Download Introduction to General Topology

Dr. Davide Batic
Introduction to General Topology
with solution manual
January 20, 2016
Springer
Contents
1
Fundamental topological concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 ABC of topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Back to topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Bases of topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Continuous maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 The identification topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Projection maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 First and second-countable topological spaces . . . . . . . . . . . . . . . . . . .
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
10
12
21
21
24
26
33
33
35
37
39
2
Homeomorphisms and topological properties . . . . . . . . . . . . . . . . . . . . . .
2.1 Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Path-connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 The Hausdorff property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
41
44
48
52
53
56
56
3
Topological vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Topological vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Finite-dimensional normed spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Continuous maps between normed spaces . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
65
67
67
68
69
71
v
vi
Contents
The Continuum Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.1 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.2 Countable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.3 The Continuum Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
73
74
80
Quick review of real analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B.1 Real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B.2 Real sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B.3 Limits of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
83
84
84
85
86
Chapter 1
Fundamental topological concepts
1.1 ABC of topology
A topological space is the most general mathematical structure where a meaningful
concept of continuity can be defined. Topological spaces appear almost everywhere
in modern mathematics and are used as a unifying concept for different branches in
this discipline. The field of mathematics dealing with topological spaces is called
topology.
Definition 1.1. Let I be an indexing set. A topological space is a pair (X, O)
consisting of a set X and a family O = {Ui ⊂ X | i ∈ I} of subsets of X, called
open sets, such that
1. 0/ and X are open;
2. arbitrary unions of open sets are open, i.e.
if Ui ∈ O ∀i ∈ I =⇒
[
i∈I
Ui ∈ O;
3. finite intersections of open sets are open, i.e.
if Ui ∈ O ∀i ∈ In = {1, 2, · · · , n} =⇒
\
i∈In
Ui ∈ O.
The collection O is called a topology on X.
The requirement of taking finite intersections cannot be relaxed as the following
example shows.
Example 1.1. Let X = R and O = { − 1n , 1n | n ∈ N}. Then,
1
2
1 Fundamental topological concepts
\
n∈N
1 1
− ,
n n
= {0} ∈
/ O.
Note that Def. 1.1 does not require that O must contain all subsets of X. Finally, the
topology O does not need to be unique.
Definition 1.2. Let (X, O) be a topological space.
1. A ⊂ X is closed in X if the complement X − A ∈ O.
2. U ⊂ X is a neighbourhood of the point x ∈ X if there exists a set V ∈ O
such that x ∈ V ⊂ U.
3. Let B ⊂ X and x ∈ X.
a. x is an interior point of B if B is a neighbourhood of x. We denote the
set of all interior points of B by Int(B).
b. x is an exterior point of B if X − B is a neighbourhood of x.
c. x is a boundary point of B if both B and X −B are not neighbourhoods of
x. By ∂ B we denote the frontier of B, i.e. the set of all boundary points
of B.
d. x is in the closure of B if for each neighbourhood N of x we have N ∩B 6=
0.
/ We denote the closure of B by B.
When we say that U is not a neighbourhood of a point x ∈ X we mean that there
does not exist an open set V such that x ∈ V ⊂ U but this is equivalent to say that
every open set V containing x is such that V 6⊂ B.
Lemma 1.1. Let (X, O) be a topological space and B ⊂ X. Then, the following
statements are true
1. If x is an exterior point of B, then x ∈
/ B.
2. Int(B) ∩ ∂ B = 0.
/
Proof. 1. Let x be an exterior point of B. Then, X − B is a neighbourhood of x.
Hence, x ∈ X − B and also (X − B) ∩ B = 0.
/ Let N = X − B. Then, there exists a
neighbourhood N of x such that N ∩ B = 0.
/ Hence, it follows that x ∈
/ B.
2. By contradiction. Suppose Int(B) ∩ ∂ B 6= 0.
/ Then, there exists x ∈ X such that
x ∈ Int(B) and x ∈ ∂ B. Therefore, x must be an interior point of B, i.e. B is a
neighbourhood of x. Since x must be a boundary point of B, i.e. both B and X − B
are not neighbourhoods of x, we have a contradiction. ⊓
⊔
1.1 ABC of topology
3
Lemma 1.2. Let (X, O) be a topological space and B ⊂ X. Then
B = Int(B) ∪ ∂ B.
Proof. We show the inclusion Int(B) ∪ ∂ B ⊂ B. We have two cases: x ∈ Int(B) =⇒
x ∈ B or x ∈ ∂ B =⇒ x ∈ B.
1. We prove the contrapositive of the first case above, namely x ∈
/ B =⇒ x ∈
/ Int(B).
If x ∈
/ B, then there exists a neighbourhood N of x such that N ∩ B = 0.
/ Hence,
x∈
/ B and B cannot be a neighbourhood of x. Therefore x ∈
/ Int(B).
/ ∂ B. Let x ∈
/ B. Then, there exists a
2. We show the contrapositive x ∈
/ B =⇒ x ∈
neighbourhood N of x such that N ∩ B = 0.
/ This implies that x ∈ X − B with N ⊂
X − B and there exists a set V ∈ O such that x ∈ V ⊂ N. Hence, we constructed
an open set V contained in X − B and therefore X − B is a neighbourhood of x.
This implies that x is an exterior point and we can conclude that x ∈
/ ∂ B.
Finally, we prove the inclusion B ⊂ Int(B) ∪ ∂ B. The contrapositive of statement
1 in Lemma 1.1 implies that x is not an exterior point and therefore x must be a
boundary point or an interior point. ⊓
⊔
Theorem 1.1. Let (X, O) be a topological space and B ⊂ X. Then,
B ∈ O ⇐⇒ Int(B) = B.
Proof. =⇒ Let B ∈ O. We have to show Int(B) ⊂ B and B ⊂ Int(B).
1. Let x ∈ Int(B). Then, B is a neighbourhood of x. By definition of neighbourhood
there exists U ∈ O such that x ∈ U ⊂ B. Hence, x ∈ B.
2. Let x ∈ B. Since B open and B ⊂ B, it follows that B is a neighbourhood of x, i.e.
x ∈ Int(B).
⇐= Let Int(B) = B. B will be open if we can show that B can be written as a union
of open sets. Since B = Int(B), B is a neighbourhood of all its points. Hence, for
every x ∈ B there exists Vx ∈ O such that x ∈ Vx ⊂ B. Moreover,
[
x∈B
Vx ⊂ B.
Since Vx ⊂ B for every x ∈ B and the union is taken over all x ∈ B, the above union
must contain all points of B. Hence, we conclude that
B=
[
x∈B
Vx .
4
1 Fundamental topological concepts
Since Vx ∈ O for every x ∈ B, the above union is again in O by the second property
of topological space. Thus, B ∈ O. ⊓
⊔
Note that the above theorem implies that every open set can be written as a union of
open sets.
Lemma 1.3. Let (X, O) be a topological space and U, B ⊂ X. If U ⊂ B, then
Int(U) ⊂ Int(B).
Proof. