Download Lecture notes of Dr. Hicham Gebran

Math 201
Topology I
Lecture notes of Dr. Hicham Gebran
[email protected]
Lebanese University, Fanar, Fall 2013-2014
www.fs2-fanar.com/Math
2
What is topology? Topology is the study of continuity in a general context. In group theory
or vector space theory, a fundamental concept is that of isomorphism. Two groups are isomorphic if they have the same algebraic structure. The analogous concept in topology is that
of homeomorphism. A homeomorphism is a continuous bijection whose inverse is continuous.
For example, a circle and a square are homeomorphic as we shall see, that is, they are topologically the same. A fundamental problem of topology is the classification of spaces up to a
homeomorphism.
References
James Munkres, Topology (Pearson, 2000).
Colin Adams and Robert Franzosa, Introduction to topology, pure and applied (Pearson,
2007).
Lynn Arthur Steen and J. Arthur Seebach, Counterexamples in Topology (Dover, 1978).
Contents
0 Some fundamental properties of the real line
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1 Metric spaces and topological spaces
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1.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 Open and closed sets and related concepts (neighborhood, interior, closure, boundary, convergence, density) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.4 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.5 Product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2 Continuous functions and homeomorphisms
3 Compactness
3.1 Definitions, examples and properties . . . . . . . . . . .
3.2 Compact subspaces of IRn . . . . . . . . . . . . . . . . .
3.3 Compact metric spaces . . . . . . . . . . . . . . . . . . .
3.4 Local compactness and the Alexandroff compactification
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4 Connectedness
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4.1 Definitions, examples and properties . . . . . . . . . . . . . . . . . . . . . . . . . 41
4.2 Components and local connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 46
5 Complete metric spaces
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3
4
CONTENTS
Chapter 0
Some fundamental properties of the
real line
We start by formulating two fundamental properties of the set of real numbers that we shall
use in the sequel. This set is denoted by IR and it is naturally identified to a line. An upper
bound of a subset E ⊂ IR is a real number M such that x ≤ M for all x ∈ E. The set E is
called bounded from above if it has an upper bound. A lower bound of a subset E ⊂ IR is a
real number m such that m ≤ x for all x ∈ E. The set E is called bounded from below if it has
a lower bound.
Definition 0.1 Let E ⊂ IR be bounded from above. The number α is called the least upper
bound of E if the following hold.
i) α is an upper bound of E.
ii) If β < α, then β is not an upper bound of E.
In this case, we write α = sup E. If E is not bounded from above, we write sup E = +∞.
Proposition 0.1 Let E ⊂ IR be bounded from above. Then α = sup E if and only if the
following hold.
i) x ≤ α for all x ∈ E.
ii) ∀ε > 0 ∃y ∈ E such that α − ε < y.
Definition 0.2 Let E ⊂ IR be bounded from below. The number α is called the greatest lower
bound of E if the following hold.
i) α is a lower bound of E.
ii) If β > α, then β is not a lower bound of E.
In this case, we write α = inf E. If E is not bounded from below, we write inf E = −∞.
Proposition 0.2 Let E ⊂ IR be bounded from below. Then α = inf E if and only if the
following hold.
i) x ≥ α for all x ∈ E.
ii) ∀ε > 0 ∃y ∈ E such that α + ε > y.
Proposition 0.3 Let ∅ 6= A ⊂ B ⊂ IR. Then sup A ≤ sup B and inf A ≥ inf B.
Exercise. Let A and B be two nonempty subsets of IR. If x ≤ y for all x ∈ A and all y ∈ B,
then sup A ≤ inf B.
Now we can state the two fundamental properties of IR.
5
6
CHAPTER 0. SOME FUNDAMENTAL PROPERTIES OF THE REAL LINE
Theorem 0.1 Any nonempty subset of IR which is bounded from above has a least upper
bound. Any nonempty subset of IR which is bounded from below has a greatest lower bound.
Any proof of this theorem involves going back to the construction of the real numbers from the
rational numbers. We do not address this issue here. Note that the above theorem is not true
for the set of rational numbers Q. And this is one of the reasons we prefer to work in the IR
rathre than in Q.
Recall now that a subset E ⊂ IR is an interval if and only if it is not empty and [x, y] ⊂
E whenever x and y belong to E. An interval has thus one of the following ten forms
{a}, ]a, b[, ]a, b], [a, b[, [a, b], ] − ∞, a], ] − ∞, a[, ]a, ∞[, [a, ∞[ and IR. An interval is called
trivial if it is a singleton.
Theorem 0.2 Any nontrivial interval of the real line contains rational as well as irrational
numbers.
This theorem will be proved in the exercises. We end with a useful result.
Proposition 0.4 Let a and b be given real numbers. If a ≤ b + ε for all ε > 0 then a ≤ b.
Proof. Otherwise, taking ε =
b−a
2 ,
we reach a contradiction.
Chapter 1
Metric spaces and topological spaces
1.1
Definitions and examples
Definition 1.1 Let X be a non empty set. A metric or distance over X is a function
d : X × X → [0, ∞[ that satisfies the following properties
1. d(x, y) = 0 if and only if x = y.
2. d(x, y) = d(y, x) for all x, y ∈ X (symmetry).
3. d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X (Triangle inequality).
A metric space is a couple (X, d) where d is a distance on X.
Examples. 1) The real line with the usual distance d(x, y) = |x − y|.
2) The real line with the distance ρ(x, y) = |x3 − y 3 |.
3) The real line with the distance γ(x, y) = min(1, |x−y|). The first two properties of a distance
are clear. We prove the triangle inequality. Observe that γ(x, y) ≤ 1 and γ(x, y) ≤ |x − y|. Let
x, y, z be three real numbers. We distinguish between two cases.
Case 1. Either |x − y| ≥ 1 or |y − z| ≥ 1. Then γ(x, y) = 1 or γ(y, z) = 1. Therefore,
γ(x, y) + γ(y, z) ≥ 1. On the other hand, γ(x, z) ≤ 1. Hence the triangle inequality in this case.
Case 2. |x − y| < 1 and |y − z| < 1. Then γ(x, y) = |x − y| and γ(y, z) = |y − z|. Therefore,
γ(x, z) ≤ |x − z| ≤ |x − y| + |y − z| = γ(x, y) + γ(y, z).
¯ y) = min(1, d(x, y)). Then d¯ is a
4) More generally, let (X, d) be a metric space and set d(x,
distance on X. The proof is similar as the proof above.
5) Let a < b and X = C([a, b]) be the set of continuous functions f : [a, b] → IR. Let
d(f, g) = supx∈[a,b] |f (x) − g(x)|. Then d is a distance.
Z b
6) Again consider the set X = C([a, b]) and set d(f, g) =
|f (x) − g(x)| dx.
a
7) Let X be a set. The discrete distance d on X is defined by
(
0
if x = y
d(x, y) =
1
if x 6= y
The first two properties are clear. Let x, y, z ∈ X. To prove the triangle inequality, we distinguish between two cases.
Case 1. x = z. Then d(x, z) = 0 ≤ d(x, y) + d(y, z).
Case 2. x 6= z. Then either y 6= x or y 6= z. Therefore d(x, z) = 1 ≤ d(x, y) + d(y, z).
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8
CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
Norms and distances on IRN
Let N be a positive integer. The set of all N −tuples of real numbers (x1 , x2 , . . . , xN ) is denoted
by IRN . Otherwise stated, IRN is the cartesian product of N copies of IR. Elements of IRN are
considered as vectors with N components. The origin of IRN is the vector (0, . . . , 0) denoted
simply by 0.
A norm on IRN is a way to measure the length of a vector (x1 , . . . , xN ). It is the analog of
the absolute value of a real number. The norm of a vector x ∈ IRN is a nonnegative number
denoted by kxk that satisfies the following properties.
(i) kxk = 0 if and only if x = 0.
(ii) kλxk = |λ|kxk for any real number λ and any x ∈ IRN .
(iii) kx + yk ≤ kxk + kyk for all x, y ∈ IRN . This is also known as the triangle inequality.
The most natural norm on IRN is the Euclidean norm. Let x = (x1 , . . . , xN ) ∈ IRN . The
Euclidean norm of x is defined by
!1/2
N
X
kxk =
|xi |2
.
i=1
It is not difficult to show that kxk satisfies the two properties (i) and (ii) above. The triangle
inequality however is not trivial. To prove (iii), let us first observe that the Euclidean norm is
just the square root of the inner product of x by itself. The inner product of two vectors x and
y of IRN is
N
X
x · y = x1 y 1 + · · · + x N y N =
xi yi .
i=1
Let α be real number and consider the scalar product (x + αy) · (x + αy). We have
(x + αy) · (x + αy) = kxk2 + 2αx · y + α2 kyk2 ≥ 0.
Thus we have a second order polynomial in α which is always nonnegative. Therefore its
discriminant (x · y)2 − kxk2 kyk2 ≤ 0 (otherwise, the polynomial would have two distinct real
roots and would therefore change sign). Thus,
|x · y| ≤ kxkkyk.
This is known as the Cauchy-Schwartz inequality.
From the last inequality we deduce that
kx + yk2 = (x + y) · (x + y) = kxk2 + 2x · y + kyk2
≤ kxk2 + 2kxkkyk + kyk2
= (kxk + kyk)2 .
Since both members of the above inequality are nonnegative, the triangle inequality follows.
Other norms can also be defined on IRN . The sup norm of a vector x is defined by kxk∞ =
max(|x1 |, . . . , |xN |) . It is easy to check that this norm satisfies the three properties of a norm.
N
P
Another norm is defined by kxk1 =
|xi |. Here again it is not difficult to prove the three
i=1
properties. More generally, let p ≥ 1. Set
||x||p =
N
X
i=1
p
|xi |
!1/p
.
We shall prove in the exercises that this is norm. Note finally that all these norms coincide
when N = 1.
Now a norm on IRN defines a metric by d(x, y) = ||x − y||.
1.2. OPEN AND CLOSED SETS AND RELATED CONCEPTS (NEIGHBORHOOD, INTERIOR, CLOSURE
Balls in metric spaces
The open ball of center x and radius r is the set
B(x, r) = {y ∈ X | d(y, x) < r}.
The closed ball of center x and radius r is the set
B 0 (x, r) = {y ∈ X | d(y, x) ≤ r}.
Examples. 1) The ball B(a, r) in IR (with the usual distance) is the interval ]a − r, a + r[. In
(IR2 , || · ||2 ) the ball B(a, r) is the usual disc of center a and radius r. In (IR3 , || · ||2 ), the ball
B(a, r) is the usual geometric ball of center a and radius r.
3) In (IR2 , || · ||∞ ) the ball B(0, 1) is the square of vertices (-1,-1), (1,-1) (1,1) and (-1,1). Draw
a figure and explain. The closed ball is the same square with its boundary.
4) In (IR2 , || · ||1 ), the ball B(0, 1) is the square of vertices (1,0), (0,-1) (-1,0) and (0,1). Draw a
figure and explain. The closed ball is the same square with its boundary.
1.2
Open and closed sets and related concepts (neighborhood,
interior, closure, boundary, convergence, density)
Definition 1.2 Let X be a metric space. A subset O ⊂ X is called open if for every x ∈ O
there is an ε > 0 such that B(x, ε) ⊂ O.
Otherwise stated, a set is open if whenever it contains a point, it contains a whole ball
around this point. This is why we call it open: if one member x can enter the set, some other
members (the friends of x) can also enter the set.
Examples. ]0, 1[ is open in IR but (0, 1] is not. Any interval of the form ]a, b[, ]a, ∞[, ] − ∞, a[
is open.
Proposition 1.1 An open ball is an open set.
Remark. This proposition is not a tautology because the adjective open has different meanings
in ”open ball” and ”open set”. An open ball is a set of the form B(a, r) = {x|d(x, a) < r},
whereas an open set is a set having the property that if a point belongs to this set, then a
whole ball around this point is also in the set. So we have to prove that an ”open ball” has this
property.
Proof. Let x ∈ B(a, r). Then d(x, a) < r. Set ρ = r − d(x, a). We claim that B(x, ρ) ⊂
B(a, r). Draw a figure. Indeed, let y ∈ B(x, ρ). Then d(x, y) < ρ = r − d(x, a). Therefore
d(x, a) + d(x, y) < r. By the triangle inequality, d(y, a) < r. This means that y ∈ B(a, r). The
claim implies that B(a, r) is open.
Proposition 1.2 The following properties hold.
(i) X and ∅ are open.
(ii) An arbitrary union of open sets is open.
(iii) A finite intersection of open sets is open.
The proof is straightforward. Just right it.
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CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
Remark 1.1 An arbitrary intersection of open sets need not be open. For example,
\
n≥1
−
1 1
,
= {0}
n n
is not open.
Definition 1.3 A neighborhood of a point x in a metric space X is a set V which contains
an open set containing x. The set of neighborhoods of a point x is usually denoted by U(x).
Remark 1.2 Note that the whole space is a neighborhood of x, therefore U(x) is not empty.
Note also that if V ∈ U(x) and V ⊂ U then U ∈ U(x).
Example. [−1, 1] is a neighborhood of any point x ∈]−1, 1[. However it is not a neighborhood
of 1 nor of −1.
Proposition 1.3 A set is open if and only if it is a neighborhood of all of its points.
Proof. Suppose first that A is open and let x ∈ A. Then A contains an open set (itself)
containing x. This means that A is a neighborhood of x. Conversely, suppose that A is a
neighborhood of all its points. This means that for every x ∈ A, there exists an open set Ox
such that x ∈ Ox ⊂ A. Then, we can write A = ∪x∈A Ox . Therefore A is open as a union of
open sets.
Remark 1.3 For some mathematicians, a neighborhood of a point is an open set containing
that point. This makes a little difference because in practise we can always assume that a
neighborhood is open.
Definition 1.4 The interior of a set A is the union of all open sets that are contained in A.
◦
It is therefore the biggest open set contained in A. It is denoted by A or intA.
Examples. a) int [0,1]=]0,1[. Indeed, ]0,1[ is open and contained in [0,1]. Therefore it is
contained in the interior of [0,1]. So we have ]0, 1[⊂ int[0, 1] ⊂ [0, 1]. Therefore int[0, 1] is either
]0,1[, ]0,1] [0,1[ or [0,1]. But the only open set among these sets is ]0,1[. Hence the result.
Similarly, int[0, 1[= int]0, 1] =]0, 1[.
◦
b) Q = ∅. This is because any open interval contains points outside Q and so Q cannot
contain any open interval and therefore cannot contain any nonempty open set. Similarly
int(IR\Q) = ∅.
◦
c) Similarly, Z = ∅.
Proposition 1.4 A set is open if and only if it is equal to its interior.
◦
◦
Proof. Suppose first A = A. But A is open. Therefore A is open. Conversely, suppose that
◦
◦
A is open . But A contains A. Since A is the biggest open set containing A, We have A ⊂ A.
◦
But A ⊂ A. Hence the equality.
Definition 1.5 Let X be a metric space. A subset F ⊂ X is called closed if its complement
is open.
1.2. OPEN AND CLOSED SETS AND RELATED CONCEPTS
11
Examples. 1) [a, b] is closed because its complement ]∞, a[∪]b, ∞[ is open.
2) [a, ∞[ and ] − ∞, a] are closed.
3) In a metric space singletons are closed. Indeed, let {a} be a singleton (one point set) and let
x ∈ X − {a}. Then x 6= a and so d(x, a) > 0. We claim that B(x, d(x, a)) ⊂ X − {a}. Indeed,
let y ∈ B(x, d(x, a)). Then d(x, y) < d(x, a). This implies that y 6= a because otherwise we
would have d(x, a) < d(x, a). This means that y ∈ X − {a}. Therefore etc.
Proposition 1.5 A closed ball is a closed set.
This proposition is not a tautology.
Proof. Let B 0 (a, r) be a closed ball in a metric space X. We have to prove that X − B 0 (a, r)
is open. Let x ∈ X − B 0 (a, r). Then d(x, a) > r. Set ρ = d(x, a) − r. We claim that
B(x, ρ) ⊂ X − B 0 (a, r). Indeed, let y ∈ B(x, ρ). Then d(x, y) < ρ = d(x, a) − r. Therefore
r < d(x, a) − d(x, y). By the triangle inequality, d(x, a) − d(x, y) ≤ d(y, a). Hence d(y, a) > r
and so y ∈ X − B 0 (a, r). This proves the claim. The claim implies that X − B 0 (a, r) is open. Proposition 1.6 In a metric space X, the following properties hold.
(i) X and ∅ are closed.
