TOPOLOGICAL SPACES

1. Topologies
Definition 1. Given a set X, a topology τ on X is defined as a subset of the power
set P (X) with the following properties:
(i) X ∈ τ ;
(ii) Given elements U1 and SU2 of τ , the set U1 ∩ U2 is in τ ;
(iii) Given σ ⊂ τ , its union σ is in τ .
A topological space is defined as an ordered pair (X, τ ) where X is a set equipped
with a topology τ . An element of X is also called a point in X.
For convenience, instead of (X, τ ), we shall frequently call X the topological
space when no confusion may arise.
Given a topology τ on a set X, the members of τ are said to be τ -open. When
no confusion may arise, we shall simply call them open in SX or open. The set X is
open by definition. The empty set is open, because ∅ = ∅. The intersection of
finitely many open sets is open.
Suppose that X is a set and τ1 and τ2 are topologies on X. If τ1 ⊂ τ2 , then
we say τ1 is coarser than τ2 or τ2 is finer than τ1 . The coarsest topology on X is
the indiscrete topology {∅, X}. The finest topology on X is the discrete topology
P (X).
Definition 2. Let X be a topological space. Given x ∈ X and N ⊂ X, we call N
a neighbourhood of x and x an interior point of N if there is an open set U such
that x ∈ U and U ⊂ N . The neighbourhood system of x is defined as the set of all
neighbourhoods of x.
Theorem 1. Let X be a topological space. Given U ⊂ X, the set U is in τ if and
only if U is a neighbourhood of each of its elements.
Proof. Every open set is a neighbourhood of its elements. Conversely, if U is a
neighbourhood of all its elements, then each x ∈ U has an open neighbourhood Vx
such that Vx ⊂ U . Hence
[
U= Vx
x∈U

is open. 

Given a topological space X, a set F ⊂ X is said to be closed if F c is open. We

observe that ∅ and X are closed sets. By De Morgan’s laws,
∩ F2 is closed, and
(i) given closed subsets F1 and F2 of X, the set F1S
(ii) given a set F of closed subsets of X, its union F is closed.
These properties propose an equivalent definition of topological spaces in term of
closed sets.
1
2 TOPOLOGICAL SPACES

A point x ∈ X is called a limit point or an accumulation point of a set E ⊂ X

if every neighbourhood of x contains a point in E different from x. We denote the
set of limit points of E by Ẽ.
Theorem 2. Given a subset F of a topological space X, the set F is closed in X
if and only if F̃ ⊂ F .
Proof. Put U = F c . Suppose that F is closed and x ∈ U . Then U is a neighbour-
hood of x disjoint from F . Conversely, if F̃ ⊂ F , then any x ∈ U has an open
neighbourhood Vx disjoint from F . Hence
[
U= Vx
x∈U
is open. 

2. Interiors and closures

Definition 3. Given a set X, an interior operator on X is defined as a function
f : P (X) → P (X) with the following properties:
(i) f (X) = X;
(ii) f (E) ⊂ E for all E ∈ P (X);
(iii) f (E) = f (f (E)) for all E ∈ P (X);
(iv) f (E1 ) ∩ f (E2 ) = f (E1 ∩ E2 ) for all E1 and E2 in P (X).
Theorem 3. Suppose that f is an interior operator on X. Then τ = ran(f ) defines
a topology on X.
Proof. That (i) and (iii) in Definition 1 hold for τ is by its
S own definition. To prove
that (ii) holds for τ , pick a subset σ of τ and put U = σ. Then for any V ∈ σ,
V ∩ f (U ) = f (V ) ∩ f (U ) = f (V ∩ U ) = f (V ) = V .
Hence V ⊂ f (U ). Since V is an arbitrary member of σ, the set U is a subset
of f (U ). Combined with property (ii), we have U = f (U ). 
Definition 4 (The Kuratowski closure axioms). Given a set X, a closure operator
on X is defined as a function g : P (X) → P (X) with the following properties:
(i) g(∅) = ∅;
(ii) E ⊂ g(E) for all E ∈ P (X);
(iii) g(E) = g(g(E)) for all E ∈ P (X);
(iv) g(E1 ) ∪ g(E2 ) = g(E1 ∪ E2 ) for all E1 and E2 in P (X).
Theorem 4. Given a set X, let f : P (X) → P (X) and g : P (X) → P (X) be such
that g(E)c = f (E c ) for all E ∈ P (X). Then
(a) (E)c = g(E c ) for all E ∈ P (X);
(b) if f is an interior operator, then ran(g) is the set of sets closed in the
topology induced by f ;
(c) the function g is a closure operator if and only if f is an interior operator;
Proof. (a) Since E = (E c )c ,
f (E)c = f ((E c )c )c
= (g(E c )c )c
= g(E c )
 TOPOLOGICAL SPACES 3

