AN ABSTRACT
OF THE THESIS OF
for the
MU -LO WANG
(Name)
in
MASTER OF SCIENCE
(Degree)
presented on
MATHEMATICS
(Major)
August 10, 1967
(Date)
Title: RELATIONS AMONG BASIC CONCEPTS IN TOPOLOGY
Abstract approved:
Redacted for privacy
Dr.
B-. H.
Arnold
It is well -known that a topology for a space can be described in
terms of neighborhood systems, closed sets, closure operator or con-
vergence as well as open sets. In fact, it is also possible to describe
a topology in
terms of interior operator or boundary operator. This
paper is devoted to these descriptions and the interrelations between
these basic topological notions.
Relations Among Basic Concepts in Topology
by
Mu -Lo Wang
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Master of Science
June 1968
APPROVED:
Redacted
Professor
for privacy
of Mathematics
in charge of major
Redacted
for privacy
Head of Department of Mathematics
Redacted
for privacy
Dean of Graduate School
Date thesis is presented
Typed by Clover Redfern for
August 10, 1967
Mu -Lo Wang
ACKNOWLEDGMENT
I
would like to take this opportunity to express
my deep appreciation to Dr. B.H. Arnold for not only
his suggestions and encouragement during the writing
of this thesis but also his fatherly care and guidance
for the past two years.
TABLE OF CONTENTS
Page
INTRODUCTION
1
TOPOLOGICAL SPACES
2
SOME EQUIVALENT WAYS TO DEFINE A TOPOLOGY
16
INTERRELATIONS OF THE BASIC CONCEPTS
32
Neighborhood Space
Closed Set Space
Closure Space
Interior Space
Boundary Space
Convergence Space
33
35
36
38
42
46
EXAMPLES
49
BIBLIOGRAPHY
52
RELATIONS AMONG BASIC CONCEPTS IN TOPOLOGY
INTRODUCTION
Although the usual way to define a topology for a set is to de-
fine a family of open sets, yet it is not the only way.
There are other
equivalent ways. that is by using other concepts. What we are going
to do in this paper is to define a topology by other basic concepts, and
find the interrelations between these basic concepts.
First,
we will define topological spaces by open
a topological space we give definitions for the
sets, then, in
six basic concepts:
neighborhoods, closed sets, closure operator, interior operator,
boundary operator and convergence; moreover we will give character-
istic properties for each of these concepts. Secondly, we will define
a topology
from each of these six concepts and prove that each one of
them is equivalent to the topology obtained from open sets.
we will discuss the
relations between these basic concepts. After the
theoretical discussion is completed, we will give
lit our interest.
Finally,
a few
examples to
2
TOPOLOGICAL SPACES
The mathematical system, topological space, with which we
will deal is a set with a structure subject to a few simple axioms.
The main point we will discuss here is that there are many equivalent
ways to give the axioms. Starting with the most commonly used way;
that is, given a family of open sets, we will define from that some
other basic concepts and then prove some theorems about these concepts in the mathematical system we define.
Definition
0: A
f
er with a family
J
(0.1)
X
(0. 2)
If
G1, G2
(0. 3)
If
S
E
topological space
of subsets of
and
=
E
.4,
J
{Gala
a
E
J
,
points of
or
X.
f
is a set
)
X
togeth-
the topology, such that:
X,
sr.
then
,
EA}Ç
G1
,
The members of the topology
to
(X,J°
rm
G2
then
f.
E
_.)
G
a a
3° are
E
f.
called open sets relative
-open sets, and the elements of
In any topological space
(X,
f
),
X
are called the
we may define
many other terms besides "open set ".
Definition
a
1: In a
topological space
neighborhood of a point
x
(X,
7 ),
a subset
if and only if there is an open set
is
U
G
3
such that
x
Gç
E
U.
(X,7 )
It is clear that a set is open in
if and only if it con-
tains a neighborhood of each of its points.
collect all the neighborhoods of a point x in
We
and
X,
call this collection the neighborhood system 71 x of x.
Theorem
(X,
3 ),
(N. 1)
is the neighborhood system of
1: If 71
x
in
then:
/)1 x
/
for all
¢
N. 2)
If
N
tN. 3)
If
N1, N
(N.4)
If
M
N
(N. 5)
If
N1
Enx,
all
y
E
(¡
then
x
in
x
Enx,
and
in
x
E
X.
N.
then
N1
(Th
N2 E'Y'x.
then M ET
N
then there exists an
N2, N1
x
.
N2 Ernx
such that for
.
E
Y
Proof:
(N. 1):
in
x
x
E
X.
X,
Since
so
is open, it is a neighborhood of every point
X
X Ern
x
for all
x
E
X;
i.e.,
%1
x
for all
4
If
(N. 2):
'fl,
N
then
E
there exists an open set
(N. 3): If
and
x
E
G2
G1
N
such that
n G2 Ç
G1
n N2,
N1
(N.4): If
such that
M
x
E
XEGÇ N1.
G
E
%t
(N. 5):
G
E
n
E
M
Ç
x
EG
ç
so
N,
x. Hence
x
E
N.
then there are two open sets
,
Ç N1
and
and, since
is a neighborhood of
N1 r-, N2
hence
x
such that
G
N2
,
is a neighborhood of
N
x
G2
G1 rm G2
i. e.
X;
E
2 N and N l x
For this
N c M.
E
N1
,
ç N2.
G1
1
Hence
is open by (0. 1),
rm
N2 ErTlx.
then there is an open set
,
G,
we have
x
E
G G M,
x.
If N1
E
Take
is open so,
N2
N1
E
then there is an open set
%1x,
=G,
then for all
for all
71
y
E
G.
G
such that
yEG, yEGG N1.
And
The existence of such an
Y
N2
G
is then proved.
Definition 2: In a topological space
-F
is closed if and only if its complement
all closed sets in
X
It is obvious that
5
is denoted by
e
and
X
(X,
satisfies the following conditions:
),
is open.
a
subset
F
The family of
.
are open and also closed.
there might be other such sets, thus the family
Theorem2: The family 5'
°
of
J
And
is not empty.
all closed sets in
(X,
J
)
5
(C.1)
X
(C. 2)
If
(C.
If
3)
g'
E
and
F1, F2
9`,
E
{Faros
a
=
3°.
E
(13,
11 }
then
ç
,
F1 _, F2
then
3°.
E
F
rTh
CL
a
41
E
.
Proof:
(C.1): It is obvious.
(C.2): It follows from the fact that if
--F1, ß--F2
E
-(-F1
therefore
2),
(C. 3):
If
By (0. 3),
F
a a
hence -'F1
39;
E
To
y5 ={
a
a
)
á la
E
E
F2
E
F1
F2
E
A}
G5,
and
r-,
r-
Since
_
then
F
a a
=
-..
a
F1
every subset
3:
E
of
(X,
7
The closure
n {FIEç
F
)
of a
E
the intersection of all closed sets of
j
then
F2
=
{"Tala
(-F a ),
11}
E
thus
we assign a set
and
F
E
(X,
subset
°
)
E
closed set
E,
in
containing
in
E
4',
(X,
7
)
is
E:
}.
Since the intersection of closed sets is closed,
a
3.,
by the following definition.
E,
Definition
For
E
.
the closure of
E
.
7.
F1, F2
we have
E
=
E.
E
is closed.
6
Proposition
of a set is
E
E
E
=
{xI
is closed then
E.
S D
x
E
4
N ,-,
,
the closure
),
E/
r//x, N r-, E
}.
/
xEGc.
and obviously
= cl),
G
E
then there is an
S,
Since
= (I).
G
open; hence
S
x
x
F
E
that
x
for some closed set
F
4
G
Thus
-F ç
S
G
S
S E,
Theorem
^-E.
x,
(1)
(K.2)
ESE.
(K.3)
E
(K.4)
A
_
=
3:
x
such that
rm S =
dp
is
S
E;
G
C E
=
E
S,
therefore
= (I).
which contradicts the fact that
S
x, x
G
such
But
G
x
E
S.
E.
The closure operator satisfies the following condi-
el)
E.
B
This implies that
and consequently
tions:
(K.1)
containing
F
is an open set, so there is an open set
and ^-F
E
frt
E
is closed. On the other hand, we have to show that
S
is a neighborhood of
Hence
x.