(ii) An arbitrary intersection of closed sets is closed.
(iii) A finite union of closed sets is closed.
Proof. Take complements and use the properties of open sets.
Corollary 1.1 In a metric space every finite set is closed.
Remark 1.4 An arbitrary union of closed sets need not be closed. For example for each n ≥ 2,
the set [ n1 , 1 − n1 ] is closed. However
is not closed.
∞
[
1
1
[ , 1 − ] =]0, 1[
n
n
n=2
Remark 1.5 A set which is not open is not necessarily closed. For example ]0,1] is neither
open nor closed. Also a set can be both open and closed. Indeed, in any metric space X, ∅
and X are closed an open. Here is another example. Let d be the discrete distance on a set X
containing more than one point. Then B(a, 1) = {a}. Hence {a} is open. But we know that
{a} is also closed. In fact every subset of X is open (because a subset is a union of singletons).
But if every subset of X is open then every subset is also closed. A subset which is both closed
and open is sometimes called clopen.
Definition 1.6 The closure of a set A is the intersection of all closed sets that contain A. It
is therefore the smallest closed set containing A. It is usually denoted by Ā and sometimes by
cl(A).
Examples. a) ]0, 1[ = [0, 1]. Indeed, [0,1] is a closed set containing ]0,1[, therefore it contains
]0, 1[. Thus ]0, 1[⊂ ]0, 1[ ⊂ [0, 1]. Therefore ]0, 1[ is either ]0,1[, ]0,1], [0,1[ or [0,1]. However, the
only closed set among these four sets is [0,1]. Hence the result. Similarly, ]0, 1] = [0, 1[ = [0, 1].
b) We shall see below that Q = IR . We say that Q is dense in IR. We shall also prove that the
irrationals are dense in IR.
c) We shall see in the exercises that if A is the open disk x2 + y 2 < 1, then the closure of A is
the closed disk x2 + y 2 ≤ 1.
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CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
Proposition 1.7 A set is closed if and only if it is equal to its closure.
Proof. Suppose first that A = Ā. Since Ā is closed, A is closed. Conversely, suppose that
A is closed. Since A contains A and Ā is the smallest closed set containing A, we have Ā ⊂ A.
But A ⊂ Ā. Hence the equality.
Definition 1.7 Let (xn ) be a sequence in a metric space (X, d). We say that (xn ) converges
to a point x if d(xn , x) converges to 0 in IR. In terms of quantifiers
∀ > 0 ∃n0 ∈ IN ∀n ≥ n0 d(xn , x) < .
Proposition 1.8 Let X be a metric space. Let A ⊂ X and let x ∈ X. Then, the following
conditions are equivalent.
(i) x ∈ Ā.
(ii) Every neighborhood of x intersects A.
(iii) There is a sequence in A which converges to x.
Proof. Let B denote the set of points t satisfying property (ii), i.e., satisfying that every
neighborhood of t meets A. Proving that (i)⇔(ii) is equivalent to proving that Ā = B
So let us prove first that Ā ⊂ B. We claim that B is closed. Let t ∈
/ B, then there is an open
neighborhood U of t not intersecting A. Consider an element y ∈ U . Since U is a neighborhood
of y not intersecting A, we deduce that y ∈
/ B. This means that U ⊂ {B and accordingly {B
is open. Therefore B is closed as claimed. Observe now that B contains A. Since Ā is the
smallest closed set containing A, we get Ā ⊂ B.
Let us prove next that B ⊂ Ā. Let t ∈
/ Ā. Then t ∈ {Ā . But {Ā is an open set not intersecting
Ā
A
A since { ∩ A ⊂ { ∩ A = ∅. This means that t ∈
/ B.
(ii)⇒(iii). For every positive integer n the open ball B(x, n1 ) intersects A. Choose accordingly
for each n, an element xn in the intersection. This defines a sequence of points in A which
clearly converges to x.
(iii)⇒(ii). Let U be a neighborhood of x. Since {xn } converges to x, we have that xn ∈ U for
n large enough. This means that U intersects A.
Corollary 1.2 Let X be a metric space. Let A ⊂ X and let {xn } be sequence in A which
converges to some x. Then x ∈ Ā.
Definition 1.8 Let X be a metric space. A subset A ⊂ X is called dense if Ā = X.
Corollary 1.3 Let X be a metric space and A ⊂ X. Then the following are equivalent
(i) A is dense in X.
(ii) Every nonempty open set of X meets A.
(iii) For every x ∈ X, there exists a sequence in A that converges to x.
Corollary 1.4 The set of rational numbers and the set of irrational numbers are dense in IR.
Proof. Let O be an open nonempty subset of IR. Then O contains an open interval I. But
any open inerval of R contains rational and irrational numbers. Therefore O meets Q as well
as IR\Q.
Other examples. a) IR∗ = IR − {0} is dense in IR.
b) IR − Z is dense in IR. This is because Z has an empty interior.
1.3. TOPOLOGICAL SPACES
13
Definition 1.9 Let A be a subset of a metric space X. A point x ∈ X is called a limit point
of A if every neighborhood of x meets A at a point different from x. The point x is called an
isolated point of A if x ∈ A and x is not a limit point of A.
It is clear that a limit point of A belongs to Ā and an isolated point of A belongs to A. So a
point in Ā is either a limit point of A or is isolated in A.
Examples. a) Let A =]0, 1[∪{2}. Then any point in [0, 1] is a limit point of A, whereas 2 is
an isolated point of A.
b) Every finite set in a metric space consists of isolated points. Indeed, let A = {x1 , . . . , xn }
be a finite set of n distinct points. For each i = 1, . . . , n,.let ri = minj6=i d(xi , xj ). Then
B(xi , ri ) ∩ A = {xi }.
c) Let A = { n1 | n = 1, 2, 3, . . .}. Then 0 is a limit point of A, and every point in A is isolated.
The following proposition will be proved in the exercises.
Proposition 1.9 Let A be a subset of a metric space X and let x be a limit point of A. Then
every neighborhood of x contains infinitely many points of A.
Definition 1.10 Let A be a set in a metric space X. The boundary of A denoted by ∂A is
◦
∂A = Ā − A.
Examples. 1) ∂[0, 1] = [0, 1]−]0, 1[= {0, 1}. Similarly, ∂]0, 1] = ∂[0, 1[= {0, 1}
◦
2) ∂Q = Q̄ − Q = IR.
3) We shall see in the exercises that if A is the open disk x2 + y 2 < 1, then ∂A is the circle
x2 + y 2 = 1.
Remark 1.6 The boundary of a set is always a closed set. Why?
Remark 1.7 ∂A = ∅ ⇔ A is clopen.
1.3
Topological spaces
Now that we have enough examples of metric spaces, we can move to the next level of abstraction.
Definition and examples
Definition 1.11 Let X be a set. A family T of subsets of of X is called a topology on X
provided the following conditions hold.
(i) ∅, X ∈ T .
(ii) An arbitrary union of elements of T belongs to T .
(iii) A finite intersection of elements of T belongs to T .
The elements of T are called open sets of X. A topological space is a couple (X, T ).
Remarks. 1) If T is a topology on X, then T ⊂ P(X).
2) To prove property (iii), it is enough to prove that the intersection of two elements of T is
also an element of T because then the result follows by induction.
14
CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
3) Suppose that T satisfies property (i). To prove (ii), we may assume that the sets are
nonempty, because the empty sets do not change the union and so we can remove them. To
prove (iii), we may assume that none of the sets is empty. Because otherwise the intersection
would be empty and so belongs to T by (i).
Examples.
1) A metric space is a topological space. This follows from Proposition 1.2.
But the converse is not true as we shall see. The topology defined from a distance is called
the topology generated by a distance. Let (X, T ) be a topological space. We say that T is
metrizable if T is generated by a distance.
2) Let X be a set. The collection {∅, X} is a topology called the trivial topology or the
indiscrete topology. This topology is not metrizable (see below).
3) Let X be a set. The collection P(X) of all subsets of X is a topology called the discrete
topology. This topology is metrizable because it is generated by the discrete distance. Observe
also that any topology on X, satisfies {∅, X} ⊂ T ⊂ P(X). The trivial topology is the smallest
(coarsest) topology and the discrete topology is the biggest (finest) topology on a set.
4) Let X be a set and A ⊂ X. Then {∅, A, X} is a topology on X. This topology is not
metrizable.
5) Let X = {a, b, c} be a three point set. Then {∅, {a}, {a, b}, X} is a topology on X. This
topology is not metrizable.
6) Let X be a set. Let T be the collection of all subsets of U ⊂ X such that X − U is finite or
equal to X. Then T is a topology on X called the finite complement topology. We prove
this.
(i) ∅ ∈ T because X − ∅ = X and X ∈ T because X − X = ∅ is finite.
(ii) Let Oα be a collection of nonempty elements of T . Then X − ∪α Oα = ∩α (X − Oα ) This
set is finite because it is an intersection of finite sets.
(iii) Let O1 and O2 be nonempty elements of T . Then X − (O1 ∩ O2 ) = (X − O1 ) ∪ (X − O2 )
is finite as a union of two finite sets.
Closed sets and related concepts
All the concepts that we learned in the previous section can be defined only in terms of open
sets. A set is closed if its complement is open. A neighborhood of a point is a set containing
an open set containing the point. The interior of a set is the biggest open set contained in the
set. The closure of a set is the smallest closed set containing the set. The boundary of a set is
the difference between the closure and the interior. Most of the results that we proved in the
metric case still hold in a general topological space. To be more precise,
Proposition 1.10 Propositions 1.3, 1.4, 1.6, 1.7 and the results in exercises 4,8,12,13,14,19,20
hold in an arbitrary topological space.
Let us now define the concept of convergence in a general topological space. In a metric
space, a sequence (xn ) is convergent to x if
∀ > 0 ∃n0 ∈ IN ∀n ≥ n0 d(xn , x) < .
This is equivalent to
∀ > 0 ∃n0 ∈ IN ∀n ≥ n0 xn ∈ B(x, ).
1.3. TOPOLOGICAL SPACES
15
Since a neighborhood of a point contains an open ball about this point, this condition is equivalent to the condition that very neighborhood U of x, contains xn for all n large enough. In
terms of quantifiers
∀U ∈ U(x) ∃n0 ∈ IN ∀n ≥ n0 xn ∈ U.
Now this condition involves only neighborhoods, so it depends only on the topology of the space
and we take as our definition of convergence in an arbitrary topological space.
Remark 1.8 In a general topological space, weird and unfamiliar situations can happen. For
example a sequence can converge to many points. Consider for instance IR equipped with the
finite complement topology and consider the sequence xn = n. Then (xn ) converges to every
x ∈ IR. Indeed, let U be neighborhood of x. Then IR − U is finite. Choose n0 > max(IR − U ).
Then xn = n ∈ U for all n ≥ n0 . This situation does not happen in metric spaces and more
generally in Hausdorff spaces (see below).
Proposition 1.8 and Corollary 1.3 remains partially true. More precisely,
Proposition 1.11 Let X be a topological space and let A ⊂ X. Consider the following
conditions.
(i) x ∈ Ā.
(ii) Every neighborhood of x intersects A.
(iii) There is a sequence in A which converges to x.
Then (i)⇔(ii) and (iii)⇒(i).
It is not true that (i)⇒(iii) in a an arbitrary topological space.
Corollary 1.5 Let X be a topological space and A ⊂ X. Consider the following conditions.
(i) A is dense in X.
(ii) Every nonempty open set of X meets A.
(iii) For every x ∈ X, there exists a sequence in A that converges to x.
Then (i)⇔(ii) and (iii)⇒(i).
Finally Proposition 1.9 does not hold in an arbitrary topological space. In the exercises
your are asked to give a counterexample.
Hausdorff spaces
Definition 1.12 A topological space X is called a Hausdorff space if every two distinct
points of X can be separated by two disjoint open sets. This means that if x 6= y then there
exists two disjoint open sets U and V such that x ∈ U and y ∈ V . A Hausdorff space is also
called a T2 space.
Proposition 1.12 A metrizable space is a Hausdorff space.
Proof. Choose a distance d that generates the topology of the space. Let x 6= y and let
r = d(x, y)/2 > 0. Then the open balls B(x, r) and B(y, r) are disjoint open sets containing x
and y respectively.
Corollary 1.6 A space which is not Hausdorff is not metrizable.
16
CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
Now you can see why some topologies defined above are not metrizable. The finite complement topology on an infinite space X is not Hausdorff and so not metrizable. In fact there are
no disjoint non empty open sets. For suppose that U ∩ V = ∅ with U, V 6= ∅. Then X − U and
X − V are finite. Therefore X − U ∪ X − V is finite but X − U ∪ X − V = X − (U ∩ V ) = X.
So X is finite contrary to our assumption. (The finite complement topology on a finite space
coincides with the discrete topology so it is metrizable).
There are however Hausdorff spaces which are not metrizable. You will encouter them in
your graduate studies.
Proposition 1.13 A finite set in a Hausdorff space is closed.
Proof.
It is enough to prove that singletons are closed. Consider a singletons {x}. Let
y ∈ X − {x}. We can separate x and y with disjoint sets U and V respectively. Then V ⊂
X − U ⊂ X − {x}. This means that X − {x} is a neighborhood of y. Since y was arbitrary,
this means that X − {x} is a neighborhood of all its points and so it is open.
Remark 1.9 A topological space set in which singletons are closed is called a T1 space. Therefore a T2 space is a T1 space. The converse is not true. For example, let X be an infinite set
equipped with the finite complement topology. Then X is a T1 space which is not T2 .
Proposition 1.14 In a Hausdorff space, a sequence converges to at most one point.
Proof. Suppose that there is a sequence (xn ) that converges to two distinct points x and y.
Let U and V be two neighborhoods of x and y respectively. Then xn ∈ U for all n ≥ n1 and
xn ∈ V for all n ≥ n2 . Then xn ∈ U ∩ V for all n ≥ max(n1 , n2 ). Contradiction.
Comparison of topologies
Definition 1.13 Let T and T 0 be two topologies on a set X. If T ⊂ T 0 , we say that T 0 is
finer or stronger than T ; we also say that T is coarser or weaker than T 0 . We say that T
and T are comparable if either T ⊂ T 0 or T 0 ⊂ T .
Examples. a) Let X = {a, b, c} and consider the following topologies T1 = {∅, {a}, {a, b}, X},
T2 = {∅, {a}, {b}, {a, b}, X} and T3 = {∅, {a}, {c}, {a, c}, X}. Then T2 is finer than T1 . However, T1 and T3 are not comparable. Also T2 and T3 are not comparable.
b) The finite complement topology on IR is weaker than the usual topology.
Suppose that we have two distances on the same set. Since each distance generates a
topology, we can ask if these topologies are comparable.
Definition 1.14 Let X be a set. Let d be distance on X that generates a topology T . Let
d0 be distance on X that generates a topology T 0 . We say that d and d0 are topologically
equivalent if T = T 0 , that is, if they generate the same topology.
We give some examples.
Proposition 1.15 Consider on IRN the three norms
kxk1 =
N
X
i=1
|xi |,
||x||2 =
N
X
i=1
2
|xi |
!1/2
,
kxk∞ = max(|x1 |, . . . , |xN |).
Let d1 , d2 and d∞ be respectively the distances associated with these norms. Then d1 , d2 and
d∞ generate the same topology on IRN .
1.3. TOPOLOGICAL SPACES
17
Proof. We prove that d2 and d∞ are equivalent. In the exercises, you are asked to prove
that d1 and d∞ are equivalent. It is not difficult to see that
√
||x||∞ ≤ ||x||2 ≤ N ||x||∞ .
It follows that
d∞ (x, y) ≤ d2 (x, y) ≤
Therefore,
Bd2 (x, r) ⊂ Bd∞ (x, r)
√
N d∞ (x, y).
r
and Bd∞ (x, √ ) ⊂ Bd2 (x, r).
N
Now let O be an open set for the topology generated by d∞ and let x ∈ O. Then there exists
r > 0 such that Bd∞ (x, r) ⊂ O. But Bd2 (x, r) ⊂ Bd∞ (x, r) and so Bd2 (x, r) ⊂ O. This means
that O is open for the topology generated by d2 .
Conversely, let O be an open set for the topology generated by d2 and let x ∈ O. Then there
exists r > 0 such that Bd2 (x, r) ⊂ O. But Bd∞ (x, √rN ) ⊂ Bd2 (x, r) and so Bd∞ (x, √rN ) ⊂ O.
This means that O is open for the topology generated by d∞ .
