TOPOLOGY

1.5 Interior, Exterior and Boundary of a set

Definition: Let X be a topological space and A Ì X . p Î A is an interior point of A if p
belongs to an open set contained in A. i.e. p Î G Ì A, G open.

Definition:
o
(a) The interior of A, denoted by int(A) or A , is the set of all interior points of A.

( )
(b)The exterior of A or ext (A) is the interior of Ac i.e. ext ( A) = int Ac .
(c)The boundary of A or b(A) is set of points not in int(A) or ext(A).
i.e. b(A)=X-[int(A) È ext(A)]

Example 1.5.1:
Consider the four intervals [a,b], (a,b), (a,b], and [a,b) whose endpoints are a and b.
The interior of each is the open interval (a,b) and the boundary of each is the set of endpoints,
i.e. {a,b}.

Example 1.5.2:
Let ¤ and ¡ be the sets of rational and real numbers respectively. Show that b(¤) = ¡ .

Solution:
Since every open subset of ¡ contains both rational and irrational points, there are no interior or
exterior points of ¤ .
Thus int(¤) = f and ext (¤) = int(¤c ) = f . Hence
b(¤) = ¡ - [int(¤) È ext (¤)] ,
b(¤) = ¡ .
In other words, the boundary of the set ¤ of rational numbers is the entire set ¡ of real numbers.
Example 1.5.3:
Let t = { X , f , {1} , {3, 4} , {1, 3, 4} ,{2,3, 4,5}} be a topology on X = {1, 2,3, 4,5} . If A = {2,3, 4} ,
find:
(a) Int(A)
(b) Ext(A)
(c) b(A)

Solution:
(a) 3 and 4 are interior points of A since 3 Î{3,4} Ì {2,3,4} and 4 Î {3,4} Ì {2,3,4}
Thus int(A)={3,4}.

(b) Ac = {1,5} .
1 is an interior point of Ac = {1,5} since 1 Î{1} Ì {1,5} .
5 is not an interior point of Ac = {1,5} since there does not exist an open set G such that

1

, 5 ÎG Ì {1,5}
Thus ext ( A) = int ( A c ) = {1}.

(c) b(A)=X-[int(A) È ext(A)]
= X-[{3, 4} È {1}]
=X-{1, 3, 4}
={2, 5}

Example 1.5.4:
Let t = { X , f , {a} , {a, b} ,{a, c, d } , {a, b, c, d } , {a, b, e}} be a topology on X = {a, b, c, d , e} .
Let A = {a, b, c} .Find
(a) int A
(b) ext A
(c) b(A)

Solution:
(a) a and b are interior points of A={a,b,c} since a, b Î {a, b} Ì A = {a, b, c} .
c is not an interior point of A={a,b,c} since there does not exist an open set G such that
c Î G Ì A = {a, b, c}
Thus intA={a,b}

(b) A c ={d,e}
There does not exist an open set G such that either d Î G Ì A c = {d , e} or e Î G Ì A c = {d , e}
Thus d , e Ïint A c = ext A .
Hence ext (A)=int( A c ) = f

(c) b(A)=X-[int(A) È ext(A)]
=X-[(a,b} È f ]
=X-{a,b}
={c,d,e}

Note:
1. The interior of a set A is the union of all open subsets of A. It is open and it’s the largest open
subset of A.
( )
2. A is open iff int Ac = A

Theorem 1.5.1: Let A be any subset of a topological space X. Then the closure of A is the union
of the interior and the boundary of A, i.e. A = A0 U b ( A ) .
Thus A = A U A¢ = A0 U b ( A )

Definition: A subset A Ì X (X a topological space) is nowhere dense in X if int ( A ) = f

2

,Example 1.5.3:
1 1 1 1
Let A = { , , ,... 2 ,...} Ì Â with the topology of open sets ( a, b ) .
2 4 16 2
0 is the only limit point of A
ì 1 1 1 ü
Hence A = í0, , ,..., n ,...ý .
î 2 4 2 þ
Since no open set {a, b} is contained in A, then A does not have any interior point.
Thus Int ( A ) = f
Therefore A is nowhere dense in  .

1.6 Neighbourhoods and Neighbourhood systems

Definition:
Let p be a point in a topological space X. A subset N of X is a neighbourhood of p iff N is a
superset of an open set containing p. i.e. p Î G Ì N .

Note that “N is a neighbourhood of a point p” is the inverse of “p is an interior point of N”.

The class of all neighbourhoods of p Î X is called the neighbourhood system of p and is denoted
by N p .

Example 1.6.1:
Let t = { X , f , {a} , {a, b} ,{a, c, d } , {a, b, c, d } , {a, b, e}} be a topology on X = {a, b, c, d , e} .
Find the neighbourhood system of:
(i) the point e
(ii) the point c

Solution:
(i) The neighbourhood of e is any superset N of G Î t such that e Î G Ì N .
Open sets containing e are {a,b,e} and X.
Supersets of {a,b,e} are {a,b,e}, {a,b,c,e}, {a,b,d,e} and X.
The only superset of X is X.
Thus, the class of neighbourhoods of e, the neighbourhood system of e, is given by:
N e ={{ a , b , e}, {a , b , c , e}, {a , b , d , e}, X }}

(ii) Open sets containing c are {a,c,d}, {a,b,c,d} and X.
Supersets of {a,c,d} are {a,c,d},{a,b,c,d}, {a,c,d,e} and X.
Supersets of {a,b,c,d} are {a,b,c,d} and X.
The only superset of X is X
Thus, the class of neighbourhoods of c, the neighbourhood system of c, is given by:
N c ={{ a , c , d }, {a , b , c , d }, {a , c , d , e}, X }}


3

, Example 1.6.2:
In Â, a Î Â, [a - d , a + d ] is a neighbourhood of a " d > 0 .

æ d dö
Since a Î ç a - , a + ÷ Ì [ a - d , a + d ]
è 2 2ø

In  2 , a Π 2 , N p (r ), r > 0 is a neighbourhood of a.

1.7 Neighbourhood axioms

The following four properties, known as the neighbourhood axioms, may be also be used to
define a topology:

i. N r is non empty and p Î N , " N Î N r
ii. The intersection of any two members of N r belongs to N r .
iii. Every superset of a member of N r belongs to N r .
iv. Every member N Î N p is a superset of a member G Î N p where G is a neighborhood of each
of its points. i.e. G ÎN g " g Î G

1.8 Coarser and Finer Topologies

Definition: Let t 1 and t 2 be topologies on a non-empty set X. If every t 1 -open set is a t 2 -open
subset (t 1 Ì t 2 ), we say t 1 is coarser than t 2 (or smaller, weaker) and t 2 is finer (larger) than
t 1 . t 1 and t 2 are comparable if t 1 Ì t 2 or t 2 Ì t 1 . They are not comparable if neither is coarser
that the other.

Example 1.8.1:
Let l , t and D be the indiscrete, any topology and discrete topologies respectively, then
l Ìt Ì D

1.9 Subspaces, Relative Topologies

Let ( X ,t ) be a topological space, A Ì X . t A = { A I B : B Ît } is a topology on A, called the
relative topology of A. (t A , A ) is a subspace of ( X ,t ) . H Ì A , is a t A -open set iff $ a t -open
subset G such that H = A I G .

Example 1.9.1:
Let, t = { X , f , {1} , {1, 2} , {1,3, 4} , {1, 2,3, 4} , {1, 2,5}} be a topology on X = {1, 2,3, 4,5} . Find t A ,
the relative topology on A={1,3,5}
.
Solution:
4