for all E ∈ P (X).

(b) Suppose that f is indeed an interior operator. Equip X with the topology
induced by f . Given any E ∈ X, since the set g(E)c = f (E c ) is open, g(E)
is closed. Conversely, suppose that E c = f (F ) for some F ∈ P (X). Then
E = g(F c ) is an element of ran(g).
(c) It follows from part (a) that
(i) g(∅) = ∅ if and only if
f (X) = g(∅)c = X .
(ii) Given any E ∈ P (X),
E ⊂ g(E) ↔ g(E)c ⊂ E c
↔ f (E c ) ⊂ E c .
(iii) Given any E ∈ P (X),
g(g(E)) = g(E) ↔ g(g(E))c = g(E)c
↔ f (g(E)c ) = f (E c )
↔ f (f (E c )) = f (E c ) .
(iv) Given any E1 and E2 in P (X),
g(E1 ) ∪ g(E2 ) = g(E1 ∪ E2 ) ↔ (g(E1 ) ∪ g(E2 ))c = g(E1 ∪ E2 )c
↔ g(E1 )c ∩ g(E2 )c = f ((E1 ∪ E2 )c )
↔ f (E1c ) ∩ f (E2c ) = f (E1c ∩ E2c ) .

Theorem 4 proves the duality between interior operators and closure operators
on a set X.
Definition 5. Let X be a topological space. Given any subset E of X,
(i) the interior of E, denoted by E o , is defined as the union of all open subsets
of E.
(ii) the closure of E, denoted by E, is defined as the intersection of all closed
sets containing E.
(iii) the boundary of E, denoted by ∂E, is defined as the set E − E o . Each point
in ∂E is called a boundary point of E.

c o
Theorem 5. Let X be a topological space. Then E = (E c ) and (E o )c = E c
for all E ⊂ X.
Proof. Given F ⊂ X, the set F is a
closed set containing E if and only if F c is an
c o
open set disjoint from E. Hence E = (E c ) . The rest follows from Theorem 4(a).

The following theorem justifies the names interior operator and closure operator.
Theorem 6. Given a topological space X, the functions E 7→ E o on P (X) is the
interior operator that induces the topology on X. Moreover, the function E 7→ E
on P (X) is its dual closure operator.
Proof. The interior of any subset of X, as a union of open sets, is open. Con-
versely, given any open subset U of X, we have U = U o . Hence the range of the
function E 7→ E o is the topology on X. The rest follows from Theorem 4 and 5. 
4 TOPOLOGICAL SPACES

Theorem 7. Given a subset E of a topological space X,

(a) the set E o is the set of all interior points of E;
(b) E = E ∪ Ẽ.
(c) ∂E = ∂(E c ).
Proof. (a) If x is an interior point of E, then x belongs to some open subset U
of E. Since U ⊂ E o , the point x is in E o . Conversely, for any x ∈ E o , by
definition of E o there is an open subset U of E such that x ∈ U . Hence E
is a neighbourhood of x.
(b) It follows from Theorem 5 that x ∈ / E if and only if x belongs to some open
set U disjoint from E. Hence x ∈ / E if and only if x ∈ / E ∪ Ẽ.
(c) By definition and Theorem 5,
o c
c o
∂E = E − E o = ((E c ) ) − E c = E c − (E c ) = ∂(E c ) .


