Book by Dr. Lutfi

.
I
.
II
Introduction to General
Topology
Dr. LUTFI N. KALANTAN
Faculty of Sciences
King Abdulaziz University
III
Acknowledgement
This project was funded by the Deanship of Scientific Research
(DSR), King Abdulaziz University, Jeddah, under grant no. (29/130/1431).
The auther, therefore, acknowledge with thanks DSR technical and fi-
nancial support.
IV
Introduction
This book is an introduction to the basics of General Topology. It
contains the syllabus of the course math 464, Introduction to General
Topology, of the B. Sc. level at Mathematics department, Faculty of
Sciences, King Abdulaziz University. The book is assumed to be taught
in one semester. The book can be self-study by a regular student. It
contains many examples, exercises, and hints for many exercises. We
encourage the student to read carefully the subjects and try to solve all
of the exercises. Also, we recommend the teacher to leave proof of some
theorems for students to read and assign lot of exercises to students to
solve. This is the first edition of the book and I will appreciate all of
the comments and observations.
Lutfi Kalantan.
2012.
V
Contents
Chapter 1: Set Theory 1
1.1 Sets 1
1.2 Functions 9
1.3 Infinite and Uncountable Sets 15
Chapter 2: Topological Spaces 23

2.1 Topology and Open Sets 23
2.2 Closed sets 29
2.3 Sequences 35
Chapter 3: Bases for a Topology 41
Chapter 4: New Spaces from Old Ones 47

4.1 Subspaces 47
4.2 The Product Topology 48
Chapter 5: Continuous Functions 55

5.1 Continuity 55
5.2 Homeomorphisms and Homeomorphic Spaces 60
Chapter 6: Separation Axioms 67
Chapter 7: Metric Spaces 73
Chapter 8: Compact Spaces 81
Chapter 9: Connected Spaces 87
References 91
Appendix A: Greek Letters 93
Appendix B: Hints and Solutions 95
VI
1 Set Theorey
In this chapter, we present some basic notions and facts of set theory
used throughout the book.
Section 1.1: Sets
A set is a collection of objects. These objects are called points,

numbers, or elements. Sets are denoted by uppercase letters such as
A, X, Λ, ..., and their elements by lowercase letters such as a, x, α, β, ...
The object x is a member of a set S is symbolized by x ∈ S. If x is
not a member of a set S, then it is denoted by x ∈ / S. The simplest
two ways of representing a set are by listing all of its elements and by
characterize all of its elements.
1.1 Definition: A set X is a subset of another set Y , denoted by

X ⊆ Y , if and only if ∀x ∈ X we have x ∈ Y . X = Y if and only if
X ⊆ Y and Y ⊆ X. We say that X is a proper subset of Y , denoted by
X ⊂ Y , if X is a subset of Y and X %= Y . The set which contains no
elements is called the empty set and it is denoted by ∅. The set of all
possible subsets of a set X is called the power set of X and is denoted
by P(X). A set consisting only one element is called singleton.
For two sets X and Y , we define:
1. X \ Y = {x : x ∈ X and x ∈ / Y }, called the complement of Y

with respect to X, or the difference of two sets X and Y .
2. X ∩ Y = {x : x ∈ X and x ∈ Y }, called the intersection of X

and Y .
3. X ∪ Y = {x : x ∈ X or x ∈ Y }, called the union of X and Y .
Main Sets of Numbers
1. The set of Natural Numbers N = {1, 2, 3, ...}
2. The set of Integers Z = {..., −3, −2, −1, 0, 1, 2, 3, ...}
3. The set of Rational Numbers Q = { ab : a, b ∈ Z and b %= 0}
1
2 Introduction to General Topology
4. There are numbers, we use √ in our daily life, cannot√be expressed

as a fraction such as π, 2. If q ∈ Q, then q + 2 cannot be
expressed as fraction. Also, any root with even order of a prime
positive number cannot be expressed as a fraction. So, there are
infinitely many such numbers. Such a number is called irrational.
The set of all irrationals is denoted by P. We have the following
fact: If x ∈ Q and y ∈ P, then x + y, x − y ∈ P.
5. The set of real numbers R = Q ∪ P. Note that Q ∩ P = ∅.
For two distinct real numbers a and b, the inequality a < b means
b−a is positive, i.e., b−a > 0. The set of real numbers has the property
that for any two distinct real numbers a and b, either a < b or b < a.
This property enable us to describe the set of real numbers R as a line
where each point in this line will present a unique real number and any
real number can be presented as a unique point in this line.
1.2 Theorem: Let X be a set and A, B, C ∈ P(X), then
1. A ∪ ∅ = A, A ∩ X = A, A ∩ ∅ = ∅, A ∪ X = X.
2. A ∪ A = A , A ∩ A = A.
3. A ∪ B = B ∪ A, A ∩ B = B ∩ A.
4. (A ∪ B) ∪ C = A ∪ (B ∪ C), (A ∩ B) ∩ C = A ∩ (B ∩ C).
5. A∪(B∩C) = (A∪B)∩(A∪C) and A∩(B∪C) = (A∩B)∪(A∩C).
6. X \ (X \ A) = A.
7. A ⊆ B if and only if X \ B ⊆ X \ A.
8. A ⊆ X \ B if and only if A ∩ B = ∅ if and only if B ⊆ X \ A.
9. A ∩ B = B if and only if B ⊆ A.
10. If X ⊆ Y , then P(X) ⊆ P(Y ).
Proof: We prove part (8) and leave the rest as an exercise. Let’s show
that A ∩ B = ∅ if and only if B ⊆ X \ A. Assume that A ∩ B = ∅.
If B = ∅, then ∅ ⊆ X \ A. Suppose that B %= ∅. Let b ∈ B be
arbitrary. Since A ∩ B = ∅, then b %∈ A, thus b ∈ X \ A. Therefore,
B ⊆ X \ A. Now, assumes that B ⊆ X \ A. Suppose that A ∩ B %= ∅.
Set Theorey 3
Pick x ∈ A ∩ B, then x ∈ A and x ∈ B. Since B ⊆ X \ A and x ∈ B,

then x ∈ X \ A. We get now that x ∈ A and x ∈ X \ A, which is a
contradiction. Therefore, A ∩ B = ∅.
1.3 De Morgan’s Theorem: Let X, Y and Z be sets, then
1. X \ (Y ∪ Z) = (X \ Y ) ∩ (X \ Z);
2. X \ (Y ∩ Z) = (X \ Y ) ∪ (X \ Z).
1.4 Definition: Let X and Y be sets. The Cartesian Product of

X and Y is the set of all possible ordered pairs {(x, y) : x ∈ X and
y ∈ Y } and it is denoted by X × Y . Observe that
(x1, y1 ) = (x2 , y2) if and only if x1 = x2 and y1 = y2 .
X × Y = ∅ if and only if either X = ∅ or Y = ∅.

We conclude that
(x1, y1) %= (x2, y2 ) if and only if x1 %= x2 or y1 %= y2 .
1.5 Theorem: If X, Y and Z are sets, then
1. X × (Y ∪ Z) = (X × Y ) ∪ (X × Z);
2. X × (Y ∩ Z) = (X × Y ) ∩ (X × Z);
3. X × (Y \ Z) = (X × Y ) \ (X × Z).
Indexed Sets
Let Λ be a given nonempty set. Suppose that with each element

α ∈ Λ there is identified a particular set denoted by Xα . Then the
collection of sets {Xα : α ∈ Λ} is called an indexed family of sets and
the set Λ is called an indexing set.
Examples:
1. For each n ∈ N, define
Xn = {(x, y) ∈ R2 : 0 ≤ x ≤ n, 0 ≤ y ≤ n} = [0, n] × [0, n].

So, for each n ∈ N, Xn is the square in the plane having [0, n] as

an edge on both axis. See figure 1.1 below.
Figure 1.1.
2. Let Λ = R and for each α ∈ R, define Xα = {(α, y) ∈ R2 : 0 ≤

y ≤ 1} = {α}×[0, 1]. So, for each α ∈ R, Xα is the line segment in
the plane which is perpendicular to the x-axis, having the points
(α, 0) and (α, 1) as its edges. See figure 1.2.
Figure 1.2.
Set Theorey 5
1.6 Definition: Let {Xα : α ∈ Λ}!be an indexed family of sets.

The union of this family is denoted by α∈Λ Xα and is defined by
"
Xα = {x : x ∈ Xα for some α ∈ Λ}.
α∈Λ
Similarly, #
Xα = {x : x ∈ Xα for all α ∈ Λ}.
α∈Λ
Note that if {Xα $

: α ∈ Λ} is an indexed
! family of sets, then for any
β ∈ Λ we have that α∈Λ Xα ⊆ Xβ ⊆ α∈Λ Xα .
Examples:
1. If Xn is as in example (1) above, then
$ !
n∈N Xn = X1 = [0, 1] × [0, 1], and n∈N Xn = [0, ∞) × [0, ∞).
2. If Xα is as in example (2) above, then

$ !
α∈R Xα = ∅, and α∈R Xα = R × [0, 1].
1
$
3. n∈N (−∞, n ) = (−∞, 0].
Proof: Let x ∈ (−∞, 0] be arbitrary, then x ≤ 0, so x < n1 , ∀ n ∈

N because n1 is positive for each n ∈ N, thus x ∈ (−∞, 1
n
), ∀ n ∈
N, therefore, x ∈ n∈N (−∞, n1 ). Thus (−∞, 0] ⊆ n∈N (−∞, n1 ).
$ $
Now, suppose x %∈ (−∞, 0], then x > 0, so by Archimedean prop-

erty (see Theorem 1.33), ∃ m ∈ N such that m1$< x, thus x %∈
(−∞, m1 ), thus x %∈ n∈N (−∞, n1 ). Therefore, n∈N (−∞, n1 ) ⊆
$
(−∞, 0]. So equality holds.
Figure 1.3.
Exercise: Calculate each of the following:

1. n∈N (−∞, n1 ].
$
2. n∈N (−∞, n1 ].
!
n
$
3. n∈N (−∞, n+1 ).
n
!
4. n∈N (−∞, n+1 ).
$
5. n∈N (−n, n).
!
6. n∈N (−n, n).
1.7 Definition: Let {Xk : k ∈ Λ} be an indexed family of sets

indexed by Λ = {1, 2, ..., n} where n ∈ N. Then the product of {Xk :
k ∈%Λ} is the set of all ordered n-tuples and denoted by X1 ×X2 ×...×Xn
or nk=1 Xk . That is,
n
&
Xk = {(x1, x2 , ..., xn) : xk ∈ Xk ∀ k = 1, 2, ..., n}.
k=1
xk is called the k th -coordinate of the point (x1, x2, ..., xn) and Xk is
called the k th -factor of the product.
(x1 , x2, ..., xn) = (y1, y2, ..., yn) if and only if xi = yi for each i =
1, 2,%..., n.
n
i=1 Xi = ∅ if and only if there exists at least one j, 1 ≤ j ≤ n
such that Xj = ∅.
1.8 Theorem:
1. ( ni=1 Xi )
% $ %n %n
( i=1 Yi ) = i=1 (Xi ∩ Yi );
2. ( ni=1 Xi )
% ! %n %n
( i=1 Yi ) ⊆ i=1 (Xi ∪ Yi ).
Exercise: Give an example to show that the reverse inclusion of

(2) above is not true.
1.9 Theorem: Let X be a set and {Aα : α ∈ Λ} be an indexed

family of subsets of X, then
! $
1. X \ α∈Λ Aα = α∈Λ (X \ Aα );
$ !
2. X \ α∈Λ Aα = α∈Λ (X \ Aα ).
Set Theorey 7
Problems
1. For any two subsets A and B of X, prove
(a) A ∩ B = B if and only if B ⊆ A.
(b) (A \ B) ∪ B = A if and only if B ⊆ A.
2. For any two sets A and B, prove that A ∩ B and A \ B are
disjoint(i.e., their intersection is empty.), and that A = (A ∩ B) ∪
(A \ B).
(This gives a way of representing A as a disjoint union.)
3. Prove, or give a counterexample to disprove, each of the following
statements:
(a) If D ⊆ X × Y , then there are subsets A ⊆ X and B ⊆ Y
such that D = A × B.
(b) If X ⊆ Z, then for any set Y , X × Y ⊆ Z × Y .
4. Assume X %= ∅ =
% Y . Prove that X × Y = Y × X if and only if
X = Y.
5. Prove that for any four sets X, Y, Z and W ,
(a) (X × Y ) ∩ (Z × W ) = (X ∩ Z) × (Y ∩ W ).
(b) (X × Y ) ∪ (Z × W ) ⊆ (X ∪ Z) × (Y ∪ W ).
(c) Give an example to show the reverse inclusion of part (b)
need not be true.
6. Let {Aα : α ∈ Λ} be an indexed family of sets indexed by Λ.
Prove
$ !
(a) α∈Λ Aα ⊆ Aβ ⊆ α∈Λ Aα, for each β ∈ Λ.
! !
(b) For any set B we have B ∪ α∈Λ Aα = α∈Λ (B ∪ Aα).
7. Let An = (− n1 , n1 ) = {x ∈ R : − n1 < x < n1 }, n ∈ N be an indexed
family of subsets of R.
$
(a) Find n∈N An and prove that your result is correct.
!
(a) Find n∈N An and prove that your result is correct.
8. Give a decreasing family {An ⊂ R : n ∈ N} of nonempty subsets
of R (decreasing means An+1 ⊆ An for each n ∈ N) such that
$
n∈N An = ∅.
9. Let {Aα : α ∈ Λ} be an indexed family of sets.

$
(a) Prove that B ⊆ α∈Λ Aα if and only if B ⊆ Aα for every
α ∈ Λ.
!
(b) Prove that α∈Λ Aα ⊆ B if and only if Aα ⊆ B for every
α ∈ Λ.
Set Theorey 9
Section 1.2: Functions
1.10 Definition: A function f from a nonempty set X to a nonempty

set Y , denoted by f : X −→ Y , is a rule that assign to each element
x in X a unique element in Y denoted by f(x). In other words, f is
a subset of X × Y such that for each x ∈ X there is a unique element
y = f(x) such that (x, y) ∈ f. X is called the domain of f and Y is
called the codomain of f. The subset f(X) = {f(x) : x ∈ X} of Y is
called the range of f. Two functions f : X −→ Y and g : X −→ Y are
equal, denoted by f = g, if and only if f(x) = g(x), ∀x ∈ X.
Image and Preimage
If f : X −→ Y is a function, then f introduces a function from

P(X) into P(Y ). Let A ⊆ X. The image of A under f, denoted by
f(A), is defined by f(A) = {f(a) ∈ Y : a ∈ A}. Now, for a subset
B ⊆ Y , the subset of all members of X whose image under f lies in B
is called the preimage of B. That is, f −1 (B) = {x ∈ X : f(x) ∈ B}.
Figure 1.4.
Note that, by above definitions we have:

(a) x ∈ A =⇒ f(x) ∈ f(A).
(b) y ∈ f(A) =⇒ ∃ x ∈ A such that f(x) = y.
(c) x ∈ f −1 (B) =⇒ f(x) ∈ B.
(d) f(x) ∈ B =⇒ x ∈ f −1 (B).
Examples:
1. Define f : R −→ R by f(x) = 2x + 1. Then

f([1, 3)) =
= {f(x) : x ∈ [1, 3)}
= {2x + 1 : x ∈ [1, 3)}
= [3, 7).
f −1 ([5, 9]) =
= {x ∈ R : f(x) ∈ [5, 9]}
= {x ∈ R : 2x + 1 ∈ [5, 9]}
= {x ∈ R : 5 ≤ 2x + 1 ≤ 9}
= {x ∈ R : 4 ≤ 2x ≤ 8}
= {x ∈ R : 2 ≤ x ≤ 4}
= [2, 4].
2. Let f : R −→ R be defined by f(x) = x2.

Then f([−2, 2)) = {x2 : x ∈ [−2, 2)} = [0, 4].
f −1 ((0, 9)) = {x ∈ R: x2 ∈ (0, 9)} = (−3, 0) ∪ (0, 3).
f −1 ((−2, 9)) = (−3, 3).
Exercise:
1. Define g : R −→ R by g(x) = 3x.

Find g((4, 7]) and g −1 ((−3, 27]).
2. Define h : R −→ R by h(x) = x5 .
Find h([5, 25)) and h−1 ((−2, 3]).
1.11 Theorem: (Image and Preimage Properties)

Let f : X −→ Y be a function. A, Aα, C ⊆ X and B, Bα, D ⊆ Y .
Then
1. A ⊆ f −1 (f(A)). [if f is 1-1, then equality holds]

Set Theorey 11
2. f(f −1 (B)) ⊆ B. [if f is onto, then equality holds]

! !
3. f( α Aα) = α f(Aα ).
4. f −1 ( α Bα ) = α f −1 (Bα ).
! !
$ $
5. f( α Aα) ⊆ α f(Aα ).
6. f −1 ( α Bα ) = α f −1 (Bα ).
$ $
7. f(C \ A) ⊇ f(C) \ f(A).
8. f(X \ f −1 (B)) ⊆ Y \ B.
9. f −1 (D \ B) = f −1 (D) \ f −1 (B).
Proof: We prove (1) and (8), and leave the rest as an exercise.
(1) Let A be an arbitrary subset of X. We have to show that A ⊆

f −1 (f(A)). If A = ∅, we are done as the empty set is a subset of
any set. So, assume that A %= ∅ and pick an arbitrary element a ∈
A. By (a), we have that f(a) ∈ f(A). Put f(A) = B. Then, by
(d), we get a ∈ f −1 (B) = f −1 (f(A)). Therefore, A ⊆ f −1 (f(A)).
Now, assume that f is 1-1 (see Definition 1.13) and we have to
show that A = f −1 (f(A)). Note that we just have to show that
f −1 (f(A)) ⊆ A. Let a ∈ f −1 (f(A)) be arbitrary. By (c), we have
that f(a) ∈ f(A). Put y = f(a). Then, by (b), ∃ a# ∈ A such
that f(a#) = y = f(a). This implies that a = a# ∈ A as f is 1-1.
Therefore a ∈ A and equality holds.
(8) if y %∈ Y \ B, then y ∈ B. Thus ∃ x ∈ f −1 (B) such that f(x) = y.

Hence x %∈ X \ f −1 (B). Therefore, f(x) = y %∈ f(X \ f −1 (B)).
1.12 Definition: Let X1 × X2 × ... × Xn be the product of the

family {Xk : k = 1, 2, ..., n}. For each k, 1 ≤ k ≤ n, there is a
projection function pk : X1 × X2 × ... × Xn −→ Xk defined by
pk (x1, x2, ..., xn) = xk , ∀ (x1, x2, ..., x2) ∈ X1 × X2 × ... × Xn .
1.13 Definition: Let f : X −→ Y be a function. f is injective

or one-to-one, denoted by 1-1, if and only if x1 %= x2 =⇒ f(x1 ) %=
f(x2 ), ∀x1, x2 ∈ X. Equivalently, if f(x1) = f(x2 ), then x1 = x2. And

f is surjective or onto if and only if f(X) = Y . That is, ∀ y ∈ Y, ∃ x ∈
X such that f(x) = y.
If f : X −→ Y is a 1-1 function, then f −1 : f(X) ⊆ Y −→ X
defined by
f −1 (y) = x ⇐⇒ f(x) = y, ∀ x ∈ X ∀ y ∈ f(X),
In this case, f −1 is called inverse function of f.
1.14 Definition: Let f : X −→ Y be a function and A ⊆ X.

The restriction of f to A is the function g : A −→ Y defined by
g(x) = f(x), ∀x ∈ A. Usually, we denote this restriction g by f|A .
1.15 Definition: Let f : X −→ Y and g : Y −→ Z be functions.