Remark 1.10 Let d and d0 be two two distances on the same set X. Then d generates a finer
topology than d0 if the following condition holds:
∀x ∈ X ∀r > 0 ∃δ > 0 Bd (x, δ) ⊂ Bd0 (x, r).
Proposition 1.16 Let (X, d) be a metric space. Set ρ(x, y) = min(1, d(x, y)). Then ρ and d
are topologically equivalent.
Proof. We know that ρ is a metric. Let x ∈ X. Observe that Bd (x, r) ⊂ Bρ (x, r) since
ρ(x, y) ≤ d(x, y). By the previous remark, d generates a finer topology than ρ.
Conversely, let O be open for d and let x ∈ O. Then there exists r < 1 such that Bd (x, r) ⊂
O. But then Bρ (x, r) ⊂ Bd (x, r). Indeed, let y ∈ Bρ (x, r). If d(x, y) > 1, then ρ(x, y) = 1 < r,
contradicting the choice of r. Therefore ρ(x, y) = d(x, y) < r. Conclusion: Bρ (x, r) ⊂ O and so
O is open for ρ.
Basis for a topology
A general way of defining a topology is by specifying a basis.
Definition 1.15 A subset B ⊂ P(X) is called a basis provided the following hold.
1. For all x ∈ X, there exists B ∈ B such that x ∈ B.
2. If B1 , B2 ∈ B and x ∈ B1 ∩ B2 , then there exists B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2 .
The elements of B are called basis elements. The topology T generated by B is defined as
follows: O ∈ T if whenever x ∈ O, there exists B ∈ B such that x ∈ B ⊂ O.
We need to check that T defined in this way is indeed a topology.
1. The empty set satisfies the condition vacuously. The first condition of a basis implies
thatX ∈ T .
2. Let (Oα )α∈A be a collection of elements of T . Let x ∈ ∪Oα . Then there is β ∈ A such
that x ∈ Oβ . Since Oβ ∈ T , there exists B ∈ B such that x ∈ B ⊂ Oβ and so B ⊂ ∪Oα .
This means that ∪Oα ∈ T .
18
CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
3. Let O1 , O2 ∈ T and let x ∈ O1 ∩ O2 . Then there exists B1 , B2 ∈ B such that x ∈ B1 ⊂ O1
and x ∈ B2 ⊂ O2 . The second condition of a basis implies that there exists B3 ∈ B such
that x ∈ B3 ⊂ B1 ∩ B2 . Therefore x ∈ B3 ⊂ B1 ∩ B2 ⊂ O1 ∩ O2 .
Remark 1.11 Observe that B ⊂ T .
Examples. 1) Let X be a metric space. The collection of balls B(x, r) (x ∈ X, r > 0) is
a basis for the metric topology. In particular, the collection of open disks and the collection
of open squares are bases for the usual topology of IR2 . The first condition of a basis is
satisfied. Let now x ∈ B(a1 , r1 ) ∩ B(a2 , r2 ). Let r = min(r1 − d(x, a1 ), r2 − d(x, a2 )). Then
B(x, r) ⊂ B(a1 , r1 ) ∩ B(a2 , r2 ).
2) Let X be a set. The collection of singletons of X is basis for the discrete topology on X.
3) The collection of all open intervals ]a, b[ of IR is a basis for the usual topology of IR.
4) Let X and Y be two topological spaces. Let B be the collection of all sets of the from U × V
where U is open in X and V is open in Y . Then B is a basis. The topology generated by this
basis is called the product topology of X × Y .
Proposition 1.17 Let B be a basis for a topology T on a set. Then T equals the collection
of all unions of elements of B.
Proof. Since B ⊂ T and T is stable under unions, a union of elements of B belongs to T .
Conversely, given O ∈ T , for each x ∈ O, there exists Bx ∈ B such that x ∈ Bx ⊂ O. Then
O = ∪x Bx .
Accumulation points and subsequences
Recall that a sequence (xn ) of a topological space is convergent to a limit x if every neighborhood
of x contains xn for all n large enough, that is except for a finite set of indices. Not all sequences
are convergent. A weaker notion than that limit is that of accumulation point or cluster point.
Definition 1.16 Let (xn ) be a sequence of a topological space. We say that x is an accumulation point or a cluster point of (xn ) if every neighborhood of x contains xn for infinitely
many indices n. In symbols
∀U ∈ U (x) ∀n ∈ IN ∃m ≥ n such that xm ∈ U.
This is equivalent to saying that {n ∈ IN|xn ∈ U } is infinite.
Compare with the condition of convergence
∀U ∈ U (x) ∃n0 ∈ IN ∀m ≥ n0 ⇒ xm ∈ U.
Examples. 1) It should be clear that a limit of a convergent sequence is an accumulation
point of the sequence. The converse however is not true as testified by the next example.
2) Let xn = (−1)n . Then (xn ) is not convergent in IR but has two accumulation points 1 and
−1.
Proposition 1.18 Let A be the set of accumulation points of a sequence (xn ). Then
A=
∞
\
n=1
{xk |k ≥ n} =
∞
\
n=1
{xn , xn+1 , . . .}.
1.4. SUBSPACES
19
Proof. Let Fn = {xn , xn+1 , . . .}. Then
x ∈ A ⇔ ∀U ∈ U (x) ∀n ∈ IN∗ ∃m ≥ n xm ∈ U
⇔ ∀n ∈ IN∗ ∀U ∈ U (x) U ∩ Fn 6= ∅
⇔ ∀n ∈ IN∗ x ∈ F¯n
⇔ x ∈ ∩F¯n .
Definition 1.17 Let {xn } be a sequence of some set and let n1 < n2 < · · · < ni < · · · be an
increasing sequence of positive integers. Then the sequence {yi } defined by yi = xni is called a
subsequence of the sequence {xn }.
Examples. a) Let xn = n. Then {x2n+1 } is a subsequence of {xn }.
b) Let xn = 2n . Then the sequence 1, 4, 16, 64, . . ., i.e. {x2n } is a subsequence of {xn }.
Proposition 1.19 Let (xn ) be a sequence of a topological space X.
(i) If (xn ) has a subsequence converging to `, then ` is an accumulation point of (xn ).
(ii) Conversely, if X is metrizable and ` is an accumulation of (xn ) then there exists a subsequence of (xn ) that converges to `.
Proof. (i). Let (xnk ) be a subsequence of (xn ) that converges to `. Let U be a neighborhood
of `. Then, there exists k0 such that xnk ∈ U for all k ≥ k0 . This means that xm ∈ U for all
m ∈ {nk0 , nk0 +1 , nk0 +2 . . .} and this set of integers is infinite.
(ii). In the condition
∀U ∈ U (x) ∀n ∈ IN ∃m ≥ n such that xm ∈ U,
take first U = B(`, 1) and n = 1. Then there exists n1 ≥ 1 such that xn1 ∈ B(`, 1). Next
take U = B(`, 12 ) and n = n1 + 1. Then there exists n2 ≥ n1 + 1 such that xn2 ∈ B(`, 12 ). At
the kth step, take U = B(`, k1 ) and n = nk−1 + 1. Then there exists nk ≥ nk−1 + 1 such that
xnk ∈ B(`, k1 ). This procedure defines a subsequence (xnk ) that converges to `.
1.4
Subspaces
Proposition 1.20 Let (X, T ) be a topological space and let Y ⊂ X. The collection
TY = {Y ∩ O|O ∈ T }
is a topology on Y called the subspace topology.
Therefore, in the subspace topology, the open sets of Y are the intersection of Y with the open
subsets of X.
Proof.
(i) Observe that ∅ = Y ∩ ∅ and Y = Y ∩ X. It follows that ∅, Y ∈ TY .
(ii) Let (Oα ) be a collection of elements S
of TY . Then
S there exists
S a collection (Uα )Sof elements
of T such that
S Oα = Y ∩ Uα . Then Oα = (Uα ∩ Y ) = ( Uα ) ∩ Y . Since Uα ∈ T , it
follows that Oα ∈ TY .
(iii) Let O1 , O2 ∈ TY . Then there exists U1 , U2 ∈ T such that O1 = Y ∩ U1 and O2 = Y ∩ U2 .
Then O1 ∩ O2 = Y ∩ (U1 ∩ U2 ) ∈ TY .
20
CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
Examples. 1) Consider [0, ∞[ as a subspace of IR with the usual topology. Then although
[0, 1[ is not open in IR, it is open in [0, ∞[ because for example [0, 1[= [0, ∞[∩] − 1, 1[.
2) Consider Z as a subspace of IR with the usual topology. Then each singleton {n} is open in
Z because {n} =]n − 1, n + 1[∩Z. It follows that that every subset of Z is open in the subspace
topology. This implies that the subspace topology and the discrete topology on Z coincide.
Corollary 1.7 Let X a be a topological space, let Y be a subspace of X and let U ⊂ Y . If U
is open in Y and Y is open in X, then U is open in X.
Proof. If U is open in Y , then U = Y ∩ V where V is open in X. Since Y is also open in X,
then so is Y ∩ V .
Remark 1.12 Let X be a topological space and let Z ⊂ Y ⊂ X. The the topology on Z
inherited from X is the same as the topology on Z inhereted from Y . Indeed if O is open in
X, then O ∩ Z = (O ∩ Y ) ∩ Z.
Proposition 1.21 Let Y be a subspace of a topological space X and let A ⊂ Y then A is
closed in Y if and only if A = Y ∩ F where F is closed in X.
Proof. We have to show first that if F is closed in X, then Y ∩ F is closed in Y , that is, its
complement in Y is open in Y . Now
Y − (Y ∩ F ) = Y ∩ (X − F ).
This set is open in Y because X − F is open in X.
Now conversely, suppose that A is closed in Y . Then Y −A is open in Y and so Y −A = Y ∩O
where O is open in X. Then A = Y ∩ (X − O). But X − O is closed in X. Hence the conclusion.
Example. ]0,1] is closed in ]0, ∞[ since ]0, 1] =]0, ∞[∩[0, 1].
Corollary 1.8 Let X a be a topological space, let Y be a subspace of X and let B ⊂ Y . If B
is closed in Y and Y is closed in X, then B is closed in X.
Let (X, d) be a metric space and let Y
Then dY is a metric on Y which therefore
relate to the subspace topology? It turns
denote by BY (x, r) the ball in Y of center
⊂ X. Let dY denote the restriction of d to Y × Y .
generates a topology on Y . How does this topology
out that the two topologies coincide. If x ∈ Y , we
and radius r. It coincides with Y ∩ B(x, r).
Proposition 1.22 The metric topology induced on Y coincides with the subspace topology of
Y.
Proof. Let A be open relative to the subspace topology. Then, A = Y ∩ O where O is open
in X. Let x ∈ A. Then x ∈ O. Since O is open there exists δ > 0 such that B(x, δ) ⊂ O. Then
B(x, δ) ∩ Y ⊂ O ∩ Y = A, that is, BY (x, δ) ⊂ A. This means that A is open relative to the
metric topology on Y .
Conversely, Suppose that A is open relative to the metric topology of Y . For each x ∈ A,
set δx = dY (x, Y − A). Then δx > 0 because dY (x, Y − A) = 0 ⇒ x ∈ Y − A = Y − A since
Y − A is closed in Y . We claim that if y ∈ Y and d(x, y) < δx then y ∈ A. For if y ∈
/ A, then
d(x, y) ≥ δx by definition of δx . It follows that A = ∪x∈A BY (x, δx ). Let now O = ∪x∈A B(x, δx ).
Then O is open in X and O ∩ Y = A.
1.5. PRODUCT SPACES
1.5
21
Product spaces
If X1 , X2 . . . , Xn are sets, Q
the set of n−tuples (x1 , . . . , xn ) where each xi ∈ Xi is denoted by
X1 × X2 × · · · × Xn or by ni=1 Xi .
Suppose now that X1 , . . . , Xn are topological space. If Ui is an open set of Xi for each
i = 1, . . . , n, their product U1 × U2 × · · · × Un will be called an open rectangle. Now, itQis easy
to check that the collection B of allQ
open rectangles is a basis (indeed, first, any x ∈ ni=1 Xi
is contained in the open rectangle ni=1 Xi ; second, the intersection of two open rectangles
is an open retangle). The topology generated by this basis is called the product topology of
X1 × X2 × · · · × Xn .
This means that a set O ⊂ X1 × X2 × · ·Q
· × Xn is open if whenever x = (x1 , . . . , xn ) ∈ O,
there exist open sets Ui ⊂ Xi such that x ∈ ni=1 Ui ⊂ O.
Proposition 1.23 The product of open sets is open. The product of closed sets is closed.
Proof. The first part is rather trivial because a product of open sets is a basis element and
therefore belongs to the product topology. For the second part, it is enough to prove that the
product of two closed sets is closed (then the result follows by induction).
Let A ⊂ X1 and
S
B ⊂ X2 be closed sets. Then X1 × X2 − A × B = (X1 − A1 ) × X2 X1 × (X2 − B) which is
the union of two open sets.
In the exercises, you are asked to prove the following proposition.
Proposition 1.24 Let (X1 , d1 ), . . . , (Xn , dn ) be n metric spaces. Let X = X1 × · · · × Xn . For
x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ), define
ρ1 (x, y) =
n
X
di (xi , yi )
i=1
ρ2 (x, y) =
n
X
di (xi , yi )2
i=1
!1/2
ρ∞ (x, y) = max di (xi , yi ).
i=1...,n
Then ρ1 , ρ2 and ρ∞ are metrics on X that generate the product topology on X.
22
CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES
Chapter 2
Continuous functions and
homeomorphisms
Let us start by defining continuity in metric spaces. Let f : (X, dX ) → (Y, dY ) be a function
between two metric space and let x ∈ X. We say that f is continuous at x if f (y) can be made
arbitrarily close to f (x) for y sufficiently close to x. In metric spaces the concept of a distance
gives a meaning to closeness. This leads to the following definition. We can think of ε as the
accuracy in the approximation.
Definition 2.1 Let f : (X, dX ) → (Y, dY ) be a function between two metric spaces and let
x ∈ X. f is said to be continuous at x if the following condition holds.
∀ε > 0 ∃δ > 0 such that dX (x, y) < δ ⇒ dY (f (x), f (y)) < ε.
Remark 2.1 In the definition above, one can replace ”<” by ”≤”.
Examples. 1) A constant function is continuous.
2) The identity function is continuous.
3) Let (X, d) be a metric space and a ∈ X. The function x 7→ d(x, a) is continuous at every point
because |d(x, a) − d(y, a)| ≤ d(x, y). More generally, if A ⊂ X, then the function x 7→ d(x, A)
is continuous at every point because |d(x, A) − d(y, A)| ≤ d(x, y).
4) All the elementary functions of calculus are continuous on their domains. The elementary
functions are the rational functions (ratio of polynomials), the logarithm, the exponential, the
trigonometric and inverse trigonometric functions.
Now there are several equivalent characterizations of continuity and we shall bring out a
condition which is independent of the metrics so that it will serve as a definition of continuity
in general topological spaces.
Proposition 2.1 Let f : (X, dX ) → (Y, dY ) be a function between two metric spaces and let
x ∈ X. Then the following are equivalent.
(i) f is continuous at x.
(ii) ∀ε > 0, ∃δ > 0 such that f (B(x, δ)) ⊂ B(f (x), ε).
(iii) ∀V ∈ U(f (x)), ∃U ∈ U(x) such that f (U ) ⊂ V .
(iv) If {xn } is a sequence of X that converges to x then {f (xn )} converges to f (x).
23
24
CHAPTER 2. CONTINUOUS FUNCTIONS AND HOMEOMORPHISMS
(v) If {xn } is a sequence of X that converges to x then there exists a subsequence (xnk ) of
(xn ) such that {f (xnk )} converges to f (x).
Proof. We prove that (i)⇒(ii)⇒(iii)⇒(iv)⇒(v)⇒(i).
(i)⇒(ii). (ii) is just a reformulation of (i). Indeed, let ε and δ be as in the definition of continuity.
Let y ∈ f (B(x, δ)). Then y = f (z) for some z ∈ B(x, δ). This means that d(z, x) < δ and
so by the condition of continuity, d(f (z), f (x)) < ε, that is, d(y, f (x)) < ε. This means that
y ∈ B(f (x), ε).
(ii)⇒(iii). Let U be a neighborhood of f (x). Then there exists a ball B(f (x), ε) ⊂ V . Take now
U = B(x, δ). Then U is a neighborhood of x. Now (ii) can be written as f (U ) ⊂ B(f (x), ε) ⊂ V .