S
is open, it is a
Suppose that it isn't true, then there is a point
Ç E.
but
is a neighborhood of
^- S
N
G
neighborhood of each of its points, which implies that
and furthermore
to show
},
and if we can show that
E,
x
(X,
therefore there is an open set
N m E
N
N
S D
Now, if
such that
.
for all
It is clear that
S.
=
S
NEr4
for all
{xl
=
Proof: Let
that
In a topological space
1:
=A
_.)
B.
7
Proof: Since for any set
containing
and
E,
This proves (K.
tion. Finally,
S
A w, B
A
A w. B
the fact that
A
`j
c
A
B
and
B
and
i
B
=
A
A
i
and
cl)
E
=E.
(K.2) follows directly from the defini-
.)
B
B
2
A S. A
v
c
B,
A
_
(12.
can be proved by showing that
,
A
the first inclusion follows
B,
is a closed set containing
B
the second, we use
A
is closed, therefore,
cl)
and(K. 3).
1)
is the smallest closed set
E, E
B
and
so
B
A
A
S
A
B
B
S A
j
v
To see
B.
to get
B.
Besides the closure operator, we have two other operators, the
interior operator and the boundary operator, their definitions and
characteristic properties are given below.
Definition 4: The interior
the union of all open sets in
Eo= v
{GIG
(X,
E°
7)
of a set
E
in
(X,
7
)
is
contained in E:
çE, GEy}.
Since any union of open sets is open, therefore
For an open set
E,
we have
closure of the complement of
Eo
E
=
E,
E°
and for any set
is the complement of
is open.
E,
the
Eo.
Theorem 4: The interior operator satisfies the following conditions:
8
(I. 1)
X°
(I. 2)
E° c- E.
(I. 3)
(E°)°
(I.4)
(A
X.
=
(Th
=
E°.
B)o =Ao
Proof: Since
open, therefore
(I. 2) follows
(I. 3).
n B)o
(A
=
set contained in
A?ArThB 2
and
Bo
2
(A
=
(A
n
B)o,
to show
n
(A
Definition 5: A point
in
3-
(X,
both
),
the boundary of
Every open
x
E
E
only if
E
-E,
Ao m Bo
B)o
x
E
=
X
n
are
r,
B)o,
so,
(A
(Th
is an open
Bo
Bo S (A
2
n
Ao
Ao
Ao2
G
x
containing
G
x
(Th
B)°
Consequently,
B)o.
Bo.
B
is a boundary point of a subset
G
containing
x
intersects
x
is called
E
Eb.
and denoted by
E
and that
B)o
The set of all boundary points of
and every open
x
n
Ao
if every open set
and --E.
E
E
these prove (I.1) and
Eo,
=
2A(ThB?(A
B
hence
B)o,
by the two inclusions,
E
(E )o
we use the facts that
A r,. B
(Th
0
and
X
of any set
Eo
directly from the definition. To prove
n Bo,
Ao
and the interior
X
Xo
Bo.
(Th
x
containing
intersects
x
E
if and only if
intersects
thus we have the following proposition.
E
if an
9
Proposition
E
b
of a set
E
Theorem
2: In a
topological space
is
=
Eb
(X,
7 ),
the boundary
E Km^-E.
The boundary operator satisfies the following con-
5:
ditions:
b
(B.1)
=4;.
(-E)b.
(B.2)
Eb
(B.3)
(Eb)b ç Eb.
(b.4)
A
=
B
r
(A
B)b=A
(Th
B
n(Ab
Bb).
Proof:
(B. 1): Since every neighborhood of every point in
b
intersect
=
.
(B. 2):
It is clear from the definition.
(B. 3):
To prove
we have every open set
-E b .
Suppose
y
(Eb)b S Eb,
G
x
this implies
x
Since
E
Eb,
y
G
E
x
intersects
x
is one of the points of
y.
and ^-E,
we notice that, if
containing
also an open set containing
E
fails to
X
x
n
Eb,
therefore
G
Eb,
X
E
Eb
then
(Eb)b,
and
G
x
is
intersects both
(Eb)b
Eb.
(B.4): By Proposition
prove
A
n
B
n
(A
n
B)°
=
2, we
A
n
B
have
Eh
n
v
(Ab
=
E
Bb)
10
, -E.
We will
by considering the
following equations.
A r,B
n(A nB)b
=
nB n[A nB n-(A nB)]
A
=
A
nB n~(A r,B).
The last equality holds by (K.2).
A
nB'Ab
_
(A
=
A
B b)
(ThB nj-^s7
=
A nB
n[(A n^'A)
n A)
nB n(--A
B
=
A
nB
L)(B
(Th--B) =
n B)
(A
nB (Th^-A) J(A r,B n"-B)
nB n--(A nB)
The last equality holds by (K.4).
The simplest form of convergence is that of a sequence of real
numbers. Using the so- called
precisely what
E
definition of a limit we can tell
we mean by convergence and the
limit of
a
sequence.
But, as we will see below, convergence and limit are also purely
topological notions and hence can be described by using only topologi-
cal concepts.
First
we introduce nets which
are
a
generalization of sequences.
The definition of a net involves a directed set, hence we define directed set
first.
11
Definition 6: A directed set
with a binary relation
m, n
If
(a)
m
fib)
if
(c)
If m
m
and p
E
set
and
>
are
n
>
A
net
there is
p
then there is
E
(S, >)
n
E
clement of
7: A
net
there is
E
E
such that
D
p > m
is frequently in
A
such that
a
set
{Sn, n
e
is a function
D
defined on
belong to the set
S(n),
E
S
n>
and
p,
A
and
S
n
e
A.
D
X.
if and only if
then
if and only if for each
n > m
direct-
on a
D, > },
S
n
m
Clearly,
E
A.
E
D
A
a net is
if and only if it is not eventually in the corn -
A
A.
(S, >)
and
D
p
p
is eventually in a set
D
n
D
X
or
Sn,
on
or
(S, >),
such that if
D
If a net
{non
then
n > p,
now ready to define a net.
(S, >)
frequently in
_
and
> n
n.
--
E
m
m.
with values in a set
D
there is
m
D,
e
whose function values
net
such that
D
E
then
D,
Definition
ed
such that
>
p
D
p.
>
We
and
is a nonempty set
(D,>)
E
D
S(n)
is frequently in a set
E
A}
such that
A,
then the set
has the property that for each
p >
m.
Such subsets of
D
m
are
E
E
12
called cofinal subsets of D.
a
set
A net
into the set
8: A
net
converges to a point
s
in a topological space
(S, >)
relative to
X
E
to mean that
7
s.
i.e.,
It is obvious that a constant net coverges to this constant;
if
is a net such that
S
es to
for this net
s,
Sn
n
S
=
for all
s
)
We will use the notation
converges to
S
(X,
if and only if the net is
34.
s.
eventually in each neighborhood of
s
which maps
D
A.
Definition
=
is frequently in
D
if and only if there is a cofinal subset of
A
lint
Sn
n
on
(S, >)
n
E
then
D,
converg-
S
is entirely in every neighborhood of
s.
Just as subsequences play an important role in the discussion
of the convergence of sequences, subnets will be important in the con-
vergence of nets. Our definition is given below.
Definition 9:
{S
n'
in
(a)
n
E
E}
A
net
{T
m
,
m
E
is a subnet of a net
D}
if and only if there is a function
N
on
D
with values
such that
E
T
=
SoN,
or equivalent
Ti
=
for each
SN
i
E
D.
i
(b)
For each
N
>
p-
m.
m
E
E
there is
n
E
D
such that if
p > n,
then
13
The second condition states, intuitively, "as
larege, so does
ly in
P
an
m
this
T
c
=
S
P
there is
Np
E
n
,
n
n
q
N
such that if p
E
p
and
E
then
D, p > n,
E
of
then
then there is
-> m; i.e.,
e
A,
E
A,
P
-
S
then for
q
then
> n
is
S
is also eventual-
S
A,
P
N
T
N
>
m,
there is
hence
n
D
E
is therefore eventually
T
P
This result proves that if
A.
q> m
and
E
E
T
is eventually in
E}
E
D
e
since
A
such that if p
in
{S
such that if
E
m
then any subnet
A,
For if
A.
and this enables us to prove: if a net
N ",
eventually in a set
becomes
p
converges to
S
s
in
7 ),
(X,
then so does each subnet of S.