The composition function of f and g, denoted by g ◦ f, is a function
from X into Z defined by (g ◦ f)(x) = g(f(x)).
For C ⊆ Z we have that (g ◦ f)−1 (C) = f −1 (g −1 (C)).
Set Theorey 13
Problems
1. Give an example of a function f : X −→ Y and two subsets A
and B of X which show that
(a) f(A ∩ B) %= f(A) ∩ f(B).
(b) f(A \ B) %= f(A) \ f(B).
2. Give examples of functions from the reals R to the non-negative

reals R+ that are
(a) Injective (1-1).
(b) Injective but not surjective.
(c) Surjective (onto) but not injective.
3. Let X be the set of all two by two matrices whose elements are
real numbers. Define f : X −→ R by f(A) = det(A) for all
A ∈ X.(det(A) is the determinant of A.) Prove or disprove that
f is 1-1. Prove or disprove that f is onto.
%n
4. Prove
%n that if k=1 Xk %= ∅, then every projection function
pj : k=1 Xk −→ Xj , 1 ≤ j ≤ n, is surjective.
5. Give an example of a projection function that is injective and of

one that is not injective.
6. Prove that the function f : X −→ Y is injective (1-1) if and only

if for every subset A ⊆ X we have f(X \ A) ⊆ Y \ f(A).
7. Let X = Y = R and consider the following functions:

(1) f : X −→ Y given by f(x) = 2x2 − 1 for each x ∈ X.
(2) f : N −→ N given by f(n) = 2n + 3 for each n ∈ N.
(3) f : X −→ Y given by f(x) = ex for each x ∈ X.
(4) f : X −→ Y given by f(x) = |x|.
(5) f : X −→ Y given by
'
1 ; if x ≥ 0
f(x) =
−1 ; if x < 0
(a) Describe f −1 for those functions f having f −1 as an inverse

function.
(b) For those functions f not having f −1 as an inverse function,

restrict f to some subset of the domain X so that f will
be injective and find the function inverse to your restricted
function.
(c) For the functions (1), (3) and (5), let B = [−1, 1]. Find
f −1 (B) in each case.
8. Let f : R2 −→ R2 be given by f(x, y) = (x + y, y − 2x) for each

(x, y) ∈ R2. Prove f is bijective (1-1 and onto) and find the
inverse function of f.
9. If f : X −→ Y is a function and B1 and B2 are subsets of Y

where B1 ⊆ B2 , prove f −1 (B1) ⊆ f −1 (B2 ).
10. If f : X −→ Y is a function and a, b ∈ Y where a %= b,

prove f −1 (a) ∩ f −1 (b) = ∅.
11. Let f : X −→ Y be a function, A ⊆ X and B ⊆ Y .

(a) Prove A ⊆ f −1 (f(A)).
(b) Prove f is 1-1 if and only if f −1 (f(A)) = A for each A ⊂ X.
(c) Prove f(f −1 (B)) ⊆ B.
(d) Prove f is onto if and only if f(f −1 (B)) = B for each B ⊂ Y .
Set Theorey 15
Section 1.3: Infinite and Uncountable Sets
Finite and Infinite Sets
1.16 Definition: A set X is infinite if and only if there exists a 1-1

function f : X −→ X such that ∅ = % f(X) ⊂ X, i.e. f(X) is a proper
subset of X. A set is finite if and only if it is not infinite.
From the definition it follows that every set is either infinite or
finite, but not both. Since the empty set and a set consisting of a
single element have no nonempty proper subset, both must be finite.
Examples:
1. Let X = {0, 1, 2}. Then any injective function f : X −→ X must

have f(0) %= f(1), f(0) %= f(2) and f(1) %= f(2). So
{f(0), f(1), f(2)} = X. Therfore, it is impossible to find injective
f : X −→ X such that f(X) ⊂ X. Thus X is finite.
2. N is infinite for the function g : N −→ N defined by g(n) = n + 1

is 1-1 and g(N) is a nonempty proper subset of N. (Another
function is g(n) = 2n.)
1.17 Theorem: If X ⊆ Y and Y is finite, then X is finite.

Proof: exercise.
Hint: If you assume X is infinite you will get a contradiction.
1.18 Theorem: If X is infinite and a ∈ X, then X \ {a} is infinite.

Proof: exercise.
Hint: ∃ a 1-1 function f : X −→ X such that f(X) is a proper subset
of X. There are only two cases for the point a:(1) a ∈ X \ f(X) or
(2) a ∈ f(X). In either case you must exhibit an injective function
g : X \ {a} −→ X \ {a} such that g(X \ {a}) is a proper subset of
X \ {a}.
1.19 Definition: For each n ∈ N, let In = {k ∈ N: k ≤ n}. Thus

I1 = {1}, I3 = {1, 2, 3}.
1.20 Theorem: For each n ∈ N, In is finite.

Proof: By Mathematical Induction. I1 is finite since it is a singleton.
Now, assume that Ik , k ≥ 2, is finite and we have to show that Ik+1 is
finite. Suppose not, then Ik+1 \ {k + 1} = Ik is infinite,(by Theorem

1.18), which is a contradiction.
1.21 Corollary: A subset A ⊆ N is finite if and only if there is an

n ∈ N such that A ⊆ In .
1.22 Theorem: Let f : X −→ Y be a function.

(1) If f is 1-1 and X is infinite, then Y is infinite.
(2) If f is onto and X is finite, then Y is finite.
Proof: exercise.
Countable and Uncountable Sets
1.23 Definition: A set X is countable if and only if either X = ∅

or there exists an onto function f : N−→ X. If X is not countable,
then it is called uncountable.
Examples:
1. X = {0, 1, 2} is countable. Indeed define f : N−→ X by f(1) =

0, f(2) = 1 and f(n) = 2 ∀n ≥ 3, then f is onto.
2. X = {0, 1, 2, 3, 4, ...} is countable. Indeed define f : N−→ X by

f(n) = n − 1 ∀n ∈ N, then f is onto.
Any finite set is countable. The converse is not always true as

example(2) above shows. Any uncountable set must be infinite. An
infinite set could be countable (as Example (2) above) and could be
uncountable as shown the following theorem.
1.24 Theorem:
X = {x ∈ R: 0 < x < 1} = (0, 1) is uncountable.
Proof: Suppose there exists an onto function f : N−→ (0, 1). Thus,
for each n ∈ N, f(n) is a real number between 0 and 1. Therefore,
f(n) has a decimal representation denoted by f(n) = 0.xn1 xn2 xn3 xn4...
where xij is an integer between 0 and 9 inclusive. That is,f(1) =
0.x11x12x13x14x15... , f(2) = 0.x21x22x23x24 x25..., ...,
f(n) = 0.xn1 xn2 xn3xn4 xn5 ..... Now, we are going to define a number y =
0.y1y2 y3... between 0 and 1 such that y ∈ / f(N) to get a contradiction.
Let y1 = 7 if x11 %= 7 and y1 = 3 if x11 = 7. Let y2 = 7 if x22 %= 7
and y2 = 3 if x22 = 7 .......... etc. Then y = 0.y1 y2y3 ... ∈ (0, 1) = X
Set Theorey 17
does not have two decimal representation because for all n ∈ N we

have 0 %= yn %= 9. Also, y %= f(1) because y1 %= x11, y %= f(2) because
y2 %= x22, .... , y %= f(n) because yn %= xnn , ... etc. Therefore y ∈
/ f(N),
hence f is not onto, a contradiction. Therefore, there is no surjective
f : N−→ (0, 1). Thus (0, 1) is uncountable.
1.25 Theorem: Let X ⊆ Y . If Y is countable, then X is countable.

Proof: We may assume X %= ∅. Let f : N−→ Y be an onto function.
Define h : Y −→ X as follows: If x ∈ X, let h(x) = x. If x ∈ Y \ X,
let h(x) = a where a is a fixed element in X. It is clear that h is onto.
Let g = h ◦ f : N−→ X be the composition function. Then we
have (h ◦ f)(N)= h(f(N))= h(Y ) = X, i.e., g(N)= X, so g is onto.
Therefore, X is countable.
1.26 Corollary: Let X ⊆ Y . If X is uncountable, then Y is

uncountable.
1.27 Theorem: R is uncountable.

Proof: Follows from Theorem 1.24 and Corollary 1.26.
!
1.28 Theorem: If An is countable for each n ∈ N, then n∈N An
is countable.
1.29 Corollary: Q is countable while P is uncountable.
We end this chapter with an important theorem about the cardi-

nality (the number of the elements in a set) of two sets.
1.30 Theorem: Two sets X and Y are having the same cardinality
if and only if there exists a 1 − 1 and onto function f : X −→ Y .
1.31 Theorem: The Principle of Mathematical Induction

Suppose that to each natural number n ∈ N there is associated a
certain statement Pn . Suppose that the statement P1 is true; and that
for every natural number n for which the statement Pn happens to be
true, the statement Pn+1 is also true. Then the statement Pn must be
true for every natural number n ∈ N.
1.32 Example: If x ∈ R with x > 0 and n ∈ N, then
(1 + x)n ≥ 1 + nx.
(This fact is known as Bernoulli’s inequality.)

Proof: Let x ∈ R be arbitrary with x > 0. For each natural number

n ∈ N let Pn be the statement that (1 + x)n ≥ 1 + nx. The statement
P1 is true because when n = 1 we have (1+x)1 = (1+ x) = 1+ (1)(x) ≥
1 + x. Now, suppose n is any natural number for which the statement
Pn ≡ (1 + x)n ≥ 1 + nx happens to be true. Then we see that
(1 + x)n+1 = (1 + x)n (1 + x)
≥ (1 + nx)(1 + x) by the inductive hypothesis
= 1 + (n + 1)x + nx2 ≥ 1 + (n + 1)x.
Which establishes the truth of the statement Pn+1 . Thus by mathe-
matical induction the inequality is true for any n ∈ N.
1.33 Theorem:1 If x ∈ R with x > 0, then there exists a natural

number n ∈ N such that n1 < x.
Proof: Let x ∈ R be arbitrary with x > 0. Since the set of natural
number N is not bounded above, x1 is not an upper bound of N. There-
fore we can choose a member n ∈ N such that n > x1 and we see at
once that n1 < x.
1.34 Definition: A real number α is an upper bound of a subset

A of R if x ≤ α for all x ∈ A. A real number β is a lower bound of a
subset B of R if x ≥ β for all x in B. A subset C of R is called bounded
above if it has an upper bound, and called bounded below if it has a
lower bound, and called bounded if it is bounded above and bounded
below.
Notice that if α is an upper bound of a set A, then every number
greater than α must also be an upper bound of A. Similarly, if β is a
lower bound of a set B, then so is every number less than β.
1.35 Examples:
(a) 6 is an upper bound of (0, 1) while -2 is not.
(b) 1 is an upper bound of [0, 1) while 0 is not.

3
(c) 1 is an upper bound of [0, 1] while 4
is not.
(d) 11 is an upper bound of −3, 2, 5 while 4 is not.

1
This theorem comes from Archimedean property which is the following theorem:
If a > 0 and b > 0 where a, b ∈ R, then for some natural number n, we have na > b.
This tells us that even if a is quite small and b is quite large, some natural multiple
of a will exceed b. Or, we can say, “Given enough time, one can empty a large
bathtub with a small spoon.”
Set Theorey 19
1.36 Definition: If a set of real numbers has a largest member,

then we call that largest member the maximum of the set. If a set of
real numbers has a smallest member, then we call that smallest member
the minimum of the set. Thus
α = max A if and only if x ≤ α for all x ∈ A and α ∈ A.
β = min B if and only if x ≥ β for all x ∈ B and α ∈ B.

For example, if A = [0, 1], the maxA = 1 and minA = 0. But for
A = (0, 1], then maxA = 1 but A has no minimum. If B = (2, 3]∪[4, 5),
then B has no maximum and no minimum while B is a bounded set.
1.37 Definition: If a number α is an upper bound of a set A and

if no number smaller than α is an upper bound of A, then α is called
the least upper bound of A or the supremum of A and denoted by either
α = lub A or sup A. Thus α = supA if and only if x ≤ α ∀ x ∈ A and
if γ < α, then γ is not an upper bound of A.
If a number β is a lower bound of a set B and if no number greater
than β is a lower bound of B, then β is called the greatest lower bound
of B or the infimum of B and denoted by either β = glb B or sup B.
For example, if A = [0, 1], the maxA = 1 =supA and minA =

0 =infA. But for A = (0, 1], then maxA = 1 =supA but A has no
minimum while infA = 0. If B = (2, 3]∪[4, 5), then B has no maximum
and no minimum but supB = 5 and infB = 2.
1.38 The Axiom of completeness: Every set of real numbers

which is nonempty and bounded above has a least upper bound.
1.39 Corollary: Suppose A is a set of real numbers which is

nonempty and bounded below. Then A has a greatest lower bound.
Problems
1. (a) If X and Y are both finite, prove that X ∪ Y is finite.
(b) Use part (a) and mathematical induction to prove that a
finite union of finite sets is a finite set. That
! is, if Xi is finite
for each i ∈ {1, 2, ..., n} where n ∈ N, then ni=1 Xi is a finite
set.
2. If X is infinite and X = A∪B, prove that A or B must be infinite.
3. Prove that if X is infinite and A ⊂ X is finite, then X \ A is

infinite.
4. Prove that the set of all integers is countable
5. Give an example of a countable collection of sets each of which is

finite, but their union is infinite.
6. Give an example of a countable infinite collection of countable

subsets of R such that each pair of these sets are disjoint.
7. Prove that the union of two uncountable sets is uncountable.
8. Prove that Q×Q is countable.
9. Prove that if X is countable and f : X −→ Y is onto, then Y is

countable.
10. (a) Prove that any finite set is countable.

(b) Prove that any uncountable set is infinite.
11. Prove that if there exists a 1-1 function f : N−→ X, then X is

infinite.
12. Use the mathematical induction to prove the following:

n(n+1)
(a) For any n ∈ N, we have 1 + 2 + 3 + ... + n = 2
.
(b) For any n ∈ N and x ∈ R, we have |sinnx | ≤ n|sinx |.
(c) All numbers of the form 7n − 2n are divisible by 5.
13. Prove that if a set A has a largest member α, then α is the least
upper bound of A.
Set Theorey 21
14. Show that if a set A has an upper bound α and α is a member of

A, then α is the largest member of A.
15. Prove that if α is an upper bound of A and β is an upper bound

of B, then the largest of the two numbers α and β is an upper
bound of A ∪ B.
16. Let ∅ %= A ⊆ B ⊂ R and B is bounded above. Prove that supA ≤

supB.
.
2 Topological Spaces
Section 2.1: Topology and Open Sets
2.1 Definition: Let X be a nonempty set. A topology on X is a

subset τ of P(X), the power set of X, satisfying the following condi-
tions:
1. The ground set X and the empty set ∅ are in τ;

2. The intersection of any two members of τ is in τ ; and
3. Any union of members of τ is in τ .
In notations:
1. X, ∅ ∈ τ ;
2. U1 , U2 ∈ τ =⇒ U1 ∩ U2 ∈ τ ; and
!
3. Uα ∈ τ , ∀ α ∈ Λ =⇒ α∈Λ Uα ∈ τ .
If τ is a topology on a nonempty set X, then the order pairs (X, τ )

is called a topological space2. If τ is understood, we just say that X
is a topological space, or shortly, X is a space. Any member of the
topology is called an open subset, or shortly open set.
2.2 Examples:
1. Let X = {a, b, c}. Let τ 1 = {∅, X, {a}, {a, b}},

τ 2 = {∅, X, {a}, {b}, {a, c}}, and τ 3 = {X, {a, b}, {a}}.
τ 1 is a topology on X because ∅ and X ∈ T1 , by the definition of
τ 1, so the first condition holds. Since ∅ ∩ Y = ∅, where Y is any
set, and also if A ⊆ X, then A ∩ X = A, then the intersection of
any two members of τ 1 is in τ 1 if one of them is ∅ or X. It remains
to show that {a} ∩ {a, b} is in τ 1. But, {a} ∩ {a, b} = {a} ∈ τ 1.
Thus, the second condition holds. Since the empty set does not
affect any union and since the union will be X (the ground set)
if X is a member of the union, it remains, to complete the proof
2
Abstract spaces with a topological structure were first introduced by F réchet in
1906 and by Riesz in 1907. The definition of a topological space was first formulated
by Kuratowski in 1922 in terms of a closure operator. See [Engelking, 1977].
of the third condition, to show that {a} ∪ {a, b} is in τ 1 . But,

{a} ∪ {a, b} = {a, b} ∈τ 1. Thus any union of members of τ 1 is
in τ 1, so the third condition holds. Thus τ 1 is a topology on X.
τ 2 is not a topology on X because {a}, {b} ∈ τ 2, but {a} ∪{b} =
{a, b} %∈τ 2. So, τ 2 does not satisfy the third condition. So, it is
not a topology on X.
τ 3 is not a topology on X because ∅ %∈τ 3.
2. Define τ ⊂ P(R) as follows: For a subset U ⊆ R,
U ∈ τ if and only if U = ∅ or U = R or U = [a, b] where a, b ∈
R with a < b.
Recall that [a, b] = {x ∈ R : a ≤ x ≤ b}.
τ is not a topology on R. The first and the second conditions
are satisfied, but the third one is not. To see that, observe that
[0, 1] ∈ τ and [2, 3] ∈ τ , but their union is [0, 1] ∪ [2, 3] %∈ τ .
Here is another!counterexample: for each n ∈ N we have that
[ n1 , 2] ∈ τ , but n∈N [ n1 , 2] = (0, 2]!%∈ τ . Since for each n ∈ N,
we have that [ n1 , 2] ⊂ (0, 2], then n∈N [ n1 , 2] ⊆ (0, 2]. For the
reverse inclusion, pick an arbitrary x ∈ (0, 2], then 0 < x ≤ 2,
1
thus there exists an m ∈ N such that 0 < m < x ≤ 2. Thus
1 1
!
x ∈ [ m , 2] ⊂ n∈N [ n , 2]. So, equality holds.
2.3 The Discrete Topology: For any nonempty set X, the dis-
crete topology on X is D= P(X). That is, D is the set of all subsets of
X. So, in a discrete space any subset of the space is an open set.
2.4 The Indiscrete Topology: For any nonempty set X, the

indiscrete topology on X is I= {∅, X}. So, in an indiscrete space there
are only two open sets and they are the empty set and the whole space.
Remark: We conclude that any set has more than one element has
at least two topologies on it. What about the singleton?
2.5 The One-Point Topology: Let X be a set having more

that one element. Fix c ∈ X. The one-point topology on X is T =
{∅, X, {c}}.
2.6 The Left Ray Topology on R:

Let L = {∅ , R} ∪ {(−∞, a) : a ∈ R} ⊆ P(R).
Topological Spaces 25
Let us verify that L is indeed a topology on R. Condition (1) is satisfied

from the definition of L. For condition (2), let A, B ∈ L be arbitrary. If
either A or B is empty, then A∩B = ∅ ∈ L. So, assume now A %= ∅ %= B.
If either A or B is the whole set R, then A ∩ B = A or B ∈ L. So,
assume now A %= R%= B. Then ∃ a, b ∈ R such that A = (−∞, a)
and B = (−∞, b). Hence (−∞, a) ∩ (−∞, b) = (−∞, c) ∈ L where
c = min{a, b}. Therefore (2)!is satisfied. Now, let {Aα : α ∈ Λ} ⊆ L.
If Aα = ∅, ∀ α ∈ Λ, then α∈Λ Aα = ∅ ∈ L. So, assume that some
member is nonempty. But since the empty set does not affect any union,
we may assume, without loss of generality, that ! Aα %= ∅, ∀ α ∈ Λ. If
there exists a β ∈ Λ such that Aβ = R, then α∈Λ Aα = R ∈ L. So,
assume now that Aα %= R ∀α ∈ Λ. Thus for each α ∈ Λ there exists
aα ∈ R such that Aα = (−∞, aα). Consider the subset {aα : α ∈ Λ} ⊆
R. There are exactly two cases.
Case 1: {aα : α ∈ Λ} is bounded above. By the completeness axiom of
R (Every set of real numbers which is nonempty and bounded
! above has
a least upper bound.), let c = sup{aα : α ∈ Λ}. Then α∈Λ(−∞, aα) =
(−∞, c) ∈ L. !
Case 2: {aα : α ∈ Λ} is not bounded above. Then α∈Λ (−∞, aα) =
R∈ L. Thus, condition (3) is also satisfied.
Hence L is a topology on R called the left ray topology on R.
Exercise:
Mimic the above example to define the right ray topology R on R.
2.7 Definition: A subset of R having the following forms is called

an Open Interval.
(1) (a, b) = {x ∈ R: a < x < b};
(2) (a, ∞) = {x ∈ R: a < x};
(3) (−∞, a) = {x ∈ R: x < a}; and
(4) (−∞, ∞) = R.
A subset of R having the following forms is called a Closed Interval.
(1) [a, b] = {x ∈ R: a ≤ x ≤ b};
(2) [a, ∞) = {x ∈ R: a ≤ x};
(3) (−∞, a] = {x ∈ R: x ≤ a}; and
(4) (−∞, ∞) = R.
A subset of R is called an Interval if it is either an open interval,
closed interval, has the form {x ∈ R: a ≤ x < b} = [a, b), or has the
form {x ∈ R: a < x ≤ b} = (a, b].
2.8 The Usual Topology on R: We define the usual topology U

⊆ P(R) on R as follows:
U ∈ U ⇐⇒ either U = ∅, or ∀ x ∈ U ∃ an open interval of the
form (a, b), where a, b ∈ R with a < b such that x ∈ (a, b) ⊆ U.
Let us prove that U is indeed a topology on R.
From the definition of U we do have that ∅ ∈ U. Let x ∈ R, then
(x − 1, x + 1) is an open interval such that x ∈ (x − 1, x + 1) ⊆ R, thus
R ∈ U. Hence (1) is satisfied. Now, let U1 , U2 ∈ U. If U1 ∩ U2 = ∅,
then U1 ∩ U2 ∈ U. So, assume now U1 ∩ U2 %= ∅. Let x ∈ U1 ∩ U2 be
arbitrary, then x ∈ U1 and x ∈ U2 . Now, x ∈ U1 ∈ U =⇒ ∃ (a, b) such
that x ∈ (a, b) ⊆ U1 and x ∈ U2 ∈ U =⇒ ∃ (c, d) such that x ∈ (c, d) ⊆
U2 . Thus the open interval (max{a, c}, min{b, d}) is containing x and
contained in U1 ∩ U2 . Thus
! U1 ∩ U2 ∈ U. So, ! (2) is satisfied. Now, let
{U
! α : α ∈ Λ} ⊆ U. If! α∈Λ Uα = ∅, then α∈Λ Uα ∈ U. So, assume
α∈Λ Uα %= ∅. Let x ∈ α∈Λ Uα be arbitrary. Then there exists a β ∈ Λ
such that x ∈ Uβ . But Uβ ∈ U,!thus there exists! an open interval (a, b)
such that x ∈ (a, b) ⊆ Uβ ⊆ α∈Λ Uα . Thus α∈Λ Uα ∈ U. So (3) is
also satisfied. Therefore U is a topology on R.
Some √ examples of open subsets of R in the usual topology U are:

(1, 3), ( 2, π), (1, 2) ∪ (5, 6), R\{−1, 3, 5}. Indeed, any open interval
belongs to the usual topology.
2.9 Definition: Let (X,τ ) be a topological space and let x ∈ X.