3. Bases and subbases

Definition 6. Given a topology τ on a set X, a subset B of τ is called a basis for
τ if each open set is the union of members of B.
Each topology is a basis for itself. One topology can have more than one distinct
bases. On the other hand, if two topologies τ1 and τ2 have the same basis, then
by definition each τ1 -open set is also τ2 -open. Different topologies have different
bases.
Theorem 8. A subset B of a topology τ on a set X is a basis for τ if and only
if for each x ∈ X and each neighbourhood N of x, there exists V ∈ B such that
x ∈ V and V ⊂ N .
Proof. Suppose that B is a basis for τ . Pick x ∈ X and a neighbourhood N of x.
Let U be an open neighbourhood of x such that U ∈ N . Since U is non-empty, U is
the union of an non-empty subset of B. Hence there exists V ∈ B such that x ∈ V
and V ⊂ U .
Suppose that the converse is true. Then given an open set U , to each x ∈ U
corresponds a set Vx ∈ B such that x ∈ Vx and Vx ⊂ U . It follows that
[
U= Vx ,
x∈U
so U is the union of members of B. 
Theorem 9. Given a set X and a subset B of P (X) such that X = B, let τ be
S
the collection of unions of members of B. Then τ is a topology on X if and only if
for every two members V1 and V2 of B and each x ∈ V1 ∩ V2 , there exists U ∈ B
such that x ∈ U and U ⊂ V1 ∩ V2 .
Proof. Suppose that τ is a topology on X. Since V = {V } for all V ∈ B, the
S
collection B is a subset of τ . It follows from definition that B is a basis for τ . Let
V1 and V2 be two members of B. Since V1 and V2 are open, V1 ∩ V2 is open. By
Theorem 8, there exits U ∈ B such that x ∈ U and U ⊂ V1 ∩V2 for each x ∈ V1 ∩V2 .
Suppose that the converse is true. Since each member of τ is the union of subsets
of X, the collection τ is a subset of P (X). In particular, X ∈ τ because X = B.
S
Given members W1 and W2 of τ , since both are unions of members of B, for
 TOPOLOGICAL SPACES 5

each x ∈ W1 ∩W2 we can pick members V1 and V2 of B such that x ∈ V1 and x ∈ V2 .

By assumption there exists Ux ∈ B such that x ∈ Ux and Ux ⊂ V1 ∩ V2 ⊂ W1 ∩ W2 .
It follows that
[
W1 ∩ W2 = {Ux : x ∈ W1 ∩ W2 } ,
so W1 ∩ W2 is the union of members Sof B. Given W ⊂ τ , since each member of W
is a union of members of B, the set W is itself the union of members of B. We
conclude that τ is a topology on X. 

Given a set X, Theorem 9 characterises all subsets of P (X) that can be a basis
for some topology on X. Given a more general subset S of P (X), the following
theorem describes the coarsest topology on X that contains S .
Theorem 10. Given a set X and a subset S of P (X) such that X = S , let B
S
be the set of all finite intersections of members of S . Then B is a basis for some
topology on X.
Proof. Given two members V1 and V2 of B, each is a finite intersection of members
of S . Hence V1 ∩ V2 is a finite intersection of members of S . The rest follows from
Theorem 9. 

Definition 7. Given a topology τ on a set X, a subset S of P (X) is called a

subbasis for τ if the set of all finite intersections of S is a basis for τ .
In view of Theorem 10, different topologies have different subbases.

4. The countability axioms

Definition 8. Given a topological space X, a local basis at x ∈ X is defined as
a collection of neighbourhoods of x such that every neighbourhood of x contains a
member in that collection. A local subbasis at x ∈ X is defined as a collection of
subsets of X such that the set of finite intersections of members of the collection is
a local basis at x.
Theorem 11. Let X be a topological space and pick x ∈ X. Suppose that for
some ordinal α, the set {Uξ : ξ < α} is a local subbasis at x. Then there is a local
basis {Vξ : ξ < α} at x such that Vξ ⊂ Vζ whenever ζ ≤ ξ < α.
T
Proof. For each β < α, put Vβ = ξ≤β Uξ . 

