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MSc. Pre (Topology) Unit - I Page 1
Def. : Let X be any given set and be any family of subsets of X satisfying the following axioms
[O1] Arbitrary union of members of is in
[O2] Finite intersection of members of is in
[O3]
[O4] X
Then, we say that defines a topology on X and the pair ( X, ) is called topological space. Sometimes
we denote a topological space ( X, ) by X only without referring .
Remark : We can always define two topologies on every set, namely, the indiscrete topology and discrete
topology.
Indiscrete topology : Let X be any set, then the collection {, X } consisting of the empty set and the
whole space is always a topology and is called the indiscrete topology (or trivial topology). The pair
( X, ) is called an indiscrete topological space. For example, if X a, b, c , then indiscrete topology on
X is , { a, b, c } .
Discrete topology : Let X be any set and be family of all subsets of X. Then satisfies all the for
axioms of a topological space. Since arbitrary union of subsets of X is a subset of X and hence belongs to
so [O1] is satisfied. Also finite intersection of subsets of X is a subset of X and thus a member of so
that [O2] is satisfied. Further X, X X ; so that , X and so [O3] and [O4] are satisfied. Thus
defines a topology on X and this topology is called discrete topology. For example, if X a, b, c , then
discrete topology is given by
, X, { a }, { b}, { c }, { a, b}, { a, c }, { b, c } .
Example 1 : Let X {a, b, c } . Which of the following sets is not a topology on X?
(i) , { a}, X (ii) , {a}, {b}, {a, b}, X (iii) , {a}, {b}, {c }, X
Solution : (i) Yes (ii) Yes
(iii) This collection is not a topology, since { a }, { b} belong to the collection but { a } { b} { a, b} does
not belong to the collection, so [O1] is not satisfied.
Example 2 : Let X a, b, c, d, e . Determine whether or not each of the following classes of
subsets of X is a topology on X.
(i) 1 , X, {a}, {a, b}, {a, c } (ii) 2 , X, {a, b, c }, {a, b, d}, {a, b, c, d}
(ii) 3 , X,{a}, {a, b}, {a, c, d}, {a, b, c, d}
Solution : (i) 1 is not a topology on X, since { a, b}, { a, c } 1 but { a, b} { a, c } { a, b, c } 1
(ii) 2 is not a topology on X, since { a, b, c }, { a, b, d } 2 but { a, b, c } { a, b, d} { a, b} 2
(iii) Yes, 3 is a topology on X.
Example 3 : Let X be any set and let {, A, B, X} where A and B are nonempty distinct proper
subsets of X. Find what conditions A and B must satisfy in order that may be a topology on X.
Solution : If is a topology on X, then A B must belong to , there are two possibilities :
Case I : A B , then A B can not be A or B , hence A B X .
, 2
Case II : A B A or A B B . In either situation one of the sets is a subset of the other, so that in
this case, we must have A B X or B A X . (It should be noted that A B can not be
equal to X, since A, B are proper subsets of X.
Example 4 : Use example 3 to list all topologies for X {a, b, c } , which consists of exactly four
members.
Solution : The topologies covered under case-I of example 3 are
1 , { a }, { b, c }, X
2 , {b}, { a, c }, X
3 , { c }, { a, b}, X
and the topologies covered under case-II of example 3 are
4 , { a }, { a, b}, X
5 , { a }, { a, c }, X
6 ,b , { a, b}, X
7 , {b}, {b, c }, X
8 , { c }, { a, c }, X
9 , { c }, {b, c }, X
Example 5 : (i) List all the topologies on the set of two elements {a, b}.
(ii) List all the topologies on the set of three elements {a, b, c } . Is there any collection of subsets
of X that is not a topology on X?