An open neighborhood of x in X is any open set U ∈ τ such that x ∈ U.
The family {U ∈ τ : x ∈ U} is called the set of all open neighborhoods
of x.
2.10 Theorem: A nonempty subset V of a space X is open if and

only if ∀ x ∈ V, ∃ an open neighborhood Ux of x such that Ux ⊆ V .
Proof: (=⇒) Clear, as V itself works.
(⇐=) Suppose that ∀x ∈ V, ∃ an open neighborhood Ux of x such
that Ux ⊆ V . We need to show that V is open, i.e., V ∈ τ where τ
denotes the topology on X. For each x ∈ V fix an open neighborhood
!
Ux ∈ τ of x! such that x ∈ Ux ⊆ V . We show that V = x∈V Ux .
First, V ⊇ x∈V Ux holds because U !x ⊆ V for each x ∈! V . Now, if
x ∈ V is arbitrary, then x ∈ Ux ⊆ x∈V Ux , thus V ⊆ x∈V Ux . So,
equality holds. Therefore, V is a union of members of τ . Thus, by the
definition of a topology, V ∈ τ . Thus V is open.
2.11 Definition: Let τ 1 and τ 2 be two topologies for a set X. If

τ 1 ⊆ τ 2 , i.e., if each open set of τ 1 is also an open set of τ 2, then τ 1

is said to be smaller than τ 2. (or τ 2 is larger than τ 1.)
If neither τ 1 ⊆ τ 2 nor τ 2 ⊆ τ 1 is true, then τ 1 and τ 2 are said to
be not comparable.
Note that if τ is a topology on X, then it is always true that I ⊆

τ ⊆ D. On R, we have L ⊆ U.
2.12 Theorem: A finite intersection of open

$ sets is open. That is,
if U1 , U2, ..., Un are open, where n ∈ N, then ni=1 Ui is open.
Proof: By Mathematical Induction. For n = 1, there is nothing to
prove. For n = 2, U1 ∩ U2 is open from the definition of the topology.
Assume now the statement is true for n = k ≥ 2, and we have to prove
it for n = k +1. Note that U1 ∩U2 ∩...∩Uk+1 = (U1 ∩U2 ∩...∩Uk )∩Uk+1
and U1 ∩ U2 ∩ ... ∩ Uk is an open set by the inductive hypothesis. Thus
U1 ∩ U2 ∩ ... ∩ Uk ∩ Uk+1 is open by the definition of the topology.
Exercise: Give an example of an infinite collection of open sets in

a topological space such that its intersection is not open.
2.13 The Cofinite Topology: Let X be any infinite set. The

cofinite topology CF on X is defined as follows: For a subset U of X,
U ∈ CF ⇐⇒ U = ∅ or (X \ U) is finite.
Exercise:
1. Show that CF is a topology on X.
2. If X is a finite set and we define the cofinite topology CF on it.
What will we get?
2.14 The Cocountable Topology: Let X be any uncountable

set. The cocountable topology CC on X is defined as follows: For a
subset U of X,
U ∈ CC ⇐⇒ U = ∅, or (X \ U) is countable.
Exercise:
1. Show that CC is a topology on X.
2. If X is a countable set and we define the cocountable topology
CC on it. What will we get?
Problems
1. List all topologies for a set containing three distinct elements.
2. Give an example of a topology on an infinite set which has only a

finite number of elements. (Do not use the indiscrete topology.)
3. Give an example of a set X and two topologies τ 1 and τ 2 on X

such that τ 1∪ τ 2 is not a topology for X.
4. Let X be a non-empty set and τ 1 and τ 2 be two topologies on

X. Prove that τ 1∩ τ 2 is a topology on X.
5. Prove that τ is the discrete topology for X if and only if each

singleton is open (i.e. {x} ∈ τ , for each x ∈ X.).
6. For each n ∈ N, define Un = {n, n + 1, n + 2, ...}. So, U1 =

{1, 2, 3, 4, ...}, U2 = {2, 3, 4, 5, ...}, ... , U98 = {98, 99, 100, 101, ...}.
Let τ = {Un : n ∈ N} ∪ {∅}. Prove that τ is a topology on N.
7. Let X be a non-empty set. Fix an element q ∈ X. Define Eq ⊂

P(X) as follows:
U ∈ Eq if and only if either U = X or q %∈ U.
Prove that Eq is a topology on X.

This topology is called the excluded point topology.
8. For R, compare the cofinite topology with the usual topology?

With the left ray topology? With the co-countable topology?
9. Which of the following is a topology on R, justify your answer:
(a) τ = {∅ , R} ∪ {[−π, a) : a > 0}

(b) τ = {∅ , R} ∪ {[−π, a] : a > 0}
(c) τ = {∅ , R} ∪ {U ⊂ R : 2 ∈ U and 3 %∈ U}
(d) τ = {∅} ∪ {U ⊂ R : [0, 2] ⊆ U}
Section 2.2: Closed Sets
Now we are ready to begin investigating the structure of topological

spaces.
2.15 Definition: Let (X,τ ) be a topological space. A subset A ⊆

X is called closed subset3 , or shortly closed set, if and only if X \ A is
an open set.
From the definition we conclude that if a subset B in a space X is

open, then X \ B must be closed
Examples:
1. For any space X, the subsets X and ∅ are always closed. (Note
that X and ∅ are also open.)
2. In (R, U), any closed interval is closed. But (0, 1] is neither open
nor closed.
3. Let us describe all closed sets in (R, CF ). Since a subset U of

R is open in the cofinite topology CF if and only if U = ∅ or
the complement R \ U is finite, then a subset F of R is closed in
(R , CF ) if and only if F = R or F is a finite set.
Remark:
1. A subset in a topological space could be only open, only closed,

closed-and-open (clopen), or neither open nor closed.
2. The most important observation is that if we know that a subset

B in a space X is not open, then this does not mean that B must
be closed, as example (2) above shows.
3. The closeness of a set depends on the topology. For example,

[0, 1] is closed in (R, U). But [0, 1] is not closed in (R, L).
3
The notions of open and closed sets as well as those of closure and interior
were introduced and studied by Cantor in the class of subsets of Euclidean spaces.
Hausdorff generalized them to abstract spaces in 1914. See [Engelking, 1977].
2.16 Theorem: Let (X,τ ) be a topological space. Then

(1) The intersection of any family of closed sets is closed.
(2) The union of any finite number of closed sets is closed.
Proof: We prove (1) and leave (2) as an exercise. Let {Aα$: α ∈ Λ}
be any family of closed subsets of X.$Need to show that α∈Λ Aα is
closed. So, we$have to show! that X \ α∈Λ Aα is open in X.
But, X \ α∈Λ Aα = α∈Λ (X \ Aα ). Since Aα is closed for each
α ∈ Λ, then X \ Aα is open for each α ∈ Λ.$ Thus, by the definition of
a topology, their union is open. Therefore α∈Λ Aα is closed.
Exercise: Give an example of an infinite collection of closed sets
such that its union is not closed.
2.17 Definition: Let (X,τ ) be a topological space, and A ⊆ X.
1. The interior of A in (X,τ ), denoted by int(A) or intA, is the set
of all points x ∈ X such that there exists an open neighborhood
Ux of x with Ux ⊆ A. That is,
int(A) = {x ∈ X : ∃ open neighborhood Ux of x with Ux ⊆ A}.
From the definition we conclude that
x ∈ int(A) ⇐⇒ ∃ Ux ∈ τ such that x ∈ Ux ⊆ A.
2. The closure of A in (X,τ ), denoted by A or CL(A), is the set of
all points x ∈ X such that Ux ∩ A %= ∅ for any open neighborhood
Ux of x. That is,
A = {x ∈ X : Ux ∩ A %= ∅ for any open neighborhood Ux of x}.
x ∈ A ⇐⇒ Ux ∩ A %= ∅ ∀ Ux ∈ τ such that x ∈ Ux .
3. The boundary of A in (X,τ ), denoted by ∂A, is the set of all
points x ∈ X such that Ux ∩ A %= ∅ %= Ux ∩ (X \ A) for any open
neighborhood Ux of x. That is,
∂A = {x ∈ X : Ux ∩ A %= ∅ = % Ux ∩ (X \ A) for any open
neighborhood Ux of x}.
x ∈ ∂A ⇐⇒ Ux ∩ A %= ∅ = % Ux ∩ (X \ A), ∀ Ux ∈ τ such that
x ∈ Ux .
From definition 2.17, for any subset A of a space X, we always have

that
int(A) ⊆ A ⊆ A .
Example: Consider (R, U). Let A = [1, 2). Then int(A) =

int([1, 2)) = (1, 2) because if x ∈ (1, 2), then (1, 2) is an open neighbor-
hood of x with (1, 2) ⊂ [1, 2). Since int(A) ⊆ A, it remains to check
the number 1. If U is an arbitrary open neighborhood of 1, then there
are a, b ∈ R with a < b and 1 ∈ (a, b) ⊆ U. Thus a < 1 < b. So, there
are infinitely many numbers between a and 1 which are not in [1, 2).
Thus U %⊆ [1, 2). Hence 1 %∈ int([1, 2)).
[1, 2) = [1, 2]. To see this, first note that [1, 2) ⊂ [1, 2). Now, if U is
an arbitrary open neighborhood of 2, then there are two real numbers
a and b, with a < b and 2 ∈ (a, b) ⊆ U. Thus a < 2 < b, hence
(a, b) ∩ [1, 2) %= ∅. Thus U ∩ [1, 2) %= ∅. So, 2 ∈ [1, 2). To complete
the proof that [1, 2) = [1, 2], it remains to show that if x %∈ [1, 2], then
x %∈ [1, 2). Let x ∈ R \ [1, 2] be arbitrary. If x > 2, let U = (2, x + 1).
If x < 1, let U = (x − 1, 1). Then, in both cases, U is an open
neighborhood of x with U ∩ [1, 2) = ∅.
∂[1, 2) = {1, 2}. (Check that)
Exercise: In (R, L), find int([1, 2)), [1, 2), and ∂[1, 2).
2.18 Theorem: Let (X,τ ) be a topological space and A ⊆ X.

Then
1. If U ⊆ A and U is open, then U ⊆ int(A);
2. int(A) ⊆ A and int(A) is open.
Proof:
1. Suppose that U ⊆ A and U is open. Need to show that U ⊆
int(A). Let x ∈ U be arbitrary. Then U is an open neighborhood
of x with U ⊆ A which is the definition of x ∈ int(A). Therefore,
U ⊆ int(A).
2. We have x ∈ intA ⇐⇒ ∃ Ux ∈ τ such that x ∈ Ux ⊆ A. So,
it is clear that int(A) ⊆ A. To show that int(A) is open, we use
theorem 2.10. First, if int(A) = ∅, we are done. So, assume now
that int(A) %= ∅. Let x ∈ int(A) be arbitrary. Need an open
neighborhood Ux of x such that Ux ⊆ int(A). But x ∈ intA ⇐⇒
∃ Ux ∈ τ such that x ∈ Ux ⊆ A. By part (i), Ux ⊆ int(A). So
int(A) is open.
2.19 Corollary:
!
1. int(A) = {U ⊆ X : U is open and U ⊆ A}. That is, int(A) is
the union of all open sets which contained in A. Thus int(A) is
the largest open set contained in A.
2. A is open ⇐⇒ A = int(A).
2.20 Theorem: Let (X,τ ) be a topological space and A ⊆ X.

Then
1. If A ⊆ F and F is closed, then A ⊆ F ;
2. A ⊆ A and A is closed.
Proof:
1. Suppose that A ⊆ F and F is closed. Need to show that A ⊆ F .
We do this by contradiction. Suppose that A %⊆ F . Then there
would be an x ∈ A such that x %∈ F , hence x ∈ X \ F which is
open. Now, A ⊆ F ⇐⇒ X \ F ⊆ X \ A. Thus (X \ F ) ∩ A = ∅
and this contradicts that x ∈ A. Therefore A ⊆ F .
2. If x ∈ A, then for any open neighborhood Ux of x we have that

x ∈ Ux ∩ A. So, Ux ∩ A %= ∅. Thus x ∈ A Therefore, A ⊆ A. Now,
to show that A is closed, we show that X \ A is open. We use
theorem 2.10 to do that. Let x ∈ X \ A be arbitrary (if X \ A = ∅,
we are done). Then x %∈ A. So there exists an open neighborhood
Ux of x such that Ux ∩ A = ∅..........(&).
Claim: Ux ∩ A = ∅.
Suppose not, then ∃ some y ∈ Ux ∩ A. Since y ∈ Ux which is
open and y ∈ A, we get Ux ∩ A %= ∅ which contradicts (&), and
hence Ux ∩ A = ∅. Thus we conclude that x ∈ Ux ⊆ X \ A. So,
X \ A is open, hence A is closed.
2.21 Corollary:
$
1. A = {F ⊆ X : F is closed and A ⊆ F }. i.e., A is the intersec-
tion of all closed sets containing A. So A is the smallest closed
set containing A.
2. A is closed ⇐⇒ A = A.
Examples:
1. In any space X, we always have X = X = int(X), ∂X = ∅
and ∅ = ∅ = int(∅) = ∂∅.
2. If A ⊆ X where the topology on X is I, the indiscrete topology,
then
'
∅ ; if A %= X
int(A) =
X ; if A = X
'
∅ ; if A = ∅
A=
% ∅
X ; if A =

 ∅ ; if A = ∅
∂A = X ; if ∅ %= A %= X
∅ ; if A = X

Don’t forget that in an indiscrete space X, the only nonempty

open set is X itself.
3. If A ⊆ X where the topology on X is D, the discrete topology,
then
int(A) = A = A and ∂A = ∅.
Don’t forget that in a discrete space, each subset is clopen.
4. Let X be an infinite set and consider (X, CF ), the cofinite topol-
ogy. Let A be an arbitrary subset of X.
Let us determine int(A). We will use the fact: if V and B are
subsets of a set X, then V ⊆ B if and only if X \ B ⊆ X \ V .
Now, for the complement X \ A, there are only two cases. Either
X \ A is infinite or it is finite. If X \ A is infinite, then for any
a ∈ A and any open neighborhood U of a, we have that X \ U
is finite, hence X \ A %⊆ X \ U, thus U %⊆ A. Hence, in this
case, int(A) = ∅. Now, if X \ A is finite, then A is open, hence
int(A) = A. So, we have
'
∅ ; if X \ A is infinite
int(A) =
A ; if X \ A is finite
Let us, now, determine A. There are only two cases. Either A
is finite or A is infinite. If A is finite, then A is closed, hence
A = A. If A is infinite, we will show that A = X. (so, any

infinite set in a cofinite space is dense, see definition 2.23 below)
It is clear that A ⊆ X. For the reverse inclusion, let x ∈ X be
arbitrary. To show that x ∈ A, let U be any open neighborhood
of x, then X \ U is finite. Since A is infinite, then A %⊆ X \ U.
Thus A ∩ U %= ∅. Therefore, x ∈ A. Thus X = A. So, we have
'
A ; if A is finite
A=
X ; if A is infinite
Check that

 A ; if A is finite
∂A = X \ A ; if X \ A is finite
X otherwise

Exercise: Prove that A \ int(A) = ∂A.
2.22 Theorem: Let (X,τ ) be a space and A ⊆ X, then
1. X \ int(A) = X \ A.
2. X \ A = int(X \ A).
Proof: This is clear from the following negations.
1. x ∈ int(A) ⇐⇒ ∃ Ux ∈ τ such that x ∈ Ux ⊆ A. Thus

x∈/ int(A) ⇐⇒ ∀ U ∈ τ , x ∈ U =⇒ U ∩ (X \ A) %= ∅.
2. We have x ∈ A ⇐⇒ ∀ U ∈ τ with x ∈ U we must have U ∩A %= ∅.

Thus x ∈
/ A ⇐⇒ ∃ U ∈ τ with x ∈ U such that U ∩ A = ∅ ⇐⇒
U ⊆ X \ A.
Exercise: Write a complete proof of Theorem 2.22.
2.23 Definition: A subset D of a topological space (X, τ ) is called

dense4 in (X, τ ) if and only if D = X.
Example: Q and P are both dense in (R, U). So a subset and its
complement could be dense.
4
The notions of the boundary of a set, the derived set and of dense sets were
introduced by Cantor. See [Engelking, 1977].
2.24 Theorem: Let (X, τ ) be a topological space and D ⊆ X.