(iii)⇒(iv). Let (xn ) be a sequence of X that converges to x. Let V be a neighborhood of f (x).
Then there exists a neighborhood U of x such that f (U ) ⊂ V . Now convergence means that
xn ∈ U for all n large enough. Therefore f (xn ) ∈ f (U ) ⊂ V for n large enough. This means
that (f (xn )) converges to f (x).
(iv)⇒(v). Is clear.
(v)⇒(i). Suppose that f is not continuous at x. Then there exists ε > 0 such that for all δ > 0
there exists y with d(y, x) < δ and d(f (y), f (x)) ≥ ε. Taking δ = n1 we get a sequence (yn )
converging to x and such that no subsequence of (f (yn )) converges to f (x), a contradiction. The only condition which does no involve metrics is condition (iii) and this will be our
definition of continuity in a general topological space.
Definition 2.2 A function f : X → Y between two topological space is said to be continuous
at x ∈ X, if for every neighborhood V of f (x), there exists a neighborhood U of x such that
f (U ) ⊂ V .
Remark 2.2 We can replace ”neighborhood” by ”open neighborhood”.
Remark 2.3 If f is continuous at x and (xn ) is a sequence converging to x, then the sequence
(f (xn )) converges to f (x). This follows from the definitions. The converse is however not true,
i.e. if for every sequence (xn ) converging to x we have (f (xn )) converges to f (x), it does not
follow that f is continuous at x. The converse is true if X is metrizable.
Definition 2.3 A function f : X → Y between two topological spaces is called continuous if
it is continuous at every point of X.
Theorem 2.1 Let f : X → Y be a function between two topological spaces. Then the following
are equivalent.
(i) f is continuous.
(ii) The inverse image under f of any open subset of Y is open in X.
(iii) The inverse image under f of any closed subset of Y is closed in X.
Proof. (i)⇒(ii). Let O ⊂ Y be open and let x ∈ f −1 (O) (so that f (x) ∈ O). Since O is a
neighborhood of f (x), there exists according to the previous proposition, a neighborhood U of
x such that f (U ) ⊂ O. Therefore U ⊂ f −1 (f (U )) ⊂ f −1 (O). This means that f −1 (O) is open.
(ii)⇒(i). Let x ∈ X and let V be an open neighborhood of f (x). Set U = f −1 (V ). Then U
is a neighborhood of x (being an open set containing x) and f (U ) = f (f −1 (V )) ⊂ V . According
to the previous proposition, f is a continuous at x. Since x was arbitrary, this proves that f is
continuous.
The equivalence (ii)⇔(iii) follows from the fact that f −1 (Y − O) = X − f −1 (O).
25
Corollary 2.1 Let X be a topological space and f : X → IR be a continuous function. Let
a ∈ IR. Then
(i) The sets {x ∈ X|f (x) < a} and {x ∈ X|f (x) > a} are open.
(ii) The sets {x ∈ X|f (x) ≤ a}, {x ∈ X|f (x) ≥ a} and {x ∈ X|f (x) = a} are closed.
Example. Since in a metric space x 7→ d(x, a) is continuous, an open ball is open and a closed
ball is closed. The set {x ∈ X|d(x, a) > r} is open, the set {x ∈ X|d(x, a) ≥ r} is closed. The
sphere {x ∈ X|d(x, a) = r} is closed.
Proposition 2.2 The following are true.
(i) The composition of two continuous functions is continuous.
(ii) The restriction of a continuous function is continuous.
(iii) If A ⊂ X, the inclusion map j : A → X is continuous.
(iv) If f : X → Y is continuous, then f : X → f (X) is also continous.
Proof. (i). Let f : X → Y and g : Y → Z be two continuous functions. We have to prove
that g ◦ f : X → Z is continuous. Observe that (g ◦ f )−1 (O) = f −1 (g−1 (O)) for every subset
O ⊂ Z. In particular, if O is open, then g−1 (O) is open in Y because g is continuous. Since f
is continuous, we have f −1 (g−1 (O)) is open. Hence the result.
(ii). Let f : X → Y be continuous and let A ⊂ X. We have to prove that f|A : A → Y is also
−1
continuous. Let O be open in Y . Then f|A
(O) = A ∩ f −1 (O) is open in A.
(iii). Let O be open in X. Then j −1 (O) = A ∩ O is open in A.
(iv). Let O be open in f (X). Then O = O0 ∩f (X) where O0 is open in Y . But f −1 (O) = f −1 (O0 )
and f −1 (O0 ) is open in X.
Proposition 2.3 For real valued functions, the sum and product of continuous functions are
continuous. If f : X → IR is continuous, then 1/f is continuous on its domain, i.e. on the set
where f does not vanish.
Proof. (i). Let f, g : X → IR be continuous at some point x0 . Then, given > 0, there exists a
neighborhood U1 of x0 such that |f (x)−f (x0 )| < /2. Also, there exists a neighborhood U2 of x0
such that |g(x)−g(x0 )| < /2. Then for all x ∈ U1 ∩U2 , we have |f (x)+g(x)−(f (x0 )+g(x0 )| < .
(ii). Write f (x)g(x) − f (x0 )g(x0 ) = f (x)g(x) − f (x0 )g(x) + f (x0 )g(x) − f (x0 )g(x0 ). Then, by
the triangle inequality
|f (x)g(x) − f (x0 )g(x0 )| ≤ |g(x)||f (x) − f (x0 )| + |f (x0 )||g(x) − g(x0 )|
Let > 0 be given. The continuity of g at x0 implies that there exists a neighborhood U1 of x0
such that |g(x)−g(x0 )| < 2(|f (x0 )|+1) for all x ∈ U1 . This implies that |f (x0 )||g(x)−g(x0 )| < /2
for all x ∈ U1 . The continuity of g at x0 also dictates that |g(x)| < |g(x0 )| + 1 for all x in
some neighborhood U2 of x0 . Finally, the continuity of f at x0 dictates that there exists a
neighborhood U3 of x0 such that |f (x) − f (x0 )| < 2(|g(x0 )|+1) for all x ∈ U3 . Therefore for
x ∈ U1 ∩ U2 ∩ U3 , we have |f (x)g(x) − f (x0 )g(x0 )| < .
(iii). Let x0 be a a point where f (x0 ) 6= 0. Then |f (x0 )| > 0. Then, there exists a neighborhood
U1 of x0 such that |f (x) − f (x0 )| < |f (x0 )|/2. Then |f (x)| > |f (x0 )|/2 and so |f (x)||f (x0 )| >
26
CHAPTER 2. CONTINUOUS FUNCTIONS AND HOMEOMORPHISMS
|f (x0 )|2 /2 for all x ∈ U1 . Given > 0 there exists a neighborhood U2 of x0 such that |f (x) −
2
f (x0 )| < |f (x2 0 )| for all x ∈ U2 . It follows that if x ∈ U1 ∩ U2 then
1
1 |f (x) − f (x0 )|
f (x) − f (x0 ) = |f (x)f (x0 )| < .
Given two sets X and Y , the projection π1 : X × Y → X is defined by π1 (x, y) = x.
The projection π2 : X × Y → X is defined by π2 (x, y) = y. Observe that if U ⊂ X then
π1−1 (U ) = U × Y and if V ⊂ Y then π2−1 (V ) = X × V . Therefore we have
Proposition 2.4 Let X and Y be topological spaces then the projections on X and Y are
continuous (X × Y is equipped with the product topology).
Remark 2.4 In fact, the product topology is the smallest topology that makes the projections
continuous. Indeed, let X and Y be two topological spaces and let T be a topology on X×Y that
makes the projections π1 and π2 continuous. We have to prove that T contains the product
topology, and for this it is enough to prove that any basis element of the product topology
belongs to T . So let U and V be open sets of X and Y respectively. We already observed
that π1−1 (U ) = U × Y . Since π1 is continuous with respect to T , it follows that U × Y ∈ T .
Similarly, X × V ∈ T . Therefore their intersection U × V also belongs to T .
Proposition 2.5 Let X, Y, Z be topological spaces and let f : X → Y × Z. Let f1 and f2
denote the components of f . Then f is continuous if and only if f1 and f2 are continuous.
Proof. Suppose first that f is continuous. Observe that if πY and πZ denote the projections
into Y and Z respectively, then f1 = πY ◦f and f2 = πZ ◦f . Since the projection are continuous,
it follows that f1 and f2 are continuous.
Suppose conversely that f1 and f2 are continuous. We show that the inverse image under f
of a basis element U × V is open in X. This follows from the fact that
f −1 (U × V ) = f1−1 (U ) ∩ f2−1 (V ).
Proposition 2.6 Let f : X × Y → Z be continuous. Then the partial map f (·, y) : X → Z
and f (x, ·) : Y → Z are continuous.
Proof. Let g(x) = f (x, y) for y fixed. Let α(x) = (x, y). Then α : X → X × Y is continuous
(because each component is continuous). Since g = f ◦ α, it follows that that g is continuous.
Similarly for the partial map f (x, ·).
Remark 2.5 If f : X × Y → Z and the partial maps f (·, y) : X → Z and f (x, ·) : Y → Z are
continuous, it does not follow that f is continuous. For example, let f : IR × IR → IR be defined
by f (x, y) = x2xy
+y 2 if (x, y) 6= (0, 0) and f (0, 0) = 0. Then the partial maps are continuous.
However f is not continuous at (0,0) because it has not limit there.
Definition 2.4 A function f : X → Y between two topological spaces is called a homeomorphism if it is bijective, continuous and its inverse is also continuous.
The concept of homeomorphism is fundamental in topology as is the concept of isomorphism
in algebra. Recall that two isomorphic groups have the same structure and therefore may be
identified. Similarly, two homeomorphic space have exactly the same topological properties and
therefore can be identified.
27
Examples.
1) The function cos : [0, π] → [−1, 1] is a homeomorphism whose inverse is
arccos : [−1, 1] → [0, π]. Continuity of these functions is proved in the first year.
2) The function sin : [− π2 , π2 ] → [−1, 1] is a homeomorphism whose inverse is arcsin : [−1, 1] →
[− π2 , π2 ].
3) The function tan :] − π2 , π2 [→ IR is a homeomorphism whose inverse is arctan : IR →] − π2 , π2 [.
4) The function f : IR → IR defined by f (x) = x3 is a homeomorphism whose inverse is given
by f −1 (x) = x1/3 .
Remark 2.6 A continuous bijection need not be a homeomorphism. Indeed, let T be the usual
topology on IR and let TF be the finite complement topology on IR. Let i : (IR, T ) → (IR, TF )
be the identity function. Then i is clearly a bijection. It is continuous because if O ∈ TF then
i−1 (O) = O ∈ T . However, i−1 : (IR, TF ) → (IR, T ) is not continuous because there exists an
element O ∈ T and O ∈
/ TF .
Here is another example. Let S 1 denote the unit circle. The function f : [0, 2π[→ S 1 given
by f (t) = (cos t, sin t) is a continous bijection. However, it is not a homeomorphism. Indeed,
one can check that its inverse is given by
(
arccos x
if y ≥ 0
−1
f (x, y) =
π + arccos(−x) if y < 0.
Then lim(x,y)→(1,0+) f −1 (x, y) = arccos(1) = 0, whereas lim(x,y)→(1,0−) f −1 (x, y) = π+arccos(−1) =
2π. So f −1 is not continuous. Note that in contrast, f :]0, 2π[→ S 1 \{(1, 0)} is a homeomorphism
(the limits above do not belong to the space ]0, 2π[).
Remark 2.7 In contrast, let I and J be intervals of IR. A continuous bijection f : I → J is
automatically a homeomorphism. This will be proved in the exercises.
Remark 2.8 Let T and T 0 be two topologies on a set X and consider the identity map i :
(X, T ) → (X, T 0 ) defined by i(x) = x. Then T = T 0 if and only if i is a homeomorphism.
Proposition 2.7 All intervals of the real line of the form [a, b] with a < b are homeomorphic.
Proof. The map x 7→ (x − a)/(b − a) is a homeomorphism between [a, b] and [0, 1].
Proposition 2.8 All open intervals of the real line are homeomorphic.
Proof. The map x 7→ (x − a)/(b − a) is a homeomorphism between ]a, b[ and ]0, 1[. Therefore
all intervals of the form ]a, b[ are homeomorphic between each other. The function x 7→ tan x
is a homeomorphism between ] − π/2, π/2[ and IR. The interval ]a, +∞[ is homeomorphic to IR
by x 7→ ln(x − a). The interval ] − ∞, a[ is homeomorphic to IR by x 7→ ln(a − x).
Other examples.
(x, y) 7→ (ax, by)).
1) A circle and an ellipse are homeomorphic (for example by the map
2) Rotations, translations, homotheties and symmetries of the plane or space are homeomorphisms.
3) The punctured disc {0 < x2 + y 2 < 1} and the annulus {1 < x2 + y 2 < 2} are homeomorphic.
X
−1 (X) = ||X||−1 X. Then f is a
For X = (x, y) let f (X) = X + ||X||
= ||X||+1
||X|| X. Then f
||X||
homeomorphism.
4) A circle and the boundary of a square are homeomorphic. To be specific consider the unit
circle S 1 defined by x2 + y 2 = 1 and the square C, defined by |x| + |y| = 1. Let f : IR2 → IR2
be given by f (x, y) = (x|x|, y|y|). Then f is continuous because its components are continuous.
28
CHAPTER 2. CONTINUOUS FUNCTIONS AND HOMEOMORPHISMS
−1 (x, y) = (g(x), g(y)) where g(x) = √x if x ≥ 0 and
f is bijective
and
its
inverse
is
given
by
f
√
g(x) = − −x if x ≤ 0. By the pasting lemma, g is continuous (draw the graph of g). It follows
that f −1 is also continuous. Thus f is a homeomorphism.
Next, we prove that f (S 1 ) = C. Indeed, let first (x, y) ∈ S 1 , then x2 + y 2 = 1 and so
|x|x|| + |y|y|| = 1. This means that f (x, y) ∈ C and so f (S 1 ) ⊂ C. Conversely, let (u, v) ∈ C.
Then there exists (x, y) ∈ IR2 such that f (x, y) = (x|x|, y|y|) = (u, v). Then x2 = |u| and
y 2 = |v|. Since |u| + |v| = 1, then x2 + y 2 = 1 i.e (x, y) ∈ S 1 .
In fact, f maps the open unit disc onto the open square, and the exterior of the disc onto
the exterior of the square.
1/2
5) Equip IR2 with the Euclidean distance d((x, y), (x0 , y 0 )) = (x − x0 )2 + (y − y 0 )2
(but
0
any other equivalent distance would do). Equipp C
l with the distance d(z, z ) = |z − z 0 | =
1/2
. Then IR2 is homeomorphic to C.
l
Indeed, consider the function
(x − x0 )2 + (y − y 0 )2
2
f : IR → C
l defined by f (x, y) = x + iy. Then f is a bijection. Next, f preserves the distances
(we say that f is an isometry). Therefore f is continuous. Since f −1 also preserves distances,
it is continuous as well.
6) A circle with a point removed is homeomorphic to IR. Let S 1 denote the unit circle in IR2
and let P denote the point with coordinates (0,1). The stereographic projection with pole P
associates to each point M ∈ S 1 − P the point N which is the point of intersection of the line
(P M ) with the horizontal line {y = 0}. If M has coordinates (x, y), then the point N has
x
coordinates ( 1−y
, 0). Indeed, let N have coordinates (X, 0). Since N belongs to (P M ), there
exists λ ∈ IR such that P~N = λP~M and so (X − 0, 0 − 1) = λ(x − 0, y − 1). It follows that
x
. This defines a continuous function F : S 1 − P → IR given by
λ = 1/(1 − y) and so X = 1−y
F (x, y) =
It inverse is given by
x
.
1−y
1
(2X, X 2 − 1).
+1
x
Indeed, if X = 1−y
, then X − x = yX. Squaring and using the fact that x2 + y 2 = 1, we get
F −1 (X) =
x=
2X
,
X 2 +1
and then y =
X 2 −1
.
X 2 +1
X2
It is clear that F −1 is continuous.