Definition 9. A topological space X is said to satisfy the first countability axiom

or to be first-countable if for each x ∈ X there is a countable local basis at x.
In view of Theorem 11, a topological space X such that there is a countable local
subbasis at each x ∈ X is first-countable.
Definition 10. A topological space is said to satisfy the second countability axiom
or to be second-countable if its topology has a countable basis.
It follows from definition that every second-countable topological space is first-
countable.
Theorem 12. Suppose that X is a second-countable topological space and E is an
uncountable subset of X. Then E ∩ Ẽ 6= ∅.
6 TOPOLOGICAL SPACES

Proof. Let B be a countable basis of X. Suppose, for contradiction, that no point

in E is a limit point of E. Then to each x ∈ E corresponds an open set Ux ∈ B
such that Ux ∩ E = {x}. Since the function defined by x 7→ Ux is an injection
from E to B, we have |E| ≤ |B|. Therefore E is countable, a contradiction. 
Definition 11. A subset E of a topological space X is said to be dense in X
if Ẽ = X.
Definition 12. A topological space is said to be separable if it has a countable
dense subset.
Theorem 13. A second-countable topological space is separable.
Proof. Suppose that X is a topological space with a countable basis B. Choose
a point from each non-empty member

c of B and let
E be the set of these points.
c
Then E is countable. Since E is open and U ∩ E = ∅ for all U ∈ B, the set

c
E must be empty. 
Definition 13. Let X be a set. Suppose that E ⊂ C for some sets E ⊂ X
S
and C ∈ P (X). Then we call C a cover of E. If a subset of C is also a cover of E,
then we call that subset a subcover. If X is a topological space and each member
of C is open, then we call C an open cover of E.
Definition 14. A topological space is called a Lindelöf space if each open cover of
the space has a countable subcover.
Theorem 14. A second-countable topological space is a Lindelöf space.
Proof. Suppose that X is a topological space with a countable basis B and C is
an open cover of X. To each member U of C there
S is a subset BU of B such that
U is the union of all members of BU . Put V = U ∈C BU . Then
[ [
V = U =X.
U ∈C
Since V ⊂ B and B is countable, there is an enumeration n 7→ Vn of V by
S ∈ N pick Un ∈ C such that Vn ∈ BUn . Put
the natural numbers. ForSeach n
U = {Un : n ∈ N}. Since V ⊂ U , the collection U covers X. 

5. Subspace topologies
Definition 15. Given a space X with a topology τ and a set Y ⊂ X, the subspace
topology τY on Y is a set defined by
τY = {U ∩ Y : U ∈ τ } .
Suppose that X is a space with topology τ and Y ⊂ X. Then τY is indeed a
topology on Y . The space Y with topology τY is called a subspace of X.
It follows from definition that if X, Y and Z are topological spaces such that Y
is a subspace of X and Z is a subspace of Y , then Z is a subspace of X.
Suppose that Y is a subspace of a topological space X. Then
(i) if Y is open X, sets open in Y are also open in X;
(ii) if Y is closed in X, sets closed in Y are also closed in X.
Theorem 15. Suppose that X is a topological space and Y is a subspace of X.
Given a subset E of Y ,
 TOPOLOGICAL SPACES 7