Then
D = X ⇐⇒ ∀ U ∈ τ , U %= ∅ we have U ∩ D %= ∅.
Proof: Exercise.
2.25 The Particular Point Topology: Let X be any non-empty

set. Fix a point p ∈ X. Define τ p ⊆ P(X) as follows:
τ p = {∅} ∪ {U ⊆ X : p ∈ U}.
Exercise: Prove that τ p is a topology on X. Then show that {p}
is dense in (X ,τ p)
2.26 Definition: A space X is called separable5 if it contains a

countable dense subset.
Examples:
1. (R, U) is separable, as Q is a countable dense subset.
2. Any particular point topological space is separable, as the single-

ton of the particular point is a countable dense subset.
3. (R, L) is separable, as {−7n : n ∈ N} is a countable dense subset.
4. If X is infinite, then (X, CF ) is separable, as any countably infi-

nite subset of X will be dense, see example 4 above.
5. (R , CC) is not separable, because any countable subset of R can-

not be dense. To see this, let A ⊂ R be any countable subset.
Then R \ A is a nonempty open subset which is disjoint from A.
Hence A cannot be dense.
Section 2.3: Sequences
2.27 Definition: A sequence in a topological space (X, τ ) is a

function a : N −→ X. For k ∈ N, the value of a at k, denoted by
a(k) = ak , is called the k th term of a; and for each m ≥ 1, the set
Am = {ak : m ≤ k} is called a tail of a. Sometimes we denote a
sequence a : N −→ X in a topological space (X, τ ) by (an )n∈N . If
the contest is clear, we just write (an ). A subsequence of a sequence
5
Separable spaces were defined by F réchet in 1906. See [Engelking, 1977].
(an )n∈N is a sequence (an )n∈E for some infinite set E ⊆ N. A sequence
(an )n∈N in a topological space X converges in X to a point x ∈ X,
denoted by an −→ x, if and only if for any open neighborhood Ux
of x, there is some tail Am contained in Ux . That is, for any open
neighborhood Ux of x there exists an m ∈ N such that ak ∈ Ux for each
k ≥ m. Let (an )n∈N be a sequence in a topological space X. The set
of all points in X that the sequence (an )n∈N converges to is called the
convergency set of (an )n∈N and it is denoted by C(an)n∈N . That is,
C(an )n∈N = {x ∈ X : an −→ x}.
Example: Consider R with the indiscrete topology I. Let (xn )n∈N

be any sequence in R, then for any x ∈ R we have that xn −→ x
because the only open neighborhood of any point in R is R itself. In
fact, any sequence (xn )n∈N in an indiscrete space (X, I) converges and
C(xn )n∈N = X.
Example: Consider R with the discrete topology D. The only

convergent sequences in (R , D) are those who are eventually constant.
That is, (xn )n∈N is a convergent sequence if and only if there exists an
m ∈ N such that xn = c, ∀ n ≥ m, where c is a constant real number.
This is because a sequence which is not eventually constant cannot
converge to any real number x, since {x} is an open neighborhood of x
which does not contain any tail of the sequence.
So, if (xn )n∈N is a sequence in a discrete space which is eventually
constant to the constant c, then C(xn )n∈N = {c}, and if (xn )n∈N is not
eventually constant, then C(xn )n∈N = ∅.
Example: Consider the sequence (− n1 )n∈N in the space (R , L), the

real numbers with the left ray topology. We will show that C(− n1 )n∈N =
[0, ∞). Let x < 0 be arbitrary. Then there exists an m ∈ N such that
x < − m1 (This follows from the fact that for any real number z greater
than zero, there exists a natural number m such that m1 < z.) Thus
1
(−∞, − m ) is an open neighborhood of x which does not contain any
tail of the sequence. Thus for x < 0, we have x %∈ C(− n1 )n∈N . Now, let
x ∈ [0, ∞) be arbitrary. So we have that x ≥ 0. An open neighborhood
of x is of the form (−∞, x + '), where ' > 0, and now it is clear that
any open neighborhood of x contains a tail of the sequence. Therefore,
C(− n1 )n∈N = [0, ∞). i.e., for any x ∈ [0, ∞), we have that − n1 −→ x.
For instance, − n1 −→ 7.
Problems
1. Give an example of a collection of closed sets in a space (X, τ)
whose union is not closed.
Exercises 2 to 7, A and B are subsets of a space X.
2. Prove that if A ⊆ B, then A ⊆ B.
3. Prove that A ∪ B = A ∪ B.
4. Prove that A = A.
5. Prove that A ∩ B ⊆ A ∩ B. Give an example to show that the

reverse inclusion is not true in general.
6. Prove that int(A ∩ B) = int(A) ∩ int(B).
7. Prove that int(A) ∪ int(B) ⊆ int(A ∪ B). Give an example to

show that the reverse inclusion is not always true.
8. (a) If A and B are subsets of a set X, prove that
A \ B = A ∩ (X \ B)
(b) Let (X, τ ) be a topological space, U ⊆ X is open and F ⊆ X

is closed. Use part (a) to prove that U \ F is open and F \ U
is closed.
9. Consider (R , R), the real numbers with the right ray topology.
So,
R = {∅, R} ∪ {(a, ∞) : a ∈ R}.
Let A = (1, 2] and B = {3} ∪(4, 5) ∪[6, ∞). Find the following:
(a) int(A). (b) A . (c) int(B). (d) B .
10. Let A = (0, 1)∪{2} ⊂ R. Find int(A), A, and ∂A in the following

topologies: (i) usual; (ii) cofinite; (iii) left ray; and (iv) discrete.
11. Let U and V be open sets such that U ∩ V = ∅. Prove that

U ∩ V = ∅.
12. Consider (R , T√2 ), the real numbers with the particular point
√
topology
√ √where the consideration point is 2. Let A = (0, 1] and
B = { 2, 3}. Find, without proof, the following:
(1) A . (2) int(A). (3) ∂(A).

(4) B . (5) int(B).
13. Prove that if A is dense in X, then for every open U ⊆ X we
have U = U ∩ A .
14. Consider (R, U). Let A ⊂ R be any nonempty subset which is

bounded above. Prove that supA ∈ A. Prove that the same is
true with inf A, if A is bounded below.
15. Definition: The exterior of a subset A of a space X, denoted by

ext(A), is defined by ext(A) = int(X \ A).
By Theorem 2.22 (2), we have ext(A) = X \ A.
(a) Prove that the three sets int(A), ∂A and ext(A) are always
pairwise disjoint. (i.e., the intersection of any two distinct
sets is empty.)
(b) Prove that X = int(A) ∪ ∂A ∪ ext(A).
16. Consider the sequence (− n1 )n∈N ⊆ R. Find C(− n1 )n∈N in (R , U)

and in (R , CF ). Prove your result.
17. Describe the convergent sequences in the particular point space.
18. Let (X, τ ) be a topological space and A ⊆ X. Let (an )n∈N be a

sequence such that an ∈ A for all n ∈ N. Prove that if b ∈ X and
an −→ b, then b ∈ A .
Use ( R , CC ) to give an example which shows that the converse of
the above statement is not always true. That is, present a subset
A ⊂ R and b ∈ R such that b ∈ A , but there is no sequence in A
which converges to b.
19. Prove that if (an )n∈N is a convergent sequence, then any subse-
quence from it is also converges.
20. Let X be an uncountable set and A be a nonempty proper subset

of X. Define τ ⊂ P(X) as follows:
U ∈ τ if and only if U = ∅ or A ⊆ U
(a) Prove that τ is a topology on X.

(b) Prove that (X , τ ) is separable
(c) If we take A = X, what will we get?
.
3 Bases for a Topology
Some times, when we want to study a property of a topological space, it

is better to reduce the topology to a smaller collection of open sets such
that any open set is a union of some members of this new collection. If
this can be done, then the new collection will be called a base for the
topological space.
3.1 Definition: Let (X, τ ) be a topological space. A base or basis

for τ is a collection B of subsets of X such that:
(1) each member of B is open.
(2) any non-empty open set is a union of members of B.
In notations:
(1) B ⊆ τ . !
(2) ∀ U ∈ τ , U %= ∅, ∃ G ⊆ B such that U = B∈G B.
Members of B are called basic open sets.
Examples:
(1) In any topological space (X, τ ), τ is a base for τ .
(2) B= {(a, b) : a, b ∈ R, a < b} is a base for (R , U).
We can reduce Theorem 2.10 to the following:
3.2 Theorem: Let (X, τ ) be a topological space and B be a base

for (X, τ ). Let U ⊆ X, U %= ∅. Then
U is open ⇐⇒ ∀ x ∈ U ∃ B ∈ B such that x ∈ B ⊆ U
Proof:
(=⇒) Assume that U is open. Since B is a base, then U is a union
of members
! of B. That is, there exists a subset G ⊆ B such that
U = B∈G B. Thus, if x ∈ U, then there exists B ∈ G ⊆ B such that
x ∈ B ⊆ U.
(⇐=) Assume that the condition!holds. For each x ∈ U, fix Bx ∈ B
such that x ∈ Bx ⊆ U. Then U = x∈U Bx . Hence, U is open being a
union of open sets.
3.3 Theorem: Let (X, τ ) be a topological space and let B ⊆ τ .

Then B is a base for τ ⇐⇒ ∀ U ∈ τ , U %= ∅ and ∀ x ∈ U ∃ B ∈ B
such that x ∈ B ⊆ U.
Proof:
(=⇒) Follows from Theorem 3.2.

(⇐=) Assume that the condition holds. We already have that B ⊆
τ . i.e., each member of B is open. From the condition, it is clear that
each nonempty open set is a union of members of B. Thus B is indeed
a base for τ .
Example: B = {(a, ∞) : a ∈ R} is not a base for (R , U). For the

open set (2, 3) does not contain any member of B.
3.4 Theorem: Let B be a nonempty collection of subsets of a

nonempty set X. Then there exists a unique topology on X for which
B is a base if and only if B satisfies:
(B1) ∀ x ∈ X ∃ at least one U ∈ B such that x ∈ U; and
(B2) if U ∈ B and V ∈ B and x ∈ U ∩ V , then ∃ W ∈ B such that
x ∈ W ⊆ U ∩V.
Proof:
(=⇒) Obvious. (use Theorem 3.3)
(⇐=) Assume that B is a collection of subsets of X satisfying B1
and B2. Need to define a topology τ on X such that B is a base for it
then show that this topology is unique.
Let τ be the set consisting of all possible unions of members of B
togother with ∅, i.e., ! !
τ = {U : U = α∈Λ Uα , Uα ∈ B, ∀ α ∈ Λ} {∅}.
Now, we show that τ is a topology on X. ∅ ∈ τ by its definition. From
B1, ∀ x ∈ X, ∃ Ux ∈ B such that x ∈ Ux and Ux ⊆ X. Therefore,
!
x∈X Ux = X ∈ τ . Now, from the definition of τ we have that any
union of members of τ is again a member of τ . It remains that the
intersection of !any two members of τ is again in τ . So, let U =
!
α∈Λ Uα , V = β∈Γ Vβ ∈ τ . Now
" " "
U ∩V = ( Uα) ∩ ( Vβ ) = (Uα ∩ Vβ ).
α∈Λ β∈Γ α∈Λ, β∈Γ
From B2, (Uα ∩ Vβ ) is a union of members of B for each α ∈ Λ and

each β ∈ Γ.(If it is empty, omit it.) Thus, the intersection U ∩ V ∈ τ .
Therfore, τ is a topology on X.
Note that by Theorem 3.3 and the definition of τ , it is clear that
B is a base for τ . To show uniqueness, let S be any topology on X
such that B is a base for S. Since B is a base for S, members of B
are open in (X, S). Let U be any open subset in (X, τ ), then U is a
union of members of B. Thus, U is open in (X, S). Therefore. τ ⊆ S.
Bases for a Topology 43
Now, let V be open in (X, S). Then V can be expressed as a union of

members of B. (because B is a base for S.) Thus, V is open in (X, τ ).
Therefore, S ⊆ τ . Hence S = τ . Therefore, τ is unique.
Theorem 3.4 gives us a very usefull tool to generate a topology on

a nonempty set X from any nonempty collection B of subsets of X
provided B satisfies B1 and B2, of the theorem even if B itself is not
a topology on X. Simply, take all possible unions and the empty set.
(if B is a topology on X, then all possible union will give the same B.)
This will give the smallest topology on X having B as a base. That is,
if B and C are satisfying B1 and B2 such that B ⊂ C, then τ B ⊂ τ C .
3.5 The Sorgenfrey Topology: Consider the set of real numbers

R. Let B= { [x, y) : x, y ∈ R and x < y}. For each x ∈ R we have that
x ∈ [x, x + 1) ⊆ R. So, B1 is satisfied by B. B2 is also satisfied by
B because for any two members of B, if their intersection is not empty,
then it must be in B. (Check it.)
The topology on R generated by B is called the Sorgenfrey topology
on R and is denoted by S. The topological space (R, S) is called,
sometimes, The Sorgenfrey Line.
So in the Sorgenfrey line S, a basic open neighborhood for any

x ∈ R is of the form [x, x + '), where ' > 0. Any open bounded interval
(x, y), which is open in (R , U), is also open in the Sorgenfrey line S
because for each z ∈ (x, y), we have z ∈ [z, y) ⊂ (x, y). In fact, any
open interval is open in S. Since the bounded open intervals form a
base for (R , U), we conclude that U ⊂ S. Any basic open set [x, y) in
S is clopen. It is closed because its complement is open.
3.6 Definition: A topological space is called second countable space

if it has a countable base.
Example: (R, U) is second countable. To prove this, we must

present a countable base for (R , U). Let B = {(a, b) ⊂ R : a, b ∈
Q with a < b}. It is clear that B ⊂ U. Now, Q is countable, hence
Q × Q is also countable, (see problem 6, page 14), thus the subset
K = {(a, b) ∈ Q × Q : a < b} ⊂ Q × Q is countable. The function
f : B −→ K defined by f((a, b)) = (a, b), ∀ (a, b) ∈ B is 1-1 and onto,
hence by theorem 1.30, B is countable.
Now, we show that B is a base for (R , U). We will use Theorem 3.3
to do that. Let U be an arbitrary nonempty open set. Let x ∈ U be
arbitrary. By the definition of U, there are two real numbers c, d ∈ R

such that x ∈ (c, d) ⊆ U. Thus c < x < d thus (c, x) and (x, d)
are nonempty open sets. Since Q is dense in (R , U), then there are
two rational numbers a, b ∈ Q such that c < a < x < b < d. Thus
x ∈ (a, b) ⊆ (c, d) ⊆ U with (a, b) ∈ B. Therefore, B is a base.
Bases for a Topology 45
Problems
1. Let X = {1, 2, 3} and B = {{1, 2}, {2, 3}, X}. Prove that B is
not a base for a topology on X.
2. Prove that (R , L) is second countable.
3. Prove that every second countable space is separable.
4. Prove that R with the cofinite topology CF is not second count-

able.
5. Find the closure, the interior, and the boundary, in the Sorgenfrey
line, of the following subsets of R: A = [0, 1], B = (1, 2) ∪ {3} ∪
(4, 5], and C = {− n1 : n ∈ N}. Is the Sorgenfrey line separable?
6. Verify that the collection B = {(a, b] ⊂ R : a, b ∈ R, a < b} is

satisfying conditions (B1) and (B2) of Theorem 3.4, then apply
its conclusion to generate a topology on R. This topology is called
right interval topology on R. Find the closure, the interior, and
the boundary, in this space, of the following subsets of R:
A = [0, 1], B = (1, 2) ∪ {3} ∪ (4, 5], and C = {− n1 : n ∈ N}.
7. Consider the sequence (− n1 )n∈N ⊆ R. Find C(− n1 )n∈N in (R , S).

Prove your result.
8. Consider the sequence ( n1 )n∈N ⊆ R. Find C( n1 )n∈N in (R , S).

Prove your result.
9. Let (X,τ ) be a topological space and Λ be an index set. Let

x ∈ X, A ⊆ X, and {Aα : α ∈ Λ} ⊆ P(X). State the definition
of the following:
(a) {Aα : α ∈ Λ} is a base for X.

(b) {Aα : α ∈ Λ} is not a base for X.
(c) x %∈ int(A), x is not an interior point of A.
(d) x %∈ A, x is not in the closure of A.
(e) A is not dense in (X,τ )
(f) (X,τ ) is not second countable.
10. Let B = { [ a , b ] : a < b and a, b ∈ R} ⊂ P(R). Prove that B

is not a base for any topology on R
11. Let B = { (−∞ , b ] : b ∈ R} ⊂ P(R). Is B a base for a topology

on R. Justify your answer.
12. Let B = { ( a , ∞ ) : a ∈ R} ⊂ P(R). Is B a base for a topology

on R. Justify your answer.
4 New Spaces from Old Ones
Section 4.1: Subspaces
4.1 Definition: Let (X,τ ) be a topological space and Y be a

nonempty subset of X. Define τ Y = {U ∩ Y : U ∈ τ }. Then (Y,τ Y )
is a topological space and it is called the subspace topology of (X,τ ).
From the definition, we conclude that if Y is a nonempty subset of

a space (X,τ ), then a subset A of Y is open in the subspace topology
τ Y if and only if there exists an open set B ∈ τ such that A = B ∩ Y .
Note that τ Y in the above definition is well defined. Check that.
Example: In (R, U), consider ( [0, 1] , U[0,1] ) as a subspace. Note

that ( 21 , 1] ⊂ [0, 1] and ( 12 , 1] = ( 12 , 2) ∩ [0, 1], thus ( 12 , 1] is open in
( [0, 1] , U[0,1] ). But ( 12 , 1] is not open in (R, U).
Now, consider ( (0, 1) , U(0,1) ) as a subspace of (R, U). Observe

that [ 21 , 1) is closed in ( (0, 1) , U(0,1) ) because its complement (0, 21 ) =
(0, 12 ) ∩ (0, 1) is open in ( (0, 1) , U(0,1) ).
N as a subspace of (R, U) is a discrete space because for each n ∈ N

we have that (n − 12 , n + 12 ) ∩ N = {n}. Thus {n} is open in (N, U N )
for each n. Thus (N, U N ) is discrete.
Thus open sets and closed sets in a subspace depend on the sets
and the original topology.
Example: Consider R with the particular point topology, see ex-

ample 2.25, where the particular point p = 0 ∈ R. So, U ⊆ R is open if
and only if U = ∅ or 0 ∈ U. Let A = (−1, 1), then A as a subspace is
the same as A itself with the particular point topology where 0 is the
particular point. Let B = (−1, 0) then B as a subspace is discrete.
4.2 Theorem: Let (X,τ ) be a topological space and Y be a

nonempty subset of X. Let B be a base for τ . Then B Y = {B ∩ Y :
B ∈ B} is a base for (Y, τ Y ).
Proof: exercise.
From Theorem 4.2, we conclude that a subspace of a second count-

able space is second countable.
4.3 Theorem: Let (A, τ A ) be a subspace of (X,τ ). A subset

C ⊆ A is closed in (A, τ A ) if and only if there exists a closed set
D ⊆ X of (X,τ ) such that C = D ∩ A.
Proof: (=⇒) Let C ⊆ A be closed in A. Then there exists U ∈ τ such
that A\C = A∩U. Thus C = A\(A\C) = A\(A∩U) = A∩(A\U) =
A ∩ (X \ U). Now, X \ U is closed in X so we may take D = X \ U.
(⇐=) Suppose C ⊆ A and ∃ D ⊆ X closed in X such that C = A ∩ D.
Then X \ D is open in X and A \ C = A \ (A ∩ D) = A ∩ (A \ D) =
A ∩ (X \ D). Thus A \ C is open in (A, τ A ). Therefor C is closed in
A.
Section 4.2: The Product Topology
4.4 Definition: Let (X,τ 1 ) and (Y, τ 2) be topological spaces and

let B= {U × V : U ∈ τ 1 and V ∈ τ 2 }. The product topology on X × Y
is the topology τ that has B as a base. (X × Y, τ ) is called the product
topology of X and Y . If U ∈ τ 1 and V ∈ τ 2, then U × V is called a
basic open set and sometimes called also open rectangle.
Example: R2 = R × R, each is with the usual topology, let us

denote this space, from now on, by (R2 , U). An open rectangle is of
the form (a, b) × (c, d). See Figure 4.1.
Figure 4.1.
New Spaces from Old Ones 49
Figure 4.2.
Figure 4.3.
Important Remarks:
Let (X,τ 1) and (Y, τ 2) be topological spaces.
1. From the definition of a base, we conclude that a subset G ⊆ X ×
Y is open if and only if for each (x, y) ∈ G, ∃ open neighborhood
Ux of x in X, ∃ open neighborhood Vy of y in Y , such that
(x, y) ∈ Ux × Vy ⊆ G.
2. If U ∈ τ 1 and V ∈ τ 2, then U × V is open in X × Y , in fact, it
is a basic open set. BUT, if G ⊆ X × Y is open, then this does
NOT mean that there must be an open set U ⊆ X and an open
set V ⊆ Y such that G = U × V .
+
Example: Consider (R2 , U). Let G = {(x, y) ∈ R2 : x2 + y 2 <
1}. Then G, which is the set of all of the interior points of the
unit circle centered at the origin, is open in (R2 , U). BUT, there
do not exist two open sets U and V of R with G = U × V . See
Figure 4.2.
3. If E ⊆ X and F ⊆ Y are closed, then E × F is closed in X × Y
(WHY?). BUT, if K ⊆ X × Y is closed, then this does NOT
mean that there must be a closed set E ⊆ X and a closed set
F ⊆ Y such that K = E × F .
+
Example: Consider (R2 , U). Let K = {(x, y) ∈ R2 : x2 + y 2 ≤
1}. Then K, which is the set of all points of the unit circle cen-
tered at the origin (the interior and the boundary points), is closed
in (R2 , U). BUT, there do not exist two closed sets E and F of
R with K = E × F . See Figure 4.3.
4.5 Theorem: Let (X,τ 1 ) and (Y, τ 2 ) be topological spaces, B 1

is a base for τ 1, and B 2 is a base for τ 2. Then B= {U × V : U ∈ B 1
and V ∈ B 2} is a base for the product topology on X × Y .
Proof: Exercise.
One can extend definition 4.4 to a finite number of products.