7) A sphere with a point removed is homeomorphic to IR2 . Let S 2 denote the unit sphere in
IR3 defined by x2 + y 2 + z 2 = 1 and let P denote the point with coordinates (0,0,1). The
stereographic projection with pole P associates to each point M ∈ S 2 − P the point N which is
the point of intersection of the line (P M ) with the plane {z = 0}. If M has coordinates (x, y, z),
y
x
, 1−z
, 0). Indeed, let N have coordinates (X, Y, 0). Since
then the point N has coordinates ( 1−z
N belongs to (P M ), there exists λ ∈ IR such that P~N = λP~M and so (X − 0, Y − 0, 0 − 1) =
y
x
λ(x − 0, y − 0, z − 1). It follows that λ = 1/(1 − z) and so X = 1−z
and Y = 1−z
. This defines
2
2
a homeomorphism F : S − P → IR given by
x
y
F (x, y, z) =
,
1−z 1−z
whose inverse is given by
F −1 (X, Y ) =
X2
1
(2X, 2Y, X 2 + Y 2 − 1).
+Y2+1
8) More generally the unit sphere S n (defined by x21 + x22 + · · · x2n+1 = 1) with the north pole
removed is homeomorphic to IRn via the stereographic projection
x1
x2
xn
F (x1 , x2 , . . . , xn+1 ) =
,
,··· ,
.
1 − xn+1 1 − xn+1
1 − xn+1
29
The inverse is given by
n
X
1
,
2X
,
·
·
·
,
2X
,
Xi2 − 1).
(2X
1
2
n
2+1
X
i=1 i
i=1
F −1 (X1 , X2 , . . . , Xn ) = Pn
Definition 2.5 Let X and Y be two topological spaces. A topological imbedding or simply
an imbedding of X in Y , is a map f : X → Y such that f : X → f (X) is a homeomorphism.
The existence of an imbedding of X in Y , means that we can place a copy of X inside Y .
Examples. 1) Let f : IR → IR2 be defined by f (x) = (x, 0). Then f is an imbedding. Here
the copy of IR is a straight line.
2) The map g : IR → IR2 defined by g(x) = (x, x3 ) is an imbedding. Here the topological copy
of IR is the curve of equation y = x3 .
3) The map g : IR → IR2 defined by g(x) = (e−x cos x, e−x sin x) is an imbedding. Here, the
topological copy of IR is an infinite spiral.
4) The map g : [0, 1] →]0, 1[ defined by g(x) =
1
3
+ 16 x is an imbedding.
30
CHAPTER 2. CONTINUOUS FUNCTIONS AND HOMEOMORPHISMS
Chapter 3
Compactness
Now we move to another important concept in topology. Compactness of in interval [a, b]
was formulated first in terms of sequences and subsequences. However, it took some time for
mathematicians to formulate the modern notion of compactness in a general topological space.
This is true for many abstract concepts of Mathematics. You usually wonder how did we arrive
at an abstract concept or a nontrivial proof. In fact, in the courses that we teach, we give you
a synthesis of a long process of distillation of mathematical ideas.
3.1
Definitions, examples and properties
Definition 3.1 Let X be
S a topological space. A collection of subsets (Oλ )λ∈L of X is called a
covering of X if X =
Oλ . The collection is called an open covering if its elements are
open.
λ∈L
Definition 3.2 A topological space X is called compact if every open
S covering of X contains
a finite subcollection that also covers X. Otherwise stated, if X =
Oλ , where each Oλ is
λ∈L
S
open, then there is a finite set J ⊂ L such that X =
Oλ .
λ∈J
If (Oλ )λ∈L is a covering of X, a subcollection (Oλ )λ∈J that also covers X is sometimes called a
subcovering of the covering (Oλ )λ∈L . Then, compactness can be also refolmulated as follows:
X is compact if from every open covering of X, we can extract a finite subcovering.
Examples. a) Any finite space is compact.
b) A space having a finite topology is compact.
c) A space X with the finite complement topology is compact. Indeed, let (Oλ ) be an open
covering of X. We may assume that X 6= ∅ (otherwise there is nothing to prove). Then there
exists an nonempty element Oλ0 . Then, by definition of open sets, X − Oλ0 is a finite set
{x1 , . . . , xn }. Now each xi belongs to an element Oλi . Therefore we have X = Oλ0 ∪ Oλ1 ∪ · · · ∪
Oλn . Thus, X is covered by finitely many elements of the original covering. Since any subspace
of X has the finite complement topology, it follows that any subspace of X is compact.
d) Consider X = {0} ∪ { n1 |n ∈ IN∗ } as a subspace of IR. Then X is compact. Indeed let (Oλ ) be
an open covering of X. Then 0 belongs to some Oλ0 . Since ( n1 ) converges to 0, Oλ0 contains all
1
1
1
1
n except finitely many terms 1, 2 , . . . , n0 . Now each i belongs to some open set Oλi . It follows
that X = Oλ0 ∪ Oλ1 ∪ · · · ∪ Oλn . The argument that we used shows that more generally, if (xn )
is a sequence of a topological space that converges to x, then the subspace {xn |n ∈ IN∗ } ∪ {x}
is compact.
31
32
CHAPTER 3. COMPACTNESS
e) The real line is not compact. Indeed, IR =
S
n∈IN
] − n, n[ and no subfamily of (] − n, n[)n can
cover IR.
S 1
f) (0, 1] is not compact. For ]0, 1] =
] n , 1] and no finite subfamily of (] n1 , 1])n can cover ]0,1].
n≥1
{ n1 |n
∗
g) The subspace Y =
∈ IN } of IR is not compact. Indeed, { n1 }n is an open covering of Y
that has no finite subcollection that covers Y . Therefore removing a point of a compact space
can destroy the property of compactness.
Remark 3.1 Following Bourbaki, most French mathematicians include the Hausdorff condition in the definition of compactness. This is not the case with American and Russian mathematicians. We follow the American and Russian defintion. In the french litterature, what we
call compact is called quasi-compact.
Compactness can also be formulated in terms of closed sets.
Definition
3.3 A collection (Fλ )λ∈L of sets is said to have the finite intersection property
T
if λ∈J Fλ 6= ∅ for every finite subcollection (Fλ )λ∈J .
has the finite intersection
Example. Set for λ ∈ IR, Fλ = [λ, ∞[. Then the collection (Fλ )λ∈IR T
property. Indeed, let J = {λ1 , . . . , λn } be a finite subset of IR. Then λ∈J Fλ = Fλ1 ∩ Fλ2 · · · ∩
F
Tλn = [t, ∞[6= ∅ where t = max(λ1 , . . . , λn ). Observe however that the whole intersection
λ∈IR Fλ = ∅ because there is no real number bigger than every other real number.
Taking complements in the definition of compactness, one can show the following (write the
proof properly as an exercise).
Theorem 3.1 Let X be a topological space. Then X is compact if and only if for every collection (Fλ )λ∈L of closed sets of X having the finite intersection property, the whole intersection
T
λ∈L Fλ is nonempty.
Corollary 3.1 Let X be topological space and let T
(Fn ) be a decreasing sequence of nonempty
closed subsets of X (that is, F1 ⊃ F2 ⊃ · · · ). Then Fn 6= ∅.
Proof. The collection (Fn ) has the finite intersection property beccause Fn1 ∩ · · · ∩ Fnk =
Fmax ni 6= ∅.
Now we prove some facts about subspaces. If Y is a subspace of aStopological space X, a
collection (Oλ ) of subsets of X is a covering of Y (or covers Y ) if Y ⊂ Oλ .
Lemma 3.1 Let Y be a subspace of a topological space X. Then Y is compact if and only if
every covering of Y by open subsets of X contains a finite subcollection that also covers Y .
Proof. This follows from the definition of the subspace topology. Write the proof properly
as an exercise.
Proposition 3.1 Every closed subspace of a compact space is compact.
Proof. Let Y be a closed subspace of a topological space X. Let (Oλ )λ∈L be a covering
of Y by open subsets of X. Add to this covering the open set X − Y . Otherwise stated let
µ be an element not in L and set Oµ = X − Y and consider the new collection (Oλ )λ∈L∪{µ} .
Then this new collection is an open covering of X. Compactness of X implies that a finite
subcollection (Oλ )λ∈J covers X and consequenlty also Y . Then the collection (Oλ )λ∈J−{µ} is a
finite subcollection of (Oλ )λ∈L that covers Y .
Theorem 3.2 Every compact subspace of a Hausdorff space is closed (in the space).
3.1. DEFINITIONS, EXAMPLES AND PROPERTIES
33
Proof. Let X be a Hausdorff space and let K ⊂ X be compact. We prove that {K is open.
Let x ∈
/ K. For any y ∈ K there is an open S
neighborhood Uy of x and an open neighborhood
Vy of y such that Uy ∩ Vy = ∅. Now K ⊂
Vy . Since K is compact K ⊂ Vy1 ∪ · · · ∪ Vyn .
y∈K
Then U = Uy1 ∩ · · · ∩ Uyn is a neighborhood of x not intersecting K, that is, contained in the
complement of K.
Remark 3.2 If the Hausdorff condition is removed in the theorem above, then the conclusion
may fail. Consider an infinite set X with the finite complement topology. Then the closed
subsets of X are X and the finite sets. However any subspace of X is compact.
Theorem 3.3 The image of a compact space under a continuous map is compact. In particular compactness is preserved by homeomorphism. We say that compactness is a topological
property.
Proof. Let f : X → Y be continuous with X compact. Let (Oλ )λ∈L be a covering of f (X)
by open subsets of Y . Then the collection (f −1 (Oλ ))λ is an open covering of X (because f
is continuous). Compactness of X implies that finitely many subsets f −1 (Oλ1 ), · · · , f −1 (Oλn )
cover X. It follows that {Oλ1 , . . . , Oλn } cover f (X).
Here is a useful criterion involving compactness for proving that a continuous bijection is
actually a homeomorphism.
Theorem 3.4 Let f : X → Y be a continuous bijection. If X is compact and Y is Hausdorff,
then f is a homeomorphism.
Proof. We have to prove that f −1 is continuous and this is equivalent to proving that f
maps closed sets into closed sets (because f = (f −1 )−1 ). So let A ⊂ X be closed. Then A is
compact. By the preceding theorem, f (A) is compact. Since Y is Hausdorff, f (A) is closed in
Y.
Our next theorem is that the product of compact spaces is compact. For this we need a
lemma.
Lemma 3.2 (The tube lemma) Let X and Y be two topological spaces with Y compact. If
N is an open set of X × Y conatining the slice {x0 } × Y , then N contains a tube W × Y where
W is a neighborhood of x0 in X.
Proof. Draw a figure and explain. If y ∈ Y , then (x0 , y) ∈ N . Then, by definition of the
product topology, there exists a basis element Uy × Vy containing (x0 , y) and contained in N .
Then the collection (Uy × Vy )y∈Y is an open covering of x0 × Y . But x0 × Y is compact being
homeomorphic to Y . Therefore, there exist finitely many basis elements U1 ×V1 , . . . Un ×Vn that
cover x0 ×SY . Let W = U1 ∩ · · · ∩ Un . Then W is an open neighborhood of x0 . We
Snclaim that
n
W × Y ⊂ i=1 Ui × Vi . Indeed, let (x, y) ∈ W × Y . The point (x0 , y) ∈ x0 × Y ⊂ i=1 Ui × Vi .
So y ∈ Vj for some j. But x ∈ W ⊂ Uj . It followsSthat (x, y) ∈ Uj × Vj and the claim is proved.
But by construction, Ui × Vi ⊂ N for all i and so ni=1 Ui × Vi ⊂ N . Consequently, W × Y ⊂ N .
Theorem 3.5 A product of compact spaces is compact.
Proof. It is enough to prove the theorem for the product of two compact spaces since the
general result follows by induction on the number of spaces. So let X and Y be two compact
spaces. Let (Oλ ) be open covering of X × Y . Given x ∈ X, the slice x × Y is compact
and therefore
can be covered by finitely many elements Oλ1 , . . . , Oλn of the covering. Let
S
N = ni=1 Oλi . By the tube lemma, N contains a tube Wx × Y containing x × Y . Now the
34
CHAPTER 3. COMPACTNESS
collection (Wx )x∈X is an open covering of X. Compactness of X implies that there exists a finite
subcollection Wx1 , . . . , Wxk that covers X. Therefore the union of the tubes Wx1 ×Y, . . . , Wxk ×Y
is X × Y . Since each of these tubes is covered by finitely many Oλ , then so is X × Y .
3.2
Compact subspaces of IRn
Theorem 3.6 Any closed interval [a, b] of the real line is compact.
S
Oλ where each Oλ is open in IR . We need to prove that [a, b] is
Proof.
Let [a, b] ⊂
λ∈L
covered by finitely many Oλ . We divide the proof into four steps.
Step 1. We show that if x ∈ [a, b[, then there is y ∈]x, b] such that [x, y] is covered by exactly
one Oµ . Thus let x ∈ [a, b[. Then x ∈ ∪λ∈L Oλ , and so there is an index µ ∈ L such that x ∈ Oµ .
But since Oµ is open, we can find an ε, 0 < ε < b such that ]x − ε, x + ε[⊂ Oµ . Choose now an
element y ∈]x, x + ε[. Then [x, y] ⊂]x − ε, x + ε[⊂ Oµ . Thus the claim is proved.
Step 2. Let C be the set of all y ∈]a, b] such that [a, y] can be covered by finitely many Oλ .
Applying Step 1 to x = a, we see that a ∈ C and so C 6= ∅. Since C is bounded from above by
b, C has a least upper bound c ∈]a, b].
Step 3. We prove that c ∈ C. There is k ∈ L such that c ∈ Ok . Since Ok is open, there is
δ > 0 such that ]c − δ, c + δ[⊂ Ok . By the property of the least upper bound, there is an element
z ∈ C such that c − δ < z ≤ c. Then [z, c] ⊂ Ok . Also, [a, z] can be covered by finitely many
Oλ since z ∈ C. Thus, [a, c] = [a, z] ∪ [z, c] can also be covered by finitely many Oλ .
Step 4. We show that c = b and the proof is finished. Suppose that c < b. Applying Step 1
to x = c, we see that there is y ∈]c, b] such that [c, y] can be covered by one element Oj . But
then, [a, y] = [a, c] ∪ [c, y] can be covered by finitely many elements of the cover. This means
that y ∈ C and accordingly y should not be greater than c. We thus got a contradiction.
Recall that we defined three topologically equivalent distances on IRn :
ρ1 (x, y) =
n
X
i=1
ρ2 (x, y) =
|xi − yi |
n
X
i=1
2
|xi − yi |
!1/2
ρ∞ (x, y) = max |xi − yi |
i=1...,n
the Euclidean metric
the square metric.
We also noticed that
ρ∞ (x, y) ≤ ρ2 (x, y) ≤
√
nρ∞ (x, y).
and
ρ∞ (x, y) ≤ ρ1 (x, y) ≤ nρ∞ (x, y).
Now a subset A of a metric space X is called bounded if it is contained in a ball, or equivalently if its diameter is finite. The inequalities between the usual distances on IRn imply that
boundedness in ρ2 is equivalent to boundednes in ρ∞ or boundedness in ρ1 .
Theorem 3.7 A subspace A of IRn is compact if and only if it is closed and bounded (in any
of the usual metrics).
3.3. COMPACT METRIC SPACES
35
Proof. Suppose first that A is compact. Then A is closed by a previous theorem. Now
consider the open covering B(0, m)m∈IN∗ of IRn . Compactness of A implies that A ⊂ B(0, m1 ) ∪
· · · ∪ B(0, mk ) = B(0, R) where R = max(m1 , . . . , mk ). This means that A is bounded.
Conversely, suppose that A is closed and bounded in ρ∞ . Then ρ∞ (0, x) ≤ M for every
x ∈ A. This means that |xi | ≤ M for every x = (x1 , . . . , xn ) ∈ A and so A ⊂ [−M, M ]n . This
last set is compact as a product of compact spaces. Being closed, A is also compact.
Remark 3.3 It is important to precise the distance in which a subspace is bounded. If a set
is bounded in d, and d0 is a distance topologically equivalent to d, then boundedness in d does
not imply boundedness in d0 . Here is an example. Define on IR, d0 (x, y) = min(1, |x − y|). Then
d0 is a distance toplogically equivalent to the usual distance d on IR. However, IR is bounded in
d0 and unbounded in d. Therefore, IR is closed and bounded in d0 , however, it is not compact
(because (IR, d) is not compact). As a convention, when we do not mention the distance on
IRn , it means that we consider any of the usual distances.
Examples. 1) The unit sphere S n−1 and the closed ball B n are compact in IRn because they
are closed and bounded.
2) The set A = {(x, x1 )|0 < x ≤ 1} is closed in IR2 . However it is not compact because it is not
bounded.
3) The set S = {(x, sin x1 )|0 < x ≤ 1} is bounded but not compact because it is not closed.