Another fact about convergence of nets is as follows: if
does not converge to
n
,
n
E
fails to converge to
D}
a
point
the complement of a neighborhood of
subset
E
D,
of
subnet of
{
subnet of
S
S
n
n
,
{S
E
n
,
E}
no subnet of
This follows by the argument that if a net
s.
which converges to
1S
then there is a subnet of S,
s,
S
n
e
E}
s
s,
then it is frequently in
and hence for a cofinal
is in the complement. Clearly no
converges to
s,
so the existence of such
is established.
With one more concept, that of product directed set, we can
then list as a theorem some conditions the convergence in a topologi-
cal space satisfies.
Suppose
(D, >a),
a
E
t
A,
_
{Da, a
a
E
A}
is a family of directed sets
the Cartesian product of C,
,
P
=
X {Da,
a
E
A },
14
is a set of all functions
all
a
is the Cartesian product
if
d, e
P,
E
such that
A
d > e
da
a
a
and the product directed set of
A,
E
on
d
,
(
=d(a))
{X {D
a
with the binary relation
P
if and only if
d
a
a
Theorem 6: Convergence in a topological space
E
E
a
for
A }, >}
such that
>
for all
-a e a
>
a
,
D
E
A.
(X,
y) satis-
fies the following conditions:
(a) If
then
is a net in
S
n
for each
s
=
(c) If
S
does not converge to s, then there is a subnet of
Let
E
D.
in
SOR
F,
D
n,
then so does every subnet of S.
converges to
m
then
Sn
S
(d)
(m, f)
such that
(b) If
no subnet of which
each
7)
s.
converges to
S
(X,
s,
converges to
s.
be a directed set, let
be the product
Let
F
let
R(m, f)
converges to
_
(m, f(m)
E
be a directed set for
m
, m
E} and for
m
then, if limlimS(m,
n) = s
m n
DXX{E
)
where
s,
S,
m
E
E
and
D
n
E
E
m
Proof: We have already shown that convergence in a topological
space satisfies (a),
(b) and (c), so, (d) is the only thing we have to
prove.
S(m, n)
Suppose lim lim
m n
of
s.
We
must find
a
=
member
s,
and
U
is an open neighborhood
(m, f)
of
F
such that if
15
then
(p, g) > (m, f)
lim S(p, n)
a
member
Ep.
If
E
U
f(p)
m,
p
(p, g) > (m, f),
g(p) > f(p),
SoR(p, g)
for each
of
E
such that
let f(p)
then
we get
be an
-m
p >
SoR(p, g)
S(p, n)
E
U
D
so that
n
S(p, g(p))
E
U.
E
p
for all n
arbitrary member
hence lim S(p, n)
=
E
and then for each such
p.> m,
P
Choose m
U.
E
U,
of
choose
-> f(p)
Ep.
and since
If
in
16
SOME EQUIVALENT WAYS TO DEFINE A TOPOLOGY
In this chapter, we will use, successively, each one of the con-
cepts neighborhood system, closed set, closure operator, interior
operator, boundary operator and convergence as the basic notion to
define a topology.
We will give a name to each type of space
indicat-
ing which concept was used as the basic one. More than that, we will
show, in each case, the concept taken as the basic notion is identical
with the corresponding concept defined by the topology. For example,
if we start with a closure operator
c
which satisfies the so- called
Kuratowski closure axioms we can get a topology
we get
7
let
,
we prove that
denote the closure of
E
c(E)
=
E
for every
E Ç X.
E
7
.
Then, after
defined from
7
,
By that we then have
shown that each type of space is equivalent to a topological space, and
therefore all the types of spaces are equivalent.
The following six theorems present these
Theorem
set
X
a
1:
(N.1)
(N. 2)
I(
If
Let there be associated with each point
collection
hood system of
F g
N
E
results.
-x
of subsets of
X,
x
of a
called the neighbor-
subject to the following neighborhood axioms:
x,
for all
,
x
then
in
x
E
X.
N.
17
If
(N. 3)
En
NI, N2
(N.4)
If
M
(N. 5)
If
N1
all
y
N2
(Th
N
r//
E
,
`
X
rn x
E
2
then
T
M
then there exists an
X,
E
N
1
and
N
E
then
x
N2, N1 E/
N2
x
E
?x
such that for
y.
Y
The members of rn
are called neighborhoods of x.
x
be the family of all subsets of
7
of their points:
topology for
{G Ix
=
J7
defined by the topology
implies
7
and, if
X,
,
Proof: We first show
'
which are neighborhoods of each
X
G
E
7
Let
G
E
`x
7
then
},
is a
is the neighborhood system of
then
7
"
ll
=
x
for all
x
i.e.,
is a topology;
x
7
x
X.
E
satisfies
(0. 1), (0.2) and (0. 3).
(0. 1): Since
then
71'x
(13.E
/
is vacuously satisfied. So,
y!` ,
x
r c for all
X
E
X.
'X
E
X
Since
then
contains no point, the condition if x
4
N
X
e
x
X,
E
ç X,
71
(0. 2): Suppose
x
E
and
G
x
E
By (N.3),
G1
(-Th
E
2
E
E
N
7,
X,
hence
E
X
(131,
By (N. 1),
fn x for each
E
7t*
that is, if
1.
and
By the definition of
G2.
G
X
7.
E
by (N.4) we find that
G1, G2
2
1
hence there is an
hence
,
X
irp
E
G1
x
E
7
G1
,
rm
G1
1
G2E.
e
G2,
then
`} X, G2
'lX
2
E
18
5
(0. 3): Suppose
X
a
G
o
G
a
CL
a.
for some
G
E
x
and, since
To show
so
,
G
a a
o
x,
then
G
E
N'
E
IN-
N
/ l
(
.
E
71
X
E
7l x
G
7,
N
E
7x
N
G G N
Now, if
y
E
such that for all
E
;
!
A
y
z EN', N
N
EI)
so we have G
topological space
(X,
E
7
a
i
E
l
0
G
x
G
E
.
'
But,
.
G
G
IN
E
7
)
E
I
l
Y.
y
E
E
x
T
Ç en x'
E
i l''`z
G
}
then
,
G
E
7,
implies
there exists an
By (N.5)
Y
which implies that
z
E
G;
:'
r/
with neighborhood systems as
(X,
,7
)
and the topol-
Next, we use closed sets.
2:
Let
9.
'1`
i.e.,
(y by (N.4). The proof is completed.
is called the neighborhood topology.
Theorem
,
be the set of all
Let
= {z
such
G
G
Thus
then
,
a a
a a
implies
G
basic concept is called a neighborhood space
ogy
G
,
E
by (N.2). If we can show
then
G,
x
by (N.4).
fl
E
x
That is, we need to show that
.
and N'
°
as a neighborhood:
N
and
then there is an open set
,
Y
y
7
Ç
}
by (N.4); hence
x
suppose
and
G
E
in
we find
points which have
x
N
E
By definition of
2 G,
N
'l
clearly,
A
E
rx
a
G Ç N.
E
la
a
By definition of
O
(7`X = 71)' If
that
= {G
be a family of subsets of a set
which satisfies the following closed sets axioms:
X
19
5
and
(C.1)
X
(C.2)
If
F1, F2
(C.3)
If
5
E
_
cp.E
7
E
{F la
a
then
y`,
v
F
y
A
E
F2
E
7 °.
the family of the complements of members of
=
{GI-G
7
E
7
then
},
eit
na F a
then
E
7
5
and if
X,
then
,
be
5*:
is a topology for
the family of closed sets defined by
7
J
Let
are called closed sets in X.
7
The members of
'`'
=
is
7.
Proof;
By (C. 1), it is
(0. 1):
(0. 2): If
(,-G1)
v
(^'G2)
G1, G2
y
E
7,
E
and
r
clear that
X
E
7
then ^G1, NG2
E
=^[( G1) v
G2
G1
and
5
it,
E
7.
and by (C.2),
,
(^'G2)],
so,
31 rThG2E1.
If
(0. 3):
By (C.3),
G
E
a a
.
S
Q
(Tha(^^Ga)
=
{Ga
E
la
1l
E
5'*,
Ç7,
then
_) aGa
and
=
61=
{G
...[(--, a(,.,Ga)].
la
E
Al Ç
Hence
7.
(5'
_
°
space
(X,
only if
' -(-F)
By the definition of a closed set in a topological
):
we have
),
=
F
E
7
.
F
So,
E
5
is and only if ^'F
7":
=
y.
E
7' if
and
20
A
topological space
(X,
is called a closed set space
J
(X,
with closed sets as basic concept
)
t) and the topology
is called
°
the closed set topology.