Exercise:
Define the product topology on X1 × X2 × X3 , and hence on X1 ×
... × Xn , n ∈ N.
4.6 Theorem: If A ⊆ X and B ⊆ Y , then

X×Y X Y
(1) A × B =A ×B ;
(2) intX×Y (A × B) = intX (A) × intY (B).
Proof: We prove part (1) and leave part (2) as an exercise.

Let (x, y) ∈ A × B be arbitrary. Need to show that (x, y) ∈ A × B.
That is, x ∈ A and y ∈ B. Let U be any open neighborhood of x and V
be any open neighborhood of y. Then U × V is an open neighborhood
of (x, y). Thus ∅ %= (U × V ) ∩ (A × B) = (U ∩ A) × (V ∩ B). Hence
U ∩ A %= ∅ and V ∩ B %= ∅. Thus x ∈ A and y ∈ B. Therfore,
(x, y) ∈ A × B. Thus A × B ⊆ A × B.
Now, let (x, y) ∈ A × B be arbitrary. This means that x ∈ A and
y ∈ B. We need to show that (x, y) ∈ A × B, so, let W be any open
neighborhood of (x, y) in X × Y , then there exists an open set U in
X and V in Y such that (x, y) ∈ U × V ⊆ W . So, U is an open
neighborhood of x in X and V is an open neighborhood of y in Y .
Thus U ∩ A %= ∅ and V ∩ B %= ∅. Thus (U × V ) ∩ (A × B) %= ∅. Thus
W ∩ (A × B) %= ∅. Thus (x, y) ∈ A × B. Therefor A × B ⊆ A × B.
Thus A × B = A × B.
Problems
1. Give an example of a topological space X, a subspace A of X and

an open subset B of A which is not open in X.
2. Give an example of a topological space X, a subspace A of X and

a closed subset B of A which is not closed in X.
3. Let I = [0, 1] and B = {1 − n1 : n ∈ N}. Find int(B), B and ∂B

in (I , LI ), (I , UI ), and (I , τ 0I ).
4. Consider A = { n1 : n ∈ N} ∪ {0} as a subspace of (R, U). In

(A, U A ), find the following:
1 1
(i) { 2n : n ∈ N} (ii) {0} (iii) int({ 2n+1 : n ∈ N}).
5. Consider A = {− n1 : n ∈ N} and B = [1, 3) ∪ (3, 5) as subsets of

R.
Without proof, find in the product space (R , L)×(R , T√2 ), where
L is the left ray topology and T√2 is the particular point topology
√
where the consideration point is 2, the following:
(a) The closure A × B ;

(b) The interior int(A × B).
6. Let (X , τ ) be a space and A ⊆ X.

(a) If A is open and B ⊆ A, then B ∈ τ A if and only if B ∈ τ .
Definition: If A is open in X, we say that A is an open
subspace of X.
(b) If A is closed and B ⊆ A, then B is closed in (A , τ A ) if and
only if B is closed in (X , τ ).
Definition: If A is closed in X, we say that A is a closed
subspace of X.
7. Let M be a subspace of a space (X , τ ) and let N be a subspace

of the space M. Prove that the two topologies defined on N,
as a subspace of M and as a subspace of X, are coincide. i.e.,
(τ M )N = τ N .
8. Let (A, τ A ) be a subspace of a topological space (X , τ ), and let

B be a subset of A. Prove that the closure of B in (A, τ A ) is
A X
A ∩ B, where B is the closure of B in X. i.e., B = A ∩ B .
9. Give an example of a topological space (X , τ ) and two subsets
B ⊆ A ⊆ X such that int(A ,τA) (B) %= int(X , τ )(B) ∩ A.
10. Let A be an open subset of a separable space (X, τ ). Prove that
(A, τ A ) is separable. Give an example to show that a subspace
of a separable space need not be separable.
Hint: Look in a suitable particular point space.
11. Let X be a topological space that contains a sequence of separable
subspaces whose union is dense in X. Prove that X is separable.
12. Let (X,τ 1) and (Y, τ 2) be topological spaces such that X1 ∩X2 =
∅. Define a topology τ 1⊕τ 2 on the set X1 ⊕ X2 = X1 ∪ X2 as
follows:
U ∈ τ 1⊕τ 2 if and only if U ∩ X1 ∈ τ 1 and U ∩ X2 ∈ τ 2 .
It is clear from the definition that an open neighborhood for an
element x ∈ X is of the form: either x ∈ U ∈ τ 1 if x ∈ X1 or
x ∈ U ∈ τ 2 if x ∈ X2 .
Prove that τ 1⊕τ 2 is indeed a topology on X1 ⊕ X2 , called the
free sum of (X,τ 1 ) and (Y, τ 2).
13. Let X and Y be infinite sets. Consider (X, D) and (Y, I). De-
scribe the product topology on X × Y .
14. Consider A = {− n1 : n ∈ N} and B = [2, 3) ∪ (3, 5) as subsets
of R. Without proof, find in the product space (R , S) × (R , U),
where S is the Sorgenfrey topology and U is the usual topology,
the following:
(a) The closure A × B ;
(b) The interior int(A × B).
15. Let X and Y be separable spaces. Prove that the product space
X × Y is separable.
16. Let X and Y be second countable spaces. Prove that the product
space X × Y is second countable.
17. Let P denote the irrationals. Prove that ( P , U P ) is separable.
.
5 Continuous Functions
Section 5.1: Continuity
Functions are one of the most powerful tool in all branches of Math-
ematics. The continuous functions play a major role in the study of
Calculus. The continuity of a function f : X −→ Y does not make
sense if X or Y is not a topological space. Note that when studied the
continuity of a real-valued function f : A ⊆ R−→ R in Calculus, the
usual topology on the domain and the codomain were considered even
though they were not mentioned explicitly. The continuity of f at a
real number a is defined as follows:
f : A ⊆ R−→ B ⊆ R is continuous at a ∈ A if and only if ∀ ' >

0, ∃ δ > 0 such that if 0 < |x − a| < δ, then |f(x) − f(a)| < '.
The statement 0 < |x−a| < δ means that x is in the open set (a−δ, a+δ)
such that x %= a. So, the terminology of open set was used. The state-
ment |f(x) − f(a)| < ' means that f(x) ∈ (f(a) − ', f(a) + ') where
the later is an open neighborhood of f(a).
The definition is, in fact, as follows:
(∀ ! > 0) Given any open neighborhood V of f (a)

(∃ δ > 0) There exists an open neighborhood U of a
(such that if 0 < |x − a| < δ) such that if x ∈ U \ {a}
(then |f (x) − f (a)| < !) then f (x) ∈ V which means f (U ) ⊆ V .
5.1 Definition: Let (X, τ ) and (Y, S) be topological spaces and

f : X −→ Y be a function. f is continuous at a point a ∈ X if and
only if given any open neighborhood V ⊆ Y of f(a), there exists an
open neighborhood U ⊆ X of a such that f(U) ⊆ V .
Figure 5.1.
f : X −→ Y is continuous on X if and only if it is continuous at

each point of X.
Discontinuity Criterion:
The function f : X −→ Y is discontinuous at a ∈ X if and only if
∃ V ⊆ Y open neighborhood containing f(a) such that ∀ open neigh-
borhood U ⊆ X of a we have that f(U) %⊆ V .
Examples:
1. Let f : (R, U)−→ (R, U) defined by f(x) = x, ∀ x ∈ R. f is
continuous on R because if a ∈ R is arbitrary and V ⊆ R is any
open set containing f(a), then let U = V , so a = f(a) ∈ V = U
and f(U) = U ⊆ V .
2. Let f : (R, CF )−→ (R, U) defined by f(x) = x, ∀ x ∈ R. Then
f is not continuous on R. In fact, f is not continuous at each
a ∈ R. For example, f is not continuous at 1 because 1 = f(1) ∈
(0, 2) ∈ U and if U is any open set in CF containing 1, then R\U
is finite. Hence f(U) = U %⊆ (0, 2).
3. Any function from a discrete space is continuous and any function
into an indiscrete space is continuous.
4. Let f : [0, 2] ∪ [3, 10] −→ (R, U), where [0, 2] ∪ [3, 10] has the
subspace topology of the usual topology, defined by

 2x ; if x ∈ [0, 2]
f(x) = 6 ; if x ∈ [3, 8]
10 ; if x ∈ (8, 10]

Continuous Functions 57
Then f is continuous at each point in its domain except at 8 for

let V = (5, 7) which is open containing f(8) = 6. Any open
neighborhood of 8 containing an open interval of the form (8 −
', 8 + ') where ' > 0. Thus containing 8 + 2$ and f(8 + 2$ ) = 10 %∈
(5, 7).
Exercise: Show that f is continuous at 2.
Proof: Let A = [0, 2] ∪ [3, 10]. Let V be any open set in (R, U)
that contains f(2) = (2).(2) = 4.
Now, what do we need? We need an open set U, (open in

A), such that 2 ∈ U and f(U) ⊆ V . By the definition of (R, U),
there exists an ' with 1 > ' > 0 such that (4− ', 4+ ') ⊆ V . Now,
since ' > 0, then 2$ > 0. Consider (2 − 2$ , 2 + 2$ ) which is an open
neighborhood of 2,(open in (R, U)). Let U = (2 − 2$ , 2 + 2$ ) ∩ A =
(2 − 2$ , 2]. Then U is open in A and 2 ∈ U. To show f(U) ⊆ V ,
let x ∈ U = (2 − 2$ , 2] be arbitrary. (i.e., f(x) ∈ f(U) and our
goal now is to show that f(x) ∈ V .) If x = 2, we are done as
f(2) = 4 ∈ V . So, assume that x %= 2, then
x ∈ (2 − 2$ , 2) =⇒ 2 − 2$ < x < 2
=⇒ 2 − 2$ < x < 2 + 2$ =⇒ − 2$ < x − 2 < 2$
=⇒ −' < 2x − 4 < ' =⇒ −' < f(x) − 4 < '
=⇒ 4 − ' < f(x) < 4 + ' =⇒ f(x) ∈ (4 − ', 4 + ') ⊆ V
Therefore f(U) ⊆ V , thus f is continuous at 2.
5.2 Theorem: Let X and Y be topological spaces and f : X −→ Y

be a function. The following are equivalent
1. f is continuous on X.
2. For each open set V ⊆ Y , f −1 (V ) is open in X.
3. For each basic open set W ⊆ Y , f −1 (W ) is open in X.
4. For each closed set M ⊆ Y , f −1 (M) is closed in X.

Figure 5.2.
Figure 5.3.
Proof:
(1)=⇒(2)
Assume f is continuous. Let V ⊆ Y be an arbitrary open set in Y . Need
to show that f −1 (V ) ⊆ X is open in X. If f −1 (V ) = ∅, we are done.
So, assume that f −1 (V ) %= ∅. We use theorem 2.10. Let x ∈ f −1 (V ) be
arbitrary, then f(x) ∈ V and V is open, so, by continuity of f, there
exists an open set Ux in X containing x such that f(Ux ) ⊆ V . Thus
Ux ⊆ f −1 (f(Ux )) ⊆ f −1 (V ), (see theorem 1.11). Therefore, f −1 (V ) is
open in X.
(2)=⇒(3)
Trivial.
(3)=⇒(4)
Assume that the inverse image of any basic open set is open. Let
M ⊆ Y be any closed set in Y . Need to show that f −1 (M) is closed
in X. If f −1 (M) = ∅ or f −1 (M) = X we are done. So, assume that
∅=% f −1 (M) %= X. We show that f −1 (M) is closed in X by showing
that X \ f −1 (M) is open in X. Let x ∈ X \ f −1 (M) be arbitrary, then
f(x) ∈ Y \ M.(for if f(x) ∈ M, then x ∈ f −1 (M) a contradiction.)
Since Y \ M is open in Y , there exists a basic open neighborhood V of
f(x) such that f(x) ∈ V ⊆ (Y \ M). Thus
x ∈ f −1 (V ) ⊆ f −1 (Y \ M) = f −1 (Y ) \ f −1 (M) = X \ f −1 (M)
By our hypothesis, f −1 (V ) is open in X. Also, x ∈ f −1 (V ). Therefore,

X \ f −1 (M) is open in X. Thus f −1 (M) is closed in X.
(4)=⇒(1)
Assume that the inverse image of any closed set in Y is closed in X
and we have to show that f is continuous on X. So, let a ∈ X be
arbitrary, and we have to show that f is continuous at a. Let V be an
arbitrary open neighborhood in Y of f(a). Need an open neighborhood
U of a in X such that f(U) ⊆ V . If V = Y , then X is an open
neighborhood of a with f(X) ⊆ Y = V . So, assume now V ⊂ Y . i.e.,
V is a proper subset of Y . Then f(a) %∈ Y \ V , hence a %∈ f −1 (Y \ V ).
By our assumption, we have that f −1 (Y \ V ) is closed in X. Let
U = X \ f −1 (Y \ V ), then U is open in X with a ∈ U. Now, U =
X \ f −1 (Y \ V ) = X \ (f −1 (Y ) \ f −1 (V )) = X \ (X \ f −1 (V )) = f −1 (V ).
Thus f(U) = f(f −1 (V )) ⊆ V . So, f is continuous at a and thus
continuous on X.
Exercise: Give others discontinuity criterion.
5.3 Theorem: If X, Y and Z are topological spaces and f : X −→

Y, g : Y −→ Z are continuous. Then g ◦ f : X −→ Z is continuous.
Proof: Exercise.
Hint: see definition 1.15.
5.4 Theorem: If f : X −→ Y is continuous and A ⊆ X, A %= ∅.

Then f|A : A −→ Y is continuous. Where A will have the subspace
topology.
Proof: Exercise.
Hint: (f|A )−1 (U) = f −1 (U) ∩ A.
5.5 Definition: Let X and Y be spaces and f : X −→ Y be a

continuous function.
f is open6 if and only if ∀ open U ⊆ X, f(U) is open in Y .
f is closed7 if and only if ∀ closed F ⊆ X, f(F ) is closed in Y .
Remark:
The term f is open (closed) does not refer to f as a subset of X × Y
being open (closed), but rather to the fact that the image under f of
each open (closed) set in X is open (closed) in Y .
Examples:
1. Let f : (R, U):−→ (I = [0, 1], U I ) defined by


 0 ; if x ≤ 0
f(x) = x ; if 0 < x < 1
1 ; if x ≥ 1

Then f is closed but not open. [f((2, 3)) = {1}]
2. Consider p1 : (R2, U)−→ (R, U), see definition 1.12. That is , p1 is

the first projection which is defined by p1 (x, y) = x, ∀ (x, y) ∈ R2.
p1 is open but not closed. For take A = {(x, x1 ) : x > 0} ⊆ R2
which is closed in R2 . But p1 (A) = (0, ∞) which is not closed in
R.
3. The function f : (R, S)−→ ({0, 1}, D) defined by

'
0 ; if x < 0
f(x) =
1 ; if x ≥ 0
is both closed and open.
Section 5.2: Homeomorphism and Homeomorphic Spaces
5.6 Definition:
A function f from a topological space X into a topological space Y
is called a homeomorphism if and only if
6
The notion of an open mapping was defined by Aronszajn in 1931. See [En-
gelking, 1977].
7
The notion of a closed mapping was introduced by Hurewicz in 1926. See
[Engelking, 1977].
(1) f is injective (1-1);

(2) f is surjective (onto);
(3) f is continuous; and
(4) f −1 as a function from Y into X is also continuous.
If f is a homeomorphism from X into Y , then X and Y are said to
be homeomorphic and denoted by X ∼ = Y.
Note that for any space X we have X ∼ = X. If X ∼
= Y , then Y ∼
= X.
∼ ∼ ∼
And if X = Y and Y = Z, then X = Z. Therefore the relation ”X
and Y are homeomorphic” is an equivalence relation.
Example 1: ( (0, 1] , U(0,1] ) ∼

= ( [1, ∞) , U[1,∞) ).
Proof: Define f : (0, 1] −→ [1, ∞) by f(x) = x1 , ∀ x ∈ (0, 1]. We prove
now that f is a homeomorphism
1. f is 1 − 1 :
1 1
Suppose that f(x1 ) = f(x2 ) where x1 , x2 ∈ (0, 1], then x1
= x2
,
hence x1 = x2 . Thus f is 1-1.
2. f is onto :
1
Let y ∈ [1, ∞) be arbitrary. Then y ≥ 1 > 0, hence 0 < y
≤ 1,
thus x = 1y ∈ (0, 1] and
1 1
f(x) = f( ) = 1 = y
y y
Thus f is onto.
3. f is continuous :
To show the continuity of f, we will use theorem 5.2, part 3. Let
W be any basic open set in [1, ∞). Note that a basic open set
in [1, ∞) is exactly a basic open set in (R, U) intersected with
[1, ∞). So it has exactly two forms: either (a, b) where 1 ≤ a < b
or [1, b) where 1 < b.
Case 1: W = (a, b) where 1 ≤ a < b. Since 1 ≤ a < b, then
0 < 1b < a1 ≤ 1, thus
1 1
f −1 (W ) = f −1 ((a, b)) = ( , )
b a
and ( 1b , a1 ) is open in (0, 1].
Case 2: W = [1, b) where 1 < b. We have f −1 (W ) = ( 1b , 1] which

is also open in (0, 1].
So, in all cases, the inverse image of a basic open set is open.
Thus f is continuous.
4. f −1 is continuous :
Exercise. Note that the definition of f −1 is
1
f −1 (x) = , ∀ x ∈ [1, ∞).
x
Example 2: If X and Y are spaces, then X × Y ∼

= Y × X.
Proof: Define f : X × Y −→ Y × X by
f((x, y)) = (y, x), ∀ (x, y) ∈ X × Y
We show that f is a homeomorphism. If f((x1, y1 )) = f(x2 , y2)), then

(y1, x1 ) = (y2 , x2). Thus (x1, y1 ) = (x2 , y2). Thus f is 1 − 1. If (y, x) ∈
Y × X, then (x, y) ∈ X × Y and f((x, y)) = (y, x). Thus f is onto.
Let V × U be any basic open set in Y × X, then f −1 (V × U) = U × V ,
where U × V is a basic open set in X × Y . Thus f is continuous. Since
f −1 : Y × X −→ X × Y is defined by f −1 ((y, x)) = (x, y). Then similar
argument will show that f −1 is also continuous. (Do it)
Thus f is a homeomorphism. Thus X × Y ∼ = Y × X.
5.7 Theorem: For a one-to-one function f of a topological space

X onto a topological space Y , the following conditions are equivalent
1. f is a homeomorphism.
2. f is closed.
3. f is open.
4. f(A) is closed in Y ⇐⇒ A is closed in X.
5. f −1 (B) is closed in X ⇐⇒ B is closed in Y .
6. f(A) is open in Y ⇐⇒ A is open in X.
7. f −1 (B) is open in X ⇐⇒ B is open in Y

Proof: Exercise.
Note that since f is 1 − 1 and onto, then we do have that A =

f −1 (f(A)) and (f −1 )−1 (A) = f(A). Also, from the above theorem we
conclude an other way to define a homeomorphism from X to Y , is to
say it is a bijection f : X −→ Y such that f(U) is open ⇐⇒ U is open.
Thus, a homeomorphism is not only a bijection between X and Y but
also between the topologies on X and Y . Therefore any property of X
that can be expressed exclusively in terms of the topology on X yields,
via the homeomorphism, the same property for Y .
5.8 Definition: A topological property is a property that if pos-

sessed by a topological space X, is also possessed by every homeomor-
phic image of X. That is, P is a topological property ⇐⇒ if X has
P and f : X −→ Y is a homeomorphism, then Y has P .
5.9 Theorem: Separability is a topological property.