Solution : (i) Let X { a, b} , then the possible topologies on X are
1 , X Indiscrete topology
2 , { a }, X
3 , { b}, X
4 , { a }, { b}, X discrete topology
(ii) Let X a, b, c , then the possible topologies on X are
1 (, X) Indiscrete topology
2 , { a }, X
3 , { b}, X
4 , { c }, X
5 , { a, b}, X
6 , { b, c }, X
7 , { a, c }, X
8 , { a }, { b, c }, X
9 , {b}, { a, c }, X
10 , { c }, { a, b}, X
11 , { a }, { a, b}, X
12 , { a }, { a, c }, X
13 , { b}, { b, c }, X
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MSc. Pre (Topology) Unit - I Page 3
14 , {b}, { a, b}, X
15 , { c }, {b, c }, X
16 , { c }, { a, c }, X
17 , { a }, { a, b}, { a, c }, X
18 , { b}, { a, b}, { b, c }, X
19 , { c }, { a, c }, { b, c }, X
20 , { a }, { b}, { a, b}, X
21 , { a }, { c }, { a, c }, X
22 , { b}, { c }, { b, c }, X
23 , { a }, { b}, { a, b}, { a, c }, X
24 , { a }, { b}, { a, b}, { b, c }, X
25 , { a }, { a, c }, { b, c }, X
26 , { a }, { c }, { a, c }, { a, b}, X
27 , { b}, { c }, { b, c }, { a, b}, X
28 , { b}, { c }, { b, c }, { a, c }, X
29 , { a }, { b}, { c }, { a, b}, { a, c }, {b, c }, X Discrete topology
There are collections of subsets of X which are not topology on X. For example,
A , { a }, b ,{ a, b}, { a, c }, {b, c }, X is not a topology on X because {b, c } A, { a, c } A .
But {b, c } { a, c } { c } A .
Example 6 : Let X = N, the set of natural numbers and let consists of , X and all subsets of X of the
form {1, 2,......, n} , n N. Show that is a topology on X. Since , X , so axioms [O3] and [O4] are
satisfied. For axiom [O1], let A be an arbitrary family of members of , then each A is of the
type A {1, 2,...., n } . If members of this family are finite and if m is maximum of all the natural
numbers n , , then clearly, A {1, 2,...., m} for all
A
1, 2,...., m
If members of this family are infinite, then A
X
For axiom O2 , let Ai i 1 be a finite family of members of , then each Ai is of the type
k
Ai {1, 2,...., ni } . If m is the minimum of natural numbers ni ’s (i = 1, 2,...., k), then clearly,
k
A {1, 2,...., m}
i 1
i
Hence is a topology on X.
Example 7 : Let X = N and let be the family consisting of , X and all subsets of the form
Gn n, n 1, n 2,..... . Show that is a topology on X.
, 4
Solution : Since , X , so axioms [O3 ] and [O4 ] are satisfied.
For axiom [O1 ] , let G be an arbitrary family of members of , where is some subset of N. If m is
the smallest positive integer of , then G : m, m 1, m 2,.... Gm
For axiom [O2] , let Gm , Gn where m, n N , then
Gm n m
Gm Gn Gm Gn
Gn m n
This process can be extended to conclude that intersection of finite members of is again in . Hence
is a topology on X.
Example 8 : (Excluded point topology) : Let X be a non-empty set and consists of X and all
those subsets of X which do not contain a particular point x0 of X. Show that is a topology on X.
Solution : Since x0 , we have . Also X (given), so axioms [O3] and [O4] are satisfied. For
axiom [O1], let {G } be an arbitrary family of members of , then
x0 G or G X for each
x0 G
or G
X
In either case, G
For axiom [O2], let G1 , G2 , then we discuss the following cases :
(i) If G1 G2 X , then G1 G2 X
(ii) If G1 X and G2 X , then G1 G2 G1 X G1
(iii) If G1 X and G2 X , then G1 G2 X G2 G2
(iv) If G1 X and G2 X , then x0 G1 , x0 G2
x0 G1 G2 G1 G2
Hence is a topology on X.
Example 9 : (Included Point Topology) : Let X be a set containing at least two points and be the
collection of all those subsets of X which contains a fixed point x0 of X together with the empty
set. Show that is a topology on X.
Solution : It is given that . Also x0 X , so x , so that axioms [O3] and [O4] are satisfied.
For axiom [O1 ] , Let G be an arbitrary family of members of , then
x0 G or G for each
x0 G
or G
In either case, G .
For axiom [O2], let G1 , G2 . If any of these two sets is , then their intersection is also , which is
clearly a member of . If G1 and G2 , then x0 G1 , x0 G2 x0 G1 G2 G1 G2 .
Hence is a topology on X.
Example 10 : (Co-finite topology) (i) Let X be an infinite set and be the family of subsets of X
consisting of , X and complements of all finite subsets of X, then show that is a topology on X.
(ii) What happens if X is finite?