Definition 1: A closure operator in a set
c
which assigns to each subset
E
of
X
a
is a mapping
X,
subset
c(E)
of
X,
and satisfies the following four axioms, the Kuratowski closure
axioms:
(K. l)
c (cp)
(K.2)
E S c(E).
(K. 3)
c(c(E))
(K.4)
c(A
=
4.
=
B)
c(E).
c(A)
=
c(B).
The following theorem will show how a closure operator deter-
mines a topology and prove that, in this topology, the operator is
topological closure.
Before proving the theorem, we notice first that, for any closure operator
c,
we can write
B
This implies
c(A)
=
implies
A
S
B
A
v
(B-A),
ç c(B).
c(A)
ç c(B).
For, if
then by (K.4), c(B)
=
c(A)
A
S
B
j c(B-A).
21
Theorem
Let
5
_
ology for
Let
3:
{Flc(F)
and, if
X,
the topology 7°
E
{GI -G
=
7
then
},
E
X.
is a top-
denotes the closure of
E
defined by
for all subsets
E
in X.
c(E)
then
,
7
and
F}
=
be a closure operator defined in a set
c
E
=
Proof:
Q0.1): By (K. 1)
which implies
X
c(--G1)
c[ (NG1)
._)
(-G1)
(-G2)
..)
G1 rm G2
=
E
G1, G2
-G1
(~G2)
4)
and we have
Ga
a
therefore
B
;
G
G1
rm
G2
(-G1)
=
=--[(-G1)
and we
that is,
(^'G2)],
6
c (- G
a
for all
c(B) S
-
) _
a,
CI,
=
B;
7
then
,
Let
B
so
i.e.,
c(B)
hence
á(^'Ga),
B
So
By (K, 3),
E Ç c(E).
=
c(B).
U G
CI,
a
c(c(E))
Hence
e
=
a
E
{FIF
E
5
(
-G a ),
=-m(G)
a
a
=
E
}SY,
11
then
and
By (K.2)
7.
E
- {FIFE
which implies
c(E),
c(E)
=
B.
G
.
'= {-Ga la
for all a,
c(-Ga) =^-Gaa
c(B)
(-G a )
ç
for all a.
-Ga
so
rm
{Ga la E11}
_
E
by (K.2),
5
E
(^,G2);
_.)
(c(E) =E): By the definition of closure,
F 2 E }.
E
By (K.4),
B
=
ck
then ^-G1, ^-G2
This implies
Ç c(B),
a a
and hence
.
X G c(X),
By (K. 2)
e7.
(0. 3): Suppose
B
9'
E
I,
E
v c(-G2)
But
.
X
E7.
X
c(-G2) =-,G2.
and
c(--G1)
] =
hence
Thus
c(X).
=
(0. 2): Suppose
have
7,
E
and
c(E)
But,
5) and
E
7,
F 2 E };
and
22
therefore
c(E) ç c(E); since
c(E)
ç
c(E)
A
On the other hand,
c(E) n E.
=
is closed,
E
E Ç E
c(E)
=
which implies
E
These two inclusions prove that
E.
topological space
(X,
7
c(E)
=
E.
with closure operator as basic
)
concept is called a closure space
and we have
(X,
7
and the topology
),
J°
is called the closure topology.
Definition 2: An interior operator is a mapping
i
of a set
of
signs to each subset
E
X
a
subset
i(E)
which asand
X,
satisfies the following four interior axioms:
X.
(I. 1)
i(X)
=
(I. 2)
i(E)
g_
(I. 3)
i(i(E)
(-°.4)
i(A
)
(-Th
E.
=
B)
i(E).
=
i(A)
Theorem 4: Let
X.
and, if
then
.
Let
E°
i(E)
=
_
{Gii(G)
rTh
i
=
i(B).
be an
G, G
ç
interior operator defined in
X },
denotes the interior of
E°
for all
E S X.
then
E
J
a
set
is a topology for
defined by the topology
X,
7
,
23
Proof:
(0.1): By (I.1),
i0)
so
i(X)
hence
= (l0;
¢
=
G1, G2
i(Gl
By (I. 3),
G2.
G2)
1
G2
G1
G
a
i, e.
;
E
i(G) ç `.
a
a
a
G
a
ç
B,
then
fore
i(A)
ç i(B).
a
)
But
a
a
A
A
=
G
=
E
we know that
a
G
A
E°
)
(`JaGa)
7,
i(E)
On the other hand,
=
By (I. 2), i(4)) ç
i(G1)
=
G1
=
a,
G
a a
)
7
,
(Th
G1
and
G2,
so
By (I. 2), we have
.
2
G
a a
by (I.4)
Ga
a
we have to show that
,
S
we notice
,
i(A)
i(A)
=
v a i(Ga
so,
a a
)
G
E
c
E}
{GIG
Eo
5
which implies
topological space
f,
E
G
S E }.
E°
E
E° Ç i(E);
(X,
a,
)
i(
ç
a s
7
(X,
i(i(E) )= i(E),
and
E
there-
i(B)
va i(Ga ç i()a Gs ).
G
=
first that,
G
a s
)
and,
.
Since
.
(Th
for all
G
a a
which implies that
G
_
A} S
E°): By the definition of interior in
=
{GIG
=
then
i(G1) n i(G2)
=
E
Now, since
for all a,
)
7,
E
VaGa
B,
(Th
i(G
a
=
i(
To see
for all
(i(E)
i(E°)
{Gala
_
5 i) naGa)
a s
consequently,
Eo
S
i(_,aGa )
,
.
if A
i(G
7.
E
7.
E
(0. 3): Suppose
a
X
7.
E
(0. 2): Suppose that
i(G2)
hence
X,
=
)
°
and
)
This implies that
I,
E°
=
E,
i(E) ç E°
i(E°) ç i(E)
so
hence
ç
i(E)
and
i(E).
with an interior operator as the
basic concept is called an interior space
(X,
7
)
and the topology
24
7
is called the interior topology.
Definition
b
boundary operator in a set
A
3:
which assigns to each subset
of
E
subset b(E)
a
X
is a mapping
X
of
X,
and satisfies the following four boundary axioms:
(B.
b(4)) _
1)
b(-E).
(B.2)
b(E)
(B.3)
b(b(E))
(B. 4)
A
r
=
b(E).
5_
B rm
b(A
(Th
B)
=
n
A
B
ç
v
(b(A)
b(B) ).
Before we state and prove the theorem about boundary operator,
we will prove
first
a
lemma.
Lemma: If b(A) r\ A
Proof: Suppose
B
2
=
then
(1),
A,
then
b(B)
A
rm
,-
A
=
A.
B
=
4
for all
Now,
r, b(A)
A
=
A
b(A)
(Th
=
(b(A)
4
b(B))
_
(A
b(A)
)
by hypothesis, therefore
(A
(, b(B)
A
2 A.
Substituting this
into the condition (B.4), we get
A
B
,
).
b(B)
=
.
25
Theorem
X.
Let
.7
for
X,
and, if
topology
1
{GIG
_
ç
Eb
then
,
be a boundary operator defined in a set
Let b
5:
X,
b (G)
nG
denotes the boundary of
b(E)
=
for all
Eb
3' is
then
},
=
a topology
defined by the
E
E S X.
Proof:
It is clear that
(0. 1):
b(cp) _
by (B. 1), so
ep.
eP
E
More-
3°.
over, by (B.1) and (B.2) we have
b(X)r-,X
Therefore
X
E
G1, G2
We have to show that
b(G1
= O (Th
X =(j).
7.
If
(0. 2):
=b(^'X)(Th X
(Th
G2)
E
G1
n
then
3°,
(Th
G2
Gi
=
1
for
i
=
1,
i.e.,
;
E
(G1 (1 G2)
n
b(Gi)
_ ci).
By (B.4) we have
(G1(ThG2)
n
b(G1
nG2)
=
G
1
nG22 (Th (b(G
)
=
[(G
_
[(G1r1b(G1))nG2]
_
($nG2) J
(ThG
1
= (I)
2)
(Th
_..)
b(G
1
b(G1)]
(G1rmci))
._}
_)
2
) )
[(G1(ThG2)
mb(G2)]
[Gln(G2r-b(G2))]
2.
26
5
(0.3): Let
vaGaE
a a
b(Ga )
i.e.