Theorem 3.8 (Extreme value theorem) Let X be a topological space and let f : X → IR
be continuous. If X is compact, then there exist points c and d in X such that f (c) ≤ f (x) ≤
f (d) for every x ∈ X. Otherwise stated, f has a maximum and minumum value on a compact
space.
Proof. Suppose that X is compact. Then f (X) is a compact subspace of IR. Therefore it is
closed and bounded. Boundedness implies that f (X) has a lowest upper bound sup f (X) and
a greateset lower bound inf f (X). But we proved in the exercises that for a subset A ⊂ IR,
sup A ∈ Ā and inf A ∈ Ā (whenver they exist). Therefore, if A is closed, it contains its sup
and inf. Consequently, sup f (X) ∈ f (X). This measn that sup f (X) = f (d) for some d ∈ X.
Similary, inf f (X) = f (c) for some c ∈ X. But inf f (X) ≤ f (x) ≤ sup f (X) for every x ∈ X.
Hence the conclusion.
Remark 3.4 sup f (X) is usually denoted by supx∈X f (x).
3.3
Compact metric spaces
Theorem 3.9 (Total boundedness) Let X be a compact space. Then for any ε > 0 there
n
S
are finitely many points x1 , . . . , xn in X such that X =
B(xi , ε). This property is called
total boundedness.
i=1
Proof. Consider the open covering (B(x, ε))x∈X . Compactness implies that there exists a
finite subcollection {B(x1 , ε), . . . , B(xn , ε)} that covers X. This is the conclusion.
Remark 3.5 The converse is not true, that is, a totally bounded space need not be compact.
For example the subspace ]0,1[ is totally bounded because it is contained in the compact space
[0,1]. However, ]0,1[ is not compact because it is not closed in IR.
Lemma 3.3 (The Lebesgue number lemma) Let A be an open covering of a metric space
(X, d). If X is compact, then there exists δ > 0 such that each subset of X having diameter
less than δ is contained in some element of A .
36
CHAPTER 3. COMPACTNESS
The number δ is called a Lebesgue number of the covering A .
Proof. If X is an element of A , then any positive number is a Lebesgue number of A . So
we assume that X ∈
/ A.
Choose a finite subcollection {A1 , . . . , An } of A that covers X. For each i, set Ci = X − Ai
and let
n
1X
f (x) =
d(x, Ci ).
n
i=1
Then f (x) > 0 for all x ∈ X. Indeed, otherwise there exists x ∈ X such that d(x, Ci ) = 0 for
all i. Since Ci is closed, we get x ∈ Ci . Thus x ∈ ∩Ci = ∩(X − Ai ) = ∅, a contradiction.
Since f is continuous, it has a minumum value δ > 0. We show that δ is the required number.
Let B be a subset of X of diameter less than δ. Choose a point x0 ∈ B. Then B ⊂ B(x0 , δ).
Now, if d(x0 , Cm ) is the largest of the numbers d(x0 , Ci ), then δ ≤ f (x) ≤ d(x0 , Cm ). It follows
that B(x0 , δ) ⊂ B(x0 , d(x0 , Cm )) ⊂ X − Cm = Am (if d(x, x0 ) < d(x0 , Cm ), then x ∈
/ Cm ).
Consequently, B ⊂ Am .
Definition 3.4 A function f from a metric space (x, dX ) to a metric space (Y, dY ) is said to
be uniformly continuous if for every > 0, there exists δ > 0 such that for every points x, y
in X, dY (f (x), f (y)) < whenever dX (x, y) < δ. In symbols
(∀ > 0) (∃δ > 0) (∀x, y ∈ X) d(x, y) < δ ⇒ dY (f (x), f (y)) < .
Compare this definition with the definition of continuity
(∀x ∈ X) (∀ > 0) (∃δ > 0) (∀y ∈ X) d(x, y) < δ ⇒ dY (f (x), f (y)) < .
The number δ in the condition of continuity depends on and x, whereas in the uniform
contiuity condition it depends only on , so it the same for all x. Hence the adjectif uniform.
Examples. 1) Consider the function f :]0, 1[→ IR defined by f (x) = x1 . We know that f is
continuous. However, it is not uniformly continuous. Let 0 < δ < 1 be given. Let x = δ and
y = δ/2. Then |x − y| = δ/2 < δ. But | x1 − y1 | = 1δ > 1 This means that x − y can be made
arbitrarily small whereas f (x) − f (y) cannot. So f is not uniformly continuous.
2) The function sin is uniformly continuous. This follows from the fact that | sin x−sin y| ≤ |x−y|
for every x, y ∈ IR.
Now we give a positive result.
Theorem 3.10 (Uniform continuity theorem) Let X and Y be metric space with X compact. Let f : X → Y be continuous. Then f is uniformly continuous.
Proof. Let > 0 be given. Consider the open covering of Y by the balls B(b, /2), b ∈ Y . Let
A be the colelction of f −1 (B(b, /2)). Then A is an open covering of X (the covering is open
because f is continuous). Let δ be a Lebesgue number of A . Let x, y be two points in X such
that d(x, y) < δ, then the set {x, y} has diameter less that δ. Therefore it is contained in some
open set f −1 (B(b, /2)). This means that f (x), f (y) ∈ B(b, /2). By the triangle inequality
d(f (x), f (y)) < .
Definition 3.5 A topological space is called limit point compact if every infinite subset of
X has a limit point. A topological space is called sequentially compact if very sequence in
X has a convergent subsequence.
It turns out that if X is metrizable then the three conditions of compactness are equivalent.
3.4. LOCAL COMPACTNESS AND THE ALEXANDROFF COMPACTIFICATION
37
Theorem 3.11 Let X be a metrizable space. Then the following conditions are equivalent.
(i) X is compact.
(ii) Every infinite subset of X has a limit point (limit point compactness).
(iii) Every sequence in X has a convergent subsequence (sequential compactness).
Proof. (i)⇒(ii). Let A ⊂ X be infinite. If no point of X is a limit point of A, then each
point x ∈ X has an open neighborhood Vx intersecting A in at most one point. Then, (Vx )x∈X
is a open covering of X. However, no finite subfamily of this covering can cover A since A is
infinite. So indeed no subfamily can cover X. A contradiction.
(ii)⇒(iii). Let {xn } be a sequence in X. Let E be the range of {xn }. If E is finite then
{xn } contains a constant and therefore a convergent subsequence. If E is infinite, then E has a
limit point x. Choose then n1 such that xn1 ∈ B(x, 1). Having chosen n1 , n2 , . . . , ni−1 , choose
ni > ni−1 such that xni ∈ B(x, 1i ). This is possible since any neighborhood of x contains
infinitely many terms of E. Then clearly, {xni } converges to x.
(iii)⇒(i). We prove this implication in three steps.
Step 1. We prove that if X is sequentially compact then the Lebesgue number lemma holds
for X. So let A be an aopen covering of X. Suppose that there is no Lebesgue number δ.
Then for each n ∈ IN∗ , there exists a set Cn of diameter less than n1 and not contained in any
element of A . Fo each n, choose a point xn in Cn . By hypothesis, the sequence (xn ) has a
subsequence (xnk ) that converges to some point x. Now x belongs to some element A of the
covering. Because A is open, there exists r > 0 such that B(x, r) ⊂ A. Now for all k large
enough, we have n1k < r/2. Then Cnk ⊂ B(xnk , r/2) (because the diameter of Cnk is less than
r/2). Also, for all k large enough, we have d(xnk , x) < r/2 and so B(xnk , r/2) ⊂ B(x, r) ⊂ A
by the triangle inequality. Thus, if k is large enough, then Cnk ⊂ A. But this contradicts our
hypothesis.
Step 2. We prove that if X is sequentially compact then it is totally bounded. So let > 0
be given. Suppose that X cannot be covered by finitely many open −balls. We construct
a sequence (xn ) of X as follows. Let x1 be an arbitrary point in X. Since B(x1 , ) does
not cover X, choose a point x2 ∈
/ B(x1 , ). Suppose that x1 , . . . , xn are contructed. Since
B(x1 , ) ∪ · · · ∪ B(xn , ) does not cover X, choose a point xn+1 not in the union of these balls.
Then d(xn , xm ) ≥ for all m 6= n. This means that (xn ) cannot contain any convergent
subsequence, contradicting our assumption.
Step 3. We prove the implication (iii)⇒(i). Let A be an open covering of X. By the first
step, there exists a Lebesgue number δ > 0 for A . Let = δ/3. By the second step, there exist
finitely many B(x1 , ) · · · B(xn , ) that cover X. Each of these balls B(xi , ) has diameter at
most 2δ/3. Therefore it belongs to an element Ai of A . Then the finite collection {A1 , . . . , An }
covers X.
Remarks. 1) Note that we did not use the metrizability of X in the implication (i)⇒(ii). It
follows that any compact space is limit point compact.
2) It follows from this implication that a compact and discrete space is necessarily finite. This
is an important fact in Analysis.
3.4
Local compactness and the Alexandroff compactification
In this section, we study the notion of local compactness and prove the fundamental theorem
that a locally compact Hausdorff space can be imbedded in a compact Hausdorff space called
its one point compactification or its Alexandroff compactification.
38
CHAPTER 3. COMPACTNESS
Definition 3.6 A space X is said to be locally compact at x if x has a compact neighborhood. If X is locally compact at each of its points, then X is said to be locally compact.
Examples. 1) A compact space is locally compact.
2) IRn is locally compact because the closed ball B 0 (x, r) is compact (closed and bounded).
3) A discrete space is locally compact because {x} is a compact neighborhood of x.
Theorem 3.12 Let X be a topological space. Then the following conditions are equivalent.
(i) X is locally compact Hausdorff.
(ii) There exists a space Y such that
(1) X is a subspace of Y
(2) Y − X is a singleton.
(3) Y is compact Hausdorff.
Furthermore, if Y and Y 0 are two spaces that satisfy the above conditions, then there is a
homeomorphism from Y to Y 0 that equals the identity map on X. Any space Y satisfying
the above conditions is called the one point compactification of X or the Alexandroff
compactification of X.
Proof. (ii)⇒ (i). Let Y be a space satisfying conditions (1)-(3). Then X is Hausdorff as a
subspace of Y . Let x ∈ X. We prove that x has a compact neighborhood. Separate x and the
point Y − X by two open sets U and V respectively. The set C = Y − V is closed and therefore
compact. It is also compact as a subspace of X. But x ∈ U ⊂ C. Therefore C is a compact
neigborhood of x.
(i)⇒ (ii). Take an object ω which is not in X and let Y = X ∪ {ω}. Define a topology on Y
as follows: a set is open in Y if either it is open in X (type 1) or it is of the form Y − C where
C is a compact subspace of X (type 2). We check that this is indeed a topology.
(a) The empty set is of type 1 and the space Y = Y − ∅ is of type 2.
(b) Let O1 and O2 be two open sets in
type 1, then O1 ∩ O2 is also of type
(Y − C1 ) ∩ (Y − C2 ) = Y − (C1 ∪ C2 )
then O1 ∩ (Y − C2 ) = O1 ∩ (X − C1 )
Y . There are three cases: 1. If O1 and O2 are of
1. 2. If O1 and O2 are of type 2, then O1 ∩ O2 =
is of type 2. 3. If O1 is of type 1 and O2 is of type 2,
is of type 1.
(c) Let (Oλ )λ∈L S
be a collection of open sets in Y , there are also three
cases:
S
S 1. If all Oλ are of
type 1, then Oλ is also of type 1. If all Oλ are of type 2, then Oλ = (Y −Cλ ) = Y −∩Cλ
is of type 2. Otherwise, let L1 be the set of λ for which Oλ is of type 1 and L2 be the set
of λ for which Oλ is of type 2. Then
[
[
[
Oλ =
Oλ ∪
(Y − Cλ ) = O ∪ (Y − C) = Y − (C − O)
λ∈L
λ∈L1
λ∈L2
is of type 2 because C − O is closed and therefore compact.
Next, we show that X is a subspace of Y . We have to show that the subspace topology
inherited from Y coincides with the original topology of X. Let O be open in Y . If O is of
type 1, then O ∩ X = O and so O ∩ X is open in X. If O is of type 2, then O = Y − C and so
O ∩ X = (Y − C) ∩ X = X − C is open in X because C is closed in X. Thus, we have proved
that the subspace topology is weaker than the original topology. Conversely, if O is open in X,
3.4. LOCAL COMPACTNESS AND THE ALEXANDROFF COMPACTIFICATION
39
then O is of type 1, and therefore belongs to the topology of Y . Now, O ∩ X = O means that
O belongs to the subspace topology.
We show now that Y is compact. Let A be an open covering of Y . The collection A
contains an element of type 2, that is, of the form Y − C, because none of the sets of type
1 contain the point ω. Take all elements of A different from Y − C and intersect them with
with X. They form a collection of open sets of X that covers C. Compactness of C implies
that finitely many of them cover C. The corresponding finite collection of A along with Y − C
covers Y .
We show that Y is Hausdorff. Let x and y be two distinct points of Y . If x and y belong to
X, then we can separate them by two open sets of X and therefore by two open sets of Y . If
x ∈ X and y = ω, we choose a compact neighborhood C of x. Then C and Y − C are disjoint
neigborhoods of x and y respectively.
We finally prove uniqueness up to an homeomorphism. Let Y and Y 0 be two spaces satisfying
condditions (1)-(3). Let {p} = Y − X and {q} = Y 0 − X. Define h : Y → Y 0 by h(x) = x if
x ∈ X and h(p) = q. It is clear that h is a bijection. We show that h is an open map. The
symmetry of the situation implies that h is an homeomorphism. Suppose first that U does not
contain p. Then h(U ) = U . Since U is open in Y and contained in X, it is open in X. But X
is open in Y 0 (because Y 0 − X is a singleton and singletons in a Hausdorff space are closed).
Therefore U is also open in Y 0 . Next, suppose that U contains p. Then C := Y −U is closed and
therefore compact. Since C ⊂ X, C is also compact as a subspace of X. But X is a subspace
of Y 0 . Therefore C is compact as a subspace of Y 0 . Since Y 0 is Hausdorff, C is closed in Y 0 .
But h(U ) = h(Y − C) = Y 0 − h(C) = Y 0 − C. Therefore h(U ) is open in Y 0 .
Remark 3.6 Using the arguments in the last part of the proof above, one can show that two
homeomorphic locally compact Hausdorff spaces have homeomorphic one point comapctifications. In the exercises, you are asked to show this.
Examples. The one point compactification of S n −P is S n . But S n −P is homeomorphic with
IRn . Therefore the one point compactification of IRn is homeomorphic with S n . In particular,
the one point compactification of C is homeomorphic to the sphere S 2 . In the theory of complex
variables, the one point compactification of C is called the Riemann sphere. The point ω that
we added to the space is usually denoted by ∞.
40
CHAPTER 3. COMPACTNESS
Chapter 4
Connectedness
4.1
Definitions, examples and properties
Connectedness is an important concept in topology. Roughly speaking, a set is connected if it
cannot be split into more than one part (it is topologically one piece). More precisely we have
the following definition.
Definition 4.1 Let X be a topological space. A separation of X is a pair {U, V } of disjoint
nonempty open subsets of X whose union is X. The space X is said to be connected if there
does not exist a separation of X.
Remarks. 1) If {U, V } is a separation of X, then V = X − U and so U and V are clopen.
2) A separation of a space is a partition of that space. However, not every partition is a
separation; we can always write X = A ∪ (X − A), but the point in the definition above is that
A is clopen.
Examples. 1) The space X = {x ∈ IR | |x| > 1} (as a subspace of IR) is not connected since
X =] − ∞, −1[∪]1, ∞[ is a separation of X.
2) The space IR∗ is not connected since IR∗ =] − ∞, 0[∪]0, ∞[ is a separation of X.
3) We shall prove that IR and all its intervals are connected.
3) A singleton is connected.
4) The only connected subsets of Q are the singletons. Indeed, let Y ⊂ Q and suppose that
Y contains two rational numbers a and b. Then there exists an irrational number c between a
and b. Then (Y ∩] − ∞, c[, Y ∩]c, ∞[) is a separation of Y .
5) Let X be a set equipped with the indiscrete topology. Then X is connected (there is no
enough open sets to separate X).
Proposition 4.1 Let X be a topological space. Then the following conditions are equivalent.
(i) X is not connected.
(ii) There exist two disjoint nonempty closed subsets of X whose union is X.
(iii) There exists a proper clopen subset of X.
(iv) There exists a non constant continuous function f : X → Z.
41
42
CHAPTER 4. CONNECTEDNESS
Proof.