Solution : For axiom [O1 ], let G be an arbitrary family of members of , then
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MSc. Pre (Topology) Unit - I Page 1
Def. : Let X be any given set and be any family of subsets of X satisfying the following axioms
[O1] Arbitrary union of members of is in
[O2] Finite intersection of members of is in
[O3]
[O4] X
Then, we say that defines a topology on X and the pair ( X, ) is called topological space. Sometimes
we denote a topological space ( X, ) by X only without referring .
Remark : We can always define two topologies on every set, namely, the indiscrete topology and discrete
topology.
Indiscrete topology : Let X be any set, then the collection {, X } consisting of the empty set and the
whole space is always a topology and is called the indiscrete topology (or trivial topology). The pair
( X, ) is called an indiscrete topological space. For example, if X a, b, c , then indiscrete topology on
X is , { a, b, c } .
Discrete topology : Let X be any set and be family of all subsets of X. Then satisfies all the for
axioms of a topological space. Since arbitrary union of subsets of X is a subset of X and hence belongs to
so [O1] is satisfied. Also finite intersection of subsets of X is a subset of X and thus a member of so
that [O2] is satisfied. Further X, X X ; so that , X and so [O3] and [O4] are satisfied. Thus
defines a topology on X and this topology is called discrete topology. For example, if X a, b, c , then
discrete topology is given by
, X, { a }, { b}, { c }, { a, b}, { a, c }, { b, c } .
Example 1 : Let X {a, b, c } . Which of the following sets is not a topology on X?
(i) , { a}, X (ii) , {a}, {b}, {a, b}, X (iii) , {a}, {b}, {c }, X
Solution : (i) Yes (ii) Yes
(iii) This collection is not a topology, since { a }, { b} belong to the collection but { a } { b} { a, b} does
not belong to the collection, so [O1] is not satisfied.
Example 2 : Let X a, b, c, d, e . Determine whether or not each of the following classes of
subsets of X is a topology on X.
(i) 1 , X, {a}, {a, b}, {a, c } (ii) 2 , X, {a, b, c }, {a, b, d}, {a, b, c, d}
(ii) 3 , X,{a}, {a, b}, {a, c, d}, {a, b, c, d}
Solution : (i) 1 is not a topology on X, since { a, b}, { a, c } 1 but { a, b} { a, c } { a, b, c } 1
(ii) 2 is not a topology on X, since { a, b, c }, { a, b, d } 2 but { a, b, c } { a, b, d} { a, b} 2
(iii) Yes, 3 is a topology on X.
Example 3 : Let X be any set and let {, A, B, X} where A and B are nonempty distinct proper
subsets of X. Find what conditions A and B must satisfy in order that may be a topology on X.
Solution : If is a topology on X, then A B must belong to , there are two possibilities :
Case I : A B , then A B can not be A or B , hence A B X .
, 2
Case II : A B A or A B B . In either situation one of the sets is a subset of the other, so that in
this case, we must have A B X or B A X . (It should be noted that A B can not be
equal to X, since A, B are proper subsets of X.
Example 4 : Use example 3 to list all topologies for X {a, b, c } , which consists of exactly four
members.
Solution : The topologies covered under case-I of example 3 are
1 , { a }, { b, c }, X
2 , {b}, { a, c }, X
3 , { c }, { a, b}, X
and the topologies covered under case-II of example 3 are
4 , { a }, { a, b}, X
5 , { a }, { a, c }, X
6 ,b , { a, b}, X
7 , {b}, {b, c }, X
8 , { c }, { a, c }, X
9 , { c }, {b, c }, X
Example 5 : (i) List all the topologies on the set of two elements {a, b}.
(ii) List all the topologies on the set of three elements {a, b, c } . Is there any collection of subsets
of X that is not a topology on X?