;
n Ga
=
= {G
a.
for all
4
n Ga
a Ga
b(,_) G
a a
A} ç
E
7
n (vaa Ga)
a
b(vaGa)
a a
b()aGa)
a a
the lemma
la
a
=
By hypothesis we have
.
for all a.
4)
a
_
vaa G a 2
But
=
We have to show that
.
{b( a G)
a
(Th
G
for all a,
a
so, by
Hence
G}=(1).
a
The proof is then completed.
(b(E)
containing
only if
b(G)
b(-F)
(--F)
rm
for any set
E
rTh
=
G
(I;
i. e.
(--E)
B
F
b(F)
(Th
b(E v b(E))
E
b(E)
Since
X.
in
E
set
a
= cl,
we have to show that
and
first that
Eb): We will show
=
is a closed set
is open if and
G
is closed if and only if
(-F)
(Th
_
cl),
.--(E
for b(-F)
J b(E))
_
4.
=
b(F).
Let
A
So
=
^-E
=-b(E): we apply (B.4) and get
(,- -b(E))
r, b( ~E,(-b(E))
_ (
-E) n (-b(E)) r- (b(-E)
v b( ~b(E)))
By DeMorgan's law and (B.2), we have
(
'E)
rm
(...b(E))
b((-E) - (- b(E)))
=
=-(E
b(E)),
b(-(E _) b(E)))
=
and
b(E v b(E)).
Substituting these into Equation (1) and again using (B.2), we get
^,(Ejb(E))
rm
b(Evb(E))
=
^'(Eub(E))
n
[b(E)vb(b(E) )]
(1)
27
Moreover, since b(b(E)) S b(E)
v b(b(E))
b(E)
by (B. 3),
=
b(E);
hence we have
(E.)b(E)) e- b(E.)b(E))
This implies that
v b(E)
E
We will prove
b(E)
=
=
...(Eb(E))(Th b(E)
is closed.
by showing that
Eb
and we shall use the fact that
Eb 2 b(E),
_
Eb
=
b(E)
Eb
E r-, --E.
and
For the
first inclusion we have
Eb
=
E ( ,-E
(E_.)b(E))
n (-- Erb( -E))
=
b(E).
To complete the proof of the second inclusion, we need only
show that
b(E)
E
v
b(E)
We will prove
E R E.
then
2
E
b(E)
rm
n (-E)
4.
and, since
Now
=
(-E)
n ('E) n
E
2 b(E),
it by contradiction. Suppose that E
X (I),
(NE)
which can be reduced to
let A
b( -E(Th-E)
and
= (
-jE)
n
B
in (B.4); we get
=
(^'E)
n
(b(-E) jb(^'E))
After simplification, it becomes
(E)E)
But, since
-E
b(E)
E B E,
=
n b(EK,E) = -(E.)E) n
E
.-E r-, (b(E)
v
E
=
E;
b(E))
=
(b(E)
b(E)).
hence
(-.E
(Th
b(E),
is closed, we have
E
^-E
=
for
b(E))
(r-E
n b(E)).
28
Now the
fact that ^-E
n
b(E)
gives --E
_ 4)
b(E)
= 4),
which is
The proof is completed.
a contradiction.
The topological space
(X,
7
with boundary operator as
)
basic concept is called a boundary space
(X,
J°
and the topology
)
.7 is called the boundary topology.
Definition 4: A convergence class for a set
of ordered pairs
where
s),
(S,
is a net in
S
X
is a class
X
and
s
point in X, which satisfies the following convergence axioms.
convenience, we say that
lim
n
(a)
S
n
=
s
If
(
S
if and only if
G )
(S, s)
is a net such that
S
()
verges
to
(6)
If
S
converges
(c)
If
S
does not converge
(d)
m
E
D
D.
(m, f)
(6 ),
E
n
=
(6)
(G)
converges
F
be the product
let R(m,
o
R
S
Con-
f)
=
converges
a
DX X {E
(m, f(m))
(CL)
s.
to
directed set for each
m
then, if
to
S.
then there is a subnet
s,
to
Let
S
then
then so does each subnet of
directed set, let Em be
then
or that
s
for all n,
s
be a
F
For
d4
s,
to
no subnet of which
S,
Let
S
E
to
a
s.
(b)
of
(5)
converges
CiL
s.
,
m
E
D}
and for
lim
lim S(m,n)
m n
=
s
29
A
for
X
Let
of
(
convergence class determines completely a unique topology
as the following theorem shows.
Theorem
6:
J
{GIG
ç
then
°
=
G },
(i)
to
Let
be a convergence class for a set
G
is a topology for
if and only if
s
-G
and no net in
X
and any net
X,
converges to
S
(a)
converges
to a point
converges
S
relative
s
X.
to ,7"
.
Proof:
Since there is no point in --X
(0. 1):
-X,
therefore
X
7.
E
(a)
can converge
(0. 2): If
to a point in
Also, no points in
to any point of
G1, G2
E
f,
G2.
We have to show that
-(G1
n
there is
converges
a
net
(5)
converges
Sn
E
{Sn, n
G1 }, D2
=
either
{Sn, n
E
E
D1
Dl}
which also converges
case, it is
a
G2
in ^-(G1
D}
s
and
or
or
(6)
therefore
0
E
1;
G1
E
Sn
D2
G2)
n
G2,
_
s
D2}
G1
(-G1)
)
G2.
v
D1
1
(CO
to a point in
n
G2
Suppose that
which
(^-G2)
=
D1
is a subnet of
G1
.
{nln
E
D2
and
D
=
D
and consequently
D,
by virtue of
contradiction, therefore
(
(Th
let
is cofinal in
E
nets
that is, no net in
we have
G2 },
E
{Sn, n
to
n
3"
E
no
converges
converges
to a point in
(Ce)
D
n
G1
to a point
{nln
therefore either
E
implies that
4
then no net in --,G1
and no net in -G2
G1
G2)
4
hence no nets in
= .4),
E
(b).
T.
{Sn, n
E
D}
In either
30
(0.3): Suppose
va Ga 3.; i.e.
to a point
^{S
ua Gs
n
n
,
=
E
then
{S
(-G
n
"
contradicts the assumption that
(5)
(Convergence
{Sn, n
n
s
E
relative
final subset
{S, n
'
E
verges
(n)
we have to show that
,
¡,
which converges
G
a a
()
{S
to
s
E
s
G
E
a
0
in
n
E
D}
s
E
vaa Gs then
,
for some a o
o
which converges to
s
a
A,
E
G
E
0
G
a a
but
s,
to
{S,
S
n
E
n
ENG
in
D}
by (b), hence
relative to .7,
s
Then by (c) there is a subnet
verges
(CO
let
=
D
m
Because
{nln
{S
m
s.
to
,
m
G
containing
e
for all
G,
n
E
s
E.
Since
this subnet con-
is not open relative to
G
(CR)
and fails to converge
m
,
m
n
D },
E
to
A
m
s;
.
and
-> m}
converges to
D}
7.
hence there is co-
(G)
s
T
s.
to
no subnet of which con -
To find a contradiction, for each
lie in the closure of each
which converges
{S
such that
D
E
e
does not converge to
S
which is a contradiction. Conversely, suppose that a net
converges to
This
.
a
e.7 for all a. Therefore
such that
G
,
to a point
is not eventually in G,
D
n
(CO
Convergence): Suppose that a net
is a subnet of
E}
a
G
(G)
D}
e
of
E
ç
Then there is an open set
3°.
to
{S, n
such that
=
converges
D}
A}
E
and, if
,
a
is a net in - -G
D}
E
la
Suppose there is a net
.
is in every NG
D}
a
which converges
)
a
a
n
va Ga
in
{G
there is no net in
,
E
=
yS
A
m
= {S
relative to
n
,
in
m
n
7
E
,
Dm}.
m
s
Hence, there is a net in each
that is, for each
m
E
D
D
must
A
m
there are
31
a directed
that
{S o
set
and a net
E
U(m, n), n
E
R(m, f)
DXX {E
m
m
then
E
E },
over, if p > m,
is
U
o
R(p, f)
>
m,
then
U
=R(p, f)
it follows that
topological space
S
(X,
o
UoR
converges
7
)
)
to
in
s. More Dm;
is a subnet of
S
(a)
to
that
which
s.
with convergence as basic con-
cept is called a convergence space (X,
called the convergence topology.
