2⇒1 is proved.
Let us prove 2⇒3
A is net open with respect to G iff AC is net closed with respect to G iff AC is net
closed with respect to T iff AC is closed with respect to T iff A is open with respect to
T iff A is net open with respect to T. Hence 2⇒3 is proved.
Proof of 3⇒2 is similar.
5.3. NET CLOSURE, NET INTERIOR:
In a generalized topological space we introduce a new closure operator and a new
interior operator using nets.
Definition 5.3.1:
Let X be a generalized topological space. Let A be a subset of X. The net closure of A
is defined to be the intersection of all net closed sets containing A. It is denoted as
Ncl(A).
Example 5.3.2:
In the example 5.1.16 A={b} , Ncl(A)={b}
Theorem 5.3.3:
Let X be a generalized topological space. A⊂X. Then Ncl(A) is the smallest net
closed set containing A.
Proof:
Ncl(A) = ∩{B/B⊃A, B is net closed}
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Since intersection of net closed sets is net closed, Ncl(A) is a net closed set. Since
Ncl(A) is the intersection of all net closed sets containing A, Ncl(A) is the smallest.
Definition 5.3.4:
Let X be a GTS. A⊂X. The net interior of A is defined to be the union of all net open
sets contained in A. It is denoted by Nint(A).
Example 5.3.5:
In the example 5.1.16 A= {a}
Nint(A) ={a}
Theorem 5.3.6:
Let X be a GTS and Let A⊂X. The Nint(A) is the largest net open set contained in A.
Proof:
Follows from definition 5.3.4.
Theorem 5.3.7:
Let X be a GTS. A⊂X. Then intA⊂NintA⊂A⊂NclA⊂clA
Proof:
A⊂X. X is a GTS. If A = φ or X then all the sets are equal and hence above is true.
Take any other A
int A = ∪{B/B⊂A, B is open} →1
Nint A = ∪{B/B⊂A, B is not open} →2
Every open set is net open but not conversely. Hence the collection 2 contains
collection 1 and hence intA⊂NintA. By definition, NintA⊂A.
clA = ∩{B/B⊃A, B is closed} →3
NclA = ∩{B/B⊃A, B is net closed} →4
Every closed set is net closed but not conversely. Hence the collection 4 contains
collection 3. Hence Ncl(A)⊂cl(A).
By definition, A⊂Ncl(A). Hence intA⊂NintA⊂A⊂NclA⊂clA.
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From the definitions and theorems, We get the following results.
Result 5.3.8:
1 A⊂B⇒NclA⊂NclB
2 NclA∪B = NclA∪NclB
3 NclA∩B = NclA∩NclB
4 A⊂B⇒NintA⊂NintB
5 NintA∪B = Nint A∪Nint B
6 Nint A∩B = NintA∩Nint B
Theorem 5.3.9:
Let X be a generalized topological space. Consider K:P(X)→P(X) defined as
K(A) = Ncl(A). Then K satisfies the following axioms. 1.K(φ)= φ
2.A⊂K(A)
3.K(K(A) = K(A) 4.K(A∪B) = K(A)∪K(B)
(ie) K is a kuratowski closure operator.
Proof:
By definition, φ is net closed and hence K(φ) = φ.
Ncl(A) is the smallest net closed set containing A and hence A⊂K(A).
Since Ncl(A) is net closed, Ncl(Ncl(A)) = Ncl(A) and hence K(K(A) = K(A).
A⊂NclA, B⊂NclB hence A∪B⊂NclA∪NclB.
Now NclA and NclB are net closed sets and hence NclA∪NclB is a net closed set
containing A∪B. But Ncl(A∪B) is the smallest net closed set containing A∪B and
hence Ncl(A∪B)⊂NclA∪NclB.
Since A⊂ A∪B, we have NclA ⊂ Ncl(A∪B)
Since B⊂ A∪B, we have NclB ⊂ Ncl(A∪B).
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Hence NclA∪NclB ⊂ Ncl(A∪B). Hence we have Ncl(A∪B) = NclA∪NclB which
implies K(A∪B) = K(A)∪K(B).
K satisfies all the four axioms and hence K(A) = Ncl(A) is a kuratowski closure
operator. The topology induced by this operator is the net topology. So we get a new
kuratowski closure operator in a GTS via nets.