Proof:
Let X be a separable space and let f : X −→ Y be a homeomor-
phism. Need to show that Y is separable. Pick D = {d1 , d2 , d3 , ...} a
countable dense subset of X. We show now that f(D) = {f(d1 ), f(d2 ),
f(d3 ), ...} is a countable dense subset of Y . It is clear that f(D) is
countable. Let V ⊆ Y be any nonempty open subset. Then f −1 (V ) is
a nonempty open subset of X. Thus f −1 (V )∩D %= ∅. Hence there exists
x ∈ f −1 (V ) ∩ D. Thus f(x) ∈ V ∩ f(D), which means V ∩ f(D) %= ∅.
Therefore, Y is separable.
Problems
1. Let Z be the integers as a subspace of (R , U). Let f : Z −→ Z
be given by f(x) = 2x for each x ∈ Z. Prove or disprove that f
is continuous.
2. Give an example of a function f : X −→ Y and a subset A ⊂ X

such that f|A is continuous but f is not continuous at any point
of A.
3. Prove or disprove:
(a) All constant functions are continuous.
(b) If the domain of the function has the discrete topology, then
the function is continuous.
(c) All injective (1-1) functions are continuous.
4. Prove that f : (R, U) −→ (R, U) defined by f(x) = x2 is contin-

uous.
5. Let (X, τ ) be a topological space. Prove that f : X −→ (R, U)

is continuous if and only if for each real number a ∈ R, both of
the sets {x ∈ X : f(x) > a} and {x ∈ X : f(x) < a} are open in
X.
6. Let X = {a, b, c}, τ = {∅, X, {a}, {b, c}} and Y = {x, y, z}. Give
a topology for Y which makes X and Y homeomorphic. Give
another topology for Y which makes X and Y non-homeomorphic.
7. Give an example of a bijection (1-1 and onto) function f : X −→
Y such that f is continuous, but f −1 : Y −→ X is not continuous.
8. Prove that f : X −→ Y is a homeomorphism if and only if

f −1 : Y −→ X is a homeomorphism.
9. Prove that (R , L) ∼
= (R , R).
10. Prove that ( (1, 5] , U(1,5] ) ∼
= ( (3, 11] , U(3,11] ).
11. Prove that ( (1, 5] , L(1,5] ) ∼
= ( (3, 11] , L(3,11] ).
12. Prove that ( (0, 12] , U(0,12] ) ∼
= ( (0, 3] , U(0,3] ).
13. Prove that ( (0, 12] , R(0,12] ) ∼
= ( (0, 3] , R(0,3] ).
14. Prove that ( (0, 12] , S(0,12] ) ∼

= ( (0, 3] , S(0,3] ).
15. Prove that [1, 3) ∼
= [7, 9), as subspaces of R with the usual topol-
ogy U.
16. Prove or disprove that [1, 3) ∼
= [7, 9), as subspaces of R with the
particular point topology τ √2.
17. Prove that ( R , U ) ∼
= ( ( 0 , ∞ ) , U( 0 , ∞ ) ).
18. Let X and Y be spaces. Let x ∈ X and y ∈ Y . Prove that
{x} × Y ∼
= Y and X × {y} ∼
= X.
19. Let X and Z be topological spaces. Prove that X × Z ∼
= Z × X.
20. Prove that the Sorgenfrey line is homeomorphic to R with the
right interval topology.
21. Prove that a continuous function f : X −→ Y is open if there is
some base B of X with f(B) is open in Y for each B ∈ B.
22. Prove that any projection function is open.
23. Let f : X −→ Y and g : Y −→ Z be continuous functions. Prove
(a) If g ◦ f is open (closed) and if f is onto, then g is open
(closed).
(b) If g◦f is open (closed) and if g is 1-1, then f is open (closed).
24. Define τ 8 ⊂ P(Q) as follows: For a subset U of Q
U∈ τ 8 if and only if U = ∅ or 8 ∈ U
(a) Prove that τ 8 is a topology on Q
(b) Prove or disprove that (R , τ √2 ) ∼
= (Q , τ 8 ).
25. Prove or disprove that (R , τ √2 ) ∼

= (R , τ √3 ).
26. Prove or disprove that ( ( 0 , ∞ ) , L( 0 , ∞ )) ∼
= (( −∞ , 0 ) , R( −∞ , 0 ) ).
27. Prove or disprove that ( ( 0 , ∞ ) , S( 0 , ∞ ) ) ∼

= (( −∞ , 0 ) , S( −∞ , 0 ) ).
28. Prove or disprove that ( R , CF ) ∼

= ( R , CC ).
.
6 Separation Axioms
Separation axioms give a way of classifying topological spaces according

to the topological distinguishability of points and subsets of the space.
6.1 Definition: Let (X, τ ) be a topological space. Then

1. The space (X, τ ) is called a T0-space8 if and only if for each pair
of distinct points x, y ∈ X, there is either an open set containing
x but not y or an open set containing y but not x.
Figure 6.1.
2. The space (X, τ ) is called a T1-space9 if and only if for each pair
of distinct points x, y ∈ X, there exist open sets U and V such
that x ∈ U, y ∈
/ U and y ∈ V, x ∈ / V.
Figure 6.2.
3. The space (X, τ ) is called a T2-space10 or Hausdorff space if and
only if for each pair of distinct points x, y ∈ X, there exist open
sets U and V such that x ∈ U, y ∈ V and U ∩ V = ∅.
Figure 6.3.
8
T0 -spaces were introduced by Kolmogoroff in 1935. See [Engelking, 1977].
9
T1 -spaces were introduced by Riesz in 1907. See [Engelking, 1977].
10
Hausdorff introduced the T2 -spaces in 1914. See [Engelking, 1977].
4. The space (X, τ ) is called a regular space11 if and only if for each
closed subset F ⊂ X and each point x ∈ / F , there exist open sets
U and V such that x ∈ U, F ⊆ V and U ∩ V = ∅.
Figure 6.4.
5. The space (X, τ ) is called a T3 -space if and only if it is T1 and

regular.
6. The space (X, τ ) is called a normal space12 if and only if for each
pair of closed disjoint subsets F1 and F2 of X, there exist open
sets U and V such that F1 ⊆ U, F2 ⊆ V and U ∩ V = ∅.
Figure 6.5.
7. The space (X, τ ) is called a T4 -space if and only if it is T1 and

normal.
WARNING:
When reading any book in Topology, you must first check the au-
thor’s definition of the separation axioms because some author defines
a T3-space as our definition of regularity and say that a regular space
space is T1 + T3, ... etc.
In this book, we will follow our definitions. Also, we will say that a
space is Ti instead of a Ti-space
It is clear from the definition that

11
Regular spaces were introduced by Vietoris in 1921. See [Engelking, 1977].
12
The class of normal spaces was defined by Tietze in 1923 and by Alexandroff
and Urysohn in 1924. See [Engelking, 1977].
Separation Axioms 69
6.2 Theorem: T2 =⇒ T1 =⇒ T0 .
The converse of Theorem 6.2 is not always true.
6.3 Example: (R,L), the left ray topology on R (see Example 2.6)
is T0. For if x < y, then (−∞, x+y2
) is an open set containing x but not
y. But, (R,L) is not T1, for if x < y, there is no way of getting an open
set containing y but not x.
Figure 6.6.
6.4 Example: Let X be any infinite set. Then (X, CF ), the cofinite
topology (see Example 2.13) is T1 but not T2. It is T1 because if x, y ∈ X
are any two distinct points, then U = X \ {y}, V = X \ {x} are
open because the complement of each is finite and x ∈ U, y ∈ / U and
y ∈ V, x ∈
/ V . (X, CF ) is not T2 because any two distinct points cannot
be separated. For if x, y ∈ X with x %= y and if we assume that there
were two open sets U and V such that x ∈ U, y ∈ V and U ∩ V = ∅, we
would get the following: U %= ∅ as x ∈ U, so X \ U is finite. Similarly,
X \ V is finite. Now, X = X \ ∅ = X \ (U ∩ V ) = (X \ U) ∪ (X \ V ).
Thus X is finite, which is a contradiction.
6.5 Example: Let X = {a, b, c} and let τ = {∅, X, {a}, {b, c}}.
Then (X, τ ) is regular but not Hausdorff. It is not T2 because b %= c
cannot be separated by disjoint open sets. It is regular because the
only closed sets are ∅, X, {a}, and {b, c}.(Observe that all of them are
also open.) Take any closed set and any point not in it, we can (after
considering all cases) separate them by disjoint open sets.
Note that (X, τ ) is not T1, thus it is not T3.
6.6 Example: Let X be any set having more than one point.
Consider the indiscrete topology I on X. The only disjoint closed sets
are ∅ and X, which are open. Thus it is normal. But X is not T1,
hence X is not T4.
6.7 Theorem: X is T1 if and only if any singleton is closed. (i.e.
∀ x ∈ X, the subset {x} is closed in X.)
Proof:
(=⇒) Suppose X is T1 . Let x ∈ X be arbitrary. Need to show that {x}
is closed in X. If {x} = X, we are done. So, assume that {x} %= X. We
show that {x} is closed by showing that X \ {x} is open in X. We use
Theorem 2.10 to do this. Let y ∈ X \ {x} be arbitrary. Then x %= y.
By T1, there exist two open sets U and V such that x ∈ U, y ∈ / U and
y ∈ V, x ∈ / V . Now, y ∈ V ⊆ X \ {x}. Hence X \ {x} is open. Thus
{x} is closed.
(⇐=) Suppose that {x} is closed in X for any x ∈ X. Need to show
that X is T1. So, let x %= y, x, y ∈ X. (if X contains only one point, we
are done). Let U = X \ {x} and V = X \ {y}. Then U and V satisfy
the condition of T1.
6.8 Corollary: In a T1 space, all finite subsets are closed.
6.9 Corollary: T4 =⇒ T3 =⇒ T2 =⇒ T1 =⇒ T0.

Separation Axioms 71
Problems
1. Verify that the discrete space satisfies all separation axioms.
2. In this problem, we will introduce a topological space which does
not satisfy any separation axiom.
Let X be an infinite set. Consider (X , CF ) and ({0, 1} , I), X
with the cofinite topology and {0, 1} with the indiscrete topology.
Prove that the product space X × {0, 1} does not satisfy any
separation axiom.
3. Prove that any subspace of a Ti space is a Ti space for i ∈
{0, 1, 2, 3}.
4. Prove that closed subspaces of a normal space are normal.
5. Prove that X × Y is a Ti space if and only if both X and Y are
Ti space, where i ∈ {0, 1, 2}
6. Prove that Ti is a topological property for i ∈ {0, 1, 2, 3, 4}.
7. Verify Ti, i ∈ {0, 1, 2, 3, 4}, of all spaces we have studied so far.
o8. Let ( X , τ 1 ) be a Hausdorff space. If τ 2 is a topology on X such

that τ 1 ⊆ τ 2, then prove that ( X , τ 2 ) is also Hausdorff.
9. Let f : X −→ Y be a continuous and injective function where Y
is Hausdorff. Prove that X is Hausdorff.
10. Let (X, τ ) be a Hausdorff topological space. Prove that the
convergency set of any convergent sequence is a singleton.
11. Let {x1 , x2, ..., xn} be a finite subset of a Hausdorff space. Prove
that there exist n open pairwise disjoint sets U1 , U2 , ..., Un such
that xi ∈ Ui , i ∈ {1, 2, ..., n}.
(pairwise disjoint means that Ui ∩Uj = ∅ whenever i, j ∈ {1, 2, ..., n}
with i %= j.)
12. Let X be a finite set. Prove that the only Hausdorff topology on
X is the discrete topology.
13. Define τ ⊂ P(R) as follows:
For a subset U ⊆ R, U ∈ τ if and only if U = ∅ or U = R or
(2 ∈ U and 3 %∈ U).
(a) Prove that τ is a topology on R.

(b) Prove that (R , τ ) is not T1.
14. Let (X , τ ) be a topological space. State the definition of the
following:
(1) (X , τ ) is not T1 . (2) (X , τ ) is not T2 .
(3) (X , τ ) is not regular. (4) (X , τ ) is not T4 .
7 Metric Spaces
Metric space was defined by F réchet on 1906, while topological space

was defined on 1920. We will see that any metric space is a topological
space, but the converse is not true in general. The class of metric spaces
was the first class of abstract spaces to which several notions and results
discovered in the infancy of general topology in the study of subsets of
the real line and of Euclidean spaces, were successfully generalized.
The class of metric spaces is sufficiently large to include many objects
studied in various branches of mathematics and thus describe them in
geometric language, and, at the same time, the spaces in this class seem
to be sufficiently simple to permit the use of geometric intuition.
7.1 Definition: A metric on a nonempty set X is a function

d : X × X −→ R that satisfies the following four conditions: (to make
our notations simple, we will write d(x, y) instead of d((x, y)).
1. d(x, y) ≥ 0, ∀ x, y ∈ X;
2. d(x, y) = 0 if and only if x = y;
3. d(x, y) = d(y, x), ∀ x, y ∈ X; and
4. d(x, z) ≤ d(x, y) + d(y, z), ∀ x, y, z ∈ X.
If d is a metric on X, then the order pairs (X, d) is called a metric
space, and if x, y ∈ X, then d(x, y) is called the distance from x to y.
7.2 Examples:
1. d : R × R −→ R defined by d(x, y) = |x − y|, ∀ x, y ∈ R is a
metric on R called the usual metric on R.
2. d : R2 × R2 −→ R defined by
+
d((x1 , y1), (x2, y2 )) = (x1 − x2 )2 + (y1 − y2)2 , ∀ (x1, y1), (x2 , y2) ∈ R2
is a metric on R2 , called the usual metric on R2 .
7.3 Example: Let X be any nonempty set. Define d : X ×X −→ R

by '
0 if x = y
d(x, y) =
1 if x %= y
Exercise: Verify that d is a metric on X. This metric is called the

discrete metric.
7.4 Theorem: Let (X, d) be a metric space. Let ∅ %= A ⊆ X. Then

d|A×A , the restriction of d to A × A (see Definition 1.14) is a metric on
A.
Proof: Exercise.
7.5 Definition: Let (X, d) be a metric space. Let ' be a positive

real number and let x ∈ X. The set
Bd (x; ') = {y ∈ X : d(x, y) < '}
is called the open ball centered at x of radius '. If d is clear, we write

B(x; ') instead of Bd (x; ').
7.6 Examples:
1. If d is the usual metric on R and ' > 0, then for each x ∈ R

B(x; ') = (x − ', x + ').
2
2. If d is the usual metric on R+ and ' > 0, then for each (x, y) ∈ R2
B((x, y); ') = {(u, v) ∈ R2 : (u − x)2 + (v − y)2 < '}, which is
the set of all “interior” points of the circle of radius ' centered at
the point (x, y).
Figure 7.1.
Metric Spaces 75
3. If d is the discrete metric on a set X, then for any x ∈ X

B(x; 1) = {y ∈ X : d(x, y) < 1} = {x}. If ' > 1, then B(x; ') =
X.
7.7 Definition: A subset U of a metric space (X, d) is said to be

open if and only if for each x ∈ U, there is an open ball Bd (x; ') such
that Bd (x; ') ⊆ U.
Figure 7.2.
7.8 Theorem: The open subsets of a metric space (X, d) have the
following properties:
1. X and ∅ are open;
2. The intersection of any two open sets is open; and
3. The union of any collection of open sets is open.
Proof:
1. For each x ∈ X and for any ' > 0, we have B(x; ') ⊆ X. Thus X
is open. ∅ is open because there is no x ∈ ∅.
$
2. Let U and V$ be any two open sets. Need to show that$U V is
open. If U$ V = ∅, we are done. So, assume that U V %= ∅.
Let x ∈ U V be arbitrary. Then x ∈ U and x ∈ V . Since U is
open and x ∈ U, then there exists an ' > 0 such that B(x; ') ⊆ U.
Since V is open and x ∈ V , then there exists a δ > 0 such that
B(x; δ) ⊆ V . Let γ = min{', δ}, then B(x; γ)$⊆ B(x; ') ⊆ U
and$ B(x; γ) ⊆ B(x; δ) ⊆ V . Thus B(x; γ) ⊆ U V . Therefore,
U V is open.
3. Let {U
!α : α ∈ Λ} be any!collection of open sets. Need to show
that !α∈Λ Uα is open. If α∈Λ !Uα = ∅, we are done. So, assume
now α∈Λ Uα %= ∅. Let x ∈ α∈Λ Uα be arbitrary. Then there
exists β ∈ Λ such that x ∈ Uβ . Since Uβ is open, ! then there
exists
! an ' > 0 such that B(x; ') ⊆ Uβ . But Uβ ⊆ α∈Λ Uα . Thus
α∈Λ Uα is open.
Now, let (X, d) be any metric space. d will generate a topology on

X as follows: Let τ d be the set of all open sets in (X, d), (open in
the sense of Definition 7.7). Then Theorem 7.8 assures that τ d is a
topology on X. This topology τ d , is called the topology induced by the
metric d.
So, we have the following result:
7.9 Corollary: Any metric space is a topological space.
7.10 Definition: Two topologies τ and S on a same set X are

said to be coincide if and only if τ ⊆ S and S ⊆ τ , i.e., U ∈ τ ⇐⇒
U ∈ S.
7.11 Definition: A topological space (X, τ ) is called metrizable

if and only if there exists a metric d on X such that the topology τ d
induced by d coincide with the original topology τ . i.e., τ =τ d .
7.12 Examples:
1. (R, U), the set of real numbers with the usual topology is metriz-
able. (Check it.)
2. R2 = R × R, where each R is considered with the usual topology,

is metrizable. (Check it.)
3. Any discrete space is metrizable. (Check it.)
Exercise: Prove that in a metric space, any open ball is open.

Conclude that the set of all open balls forms a base for a metrizable
space.
7.13 Theorem: Any mertizable space is Hausdorff.

Proof: Let (X, τ ) be a metrizable space. Then there exists a metric
d on X such that τ d = τ . We show that (X, τ d ) is Hausdorff. Let
Metric Spaces 77
x and y be any distinct points in X. By the definition of a metric,

we have d(x, y) > 0. i.e., the distance between x and y is positive.
Let ' = d(x, y), then 4$ > 0. Let U = Bd (x, 4$ ) and V = Bd (y, 4$ ),
then U is open and x ∈ U and V is open and y ∈ V . To see that
U ∩ V = ∅, suppose that U ∩ V %= ∅. We will get a contradiction as
follows: Pick z ∈ U ∩ V = Bd (x, 4$ ) ∩ Bd (y, 4$ ), then z ∈ Bd (x, 4$ ) and
z ∈ Bd (y, 4$ ). Thus d(x, z) < 4$ and d(y, z) = d(z, y) < 4$ . Now, we
have ' = d(x, y) ≤ d(x, z) + d(z, y) < 4$ + 4$ = 2$ , i.e., ' < 2$ , which is
a contradiction because ' > 0. Thus U ∩ V = ∅. Thus any metrizable
space is Hausdorff.
7.14 Corollary: Any metrizable space is T1 and hence T0.
It is natural to ask about regularity and normality of a metrizable

space. We will see that any metrizable space has all of the separation
axioms.
7.15 Theorem: Any metrizable space is T4.