(a) the set E is closed in Y if and only if E is the intersection of Y and a

closed subset of X;
(b) for any y ∈ Y , the point y is a limit point of E in Y if and only if it is a
limit point of E in X;
(c) the closure of E in Y is the intersection of Y and the closure of E in X;
(d) the interior of E in Y is the intersection of Y and the interior of E in X.
Proof. (a) By definition, Y − E is open in Y if and only if there is a set F
closed in X such that Y − E = Y ∩ (X − F ). Then E = Y ∩ F .
(b) Suppose that y ∈ Y is a limit point of E in Y . Let U be an open neighbour-
hood of y in X. Then U ∩Y is open neighbourhood of y in Y , so (U ∩Y )∩E
contains a point different from y, which is also a member of U ∩ E.
Conversely, suppose that y ∈ Y is a limit point of E in X. Let V be
an open neighbourhood of y in X. Then y has an open neighbourhood U
in X such that V = U ∩ Y , and U ∩ E contains a point x different from y.
Since E ⊂ Y , the point x is in V ∩ E.
(c) Let E denote the closure of E in X. From (a) we conclude that E ∩ Y is
closed in Y . Given any set G closed in Y , there exists a closed subset F of
X such that G = F ∩ Y . Since E ⊂ F , the set (E ∩ Y ) is a subset of G.
Hence E ∩ Y is the intersection of all sets closed in Y that contains E.
(d) Apply (c) and Theorem 5.


Theorem 16. Suppose that X is a topological space and Y is a subspace of X.

Given a set U ⊂ P (X), put
UY = {U ∩ Y : U ∈ U } .
Then
(a) if U is a basis of X, then UY is a basis of Y ;
(b) if there exists x ∈ X such that U is a local basis in X at x and x ∈ Y ,
then UY is a local basis in Y at x;
(c) if U is an open cover of X, then UY is an open cover of Y .
Proof. By definition and set algebra. 

Corollary 1. Suppose that X is a topological space and Y is a subspace of X.

Then
(a) the space Y is first-countable if X is first-countable;
(b) the space Y is second-countable if X is second-countable;
(c) the space Y is separable if X is separable;
(d) the space Y is a Lindelöf space if X is a Lindelöf space.
Proof. By definition, Theorem 15 and Theorem 16. 

6. Separated sets
Definition 16. Given a topological space X and subsets E1 and E2 of X, we say
that E1 and E2 are separated if both E 1 ∩ E2 and E2 ∩ E 2 are empty. Given a
subset E of X, a subset S of P (X) is called a separation of E if S = E and
S
any two members of S are separated.
8 TOPOLOGICAL SPACES

Suppose that E1 and E2 are separated subsets in a topological space X. If Y is

a subspace of X, then by Theorem 15(c) E1 ∩ Y and E2 ∩ Y are separated in Y .
Theorem 17 shows that being separated is a property that can pass from subspaces
of a topological space X to X itself. Hence we can talk about sets being separated
without referring to the background topology.
Theorem 17. Suppose that X is a topological space and Y is a subspace of X.
Let E1 and E2 be subsets of Y such that E1 and E2 are separated in Y . Then E1
and E2 are separated in X.
Proof. By Theorem 15(b), no point in E2 is a limit point of E1 in X. 
Theorem 18. Suppose that X is a topological space and E1 and E2 are disjoint
subsets of X. Then E1 and E2 are separated if either both are open or both are
closed.
Proof. If E1 is open, then E1c is closed. Hence E 2 ⊂ E1c and E1 ∩ E 2 = ∅ follows.
If E1 is closed, then E 1 = E1 . Hence E 1 ∩ E2 = ∅. 
The converse is also true. If S is a finite separation of X, then each member
of S is open and closed in X.
Theorem 19. Suppose that E1 and E2 are subsets of a topological space X such
that E1 and E2 are either both open or both closed. Then the sets E1 − E2 and
E2 − E1 are separated.
Proof. Put
F1 = E1 − E2 ,
F2 = E2 − E1 .
Put Y = E1 ∪ E2 and equip Y with the subspace topology inherited from X.
Then F1 and F2 are disjoint and either both open or both closed in Y . It follows
from Theorem 18 that F1 and F2 are separate in Y . By Theorem 17, they are
separated in X. 
Theorem 20. Suppose that X is a topological space and Y1 and Y2 are subsets
of X such that Y1 − Y2 and Y2 − Y1 are separated. Given E ⊂ X, the closure of E
in X is the union of the closure of E ∩ Y1 in Y1 and the closure of E ∩ Y2 in Y2 .
Proof. To avoid ambiguity, let A be the closure of A in X for all A ⊂ X. By
Theorem 15(c), the closure of E ∩ Yk in Yk is E ∩ Yk ∩ Yk for k ∈ {1, 2}. Since
E ∩ Yk ∩ Yk ⊂ E ∩ Yk for k ∈ {1, 2},