Solution : (i) Let X { a, b} , then the possible topologies on X are
1 , X Indiscrete topology
2 , { a }, X
3 , { b}, X
4 , { a }, { b}, X discrete topology
(ii) Let X a, b, c , then the possible topologies on X are
1 (, X) Indiscrete topology
2 , { a }, X
3 , { b}, X
4 , { c }, X
5 , { a, b}, X
6 , { b, c }, X
7 , { a, c }, X
8 , { a }, { b, c }, X
9 , {b}, { a, c }, X
10 , { c }, { a, b}, X
11 , { a }, { a, b}, X
12 , { a }, { a, c }, X
13 , { b}, { b, c }, X
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MSc. Pre (Topology) Unit - I Page 3
14 , {b}, { a, b}, X
15 , { c }, {b, c }, X
16 , { c }, { a, c }, X
17 , { a }, { a, b}, { a, c }, X
18 , { b}, { a, b}, { b, c }, X
19 , { c }, { a, c }, { b, c }, X
20 , { a }, { b}, { a, b}, X
21 , { a }, { c }, { a, c }, X
22 , { b}, { c }, { b, c }, X
23 , { a }, { b}, { a, b}, { a, c }, X
24 , { a }, { b}, { a, b}, { b, c }, X
25 , { a }, { a, c }, { b, c }, X
26 , { a }, { c }, { a, c }, { a, b}, X
27 , { b}, { c }, { b, c }, { a, b}, X
28 , { b}, { c }, { b, c }, { a, c }, X
29 , { a }, { b}, { c }, { a, b}, { a, c }, {b, c }, X Discrete topology
There are collections of subsets of X which are not topology on X. For example,
A , { a }, b ,{ a, b}, { a, c }, {b, c }, X is not a topology on X because {b, c } A, { a, c } A .
But {b, c } { a, c } { c } A .
Example 6 : Let X = N, the set of natural numbers and let consists of , X and all subsets of X of the
form {1, 2,......, n} , n N. Show that is a topology on X. Since , X , so axioms [O3] and [O4] are
satisfied. For axiom [O1], let A be an arbitrary family of members of , then each A is of the
type A {1, 2,...., n } . If members of this family are finite and if m is maximum of all the natural
numbers n , , then clearly, A {1, 2,...., m} for all
A
1, 2,...., m
If members of this family are infinite, then A
X
For axiom O2 , let Ai i 1 be a finite family of members of , then each Ai is of the type
k
Ai {1, 2,...., ni } . If m is the minimum of natural numbers ni ’s (i = 1, 2,...., k), then clearly,
k
A {1, 2,...., m}
i 1
i
Hence is a topology on X.
Example 7 : Let X = N and let be the family consisting of , X and all subsets of the form
Gn n, n 1, n 2,..... . Show that is a topology on X.
, 4
Solution : Since , X , so axioms [O3 ] and [O4 ] are satisfied.
For axiom [O1 ] , let G be an arbitrary family of members of , where is some subset of N. If m is
the smallest positive integer of , then G : m, m 1, m 2,.... Gm
For axiom [O2] , let Gm , Gn where m, n N , then
Gm n m
Gm Gn Gm Gn
Gn m n
This process can be extended to conclude that intersection of finite members of is again in . Hence
is a topology on X.
Example 8 : (Excluded point topology) : Let X be a non-empty set and consists of X and all
those subsets of X which do not contain a particular point x0 of X. Show that is a topology on X.
Solution : Since x0 , we have . Also X (given), so axioms [O3] and [O4] are satisfied. For
axiom [O1], let {G } be an arbitrary family of members of , then
x0 G or G X for each
x0 G
or G
X
In either case, G
For axiom [O2], let G1 , G2 , then we discuss the following cases :
(i) If G1 G2 X , then G1 G2 X
(ii) If G1 X and G2 X , then G1 G2 G1 X G1
(iii) If G1 X and G2 X , then G1 G2 X G2 G2
(iv) If G1 X and G2 X , then x0 G1 , x0 G2
x0 G1 G2 G1 G2
Hence is a topology on X.
Example 9 : (Included Point Topology) : Let X be a set containing at least two points and be the
collection of all those subsets of X which contains a fixed point x0 of X together with the empty
set. Show that is a topology on X.
Solution : It is given that . Also x0 X , so x , so that axioms [O3] and [O4] are satisfied.
For axiom [O1 ] , Let G be an arbitrary family of members of , then
x0 G or G for each
x0 G
or G
In either case, G .
For axiom [O2], let G1 , G2 . If any of these two sets is , then their intersection is also , which is
clearly a member of . If G1 and G2 , then x0 G1 , x0 G2 x0 G1 G2 G1 G2 .
Hence is a topology on X.
Example 10 : (Co-finite topology) (i) Let X be an infinite set and be the family of subsets of X
consisting of , X and complements of all finite subsets of X, then show that is a topology on X.
(ii) What happens if X is finite?
Solution : For axiom [O1 ], let G be an arbitrary family of members of , then