Proof: We have to show the normality. So, let A and B be any closed
nonempty disjoint subsets of a metric space (X , d). For each a ∈ A,
there exists an ra > 0 such that B(a; ra) ⊆ X \ B and for ! each b1 ∈ B
there exists an rb > 0 such that B(b; r b ) ⊆ X\A. Let U = {B(a; 2 ra ) :
1
!
a ∈ A} and V = {B(b; 2 rb ) : b ∈ B}. Then U and V are open with
A ⊆ U and B ⊆ V . Suppose that there exist an x ∈ U ∩ V . Then
there would exist a ∈ A and b ∈ B such that x ∈ B(a; 21 ra) ∩ B(b; 12 rb ).
Then, by the triangle inequality, we would have
1 1
d(a, b) ≤ d(a, x) + d(x, b) < ra + rb ≤ max{ra , rb }.
2 2
So either a ∈ B(b; rb ) or b ∈ B(a; ra), which is a contradiction. Thus U
and V are disjoint.
7.16 Definition: A sequence (an )n∈N in a metric space (X , d) is

called a Cauchy sequence provided that for each ' > 0 there is an N ∈ N
such that if i and j are greater than or equal N then d(xi , xj ) < '.
7.17 Theorem: Let (X , d) be a metric space and let (an )n∈N be

a convergent sequence in X. Then (an )n∈N is a Cauchy sequence.
Proof: Let (X , d) be a metric space and let (an )n∈N be any conver-
gent sequence in X. Since (X , d) is Hausdorff, then, by problem 10 of
Chapter 6, there is a unique a ∈ X such that an −→ a. Let ' > 0. Since
an −→ a there is a natural number N ∈ N such that ak ∈ Bd (a; 2$ ) for

all k ≥ N. Suppose that i, j ≥ N, then d(ai , a) < 2$ and d(aj , a) < 2$ .
Therefore d(ai , a) < d(a, aj ) + d(ai , aj ) < 2$ + 2$ = '.
The converse of the above theorem is not always true. That is, if
(an )n∈N is a Cauchy sequence in X, then (an )n∈N may not converge. For
an example the sequence ( 21n )n∈N in X = R \ {0} is a Cauchy sequence
which is not a convergent sequence in X. The metric space R \ {0} with
the usual metric is not a complete metric space as the next definition
says.
7.18 Definition: A metric space (X , d) is called complete provided

that every Cauchy sequence in X converges to a point of X.
In general, Cauchy sequence may not have a convergent subse-

quence. But if it happens in a metric space, then it should be a complete
metric spaces.
7.19 Theorem: Let (X , d) be a metric space such that every

Cauchy sequence in X has a convergent subsequence. Then (X , d) is
complete.
Proof: Let (an ) be a Cauchy sequence and let (ank ) be a subsequence
of (an ) that converges to a ∈ X. We will show that an −→ a. Let
' > 0. There is N ∈ N such that for all i, j ≥ N we have d(ai , aj ) < 2$ .
Since ank −→ a, then there is a natural number np such that np > N
and d(anp , a) < 2$ . Then for n > N we have d(an , a) ≤ d(an , anp ) +
d(anp , a) < '. Therefore an −→ a and (X , d) is complete.
The most important result in completeness, which we omit its proof

here, is that R with the usual metric is complete and a subspace of
complete metric space is complete if and only if it is closed. You will
find a wild study of this subject in more advanced books.
Metric Spaces 79
Problems
1. Let (X, d) be a metric space. Define e : X × X −→ R by
e(x, y) = min{1, d(x, y)}, ∀ x, y ∈ X
Prove that e is a metric on X
2. Let d1 , d2 : R × R −→ R be defined by
d1 (x, y) = x2 − y 2, ∀ x, y ∈ R
d2 (x, y) = |x2 − y 2|, ∀ x, y ∈ R

Show that neither d1 nor d2 are metric on R.
3. Let X be the set of all real-valued continuous functions defined

on the unit closed interval I = [0, 1]. Define d : X × X −→ R by
, 1
d(f(x), g(x)) = |f(x) − g(x)|dx. ∀ f(x), g(x) ∈ X
0
Prove that d is a metric on X.
4. Let e1 , e2 : R2 × R2 −→ R be defined by
e1((x, y), (z, w)) = |x − z| + |y − w|, ∀ (x, y), (z, w) ∈ R2
e2((x, y), (z, w)) = max{|x − z|, |y − w|}, ∀ (x, y), (z, w) ∈ R2
Prove that e1 and e2 are metric on R2 .
5. Let d denote the usual metric on R2 . Let e1 and e2 be as in the pre-

vious exercise. Describe graphically the sets Bd ((0, 0); 2), Be1 ((0, 0); 2),
and Be2 ((0, 0); 2).
6. Define d : R × R −→ R by
|x − y|
d(x, y) = ; ∀ x, y ∈ R.
1 + |x − y|
Prove that d is a metric on R.
7. Let P denote the set of irrationals. Define τ π ⊂ P(P) as follows:

U ∈ τ π if and only if U = ∅ or π ∈ U.
(a) Prove that τ π is a topology on P.

(b) Prove that ( P , τ π ) is not normal.
(c) Prove that ( P , τ π ) is separable.
(d) Is ( P , τ π ) metrizable? Justify your answer.
8. Show that R with the topologies L, CF , T 0 are not metrizable.

8 Compact Spaces
8.1 Definition: Let {Aα : α ∈ Λ} be a family of subsets of the

space X and B ⊆ X. We say that the family {Aα : α ∈ Λ} is a
cover of !B (or that the family {Aα : α ∈ Λ} covers B) if and only
if B ⊆ α∈Λ Aα . If Λ is finite and {Aα : α ∈ Λ} covers B, then
{Aα : α ∈ Λ} is called a finite cover of B. If each Aα , α ∈ Λ is open
(closed) in X and {Aα : α ∈ Λ} covers B, then {Aα : α ∈ Λ} is called
an open cover (closed cover) of B.
8.2 Example: Consider (R, U). Let B = (0, 1] and for each n ∈ N
let An = ( n1 , 2). Then An is an open set in (R, U) for each n ∈ N. Also,
" " 1
B = (0, 1] ⊆ An = ( , 2)
n∈N n∈N
n
Thus, {An : n ∈ N} is an open cover of B.
Now, for each n ∈ N, !

let Un = (−n, n). Then Un is open in (R, U)
for each n ∈ N, and R ⊆ n∈N Un . Thus {Un : n ∈ N} is an open cover
of R.
8.3 Definition: Let {Aα : α ∈ Λ} be a cover of a subset B of a

space X. Let Ω ⊆ Λ. Then the family {Aα : α ∈ Ω} is called a subcover
of the cover {Aα : α ∈ Λ} for B if and only if {Aα : α ∈ Ω} is a cover
of B.
8.4 Example: Let E= {n : n is even} ⊂ N. Then {An : n ∈ E} is

an open subcover of {An : n ∈ N} for B = (0, 1] of Example 8.2.
8.5 Definition: A topological space X is called compact13 if and

only if any open cover for X has a finite subcover for X. A subset B
of a space X is compact if and only if B is a compact topological space
with the subspace topology.
8.6 Example: (R, U) is not compact because {Un = (−n, n) : n ∈

N} is an open cover for R that does not have a finite subcover. For
13
The genesis of the notion of compactness is connected with the Borel theorem
(proved in 1894) stating that every countable open cover of a closed bounded interval
has a finite subcover, and with the Lebesgue observation that the same holds for
every open cover of a closed bounded interval. The concept of a compact space was
introduced by Vietoris in 1921. See [Engelking, 1977].
if
! D is any finite subset of N, then let m = max D. Then we have
n∈D Un = (−m, m) %⊇ R.
B = (0, 1] as a subset of (R, U) is not compact because {Vn : n ∈ N},

where Vn = ( n1 , 1] = ( n1 , 2) ∩ (0, 1], ∀ n ∈ N is an open (in the subspace
topology) cover for B which does not have a finite subcover. (check it.)
8.7 Example: Let X be any infinite set. Then (X, CF ), the cofinite
topology, is compact.
Proof: Let {Uα : α ∈ Λ} be an arbitrary open cover of X. Recall that
U is open in the cofinite topology
! if and only if U = ∅ or X \ U is
finite. Note that since X ⊆ α∈Λ Uα , we may assume, without loss of
generality, that Uα %= ∅, ∀ α ∈ Λ. To show that (X, CF ) is compact,
we must find a finite subcover of {Uα : α ∈ Λ}.
Take any α0 ∈ Λ. Then Uα0 covers all X except possibly a finite
set of points {x1 , x2, ..., xn}.(as X \ Uα0 is finite) Since {Uα : α ∈ Λ}
is a cover of X, then for each i ∈ {1, 2, ..., n}, there exists an αi ∈ Λ
such that xi ∈ Uαi . Thus {Uα0 , Uα1 , Uα2 , ..., Uαn } is a finite subcover of
{Uα : α ∈ Λ} which covers X.
8.8 Theorem: A subset A of a topological space X is compact if

! if for any family {Us : s ∈ S} of open subsets !
and only of X such that
A ⊆ s∈S Us , there exist {s1, s2 , ..., sn} ⊆ S with A ⊆ ni=1 Usi . That
is, any open cover of subsets of X has a finite subcover.
Proof: Exercise. (Recall that U ⊆ A is open in A if and only if there
exist V ⊆ X which is open in X such that U = V ∩ A.)
8.9 Theorem: If X is compact and B ⊆ X is closed, then B is

compact. That is, closed subset of compact space is compact.
Proof: Let {Uα : α ∈ Λ} be any open cover of B, where Uα ⊆ X is
open !in X, for each α ∈ Λ. Need finite α1 , α2 , ..., αn ∈ Λ such that
B ⊆ ni=1 Uαi .
Since B is closed in X, then X \ B is open in X. Thus {Uα :
α ∈ Λ} ∪ {X \ B} is an open cover for X. Since !n X is compact, then
there!exist α1, α2 , ..., αn ∈ Λ such that X ⊆ ( i=1 Uαi ) ∪ X \ B. Thus
n
B ⊆ i=1 Uαi . Thus B is compact.
Remarks:
1. [0, ∞) is a closed subset of (R, U). But it is not compact. Observe

that (R, U) is not compact.
Compact Spaces 83
2. [0, ∞) is not closed in (R, CF ). But it is compact with respect

to the cofinite topology.
8.10 Theorem: Let X be a Hausdorff space and A ⊆ X be a

compact subspace. Then for each x %∈ A there exist open subsets U
and V of X such that A ⊆ U, x ∈ V , and U ∩ V = ∅.
Proof: By T2 we have ∀ a ∈ A, ∃ open disjoint Ua and Va such that
a ∈ Ua and x ∈ Va and Ua ∩ Va = ∅.! (we cannot take U = ∪a∈A Ua
and V = ∩a∈A Va , WHY?) Since A ⊆ a∈A Ua and A ! is compact, then
..., an ∈ A such that A ⊆ ni=1 Uai . Now, let
there!exist finite a1, a2 , $
U = ni=1 Uai and V = ni=1 Vai . Then A ⊆ U and x ∈ V where both
U and V are open in X. (V is open because it is a finite intersection
of open sets.) To show that !n U ∩ V = $n∅, we assume that U ∩ V %=
∅. Pick y ∈ U ∩ V = ( i=1 Uai ) ∩ ( i=1 Vai ), then y ∈ Vai ∀ i and
∃ j ∈ {1, 2, ..., n} such that y ∈ Uaj . Thus y ∈ Uaj and y ∈ Vaj , thus
Uaj ∩ Vaj %= ∅, which is a contradiction. Thus U ∩ V = ∅.
8.11 Corollary: A compact subset of a T2-space is closed.

Proof: Let A be a compact subset of a Hausdorff space X. Need to
show that A is closed in X. If A = X, we are done. Assume that
A %= X. Pick an arbitrary x ∈ X \ A. So, x %∈ A. By Theorem 8.10,
there exist two open sets U and V such that A ⊆ U, x ∈ V , and
U ∩ V = ∅. Thus V ∩ A = ∅. therefore, x ∈ V ⊆ X \ A. Thus X \ A is
open. Hence A is closed.
8.12 Corollary: Let X be a T2 compact topological space and

A ⊆ X. Then A is compact if and only if A is closed.
8.13 Theorem: Every T2 compact space is T4.

Proof: Let X be any T2 compact space. To show that X is T4, we have
to show that X is T1 and X is normal. Since X is T2, then it is T1.
To show normality of X, let A and B be any closed disjoint nonempty
subsets of X. Need two open sets U and V such that A ⊆ U, B ⊆ V ,
and U ∩ V = ∅.
Since X is T2 compact and both A and B are closed, then, by
Corollary 8.12, both A and B are compact. Now, by Theorem 8.10, for
each a ∈ A there exist open sets!Ua and Va such that a ∈ Ua , B ⊆ Va ,
and Ua ∩ Va = ∅. Thus A ⊆ a∈A ! Ua . By compactness!of A, there
exist a$1 , ..., an ∈ A such that A ⊆ ni=1 Uai . Let U = ni=1 Uai and
n
V = i=1 Vai . Then A ⊆ U, B ⊆ V , and both U and V are open
(WHY?). Also, U ∩ V = ∅ (Check it). Thus X is normal and T1.

Hence X is T4.
8.14 Theorem:
If there exists a continuous function f : X −→ Y of a compact
space X onto a space Y , then Y is compact.
Proof: Let {Uα : α ∈ Λ} be any open cover of Y . Since f ! is continuous,
−1
then f (Uα ) is open in X for ! each α ∈ Λ. ! Since Y ⊆ α∈Λ Uα then
we have X = f −1 (Y ) ⊆ f −1 ( α∈Λ Uα ) = α∈Λ f −1 (Uα ), which means
that {f −1 (Uα ) : α ∈ Λ} is an open cover of X. So, by the compactness
of X, there exist α1 , α2, ..., αn ∈ Λ such that
f −1 (Uα1 ) ∪ f −1 (Uα2 ) ∪ ... ∪ f −1 (Uαn ) = X
=⇒ f[f −1 (Uα1 ) ∪ f −1 (Uα2 ) ∪ ... ∪ f −1 (Uαn )] = f(X)
=⇒ f(f −1 (Uα1 )) ∪ f(f −1 (Uα2 )) ∪ ... ∪ f(f −1 (Uαn )) = Y
=⇒ Uα1 ∪ Uα2 ∪ ... ∪ Uαn = Y (Note that f(X) = Y and f(f −1 (Uαi )) =
Uαi , ∀ i ∈ {1, 2, ..., n} because f is an onto function.) Thus Y is
compact.
8.15 Corollary: Compactness is a topological property.
8.16 Corollary: Continuous image of a compact space is compact.

That is, If X is compact and f : X −→ Y is continuous, then f(X) is
compact as a subspace of Y .
Compact Spaces 85
Problems
1. Consider (R,L), the real numbers with the left ray topology.
(a) Prove that (R,L) is not compact.
(b) Prove that (−∞, 0) is not compact in (R,L).
(c) Prove that (−∞, 0] is compact in (R,L).
2. Prove that R2 is not compact, where each coordinate is considered

with the usual topology U.
3. Suppose A and B are compact subspaces of a space X.

(a) Prove or disprove that A ∩ B is compact.
(b) Prove or disprove that A ∪ B is compact.
4. Let (X, τ ) be a compact topological space.

(a) Prove that if τ 1 ⊂ τ , then (X, τ 1) is compact.
(b) Give an example to show that if τ ⊂ τ 1, then (X, τ 1) need
not be compact.
5. Is (R,S), the Sorgenfrey line, compact?
6. Is (R , CC) compact?
7. Prove that if f : (X , τ ) −→ (R , U) is continuous and X is

compact, then there exist a, b ∈ X such that f(a) ≤ f(x) ≤
f(b), ∀ x ∈ X.
8. Prove that if f : X −→ Y is a continuous function, X is compact,

and Y is T2 , then f is a closed function.
9. Let J = (0, 1), the open unit interval. Define τ ⊂ P(J ) as follows:
τ = {∅, J } ∪ {(0, 1 − n1 ) : n ∈ N \ {1} }
(a) Verify that τ is a topology on J .
(b) Prove that (J , τ ) is not regular, but it is normal.
(c) Is (J , τ ) compact? justify your answer.
.
9 Connected Spaces
9.1 Definition: A topological space X is connected14 if and only if it

cannot be expressed as a union of two nonempty disjoint open subsets.
A topological space X is disconnected if and only if it is not connected.
That is, X is disconnected if and only if there exist two nonempty open
sets U and V such that U ∩ V = ∅ and U ∪ V = X. If the sets U and
V satisfy the previous conditions (i.e., if U and V are both nonempty
open sets such that U ∩ V = ∅ and U ∪ V = X.), then we say that U
and V form a separation of X. Note that if U and V form a separation
of X, then both U and V are also closed.
9.2 Examples:
1. Let X = {a, b, c}. Let τ = {∅, X, {a}, {b, c}} and

let S= {∅, X, {a}, {a, b}}.
Then (X, τ ) is disconnected, while (X, S) is connected.
So, connectedness depends on the topology and the ground set
itself.
2. Any singleton space is connected.
3. Any indiscrete space is connected.
4. Any discrete space having more than one point is disconnected.
5. Consider A = (1, 5]∪[7, 10] as a subspace of (R, U). Then (A, U A )

is disconnected, as (1, 5] and [7, 10] form a separation of A. (1, 5]
is open in A because (1, 5] = (1, 6) ∩ A where (1, 6) ∈ U, and
[7, 10] is open in A because [7, 10] = (6, 11) ∩ A where (6, 11) ∈ U.
So, in the usual topology think about a connected set as the set
consists of “one piece”.
9.3 Theorem: For a space X, the following are equivalent:
1. X is connected.
14
The present definition of connectedness was introduced by Jordan in 1893 for
the class of compact subsets of the plane; generalization to abstract spaces is due
to Riesz in 1907, Lennes in 1911, and Hausdorff in 1914. See [Engelking, 1977].
2. The only clopen (closed-and-open) subsets of X are X and ∅.
3. There is no onto continuous function f : X −→ {a, b} where

{a, b} has the discrete topology.
Proof:
(1) =⇒ (2)
If there exists a clopen set A such that ∅ %= A %= X, then A and X \ A
would form a separation of X.
(2) =⇒ (3)
If there exists a continuous function f from X onto {a, b}, where {a, b}
has the discrete topology, then f −1 ({a}) would be a clopen subset of
X such that ∅ %= f −1 ({a}) %= X.
(3) =⇒ (1)
Suppose that X is disconnected. Then there exist two open nonempty
disjoint subsets A and B of X which form a separation of X. Define
f : X −→ {a, b}, where {a, b} has the discrete topology, by f(x) = a
for all x ∈ A and f(x) = b for all x ∈ B. Then f is continuous and
onto, which is a contradiction.
9.4 Theorem: If f : X −→ Y is a continuous and onto function

and X is connected, then Y is connected.
Proof:
Suppose Y is disconnected. Then there exist two open nonempty dis-
joint open sets U and V of Y which form a separation of Y . Consider
f −1 (U) and f −1 (V ). Both of them are open in X (by continuity of f)
nonempty (as both U and V are nonempty and f is onto), and dis-
joint(why?). Also, X = f −1 (Y ) = f −1 (U ∪ V ) = f −1 (U) ∪ f −1 (V ).
Thus f −1 (U) and f −1 (V ) form a separation of X, which is a contradic-
tion. Therefore, Y is connected.
9.5 Corollary: Connectedness is a topological property.
9.6 Corollary: If f : X −→ Y is a continuous function and X is

connected, then f(X), as a subspace of Y , is connected.
The most important topological spaces regarding the applications

are (R, U) and all of its subspaces.
9.7 Theorem: Let A ⊆ R which contains more than one point. A

is an interval if and only if given any a, b ∈ A with a < b, then for any
x ∈ R with a ≤ x ≤ b we have that x ∈ A.
Connected Spaces 89
9.10 Theorem: Consider (R, U). Let A ⊆ R where A has more

than one point. A is connected if and only if A is an interval.
9.11 Theorem: (The Intermediate Value Theorem)

Let (X, τ ) be a connected topological space and let f : X −→ R
be a continuous function. Let a, b ∈ X such that f(a) < f(b) and
let d ∈ R with f(a) < d < f(b). Then there exists c ∈ X such that
f(c) = d.
Proof: By Corollary 9.6, we have that f(X) is a connected sub-
space of R. By Theorem 9.10, we have that f(X) is an interval. Thus
[f(a), f(b)] ⊆ f(X). Hence there exists c ∈ X such that f(c) = d.
9.12 Theorem: If U and V form a separation of X and A is a

connected subset of X, then either A ⊆ U or A ⊆ V .
Proof: A∩U and A∩V are open disjoint in A. If both are nonempty,
then A would be disconnected. Thus either A ∩ U = ∅ =⇒ A ⊆ V or
A ∩ V = ∅ =⇒ A ⊆ U.
9.13 Theorem: If A is a connected subspace of X and B ⊆ X

with A ⊆ B ⊆ A, then B is connected.
Proof: Suppose U and V form a separation of B. Then, by Theorem
9.12, we have either A ⊆ U or A ⊆ V . Without loss of generality,
assume that A ⊆ U, then A ⊆ U. Since U ∩ V = ∅, then A ∩ V = ∅.
Thus B ∩ V = ∅. Hence V = ∅, which is a contradiction. Thus B is
connected.
9.14 Corollary: If A is connected, then A is connected.