E ∩ Y1 ∩ Y1 ∪ E ∩ Y2 ∩ Y2 ⊂ E ∩ Y1 ∪ E ∩ Y2 = E .
Since Y1 − Y2 and Y2 − Y1 are separated and E ∩ Y1 − Y1 ⊂ Y2 − Y1 ,


E ∩ Y1 − Y1 ∩ E ∩ (Y1 − Y2 ) ⊂ (Y2 − Y1 ) ∩ Y1 − Y2 = ∅ .
Since
E ∩ Y1 − Y1 ⊂ E ∩ Y1 = E ∩ (Y1 − Y2 ) ∪ E ∩ Y1 ∩ Y2 ,
it must be that E ∩ Y1 − Y1 ⊂ E ∩ Y1 ∩ Y2 . Since E ∩ Y1 ∩ Y2 ⊂ E ∩ Y2 and
E ∩ Y1 − Y1 ⊂ Y2 ,
E ∩ Y1 − Y1 ⊂ E ∩ Y2 ∩ Y2 .
 TOPOLOGICAL SPACES 9

It follows that



E ∩ Y1 = E ∩ Y1 ∩ Y1 ∪ E ∩ Y1 − Y1



⊂ E ∩ Y1 ∩ Y1 ∪ E ∩ Y2 ∩ Y2 .
Similarly,



E ∩ Y2 ⊂ E ∩ Y1 ∩ Y1 ∪ E ∩ Y2 ∩ Y2 .
Hence



E = E ∩ Y1 ∪ E ∩ Y2 ⊂ E ∩ Y1 ∩ Y1 ∪ E ∩ Y2 ∩ Y2 .

Corollary 2. Suppose that X is a topological space and Y1 and Y2 are subsets of X
such that Y1 − Y2 and Y2 − Y1 are separated. Let E be a subset of X. Then
(a) the set E is closed in X if E ∩ Yk is closed in Yk for k ∈ {1, 2};
(b) the set E is open in X if E ∩ Yk is open in Yk for k ∈ {1, 2}.
Proof. (a) Given k ∈ {i, 1}, if E ∩ Yk is closed in Yk , then the closure of E ∩ Yk
in Yk is E ∩ Yk . By Theorem 20,
E = (E ∩ Y1 ) ∪ (E ∩ Y2 ) = E ,
where E denotes the closure of E in X. Hence E is closed.
(b) If E ∩ Yk is open in Yk for k ∈ {1, 2}, apply (a) to X − E.


7. Connectedness
Definition 17. A subset E of a topological space X is said to be connected if there
is no separation {E1 , E2 } of E such that both E1 and E2 are non-empty.
By Theorem 17, if E is a connected subset of Y , where Y is a subspace of a
topological space X, then E is a connected subset of X. Hence we can discuss the
connectedness of a set without referring to the background topology.
It follows from our discussion on Theorem 18 that a topological space X is
connected if and only if ∅ and X are the only subsets of X that are both open and
closed.
Theorem 21. The closure of a connected set is connected.
Proof. Suppose that E is a connected subset of a topological space X and {F1 , F2 }
is a separation of E. To prove the connectedness of E, it suffices to show that
either F1 or F2 is empty. Note that {E ∩ F1 , E ∩ F2 } is a separation of E. Without
loss of generality, suppose that E ∩ F1 is non-empty. Since E is connected, E ⊂ F1
and E ∩ F2 = ∅. Since F1 is closed in E and E is closed in X, the set F1 is closed
in X. Hence F1 = E and F2 = ∅. 
An immediate consequence of Theorem 21 is that if E is a connected subset of
a topological space X and E ⊂ F ⊂ E for some F ⊂ X, then F is connected. This
is because F is the closure of E in F .
Theorem 22. Suppose that E is a collection of connected subsets of a topological
space X such that no two members of E are separated. Then E is connected.
S
10 TOPOLOGICAL SPACES