(i)⇒(ii). Let {A, B} be a separation of X. Then A is also closed because its
complement B is open. Similarly B is closed.
(ii)⇒(iii). If X = A ∪ B where A and B are closed disjoint nonempty subsets, then A and B
are also open. So A is a proper clopen subset of X.
(iii)⇒(iv). Let A be a proper clopen subset of X. Let f (x) = 1 if x ∈ A and f (x) = 0 if
x ∈ X − A (f is the characteristic function of A). Then f is continuous because the inverse
image of a subset O of Z is either ∅ or A or Ac or X. In all cases, it is open. Now clearly f is
not constant.
(iv)⇒(i). Let a and b be two distinct values taken by a continuous non constant function
f : X → Z. Now {a} is open in Z because Z has the discrete topology. Therefore f −1 (a) is
open in X and nonempty. On the other hand, Z − {a} is also open and nonempty (because
it contains b). Therefore f −1 (Z − {a}) is open in X and nonempty. Now we can write X =
f −1 (a) ∪ f −1 (Z − {a}) where f −1 (a) ∩ f −1 (Z − {a}) = ∅.
Corollary 4.1 Let X be a topological space. Then the following conditions are equivalent.
(i) X is connected.
(ii) There do not exist two disjoint nonempty closed subsets of X whose union is X.
(iii) The only clopen subsets of X are X and ∅.
(iv) If f : X → Z is continuous, then f is constant.
Remark 4.1 Let X be a topological space and let Y ⊂ X. We say that Y is connected if it is
connected with respect to the subspace topology. We have also the following characterization.
Lemma 4.1 Let X be a topological space and let Y ⊂ X. Then Y is not connected if and
only if there exist two nonempty sets A and B such that A ∪ B = Y and Ā ∩ B = A ∩ B̄ = ∅.
Proof. Suppose first that Y is not connected, then there exist two disjoint nonempty sets A
and B which are clopen in Y and whose union is Y . We proved in the exercises that the closure
of A in Y is Ā ∩ Y where Ā is the closure of A in X. Since A is closed in Y , we have A = Ā ∩ Y .
Then Ā ∩ B = Ā ∩ B ∩ Y = A ∩ B = ∅. A similar argument shows that B̄ ∩ A = ∅.
Now suppose conversely that A and B are two nonempty sets whose union is Y and such
that Ā ∩ B = A ∩ B̄ = ∅. Then we claim that Ā ∩ Y = A and B̄ ∩ Y = B. Indeed, first
A ⊂ Y ∩ Ā. Second, let x ∈ Ā ∩ Y . If x ∈
/ A, then x ∈ B since A ∪ B = Y . Then x ∈ Ā ∩ B = ∅,
a contradiction. The other equality is proved similarly. The claim means that A and B are two
disjoint nonempty closed subsets of Y whose union is Y . Thus Y is not connected.
√
Example. Consider
Q as a subspace of IR. We can write Q = A ∪ B where A = √
Q∩] − ∞, 2[
√
and
√ B = Q∩] 2, +∞[. So {A, B} is a separation of Q. Note that
√ Ā =] − ∞, 2] and B̄ =
[ 2, +∞[. So Ā ∩ B = A ∩ B̄ = ∅. Note however that Ā ∩ B̄ = { 2} =
6 ∅.
The next lemma will be useful.
Lemma 4.2 If (C, D) is a separation of a topological space X and Y is a connected subspace
of X, then either Y ⊂ C or Y ⊂ D.
Proof. Since C and D are both open in X, the sets C ∩ Y and D ∩ Y are open in Y . These
two sets are disjoint and their union is Y ; if they were both nonempty, they would constitute a
separation of Y . Therefore one of them is empty. Hence Y is contained in C or D.
4.1. DEFINITIONS, EXAMPLES AND PROPERTIES
43
Theorem 4.1 Let X be topological space. The union of connected subspaces of X that have
a common point is connected.
T
Proof.
Let
(A
)
be
a
collection
of
connected
subspaces
of
X
and
let
p
∈
Aλ . Suppose that
λ
S
Y := Aλ is not connected, then there exists a separation (C, D) of Y . The point p is in one
of the sets C or D. We can assume that it is in C (otherwise relable C and D). Since Aλ is
connected then by the previous lemma, it must be contained in either C or D; and it cannot
be contained in
S D because it contains the point p of C. Hence Aλ ⊂ C. Since λ was arbitrary,
we have Y = Aλ ⊂ C, contradicting the fact that D is not empty.
Theorem 4.2 Let A be a connected subspace of a X. If A ⊂ B ⊂ Ā, then B is also connected.
Said differently: if B is obtained from a connected subspace A by adjoining some or all its limits
points, then B is connected.
Proof. Suppose that {C, D} is a separation of B. By a previous lemma, A must lie either in
C or D. Suppose that A ⊂ C. Then B ⊂ Ā ⊂ C̄. But C̄ ∩ D = ∅, therefore B cannot intersect
D. This contradicts the fact that D is a nonempty subset of B.
Theorem 4.3 The image of a connected space under a continuous map is connected. In
particular, connectedness is preserved by a homeomophism.
Proof.
Let f : X → f (X) be continuous with X connected. Suppose that {A, B} is a
separation of f (X). Then f −1 (A) and f −1 (B) are open (because f is continuous and A, B are
open) and disjoint because A ∩ B = ∅. They are also nonempty because A and B are contained
in the image of f . Therefore {f −1 (A), f −1 (B)} is a separation of X, contradicting the fact that
X is connected.
Theorem 4.4 A product of connected spaces is connected.
Proof.
We prove the theorem for the product of two spaces X and Y . The result for
an arbitrary number of spaces will follow by induction. Choose a base point (a, b) ∈ X × Y .
The ”horizontal slice” X × {b} is connected because it is homeomorphic with X. Also, each
vertical slice {x} × Y is connected, being homeomorphic with Y . As a result, the space Tx =
(X × {b}) ∪ ({x} × Y ) is connected asSa union of two connected subspaces having the point
(x, b) in common. Now form the union x∈X Tx of all these subspaces. ThisSunion is connected
because these subspaces have the point (a, b) in common. But X × Y = x∈X Tx . Therefore
X × Y is connected.
Theorem 4.5 A subset E ⊂ IR is connected if and only if it is an interval.
Proof. Let E be connected and assume it is not an interval. Then there are two points x
and y in E and z ∈]x, y[ such that z ∈
/ E. Then, E has the following decomposition
E = E ∩ ] − ∞, z[
[
E ∩ ]z, ∞[.
The two subsets are both open in E and nonempty since the first one contains x and the second
contains y. This means that E is not connected contrary to the assumption.
Conversely, let E be an interval and suppose that is not connected. Then it is the union
of two disjoint nonempty sets A and B each of which is open in E. Choose a ∈ A and b ∈ B.
We may assume that a < b (otherwise we relabel A and B). The interval [a, b] is contained in
E since E is an interval. Hence [a, b] is the union of the disjoint nonempty sets A0 = A ∩ [a, b]
and B0 = B ∩ [a, b]. Let c = sup A0 . We distinguish between two cases.
44
CHAPTER 4. CONNECTEDNESS
Case 1. c ∈ A0 . In this case, c 6= b, because otherwise c ∈ A0 ∩ B0 . So either c = a or
a < c < b. In both cases, because A0 is open in [a, b], there is some interval [c, c + ε[ contained
in A0 . In particular, this means that there is z ∈ A0 such that c < z < c + ε. The fact that
z ∈ A0 means that z ≤ sup A0 = c. Thus, we reached a contradiction.
Case 2. c ∈ B0 . In this case, c 6= a, so either c = b or a < c < b. In either case, it follows
from the fact that B0 is open in [a, b] that there is some interval of the form ]c − ε, c] contained
in B0 . But since c = sup A0 , the interval ]c − ε, c] contains an element of A0 . Again this is a
contradiction since A0 ∩ B0 = ∅.
Theorem 4.6 (Intermediate value theorem) Let X be a connected topological space and
let f : X → IR be a continuous function. If a and b are two points of X and r is a number
between f (a) and f (b), then there exists a point c ∈ X such that f (c) = r.
Proof. Since X is connected and f is continuous, then f (X) is connected and therefore an
interval. We can suppose that f (a) < f (b). Since f (a) and f (b) belong to the interval f (X),
[f (a), f (b)] ⊂ f (X). Hence any number between f (a) and f (b) is the image of an element of
X.
Now we come to a condition stronger than connectedness: path connectedness.
Definition 4.2 Given two points x and y of a topological space X, a path in X from x to y
is a continuous map f : [a, b] → X such that f (a) = x and f (b) = y. A space X is said to be
path connected if any two points of X can be joined by a path in X.
Note that since [a, b] is homeomorphic to [0, 1], the domain of any path can be taken as [0,1].
This is usually done in algebraic topology.
Examples. 1) Intervals of IR are path connected.
2) A ball B(a, r) in IRN is path connected. In fact it is convex, which means that any two points
in B can be joined by a straight line segment. Indeed let x and y be two points in B(a, r). Set
f (t) = (1 − t)x + ty for t ∈ [0, 1]. Then f is a path joining x to y which is contained in B(a, r).
Indeed ||f (t) − a|| = ||(1 − t)(x − a) + t(y − a)|| < (1 − t)r + tr = r. Similarly, a closed ball is
path connected.
3) The punctured Euclidean space is IRN − {0}. If N > 1, then this space is path connected.
Indeed, let x and y be two points in this space. If the straight line segment joining them does
not pass by 0, then this segment is a path in IRN − {0} joining x to y. Otherwise, we can choose
a point z not on the line joining x to y. Then x and y can be joined by the broken line path
from x to z and then from z to y.
4) The continuous image of a path connected space is also path connected (why?). In particular,
the unit sphere S n−1 in IRn is path connected for n > 1. Indeed, let g : IRn − {0} → S n−1 be
x
defined by g(x) = ||x||
. Then g is continuous and surjective. Therefore the result follows from
the previous example.
Proposition 4.2 A path connected space is connected.
Proof. Let X be a path connected space. Suppose that X = A ∪ B is a separation of X.
Let x ∈ A and y ∈ B. Then there exists a path f : [a, b] → X joining x to y. Now f ([a, b])
is connected as the image of a connected space under a continuous function. It follows that
f ([a, b]) is contained in either A or B, which is impossible since f (a) = x ∈ A and f (b) = y ∈ B.
The converse of this proposition is however not true as the following example shows.
4.1. DEFINITIONS, EXAMPLES AND PROPERTIES
45
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
0
0.2
0.4
0.6
0.8
1
Figure 4.1: The topologist’s sine curve
Example. Let S = {(x, sin x1 ), 0 < x ≤ 1}. Then S is connected as the image of ]0, 1] under
a continuous function. Therefore its closure S̄ is also connected. S̄ is called the topologist’s
sine curve. In fact S̄ = ({0} × [−1, 1]) ∪ S. Let us prove this. First, we claim that {0} ×
[−1, 1] ⊂ S. Indeed, if (0, u) ∈ {0} × [−1, 1], then u = sin v for some v . Then the sequence
1
( v+2nπ
, sin(v + 2nπ)) belongs to S and converges to (0, u). It follows that ({0} × [−1, 1]) ∪ S ⊂
S̄. Next, it should be clear that S̄ ⊂ [0, 1] × [−1, 1]. Finally we prove the reverse inclusion
S̄ ⊂ ({0} × [−1, 1]) ∪ S. Let (x, y) ∈ S̄ − {0} × [−1, 1]. Then x > 0. If y 6= sin x1 , separate
y and sin x1 by two neighborhoods V1 and V2 . Let g(x) = sin x1 and let U be a neighborhood
of x such that g(U ) ⊂ V2 . Then U × V1 is a neighborhood of (x, y) that does not intersect S,
contradicting the fact that (x, y) ∈ S̄.
We now show that S̄ is not path connected. Suppose there is a path f : [a, c] → S̄ joining the
origin to a point in S. The set {t ∈ [a, c]|f (t) ∈ {0}× [−1, 1]} is closed and bounded from above.
Therefore it has a biggest element b. Then f : [b, c] → S̄ is a path that maps b into the vertical
slice {0}×[−1, 1] and maps the other points of [b, c] to points of S. Let f (t) = (x(t), y(t)). Then
x(b) = 0 while x(t) > 0 and y(t) = sin(1/x(t)) for t > b. By continuity, limt→b+ x(t) = x(b) = 0
1
and so limt→b+ x(t)
= +∞. But the sine function has no limit at infinity. Therefore y(t) has no
limit as t → b. However, it should have because f is continuous. This contradiction shows that
there is no path joining the origin to a point in S, and so S̄ is not path connected.
To prove that y(t) has no limit as t → b, we proceed as follows. Let b < b0 < c. Then x([b, b0 ])
is an interval that contains 0 and a positive number x(b0 ). Therefore [0, x(b0 )] ⊂ x([b, b0 ]). Now
1
for every n large enough π +2nπ
∈ [0, x(b0 )]. Therefore there exists a sequence tn ∈ [b, b0 ] such
that x(tn ) =
1
π
+2nπ
2
2
. Then y(tn ) = 1. Similarly, there exists a sequence sn ∈ [b, b0 ] such that
1
x(sn ) = π +(2n+1)π
and so y(sn ) = −1. Passing to a subsequence, we may assume that tn → t
2
and sn → s. Then x(tn ) → x(t) = 0 and x(sn ) → x(s) = 0. Therefore t = s = b because
otherwise x(t) > 0 or x(s) > 0. So we have two sequences that converge to b, yet their images
under y converge to distinct limits.
46
CHAPTER 4. CONNECTEDNESS
4.2
Components and local connectedness
If a space is not connected, it can be divided into pieces that are connected. Indeed, given a
topological space X, define the binary relation on X by setting x ∼ y if there exists a connected
subset of X containing both x and y. This is an equivalence relation. Indeed, first, the relation
is reflexive: x ∼ x because {x} is connected. Next, symmetry is evident. Last (transitivity),
if x ∼ y and y ∼ z, then, there is a connected subset A containing x and y and a connected
subset B containing y and z. Then A ∪ B contains x and z and is connected because A an B
have y in common. Therefore x ∼ z.
Definition 4.3 The equivalence classes for the relation above are called the connected components (or sometimes just components) of X.
Now recall that given an equivalence relation on a set, the equivalence class containing an
element x is the set of points y that are in relation with x. Two equivalence classes are either
disjoint or identical and the whole set is the union of the pairwise disjoint equivalence classes. It
is also possible that there is only one equivalence class (in this case, all elements are equivalent).
Examples. 1) The connected components of IR∗ are ] − ∞, 0[ and ]0, ∞[.
2) The connected components of a discrete space are the singletons. In particular, the connected
components of Z are the singletons {n}. A space whose components are singletons is usually
called totally disconnected.
3) The connected components of Q are the singletons.
4) A connected space has only one component: the space itself.
Here are some important properties of connected components.
Theorem 4.7 Let X be a topological space. Then
(i) Two connected components of X are either disjoint or identical and X is the union of the
pairwise disjoint components.
(ii) The connected components of X containing an element x is the biggest connected subset
of X that contains x. It is usually denoted by C(x) or [x].
(iii) The components of X are closed.
(iv) Every nonempty connected subset of X is contained in exactly one connected component.
Proof. (i) is true for any equivalence relation.
(ii). Let D(x) be the union of all connected sets containing x. Then D(x) is connected as
a union of connected sets having a common point. Now if y ∈ D(x), then x ∼ y because D(x)
is a connected set containing x and y. Therefore y ∈ C(x). This proves that D(x) ⊂ C(x).
Conversely, let y ∈ C(x), then there is a connected set A containing x and y. Then A ⊂ D(x)
and so y ∈ D(x). This proves that C(x) ⊂ D(x). Now of course, the union of all connected
sets that contain x is the biggest connected set containing x.
(iii). Let C(x) be a connected component. Then C(x) is connected. By the previous point,
it is contained in C(x). This means that C(x) is closed.
(iv) We already observed this. If A is a nonempty connected set, it contains x and so
A ⊂ C(x). Since distinct components are disjoint, A cannot meet another component.
Remark 4.2 If X has finitely many components, then every connected component is also open
because its complement is a finite union of closed sets. In general, components need not be
open. For instance the components of Q are not open.
4.2. COMPONENTS AND LOCAL CONNECTEDNESS
47
Theorem 4.8 Let f : X → Y be a homeomorphism. Then for every x ∈ X, f (C(x)) =
C(f (x)). In words: a homeomorphism takes connected components into connected components.