Problems
1. Prove that the empty set is connected.
2. Prove that singletons are connected.
3. Prove that Q, as a subspace of (R, U), is disconnected.
4. Prove that (R, L) is connected.
5. Is Q, as a subspace of (R , L), connected?
6. Prove that (R, CF ) is connected.
7. Is Q, as a subspace of (R , CF ), connected?
8. Let X be an infinite set. Is (X , F ) connected? where F is the

Fort topology on X.
9. Is the Sorgenfrey line connected?
10. Let A be a connected subset of a space X. If B ⊆ X is clopen

and A ∩ B %= ∅, prove that A ⊆ B.
11. Let {Cα : α ∈$Λ} be a collection of connected

! subsets of a space
X such that α∈Λ Cα %= ∅. Prove that α∈Λ Cα is connected.
12. Use the previous problem to show the following:

Let X be a space such that each pair of points in X is contained
in a connected subset of X. Prove that X is connected.
13. Use the previous problem and Theorem 9.4 to show the following:
Let X and Y be spaces. Prove that X × Y is connected if and
only if both X and Y are connected.
References
[ 1 ] Long, Paul, An Introduction to General topology, Charles E.
Merrill Publishing Company, 1971.
[ 2 ] Patty, C. Wayne, Foundations of Topology, PWS-KENT

Publishing Company, 1993.
[ 3 ] Engelking, Ryszard, General Topology, PWN, Warszawa, 1977.

.
Appendixes
(A) Greek Letters
We used the following Greek letters as symbols.
α alpha
β beta
γ gamma
δ delta
' epsilon
ζ zeta
η eta
θ theta
λ lambda
µ mu
π pi
ρ rho
σ sigma
φ phi
ϕ varphi
χ chi
ψ psi
ω omega
τ tau
Λ capital lambda.
.
(B) Hints and Solutions
Chapter 2, page 28
3. There are many examples. On R, take L ∪ R. Verify that it is

not a topology. Look for another finite counterexample.
5. Assume that τ is the discrete topology. This means that any

subset of X is open (in fact, clopen). So, in particular, {x} is
open for any x ∈ X. Now, assume the condition holds. Let A be
any subset of X. If A is the empty set, then A is open. Assume
that A is not empty. Then A = ∪x∈A {x} and by the condition,
A is open being union of open sets. Therefore, any subset of X
is open which means that τ is the discrete topology
!
6. Hint: Observe that U1 = N and the Uns are decreasing which
means that for i, j ∈ N, we have Ui ∪Uj = Uk where k = min{i, j}
and Ui ∩ Uj = Uk where k = max{i, j}.
8. CF ⊂ U. CF and L are uncomparable. CF ⊂ CC. You should

give a proof for each one.
n
9. (b) It is not a topology, because [−π , n+1
] ∈ τ for each n ∈ N,
n
but ∪n∈N [−π , n+1 ] = [−π , 1) %∈ τ .
Chapter 2, page 37
2. If A = ∅, then we are done because the empty set is a subset of

any set. Assume that A %= ∅. Let x ∈ A be arbitrary. Let U be
any open neighborhood of x. Since x ∈ A, then U ∩ A %= ∅. Since
A ⊆ B, then U ∩ B %= ∅. Thus x ∈ B. Hence A ⊆ B.
5. Hint: Use the fact A ∩ B ⊆ A, A ∩ B ⊆ B and problem 2 above.

We may also prove it by using the definition.
9. int((1 , 2 ]) = ∅, (1 , 2 ] = (−∞ , 2 ], intB = ( 6 , ∞ ), and B = R.

The reader should prove each one.
10. (i) Usual: intA = ( 0 , 1 ) , A = [ 0 , 1 ] ∪ {2} , ∂A = {0, 1, 2 }.

(ii) Cofinite: intA = ∅ , A = R , ∂A = R.
(iii) Left Ray: intA = ∅ , A = [ 0 , ∞ ) , ∂A = [ 0 , ∞ ).
(iv) Cofinite: intA = A , A = A , ∂A = ∅ because A is clopen.
11. Hint: If you assume U ∩ V %= ∅ you will get a contradiction.
12. A = A, intA = ∅, ∂A = A, B = R, intB = B.
15. Hint: A short way to do this is by proving just two things: (1)
If x ∈ ext(A), then x %∈ int(A) ∪ ∂A. (2) If x ∈ int(A), then
x %∈ ∂A.
16. The sequence (− n1 )n∈N in (R , CF ) converges to each real number.

That is, C(− n1 )n∈N = R. To prove this, let x ∈ R be arbitrary.
We show that − n1 −→ x. Let U ∈ CF be arbitrary such that
x ∈ U. This means R \ U is finite. Suppose U has no tail of the
sequence (− n1 )n∈N . Then a tail would be inside the complement
R \ U, and this is a contradiction because R \ U is finite while any
tail of (− n1 )n∈N is infinite. Thus U has a tail and hence − n1 −→ x.
The sequence (− n1 )n∈N in (R , U ) converges to 0. That is,
C(− n1 )n∈N = {0}. The reader should prove this by using the same
idea in Real Analysis class.
Chapter 3
1. Suppose B is a base, then {2} = {1, 2} ∩ {2, 3} is open being an

intersection of two basic open sets. Now there is no B ∈ B such
that 2 ∈ B ⊆ {2}. Thus (B2) is not satisfied, a contradiction.
Thus B cannot be a base for any topology on X.
2. Let B = {( −∞ , a ) : a ∈ Q}. Then B ⊂ L. Let ( −∞ , a ),

( −∞ , b ) ∈ B be arbitrary. Let x ∈ ( −∞ , a ) ∩ ( −∞ , b ) be
arbitrary. Since ( −∞ , a ) ∩ ( −∞ , b ) = ( −∞ , c ) where c =
min{a , b}. Then the intersection itself belongs to B. Thus x ∈
( −∞ , c ) ⊆ ( −∞ , a ) ∩ ( −∞ , b ). Thus B is a base for L and B
is countable. Thus R with L is second countable.
3. Let ( X ,τ ) be any second countable space. Let B = {Bi : i ∈

N} be a countable base for τ . We may assume without loss of
generality that Bi %= ∅ for each i ∈ N. For each i ∈ N, pick
di ∈ Bi and define D = {di : i ∈ N}. Then D ⊆ X and D is
countable. Furthermore, Bi ∩ D %= ∅ for each i ∈ N. Thus D is
dense and hence ( X ,τ ) is separable.
Appendixe B 97
4. Let {Un : n ∈ N} be any countable family of open subsets in

( R , CF ). We show that {Un : n ∈ N} cannot be a base for
( R , CF ). We may assume without loss of generality that Un %= ∅
for each n ∈ N. Thus, for each n ∈ N we have R \ Un is a finite
set. Thus ∪n∈N (R\Un ) is a countable set. But ∪n∈N (R\Un ) = R\
(∩n∈N Un ). i.e., R\(∩n∈N Un ) is countable. Since R is uncountable,
then
∅ %= R \ (R \ (∩n∈N Un )) = ∩n∈N Un .
Pick y ∈ ∩n∈N Un . Then y ∈ Un for each n ∈ N. Define V =
R \ {y}, then V ∈ CF and for each n ∈ N we have that Un %⊆ V
because y ∈ Un for each n and y %∈ V . Thus {Un : n ∈ N} cannot
be a base for ( R , CF ). Since {Un : n ∈ N} was arbitrary, then
( R , CF ) is not second countable.
5. A = [ 0 , 1 ], intA = [ 0 , 1 ), ∂A = {1}
intB = ( 1 , 2 )∪( 4 , 5 ), B = [ 1 , 2 )∪{3}∪[ 4 , 5 ], ∂B = {1, 3, 4, 5}
C = C, intC = ∅, ∂C = C
(R , S ) is separable as Q is a countable dense subset. It is dense
as any basic open set [ a , b ) has to meet Q.
7. C(− n1 )n∈N = ∅. Proof: Let x ∈ R be arbitrary. If x < −1,

then x ∈ [ x , −1) ∈ S and [ x , −1) has no tail. If x ≥ 0, then
x ∈ [ x , x+1 ) ∈ S and [ x , x+1 ) has no tail. Now, if −1 ≤ x < 0,
then there exists an m ∈ N such that x ∈ [ − m1 , − m+1 1
) ∈ S and
[ − m , − m+1 ) has no tail. Thus in all cases we proved that − n1
1 1
does not converge to x.
8. C( n1 )n∈N = {0}.
9. (a) {Aα : α ∈ Λ} is a base for X if {Aα : α ∈ Λ} ⊆ τ and for

each ∅ %= U ∈ τ there exists Ω ⊆ Λ such that U = ∪α∈Ω Aα.
(b) {Aα : α ∈ Λ} is not a base for X if there exists β ∈ Λ such
that Aβ %∈ τ or there exists an open U ∈ τ and an x ∈ U
such that for each α ∈ Λ either x %∈ Aα or x ∈ Aα %⊆ U.
(c) x %∈ int(A), x is not an interior point of A if for each U ∈ τ
with x ∈ U we have U %⊆ A.
(d) x %∈ A, x is not in the closure of A if there exists U ∈ τ with
x ∈ U and U ∩ A = ∅.
(e) A is not dense in (X,τ ) if A %= X.

(f) (X,τ ) is not second countable if {Un : n ∈ N} is any count-
able family of open sets, then {Un : n ∈ N} is not a base
for τ .
Chapter 4
3. In ( I , LI ), intB = ∅, B = I, ∂B = I.
In ( I , UI ), intB = ∅, B = B ∪ {1}, ∂B = B.
In ( I , τ0 I
), intB = B, B = I, ∂B = I \ B.
10. Hint: If D is a countable dense subset in ( X , τ ), then A ∩ D

should be a countable dense subset in ( A , τ A ).
14. (a) A × [ 2 , 5 ] (b) ∅ × ( ( 2 , 3 ) ∪ ( 3 , 5 ) ) = ∅.
17. Hint: Prove that D = {π + q : q ∈ Q} is countable and dense in

( P , UP ).
Chapter 5
1. Hint: UZ is the discrete topology on Z.
9. Hint: Define f : ( R , L ) −→ ( R , R ) by f(x) = −x for each

x ∈ R.
10. Hint: Define f : ( 1 , 5 ] −→ ( 3 , 11 ] by f(x) = 2x + 1 for each

x ∈ ( 1 , 5 ].
x
13. Hint: Define f : ( 0 , 12 ] −→ ( 0 , 3 ] by f(x) = 4
for each x ∈
( 0 , 12 ].
17. Hint: Define f : R −→ ( 0 , ∞ ) by f(x) = ex for each x ∈ R.

√ √ √
25. √
Hint: Define
√ f : R −→
√ R by f(x) = x if 2 %
= x %
= 3, f( 2) =
3, and f( 3) = 2.
28. The statement is not true.

Appendixe B 99
Chapter 6
2. Hint: First prove that X × {0, 1} is not T0 . Then prove that
no two nonempty open subsets of X × {0, 1} are disjoint. Now,
complete the proof.
3. We give a proof for the T2 case.
Proof: Let (X,τ ) be any T2 topological space and let ∅ %= Y ⊆ X
be arbitrary. If Y is a singleton, we are done. Assume that Y
has more than one element. Let a, b ∈ Y be arbitrary such that
a %= b. Then a, b ∈ X and X is Hausdorff, thus there are U, V ∈
τ such that a ∈ U, b ∈ V and U ∩ V = ∅. Define G = U ∩ Y and
H = V ∩ Y . Then G and H are open in Y , a ∈ G, b ∈ H and
G ∩ H = (U ∩ Y ) ∩ (V ∩ Y ) = (U ∩ V ) ∩ Y = ∅ ∩ Y = ∅. Thus
( Y,τ Y ) is T2 .
4. Hint: Recall that if Y ⊆ X is closed and E ⊆ Y is closed in Y ,
then E is closed in X.
5. We give a proof for the T2 case:
Proof:
(⇒)
Assume X × Y is T2 . Assume that X has more than one element.
Let a, b ∈ X be arbitrary such that a %= b. Pick y ∈ Y . Then (a, y)
and (b, y) are distinct elements in the product X ×Y . By T2, there
are basic open sets U1 × V1 , U2 × V2 such that (a, y) ∈ U1 × V1 ,
(b, y) ∈ U2 × V2 and ∅ = (U1 × V1 ) ∩ (U2 × V2 ). This means
U1 ∩ U2 = ∅, a ∈ U1 , and b ∈ U2 . Hence X is T2 . A proof of Y is
T2 is similar.
(⇐)
Assume X and Y are both T2. Assume that X × Y has more
than one element. Let (x1, y1), (x2, y2 ) ∈ X × Y be any distinct
elements. There are just two cases, either x1 %= x2 or y1 %= y2.
Case 1: x1 %= x2 . Since X is T2, then there are two open sets U
and V in X such that x1 ∈ U, x2 ∈ V , and U ∩ V = ∅. Consider
U × Y and V × Y which are basic open in X × Y such that
(x1, y1 ) ∈ U × Y , (x2, y2) ∈ V × Y , and (U × Y ) ∩ (V × Y ) =
(U ∩ V ) × Y = ∅ × Y = ∅. Thus X × Y is T2 .
Case 2 can be proved by a similar way.
9. Assume that f : X −→ Y is a one-to-one continuous function
and Y is Hausdorff. To show that X is Hausdorff assume that
X has more than one element because if X is a singleton we are

done. Let a, b ∈ X be arbitrary such that a %= b. Since f is
one-to-one, then f(a) %= f(b). Since Y is Hausdorff, then there
are two open sets U and V in Y such that f(a) ∈ U, f(b) ∈
V , and U ∩ V = ∅. Since f is continuous, then f −1 (U) and
f −1 (V ) are open in X such that a ∈ f −1 (U), b ∈ f −1 (V ), and
f −1 (U) ∩ f −1 (V ) = f −1 (U ∩ V ) = f −1 (∅) = ∅. Therefore, X is
Hausdorff.
10. Let ( X , τ ) be a Hausdorff space and let (an )n∈N be any con-
vergent sequence. Suppose that the convergency set of (an )n∈N is
not a singleton, i.e., suppose that there are two elements x, y ∈ X
such that an −→ x, an −→ y, and x %= y. Since X is Hausdorff,
then there are two open sets U and V such that x ∈ U, y ∈ V ,
and U ∩ V = ∅. Since an −→ x, then U has a tail of (an )n∈N .
Since an −→ y,, then V has a tail of (an )n∈N . This is a contradic-
tion because U ∩ V = ∅. Thus the convergency set of (an )n∈N is
a singleton.
14. 1. We can state the negation of T1 in many ways. Here is one

of them: ∃ x, y ∈ X with x %= y such that ∀ open sets U, V ∈
τ either x, y ∈ U ∩ V or x, y %∈ U ∪ V .
2. ∃ x, y ∈ X with x %= y such that ∀ open sets U, V ∈ τ with
x ∈ U and y ∈ V we have U ∩ V %= ∅.
3. ∃ a closed set E ⊂ X and an element x ∈ X \ E such that
∀ open sets U, V ∈ τ with x ∈ U and E ⊆ V we have
U ∩ V %= ∅.
4. either X is not T1 or X is not normal.
Chapter 7
8. R with the left ray toplogy L is not T1 hence cannot be metrizable.
Chapter 8
1. (a) {( −∞ , n ) : n ∈ N} is an open cover which has no finite

subcover.
4. (a) Hint: Any open set in τ 1 is open in τ .

Appendixe B 101
(b) CF ⊂ U.
5. {[ n , n+1 ) : n ∈ Z} is an open cover which has no finite subcover.
6. The answer is NO. Think why.
10. (c) (J , τ ) is not compact. {(0, 1 − n1 ) : n ∈ N \ {1} } is an open

cover which has no finite subcover.
Chapter 9
3. ( −∞ , π ) ∩ Q, ( π , ∞ ) ∩ Q form a separation for Q.
4. Hint: Any two non-empty proper open subsets cannot be disjoint.
6. Hint: Any two non-empty proper open subsets cannot be disjoint.

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Chapter 2: Topological Spaces 23

Chapter 3: Bases for a Topology 41

Chapter 4: New Spaces from Old Ones 47

Chapter 5: Continuous Functions 55

Chapter 6: Separation Axioms 67

Chapter 7: Metric Spaces 73

Chapter 8: Compact Spaces 81

Chapter 9: Connected Spaces 87

Appendix A: Greek Letters 93

Appendix B: Hints and Solutions 95

Section 1.1: Sets

A set is a collection of objects. These objects are called points,

1.1 Definition: A set X is a subset of another set Y , denoted by

For two sets X and Y , we define:

1. X \ Y = {x : x ∈ X and x ∈ / Y }, called the complement of Y

2. X ∩ Y = {x : x ∈ X and x ∈ Y }, called the intersection of X

3. X ∪ Y = {x : x ∈ X or x ∈ Y }, called the union of X and Y .

Main Sets of Numbers

1. The set of Natural Numbers N = {1, 2, 3, ...}

2. The set of Integers Z = {..., −3, −2, −1, 0, 1, 2, 3, ...}

3. The set of Rational Numbers Q = { ab : a, b ∈ Z and b %= 0}

4. There are numbers, we use √ in our daily life, cannot√be expressed

5. The set of real numbers R = Q ∪ P. Note that Q ∩ P = ∅.

1.2 Theorem: Let X be a set and A, B, C ∈ P(X), then

5. A∪(B∩C) = (A∪B)∩(A∪C) and A∩(B∪C) = (A∩B)∪(A∩C).

8. A ⊆ X \ B if and only if A ∩ B = ∅ if and only if B ⊆ X \ A.

10. If X ⊆ Y , then P(X) ⊆ P(Y ).

Pick x ∈ A ∩ B, then x ∈ A and x ∈ B. Since B ⊆ X \ A and x ∈ B,

1.3 De Morgan’s Theorem: Let X, Y and Z be sets, then

1.4 Definition: Let X and Y be sets. The Cartesian Product of

(x1, y1 ) = (x2 , y2) if and only if x1 = x2 and y1 = y2 .

X × Y = ∅ if and only if either X = ∅ or Y = ∅.

(x1, y1) %= (x2, y2 ) if and only if x1 %= x2 or y1 %= y2 .

1.5 Theorem: If X, Y and Z are sets, then

Let Λ be a given nonempty set. Suppose that with each element

1. For each n ∈ N, define

Xn = {(x, y) ∈ R2 : 0 ≤ x ≤ n, 0 ≤ y ≤ n} = [0, n] × [0, n].

So, for each n ∈ N, Xn is the square in the plane having [0, n] as

2. Let Λ = R and for each α ∈ R, define Xα = {(α, y) ∈ R2 : 0 ≤

1.6 Definition: Let {Xα : α ∈ Λ}!be an indexed family of sets.

Note that if {Xα $

2. If Xα is as in example (2) above, then

Proof: Let x ∈ (−∞, 0] be arbitrary, then x ≤ 0, so x < n1 , ∀ n ∈

Now, suppose x %∈ (−∞, 0], then x > 0, so by Archimedean prop-

Exercise: Calculate each of the following:

1.7 Definition: Let {Xk : k ∈ Λ} be an indexed family of sets

Exercise: Give an example to show that the reverse inclusion of

1.9 Theorem: Let X be a set and {Aα : α ∈ Λ} be an indexed

9. Let {Aα : α ∈ Λ} be an indexed family of sets.

Section 1.2: Functions

1.10 Definition: A function f from a nonempty set X to a nonempty

Image and Preimage

If f : X −→ Y is a function, then f introduces a function from

Note that, by above definitions we have:

1. Define f : R −→ R by f(x) = 2x + 1. Then

2. Let f : R −→ R be defined by f(x) = x2.

1. Define g : R −→ R by g(x) = 3x.