S Suppose that {F1 , F2 } is a separation of E . To prove the connectedness

S
Proof.
of E , it suffices to show that either F1 or F2 is empty. For each E ∈ E , since E
is connected and {E ∩ F1 , E ∩ F2 } is a separation of E, either E ∩ F1 or E ∩ F2 is
empty. Hence {E1 , E2 } is a partition of E , where
E1 = {E ∈ E : E ⊂ F1 } ,
E2 = {E ∈ E : E ⊂ F2 } .
Suppose, for contradiction, that E1 ∈ E1 and E2 ∈ E2 . Since F1 and F2 are
separated, so are E1 and
S E2 , contradicting our assumption. Hence one of E1 and E2
is empty. Since F1 = E1 and F2 = E2 , one of F1 and F2 is empty.
S

Corollary 3. Let E be a collection of connected subsets of a topological space X.
Suppose that for any two members E0 and E of E there exists members E1 , . . . , En
of E such
S that En = E and Ek and Ek+1 are not separated for all k ∈ {0, . . . , n−1}.
Then E is connected.
Proof. If all members of E are empty, then E = ∅ is connected. Otherwise, fix
S
some non-empty E0 ∈ E . For each E ∈ E , put
EE = {Ek : k = 0, . . . , n} ,
where E1 , .S. . , En are picked as described in the corollary. By Theorem 22 and
induction, EE is connected.
Suppose that E and F are two members of E . Since both EE and EF contain
S S
the non-empty set E0 , they are not separated. Since
[ [ [ 
E = EE ,
E∈E

applying Theorem 22 to the set { EE : E ∈ E } shows that E is connected.

S S

Definition 18. Given a topological space X and a subset Y of X, a subset of Y
is called a component of Y if it is connected and not properly contained in any
connected subset of Y .
Theorem 23. Let X be a topological space. Then
(a) each connected subset of X is contained in a component of X;
(b) each component of X is closed;
(c) the set of components of X is a separation of X;
Proof. (a) Suppose that E is a connected subset of X. Let C be the union of all
connected subsets of X containing E. By Theorem 22, C is connected. If a
connected subset F of X contains C, then F also contains E. By definition
of C, the set F is a subset of C. Hence C is a component of X.
(b) Suppose that C is a component of X. By Theorem 21, C is a connected
subset of X containing C. Hence C = C.
(c) Let C be the set of components of X. Since {x} is S connected for all x ∈ X,
by (a) x belongs to a member of C . Hence X = C . Let C1 and C2 be
two components of X. If x ∈ C1 ∩ C2 , then both C1 and C2 are connected
sets containing {x}. It follows from definition that C1 = C2 . Hence C is a
partition of X. By (b) and Theorem 18, distinct components are separated.
Therefore C is a separation of X.

 TOPOLOGICAL SPACES 11

Definition 19. A topological space X is said to be locally connected at x if x ∈ X

and every neighbourhood of x contains a connected neighbourhood of X. If X is
locally connected at each of its points, then X is said to be locally connected.
Theorem 24. A topological space X is locally connected if and only if for every
open subset U of X, each component of U is open.
Proof. Suppose that X is a locally connected topological space and U is an open
subset of X. Let C be a component of U . For each x ∈ C, the set U contains a con-
nected neighbourhood N of x. By definition N ⊂ C. Hence C is a neighbourhood
of x. It follows from Theorem 1 that C is open.
Conversely, suppose that X is a topological space such that each component of
each open subset of X is open. Given x ∈ X and an open neighbourhood U of x, let
V be the component of U containing x. Then V is a connected open neighbourhood
of x. 
Corollary 4. A topological space is locally connected if and only if the collection
of connected open sets is a basis for the topology.
Proof. Suppose that B is the collection of connected open sets in a topological
space X. If B is a basis, then each neighbourhood of each point in X contains
a member of B which is connected. Hence X is locally connected. Conversely,
suppose that X is locally connected. By Theorem 24, for each x ∈ X and each
open neighbourhood U of x, the component of U that contains x is a connected
open set, hence a member of B.