Applying condi-
(m, f)
(
such
D
belongs to
U(p, f(p))
S
s.
to
converges
=
in
Em}
for each
(m, f(m))
SoUoR
is a contradiction. Hence
A
=
E
(a)
converges
Em}
tion (d), we have if
,
{U(m, n), n
7)
and the topology
°
is
32
INTERRELATIONS OF THE BASIC CONCEPTS
As we have shown in the
a
last chapter, we can get
a topology for
set by knowing first any one of the six basic concepts, and more-
over, by showing that each space is equivalent to a topological space
with open sets as basic concept, we have shown that each one is
equivalent to the others; that is, all the concepts are equivalent. But
it is not enough in some sense; we would like to know the ways to
go
from one directly to the others which are sometimes more interesting
than going around through open sets to get the equivalence. This gives
us the idea to
try to define each one of these concepts in terms of
each of the others, and that is what we are going to do in this chapter.
We will give, in each space, definitions for the other concepts to see
the ways the other concepts behave, and claim that with these defini-
tions, these notions satisfy their characteristic properties respectively.
And
more than that, the topologies which come from them are
equivalent. For example, in
a
neighborhood space
(X,
7 ),
we
give a definition for closed sets in terms of neighborhoods, then
claim that the family
of all such closed sets is actually the
7
family of all closed sets relative to the neighborhood topology
by which we mean
y
satisfies the closed sets axioms. Further-
more, we conclude that the closed set topology
.
tained from
7
=
"'
We
;
°
s
for
X,
is the same as the neighborhood topology
7;
ob-
i.e.,
claim similar conclusions for all the concepts in each
space, but, all we shall prove are just a few cases as examples. The
33
discussions will be given in several sections.
Neighborhood Space
In a neighborhood space
(X,
s°
the basic concept is neigh-
),
'`x satisfies
borhood system; the family of neighborhood systems
7
the neighborhood axioms. The topology
5=
{G Ix
E
implies
G
G
E
-n
for
} .
Definition 1.1: In a neighborhood space
is closed if and only if x
Theorem
7*
Let
1:
neighborhood space
implies
F
{
-F
E
(X,
71
7),
a subset
F
x
denote the family of all closed sets in a
7 ),
(X,
is
X
then
7
.k
satisfies the following
closed set axioms:
5
and
(C.
1)
X
(C.
Z)
If
F1, F2
(C. 3)
If
6
E
_ {F
from
7 0,
then
7".
e>.,,
a a
Moreover, if
4, e
E
l
11}
then
G
."',
F1
u
F2
then
E
(Th
7*.
a
F
a
E
3 * is the closed set topology for
7* 7
; .
X
obtained
=
Proof:
(C. 1): 4
E
7*
is clear, for every point is in the complement
34
,
of
the set
EI*
X
and also
X,
is a neighborhood of every point.
X
is also obvious, for there is no point in the complement of
'
hence the implication is true.
X,
F1, F2
(C. 2): If
Suppose
x
4
F
F2,
c
5*,
E
then
x
1
F
4
and
x
1
4
(-F1)
jF2)
(r-F2) =--(F1
E
F2
E
By (N.3),
71X.
v
F
1
F2
is
closed.
.
Let
(C.3):
(Th
F
E
a s
F
a
E
0
%'1
x
If
5
x
for
F
(Th
a
a
(7*
F,s
Al ç T '
then
we have to show that
F
X
a
for some
,
a
0
o
.
But
0
((Th
:
E
a
is closed. Now,
^-((Th
hence by (N.4)
{F la
=
a Fa )
F
a a)
E
=
(-Fa )? -Fa
a
therefore
71x,
x
,
0
r, Fa
E
CL
The following chain of equivalences proves this:
GEJ
if and only if
x
E
if and only if
x
4
if and only if '-G
if and only if
G
E
implies
G
G
implies
is closed
34)*.
G
E
7 x
-(-G)
E
en
x
.
and
F1
1
which implies that
Enx,
j
Since
F2.
are closed by hypothesis, we have -F1' ^-F22
F2
Fl
we want to show that
35
Definition 1.2: In a neighborhood space
ator
E
is:
c
c(E)
=
for all
{xl
N
71
E
x
N - E
,
/ it}
for all
X.
Definition 1.3: In a neighborhood space
operator
i
is:
i(E)
there is an
_ {xl
N
e %1
Definition 1.4: In a neighborhood space
operator
b
is:
b(E)
N E77
for all
EÇ
x
such that
x
N
SE}
}
{xIN
=
E
rm
/
4
and
verges to a point
s
the boundary
X,
N
rTh
(
-E)
/
for all
4
X.
Definition 1.5: In a neighborhood space
E
the interior
X,
EçX.
for all
N
the closure oper-
X,
if and only if
S
a net
X,
con-
S
is eventually in
for all
N
nS"
Closed Set Space
In a closed set space
set; the family
and the topology
5
7
Definition 2.
1:
xeGçN
'5
),
of all closed sets
for
X
is
7
the basic concept is closed
satisfies the closed set axioms
= {G
In a closed set space
neighborhood of a point
that
(X,
x
and -GE
HG
e
X,
5
a
} .
subset
if and only if there is a set
3.
G
N
is a
such
36
Definition 2.2: In a closed set space
tor
is
c
c(E)
,
=
? E,
{FIF
F
E
5} for all
Definition 2.3: In a closed set space
tor
is:
i
i(E)
ç
{GIG
= _.
E
and
is:
b
b(E)
n
=
{FIF
?
or
E
taining
if and only if
s
the interior opera-
X,
E
E
F
? ^-E, F
E
}
a net
for all E Ç X.
S
is eventually in every set
S
ç X.
the boundary opera-
X,
Definition 2.5: In a closed set space X,
to a point
E S X.
-G 5} for all
Definition 2.4: In a closed set space
tor
the closure opera-
X,
converges
N
con-
whose complement is closed.
s
Closure Space
In a closure space
operator
c
topology for
J
(X,
),
the basic concept is closure
which satisfies the Kuratowski closure axioms, and the
J
is
X
{G le
=
(-G) =-G} .
Definition 3.1: In a closure space
neighborhood of a point
that
X
E
G G N
and
Definition 3.
if and only if c(F)
=
c(-G)
F.
a
subset
if and only if there is a set
x
2: In a
X,
=
N
is a
G
such
F
is closed
^-G.
closure space
X,
a
subset
37
Definition 3. 3: In a closure space
i
is:
i(E)
=
-c(-E) for all
Theorem
2: In a
E
the interior operator
X,
ç X.
closure space
(X,
7 ),
the interior operator
satisfies the following interior axioms:
X.
(I. 1)
i(X)
(I.2)
i(E) S E.
(I. 3)
i(i(E))
(I.4)
i(ArB)
=
=
i(E).
=
i(A) m i(B).
7* is the
7*
7 7.
Moreover, if
from
i,
then
interior topology for
obtained
X
=
Proof:
By the definition of
(I. 1):
have
i(X)
(I. 2):
=
-c (.X) ='-c()
By (K. 2),
^-E
=
i
c(c)
and (K. 1),
=
hence we
4,
X.
ç c(-E),
hence
(I. 3): Apply the definition and (K. 3);
E
2 -c (--E)
=
it is clear that
i(i(E)) =,--c(-i(E)) =-c[4-c(-E))] =-rc(c(-E)) =-c(-E)
=
i(E).
i(E).
38
(I.4): Apply the definition and (K.4); we have
i(A(,B)=
(~c(~A))
=
(7*
c(-(AnB)) =°c (-A,,-B)
,
n
(.-c(^'B))
=
^-[c(--A).,c(^'B)]
=
n
i(A)
i(B).
equivalences proves this:
): The following chain of
=
GE7
if and only if
c(-G) =-G
(--G)
if and only if -c(-G)
if and only if
i(G)
if and only if
G
E
=
=
G
I
*.
G
Definition 3.4: In a closure space
b
is:
b(E)
=
c(E)
r, c(-E) for all
E
X,
ç X.
Definition 3.5: In a closure space
to a point
s
if and only if
which has a subset
G
the boundary operator
X,
a net
S
converges
is eventually in every subset
S
containing
s
and ^-G
N
is invariant under
c.
Interior Space
In an interior space
operator
j'
for
(X,
° ),
the basic concept is interior
which satisfies the interior axioms. And the topology
i
X
is
5 ={Gli(G)
=
G}.