In particular, the number of connected components of a space is a topological invariant.
Proof. f (C(x)) is a connected subset containing f (x). Therefore it is contained in C(f (x)).
Conversely, let y = f (z) ∈ C(f (x)). Then f (z) ∼ f (x) i.e., f (z) and f (x) belong to the
connected set C = C(f (x)). But then z and x belong to the same connected set f −1 (C). This
means that z ∼ x and so z ∈ C(x). Therefore y = f (z) ∈ f (C(x)).
Examples. 1) The spaces [0, 1] mand S 1 are not homeomorphic. Indeed, if we remove 12
from [0, 1], the resulting space is not connected; whereas if we remove any point from S 1 , the
resulting space is connected.
2) The letters X and Y (as compact subspaces of IR2 ) are not homeomorphic. Indeed, if we
remove the middle point from X, the resulting space has 4 connected components; whereas if
we remove any point from Y, the resulting space has at most 3 components.
3) The letters A and ∞ are not homeomorphic. Indeed, if we remove from A the topological
circle that it contains, the resulting space has two connected components; whereas if we remove
from ∞ any topological cicle, the resulting space is a circle (which is connected).
4) The sphere S 2 and the torus T 2 (surface of an american doughnut, chambre à air, bouée)
are not homeomorphic. Indeed, if we remove an apropriate circle from the torus the resulting
space is connected, whereas if we remove from S 2 any topological circle, the resulting space is
not connected.
Now we define another equivalence relation on a topological space X. Let x ∼ y if there is a
path in X joining x to y. Then this is an equivalence relation. Indeed, first x ∼ x because the
constant path f (t) = x joins x to itself. Next, let f : [0, 1] → X be a path joining x to y. Let
g(t) = f (1 − t). Then g is a path joining y to x. This proves the symmetry of the relation.
Finally, let f : [0, 1] → X be a path joining x to y and let g : [1, 2] → X be a path joining y to
z. Let
(
f (t)
if 0 ≤ t ≤ 1
h(t) =
g(t)
if 1 ≤ t ≤ 2.
Since f (1) = g(1) = y, it follows from the pasting lemma that h is continuous. Since h(0) = x
and h(2) = z, h is a path joining x to z. This proves the transitivity of the relation.
Definition 4.4 The equivalence classes for the relation above are called the path components
of X.
Example.
The topologist’s sine curve S̄ has two path components: the curve S and the
vertical segment V = {0} × [−1, 1]. Note that S̄ has one connected component since it is
connected.
If one removes from S̄ the points of V with rational second coordinates, the resulting space
is still connected but has infinitely many path components.
Theorem 4.9 Let X be a topological space. Then
(i) Two path components of X are either disjoint or identical and X is the union of the
pairwise disjoint components.
(ii) The path component of X containing an element x is the biggest path connected subset
of X that contains x.
(iii) Every nonempty path connected subset of X is contained in exactly one path component.
48
CHAPTER 4. CONNECTEDNESS
Proof. The proof is similar to that of Theorem 4.7. One has to observe that the union of
path connected sets having a common point is path connected.
Remark 4.3 Path components need not be closed. For instance in the topologist’s sine curve,
S is a path component which is not closed.
Connectedness is a useful property for a space to possess. But for some purposes, it is more
important that the space satisfy a connectedness condition locally. Local connectedness means
that each point has arbitrarily small neighborhoods that are connected. More precisely, we have
the following definition.
Definition 4.5 A space X is said to be locally connected at x if for every neighborhood U
of x, there is a connected neighborhood V of x contained in U . If X is locally connected at
each of its points, it is said to be locally connected. Similarly, a space X is said to be locally
path connected at x if for every neighborhood U of x there is a path connected neighborhood
V of x contained in U . If X is locally connected at each of its points, it is said to be locally
path connected.
Remark 4.4 Since path connected ⇒ connected then locally path connected ⇒ locally connected.
Examples. 1) Each interval of IR is both connected and locally connected. In fact in IR, path
connected⇔ path connected ⇔ convex ⇔ interval.
2) IRn is path connected and locally path connected. Every open subset of IRn is locally path
connected. This follows form the fact that an open ball in IRn is path connected.
3) IR∗ is not connected but locally connected.
4) The topologist’s sine curve is connected but not locally connected.
5) The set of rationals Q is neither connected nor locally connected.
Theorem 4.10 A space X is locally connected if and only if for every open set U of X, each
component of U is open in X.
Proof. Suppose first that X is locally connected. Let U be an open subset of X and let
C be a component of U . Let x ∈ C. Since U is a neighborhood of x, there exists by local
connectedness a connected neighborhood V of x such that V ⊂ U . Since V is connected and
contains x, it is contained in C. This means that C is a neighborhood of each of its points.
Therefore C is open in X.
Conversely, suppose that components of open sets are open. Given a point x ∈ X and an
open neighborhood U of x, let C be the component of U containing x. Since C is open and
connected, it is a connected neighborhood of x. This means that X is locally connected.
Corollary 4.2 The connected components of a locally connected space are clopen.
Similarly, we have the following.
Theorem 4.11 A space X is locally path connected if and only if for every open set U of X,
each path component of U is open in X.
The relation between connected components and path components is given in the following
theorem,
4.2. COMPONENTS AND LOCAL CONNECTEDNESS
49
Theorem 4.12 Let X be a topological space. Each path component is contained in exactly
one connected component. If X is locally path connected, then the connected components and
the path components are the same.
Proof.
The first statement is clear because a path component is connected and so it is
contained in exactly one component. Now assume that X is locally path connected (actually
a weaker assumption is needed). Let x ∈ X and let C(x) and P (x) denote respectively the
component and path component containing x. We need to show that C(x) ⊂ P (x), and for this
it is enough to show that C(x) is path connected. Let A be the set of points in X that can be
joined to x by a path. Let us denote this relation by ∼p . Note that A ⊂ C(x) because a path
is connected. Then x ∈ A and so A is not empty. We show that A is clopen. Let y ∈ A and
let U be a path connected neighborhood of y. If z ∈ U then z ∼p y. But y ∼p x. Therefore
z ∼p x and so z ∈ A. This proves that A is a neighborhood of each of its points and so it is
open. Now let y ∈ X − A and let V be a connected neighborhood of y. Then V ⊂ X − A
because if z ∈ V then z ∼p y; if z ∈ A, then z ∼p x and so x ∼p y, which means that y ∈ A,
a contradiction. Conclusion: A is nonempty and clopen in C(x). Since C(x) is connected, it
follows that C(x) = A and so every point of C(x) can be joined to x by a path. This means
that C(x) is path connected.
Corollary 4.3 Let O be an open subset of IRn . Then O is connected if and only if it is path
connected.
We end this chapter by a useful characterization of open sets of the real line.
Theorem 4.13 Every open set O of the real line is the union of a countable family of pairwise
disjoint open intervals.
Proof. Let (Iλ )λ∈L be the collection of connected components of O. We know that each Iλ
is an open interval (recall that in a locally connected space, a connected component of an open
set is open). Since the rational numbers are dense in IR, each Iλ contains a rational number.
By the axiom of choice, we can choose exactly one rational number in each Iλ . This defines a
function between L and Q which is one to one because two distinct components are disjoint.
Therefore L is countable. Finally we have indeed O = ∪λ∈L Iλ .
50
CHAPTER 4. CONNECTEDNESS
Chapter 5
Complete metric spaces
Let (xn ) be a sequence of a metric space (X, d) that converges to x. Recall that this means that
for any ε > 0, we can find an integer n0 such that d(xn , x) < ε for all n ≥ n0 . Therefore, by the
triangle inequality, d(xn , xm ) < 2ε for all integers n and m greater or equal to n0 . Otherwise
stated, the difference of any two terms can be made arbitrary small when their indices are large
enough. We are thus led to formulate the following definition.
Definition 5.1 Let (X, d) be a metric space. A sequence (xn ) of X is called a Cauchy sequence
if for any ε > 0 there is a number n0 such that d(xn , xm ) < ε for all integers n, m ≥ n0 .
According to what we said, any convergent sequence is a Cauchy sequence. Is the converse
true? The converse need not be true. If it is true, the space is called complete.
Definition 5.2 A metric space (X, d) is called complete if every Cauchy sequence of X is
convergent.
Counterexamples. 1) The sequence
1.4, 1.41, 1.414, 1.4142, 1.41421, . . . of rational numbers
√
is a Cauchy sequence whose limit 2 is not in Q. Thus Q is not complete.
2) Consider the sequence (xn ) of (0,2) defined by xn = n1 . This sequence is a Cauchy sequence
which is not convergent in (0,2). Thus (0,2) is not complete. This example also shows that completeness is not a topological property (that is, not invariant under homeomorphisms) because
(0,2) is not complete whereas IR as we shall see is complete.
Lemma 5.1 A Cauchy sequence is bounded.
Proof. Let (xn ) be a Cauchy sequence of a metric spacce(X, d). Taking ε = 1 in the definition
of a Cauchy sequence, we see that there exists n0 such that d(xn , xm ) < 1 for n, m ≥ n0 . Let
M = max{d(xn , xm )|n, m ≤ n0 }. Then, by the triangle inequality d(xn , xm ) ≤ d(xn , xn0 ) +
d(xn0 , xm ) ≤ M + 1 for all n, m ∈ IN∗ . This means that the sequence (xn ) is bounded.
Lemma 5.2 If a Cauchy sequence (xn ) contains a convergent subsequence, then (xn ) is convergent.
Proof. Let (xni ) be a subsequence converging to x. Let ε > 0 be given. Then, there is an
index m1 such that d(xni , x) < ε/2 for i ≥ m1 . Since {xn } is a Cauchy sequence and ni ≥ i,
there is an integer m2 such that d(xni , xi ) < ε/2 for i ≥ m2 . Therefore, d(xi , x) < ε for all
i ≥ max(m1 , m2 ).
Corollary 5.1 A compact metric space is complete.
Theorem 5.1 IRn is a complete under any one of the usual distances.
51
52
CHAPTER 5. COMPLETE METRIC SPACES
Proof. Since a Cauchy sequence is bounded, it belongs to a compact set (a closed ball).
Therefore, it contains a convergent subsequence. According to the previous lemma it converges.
Remark 5.1 Thus we have established a new criterion for convergence. Note that, this criterion permits us to decide whether a sequence is convergent without knowing its limit. This is
very useful, especially in theoretical considerations.
Remark 5.2 Here is another proof of the completeness of IR which is a direct consequence of
the sup property. Let {xn } be a Cauchy sequence and let ε > 0 be given. Let N be such that
|xn − xm | < ε for n, m ≥ N . Define
S = {y | xn < y for finitely many n} = {y | xn ≥ y for n large enough}.
Now observe that xn − ε ∈ S for all n. In particular xN − ε ∈ S. We claim that y ∈ S ⇒ y <
xN + ε. Indeed, let y ∈ S. Then xn ≥ y for n large enough. If y ≥ xN + ε, then xn ≥ xN + ε
for n large enough, which contradicts the definition of N . Therefore y < xN + ε. This means
that xN + ε is an upper bound of S. Let now b = sup S. It follows from what we have said that
xN −ε ≤ b ≤ xN +ε, that is |xN −b| ≤ ε. Now if m ≥ N , then |xm −b| ≤ |xm −xN |+|xN −b| ≤ 2ε.
Since ε is arbitrary, it follows that that xn → b.
Proposition 5.1 Let (X, d) be a complete metric space and let F be a subspace of X. Then
F is complete if and only if it is closed in X.
Proof. Suppose that F is complete (under d). Let x ∈ F̄ . Then there is a sequence (xn ) of F
that converges to x. Therefore (xn ) is a Cauchy sequence. Since F is complete, (xn ) converges
to some y ∈ F . But (xn ) converges to x. Therefore x = y ∈ F . This means that F is closed as
a subspace of X.
Conversely, suppose that F is closed in X. Let (xn ) be a Cauchy sequence of F . Since X is
complete, (xn ) has a limit x ∈ X. But closedness of F implies that x ∈ F .
Definition 5.3 Let X be a set and f : X → X. A fixed point of f is a point a ∈ X such
that a = f (a).
Definition 5.4 Let (X, dX ) and (Y, dY ) be metric space. A function f : X → Y is called
a contraction if there exists a constant k < 1 such that dY (f (x), f (y)) ≤ kdX (x, y) for all
x, y ∈ X.
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Example. The function f : [−1.1] → [−1, 1] defined by f (x) = x3 is a contraction. Indeed,
f 0 (x) = 23 x and so |f 0 | is bounded by 23 . By the mean value theorem, |f (x) − f (y)| ≤ 23 |x − y|
for all x, y ∈ [−1, 1].
The following theorem is important in analysis.
Theorem 5.2 (The Banach fixed point theorem) Let X be a complete metric space and
let f : X → X be a contraction. Then f has a unique fixed point.
Proof. Let k < 1 satisfy d(f (x), f (y)) ≤ kd(x, y) for all x, y ∈ X. We first prove uniqueness.
Suppose that we have two ficed points a and b. Then d(f (a), f (b)) ≤ kd(a, b) and so d(a, b) ≤
kd(a, b). This implies that d(a, b) = 0 because otherwise, we get 1 ≤ k.
Next, we prove existence of a fixed point. Pick a point a ∈ X and define the sequence (xn )
by x0 = a and xn+1 = f (xn ). Observe that for n ≥ 1,
d(xn+1 , xn ) = d(f (xn ), xn ) = d(f (xn ), f (xn−1 )) ≤ d(xn , xn−1 ).
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It follows by induction that d(xn+1 , xn ) ≤ kn d(x1 , x0 ). Let now p ≥ 1. Using the traingle
inequality and the last estimate, we get
d(xn , xn+p ) ≤ d(xn , xn+1 ) + d(xn+1 , xn+2 ) · · · + d(xn+p−1 , xn+p )
≤ kn d(x1 , x0 ) + kn+1 d(x1 , x0 ) + · · · kn+p−1 d(x1 , x0 )
= (kn + kn+1 + · · · kn+p−1 )d(x1 , x0 )
kn
1 − kp
d(x1 , x0 ) ≤
d(x1 , x0 ).
= kn
1−k
1−k
n
k
d(x1 , x0 ) < ε for all n large enough.
Let ε > 0 be given. Since kn → 0 as n → ∞, we have 1−k
Then we have d(xn , xn+p ) < ε for all n large enough and all p ≥ 1. This means that (xn ) is
a Cauchy sequence. Since X is complete (xn ) has a limit a. Letting n → ∞ in the relation
xn+1 = f (xn ) and using the continuity of f we get a = f (a).
We already noticed that a compact metric space is complete. The converse is not true
because IR is complete but not compact. However, it turns out that if a space is complete and
totally bounded, then it is compact.
Theorem 5.3 A metric space is compact iff it is totally bounded and complete.
Proof. If X is compact, then it is totally bounded and complete. Conversely, suppose that
X is totally bounded and complete. We prove that X is sequentially compact. So let (xn ) be
a sequence of X. First cover X by finitely many balls of radius 1. At least one of these balls,
call it B1 contains xn for infintely many indices n. Let J1 = {n ∈ IN∗ |xn ∈ B1 }. Next, cover X
by finitely many balls of radius 12 . Because J1 is infinite, at least one of these balls, call it B2 ,
contains xn for infinitely many indices n ∈ J1 . Let J2 = {n ∈ J1 |xn ∈ B2 }. At the kth step, let
Jk = {n ∈ Jk−1 |xn ∈ Bk } where Bk is a ball of radius k1 and Jk is infinite.
Choose n1 ∈ J1 . Given nk−1 , choose nk ∈ Jk such that nk > nk−1 (this is possible because
Jk is infinite). Now if i, j ≥ k, the indices ni and nj belong to Jk (because (Jk ) is a decreasing
sequence of subsets of IN∗ ). Therefore for all i, j ≥ k, the points xni and xnj belong to the ball
Bk of radius k1 . Therefore d(xni , xnj ) ≤ k2 . This means that (xni ) is a Cauchy sequence. Since
X is complete (xni ) is convergent and this means that X is sequentially compact.
Corollary 5.2 In a complete metric space, a subset has compact closure if and only if it is
totally bounded.
Proof. Let X be a complete metric space and let A ⊂ X. Suppose first that Ā is compact.
Then Ā is totally bounded. It follows that A is also totally bounded (prove this). Conversely,
suppose that A is totally bounded. Then Ā is also totally bounded (prove this). But since Ā
is closed in a complete space, it is itself complete. Being totally bounded, it is compact by the
previous theorem.
A subset with a compact closure is also called relatively compact.