5.4 FILTER CLOSURE IN GTS
Definition 5.4.1:
A filter F in a generalized topological space X is said to converge to an element x∈X
if every open neighborhood of x belongs to F.
Definition 5.4.2:
Let X be a GTS. A⊂X is called a filter closed if no filter containing A can converge to
any point outside A.
Theorem 5.4.3:
Let X be a GTS. A⊂X and A is closed then A is filter closed.
Proof:
X is a GTS.
A⊂X is closed.
F is a filter containing A
Let F→x.
Then Nx⊂F
Let U be a neighborhood of x, then U∈ Nx
U∈F and A∈F.
U∩A≠φ.
Every neighborhood of x intersects A which implies x∈Ā.
Hence x∈A
If F contains A and F→x which implies x∈A.
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A filter containing A can not converge to any point outside A.
Hence A is filter closed.
Result 5.4.4:
Converse is not true.
Example 5.4.5:
X = {a,b,c},
G = {φ,{a,b},{a,c},X}
A = {b,c},
F = {{b,c},X} is a filter containing A
It is clear that Fքa
Let A∈F’ and F’→a. Then
F’→a ⇒ {a,b},{a,c}∈ F’
⇒ {a}∈ F’
⇒ AC ∈ F’
Now A ∈ F’ and AC ∈ F’ ⇒⇐
We can not have a filter containing A which converges to point outside A.
Hence A is filter closed.
But A is not closed.
Hence a filter closed set need not be a closed set.
Theorem 5.4.6:
Let X be a GTS. If G is a topology then a filter closed set A⊂X is closed.
Proof:
In any toplogical space it is true that x∈ cl(A) iff ∃ a filter F containing A which
converges to x. Now let A be filter closed. Let x∈cl(A). Then ∃ a filter F containing
A which converges to x. Since A is filter closed, x ∈A.Hence cl(A) = A.
Hence A is closed.
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Theorem 5.4.7:
Arbitrary intersection of filter closed sets is filter closed.
Proof:
Let {Aα/α∈I} be a family of filter closed sets.
Let A = ∩ Aα
Claim: A is filter closed.
Suppose F is a filter containing A and F→x. We have to prove that x∈A.
F is a filter containing A = ∩ Aα
Now A ⊂ Aα
Hence Aα∈F for each α∈I.
Now Aα∈F and F→x and Aα is a filter closed.
Hence x∈Aα, this is true for each α∈I.
Hence x∈∩Aα ⇒ x∈A
Hence A is filter closed.
Theorem 5.4.8:
Finite union of filter closed sets is filter closed.
Proof:
Let A and B be filter closed sets.
Let C=A∪B
Let F be a filter containing C and F→x
We know that every filter is contained in an ultra filter.
Let F′ be an ultra filter containing F
Then F′→x
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Now C∈F ⇒ C∈F′
⇒A∪ B ∈F′
⇒A∈F′ or B∈F′
Case:1
A ∈F′ and F′→x
This implies x∈A since A is filter closed
Case:2
B∈ F′ and F′→x
This implies x∈B since B is filter closed.
X∈A or x∈B ⇒ x∈A∪B ⇒ x∈C
Hence C is filter closed.
Theorem 5.4.9:
The set of all complements of filter closed sets in GTS is a topology.
Proof:
Follows from Theorem 5.4.3, 5.4.7 and Theorem 5.4.8
Theorem 5.4.10:
Let X be GTS. Let A⊂X. If A is filter closed then A is net closed.
Proof:
Let A be filter closed.
Let (xλ) be a net in A and let (xλ)→x
Let G be a collection of all tails of the net (xλ)
G has finite intersection property.
Generate a filter from G.
Let it be F.
Claim: F → x
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Let U be a neighborhood of x.
Since (x λ )→x, there exist a tail tλ0, ∋ tλ0 ⊂U.
Now tλ0 ∈ F and hence U∈F.
Every neighborhood of x is an element of F.
Hence F→x
Now tλ0 ∈ F and hence A∈F
Hence F is a filter containing A and F→x.
Since A is filter closed, x∈A.
(xλ) is a net in A and (xλ)→x ⇒x∈A
Hence A is net closed.
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