39
Definition 4.1: In an interior space
neighborhood of a point
if and only if
x
x
Definition 4. 2: In an interior space
closed if and only if
F
is:
c(E)
=
'-(i(-E))
for all
b
is:
b(E) =-(i(E)
°
We will prove
first
v i (^-E))
equivalently,
i(-E)
1:
two
,i(-E)
Proof: Since
E
,-h
(Th
E
=
Lemma 2: i(AvB)
is
X,
a subset
X,
the closure operator
X,
the boundary opera-
E
3
below.
for all subset
E
in
X,
or,
E.
by (I. 2), it is clear that
---E
and
(Th
F
i(N).
properties about interior operator which
= 4
i(-E) ç
n E ç -E E = c),
-i(-E) 2 E, or -i(-E)
i(-E)
is a
for all E E X.
we will use in the proof of Theorem
Lemma
N
ç X.
E
Definition 4.4: In an interior space
tor
E
a subset
(i( --F)).
= ^-
Definition 4. 3: In an interior space
c
X,
=
i(-E)
(Th
E
= 4)
is equivalent to
E.
? i(A) v i(B)
for any sets
A, B
in X.
Proof: As we have shown before in the proof of Theorem
i(E) ç i(F).
page 23, if
E E F
then
AS A
and
ç A v B,
B
B
i(A)
4 on
Applying this we have, since
ç i(A_)B)
and
i(B)
ç i(A
3 );
40
v
therefore i(A)
i(B)
S i(AvB).
Theorem 3: In an interior space
ator
satisfies the following boundary axioms:
b
1)
b(4)
(B. 2)
b(E)
(B.
b(b(E))
(B.
3)
(B.4)
A
4.
=
b(--E).
=
nB
ç b(E).
rm
b(A(-B)
7*
1* 7.
Moreover, if
b,
the boundary oper-
(X, ,7 ),
then
=
A
(b(A)vb(B)).
rm B
is the boundary topology for
X
obtained from
=
Proof:
(B. 1): By the definition, and (I. 1),
MO
(i(4)A(X))
=
=
-(i(4)uX)
=
i(X)
=
X;
we have
4.
(B.2): By the definition, we have
b(...E)
_
^-[i(^'E)Ji(~(^'E))] =^-[i(...E)vi(E)]
(B.3): To show b(b(E)) S b(E)
b(b(E))
--b(E)
= 4).
By Lemma
1,
=
b(E).
is the same as showing
[-- i(b(E))]
From this equation and the definition, we have
r [-b(E)] =- -b(E).
41
b(b(E))
n ^-b(E)
_
,..[i(b(E))i(,,.b(E))]
n -b(E)
,i(-b(E))n-b(E).
=-i(^.b(E))r-,[^'i(b(E)),Th-b(E)]
We now have to show that
i(- -b(E))
? -b(E).
i(. -b(E))
=
-
i(tib(E))r -b(E)
By Lemma
i(i(E) ji(.E))
2
=
which is the same as
4
and (I. 3), we have
? i(i(E))vi(i( --E))
=
i(E)vi(-E)
.b(E).
The proof is then completed.
(B.4):
This can be proved by considering the following equa-
tions. By the definition and Lemma
A
n
B
n
b(AnB)
nB
1,
[-i(A(ThB) n-,.i(-(A(ThB))]
=
A
_
[(AnB) n^-i(,-(ArThB))]
=
A
n
B
(Th
n ,i(AnB)
n (i(AnB))
On the other hand, we have
AnBn(b(A)b(B))
=
AnBn[(-i(A)n-i(A))v(--i(B)(---i(-B))]
=
[AnBn(-i(A)n-i(-A))] v [ArBr1(-i(B)(Th-i(- B))]
_
[(A(Th-i(-A))nBn-i(A)]v[An(B(Th-i(- B))n--i(B)]
=
(AnBn-i(A))v(AnB(Th-ri(B))
=
AnBnLi(A)v 1(B)]
=
AnBn(,,i(A(ThB))
=
AnBn[^'(i(A)ni(B))]
42
last equality holds by (I.4). Hence we have shown that
The
b(A(ThB)
B
A
(7* _ 7):
-4i(G)
b(G) ,- G
If
E
.7*,
-b(G)
G
The
=
=
then
i(G)
7,
E
,
A
G
E
then
i(G)
=
nG
i(G) ? G,
(i(G)ii( -G)) n G
=
(i(G)nG)
G
b(G)
E
v
(i( -G)(ThG)
i(NG)
7,
n
G
s
such that
s
E
if and only if
X
i (N)
4,
ç G or,
=
= 4
i(G)
,
G.
by Lemma 1.
7
.
The
These two in-
7
7
in
=
On the other hand, if
hence 3° *c.
Definition 4.5: In an interior space
to a point
-G (Th- -i(~G) r, G
then we have
last equality holds because
above equation implies
=
7 *.
that is;
(I;
=
3° ç
so
case,
G. In this
G
1*,
b(G)
clusions prove
n(b(A)vb(B))
B
,i( ~G)] n G
which implies that
G
G
=
X,
a net
S
converges
is eventually in every set N
S
.
Boundary Space
In a boundary space
operator
7
for
b
X
(X,
1),
the basic concept is boundary
which satisfies the boundary axioms, and the topology
is
7
=
{GIb(G)
rm
G
=
1)}.
43
Definition 5.1: In a boundary space
neighborhood of a point
x
E
b(E)
and
E Ç N
,m
E=
.
Definition 5. 2: In a boundary space
is
c(E)
=
E
v b(E)
for all
is:
i(E)
=
E ^-b(E)
for all
that
s
E
s
N
if and only if
and
subset F
is
X,
a
X,
the closure operator
X,
the interior operator
E G X.
Definition 5.5: In a boundary space
to a point
such that
ç X.
E
Definition 5.4: In a boundary space
i
E
is a
N
ç F.
Definition 5. 3: In a boundary space
c
a subset
if and only if there is an
x
closed if and only if b(F)
X,
b(N) rThN
S
X,
a net
S
converges
is eventually in every set
such
N
=4.
Theorem 4: In a boundary space
(X,
7 ),
the convergence
satisfies the following convergence axioms:
(a)
If
S
is a net such that
S
n
=
s
for all
n,
then
S
Con-
verges to s.
then so does each subnet of
(b)
If
S
converges to
(c)
If
S
does not converge to s,
s,
no subnet of which converges to
S.
then there is a subnet of S,
s.
44
Let
(d)
M
E
be a directed set, let
D
Let
D.
(m, f)
E
then
S
o
be the product
F
let R(m,
F
f)
Moreover, if
7
be a directed set for each
m
D X X {E
m
m
,
E
D}
s.
is the convergence topology for
>
and for
lim
limS(m, n) =s,
m n
then, if
(m, f(m))
_
converges to
R
E
then
X,
7* _7.
Proof:
(a): This is obvious, for such a net
s.
set containing
This follows directly from the fact that if
(b):
ly in a set
tually in
N;
subset of
D
{S
n
,
n
E
{S
,
n"
containing
N
D'}
n
S
S
(d): Suppose
s
with
such that if
b(N)
s,
4,
but
S
is frequently in --N.
Let
D'
such that
of
does not converge to
D)
with
s
thus
E
S
n
E
b(N) r= N
-N for
n
=
D',
E
then
(
be a cofinal
the subnet
N
n
= (}).
(p, g) > (m, f)
hence no
s.
lim lim S(m,
(Th
then there
is not even-
is a subnet which is entirely in ^-N,
subnet of it can converge to
ing
is eventual-
S
then so is each subnet of S.
N,
(c): If a net
is an
is entirely in every
S
We
then
n)
=
s,
and
N
is a set contain-
must find a member
S
a
R(p, g)
E
N.
(m, f)
Choose
m
E
of
F
D
so
45
lim S(p, n)
that
choose a member
E
Since
(f*
7 ): If
=
then there is a net
i.e.
G;
that
b(N)
=
N
--G.
e
So,
7*,
to show
G
E
E
in
S
then
-W-G
E
E
g7
=
suppose
(p),
and
-'
.7
and we have
i.e.,
i.e.
7;
x
E
b(G)
rTh
}
,
n
G
G,
Suppose that
and
b(G)
N
E
in
x
such
G
=
4,,
is a net in
On the other hand, if
= (I).
Ç N,
=
G
7x
We have
G.
b(G)
r- -G
(m
G
/
4,
hence
there is
G
ç
N.
converges to
x
then
which is in --G,
Consequently,
1*.
7'!;
4
a
point SN
N
Because 71 x is directed by G
N.
M
5_
n
S
converges to a point in
b(G)
G
x
b(G)
but this is impossible, for
G,
contradicts the hypothesis that
7
G
which converges to a point
G
N.
E
N.
x
Since
4.
cnx such that
)7x, C
n
E
contains
=
n)
n
containing
which belongs to
M, N
lim S(p,
hence
S(p, g(p))
=
b(G)
Then, we know that for each
-G.
{SN, N
1
then no net in NG
G
of ^-G
if
E
R(p, g)
arbitrary member
be any
p > m;
p
for all
N
E
N
then there is some
X
G
o
S
S(p, n)
is eventually in every set
is eventually in
S
G
S
,
let f(p)
then
we get
g(p) > f(p),
and then for each such
m,
such that
E
(p, g) > (m, f),
p
p >
p it m,
If
E
If
.
of
f(p)
E.
in
n > f(p)
of
for each
N
E
n
E
7*.
=7.
SM
So
E
M
G
E
7,
But then net
E
G.
and
This
46
Convergence Space
In a convergence space
(X,
u' )
,
the basic concept is con-
vergence which satisifes the convergence axioms, and the topology
7
for
is
X
point of
G)
7
=
{GIG
çX
.
Definition 6.1: In a convergence space
neighborhood of a point
x
E
M
ç
x
closed if and only if no net in
F
s)
is
for all E
c(E)
= {s
for some net
I
S
in
all
i
is:
such that
X,
a
subset
F
X,
i(E)
= {s
Is
E
E,
no net in
-E
Theorem 5: In a convergence space
i
is
E,
the closure operS
converges to
X,
the interior oper-
converges to
s}
EGX.
operator
is a
X.
Definition 6.4: In a convergence space
ator
N
converges to a point of --F.
Definition 6.3: In a convergence space
c
a subset
converges to a point of M.
Definition 6.2: In a convergence space
ator
X,
if and only if there is an M
and no net in --M
N
converges to a
and no net in NG
satisfies the interior axioms:
(X,
7 ),
the interior
for
47
(I.
i(X)
1)
X.
=
(I. 2)
i(E) Ç E.
(I. 3)
i(i(E))
(I.4)
i(AnB)
i(E).
=
i(A)
=
r, i(B).
Moreover, if .7
from
f.
then
i,
is the interior topology for
X
obtained
_
.
Proof:
(I. 1):
no net can be in
Since there are no points in --X,
---X,
therefore every point satisfies the condition to get into i(X).
So,
i(X)
=
X.
(I. 2):
This is obvious from the definition.
(I. 3): If we can show that
i(i(E))
(I. 2),
s
E
i(E),
but
s
{S(m, n), n
by the
E
then there is a net
i(i(E))
4
E
G
m
}
s.
i(E).
Thus,
s.
i(E)
For each
in ^-E
last convergence axiom
which converges to
s
we get what we want.
which converges to
-i(E)
net
ç i(E),
then together with
i(E) e- i(i(E)),
Now, suppose that
{T
m
E
S(m, n)
D}
T
m.
a
Then,
is a net in
This contradicts the hypothesis that
ç i(i(E)).
in
there is
Tm ..i(E),
which converges to
(d) we have
m
-'E
48
(I. 4): Suppose that
in --.(A(ThB)
=
-A
-A converges
x
E
n
i(A)
-B
to
i(B).
x
converges to
By this we have
a
to
Let
x.
D
{nIn
=
final in
Sn
net
E
=
and
D
n
{nIn
e
Sn,
n
n
E
G
e
7
=
):
Sn
E
_
x
-A
ì -B
-A}
{Sn, n
E
x
i(A)
E
r i(B).
rm
i(B).
if and only if
i (G)
if and only if
G
to
s
E
X.
E
is a cois a
D2}
In either case it
Hence
x
i(A(ThB)
-G
E
=
i(A(B),
i(A)
n
e
=
G
converges to a point in
belongs to
G
(since i(G) S
and
i(B).
b(E)
G
G
i(G)
is always true)
7 *.
Definition 6. 5: In a convergence space
is:
D2
The following chain of equivalences proves it.
if and only if every point in
b
then
which converges
{Sn, n
x.
Therefore
if and only if no net in
operator
On the
i(A(B),
or
D1
which converges to
D)
4
hence
x;
and
or
D1}
to
- i(B).
and
then either
Sn EBB },
i(A(B) 2 i(A)
(7*
and
D
contradicts the fact that
we get
r i(B)
and no net
B
(Th
-B converges
-(A(- B)
in
D}
E
A
e
i(A(ThB) S i(A)
i(A)
e
hence either
D;
subnet of
D
,
x
x
which implies no net in
x,
and no net in
x,
other hand, suppose that
there is
then
i(A(ThB),
E
= {s1
and there is a net in
there is
a
net in
X,
E
-E which converges
the boundary
which converges
to
s}
for all
49
EXAMPLES
The reason we discussed some equivalent ways to define a
topology is that in some cases it is more convenient to define a topology from other concepts than open
sets.
We will give below some
examples to demonstrate this and also some trival examples.
1.
Neighborhood space
It is well -known that neighborhoods are a commonly used and
convenient way to get a topology in any metric space. A special example will be as follows:
Example
and let
X
Let
then
x,
x
is a neighborhood system of
then
X
has the discrete topology.
?,
Closed set space
Example
be a set with
Let
X,
X
7i
Z:
in
X
be the real line with the usual metric
x,
interval
which
the usual topology.
Example
be
X
be the family of all sets containing an open
71 x
containing
gives
1:
ï_
X
Let
X
rrt
x
=
{E Ix
E
T1
for all
be a set and let the family of closed sets
together- with the family of all finite subsets of
has the minimal
E}
X,
then
topology. This is a well -known example,
ana hence needs no more discussion.
x
50
Closure space
3.
Example
a
c(0
by
c
1:
= cI),
be the real line, and define an operator
Let
X
and
c(E)
closure operator in
X,
=
E
v {O}
for
O.
Also, we know that
ty subset of
(ch) _
c(E)
and
(I),
Then
O
is
c
In
X.
and any set not contain-
X
is in the closure of every nonemp-
X.
Example 2: Let
c
, 4.
hence it gives a topology for
this topological space the open sets are
ing
E
operator in
=
X
be a set, we define an operator
X
for all
which gives
X,
Then
E Ç X.
c
by
c
is a closure
the indiscrete topology.
X
4. Interior space
Example
i
1:
Let
as follows: If
O
4
X
be the
real line;
then i(E)
E,
Then
i
is an
interior operator in
in which the open sets are
Example
i(X)
an
5.
=
X
and
2:
Let
i(E)
= 4)
interior operator in
X
4)
X
O
E
operator
then
E,
i(E)
E.
and contained in
O
and it gives
X
a topology
and open intervals containing
O
.
be a set and we define an operator
for all
X
if
= (1):
is the largest open interval containing
we define an
E
çX
which gives
and
X
E
/
X.
Then
i
i
by
is
the indiscrete topology.
Boundary space
Example
1:
Let
X
be the real line and define an operator
b
51
b(E)
by
meets
= {O}
E
if and only if every open interval containing
'E,
and also meets
is a boundary operator in
b
that all sets not containing
X,
O
b(E)
and
otherwise. Then
= 4)
which gives
O
a topology such
X
are open, all sets containing
as
O
an isolated point are open, and no other sets are open.
Example
Let
2:
X
be a set and let
b(E)
_
cp
for all E
Ç X,
then clearly the topology is the discrete topology.
6. Convergence space
The most well -known example for a convergence space is that
of a function space.
Example
to a
X
ff
n
,
n
E
F
Let
F
topological space
D}
converges to
for
1:
in
F
g(x)
be a family of functions, each on a set
and define convergence so that a net
Y,
converges to
for each
x
g
if
in
X.
and only if
{fn(x), n
E
D}
Then clearly the topology
is the topology of pointwise convergence.
52
BIBLIOGRAPHY
Gaal, S.A. Point set topology. New York, Academic Press, 1964.
317 p.
Kelley, J.L. Convergence in topology. Duke Mathematical Journal
17:277 -283. 1950.
Kelley,
J.L. General topology.
New York, Van Nostrand, 1955.
298 p.
Pervin,
ic
W.
J. Foundations of general topology. New York, Academ-
Press,
1964.
209 p.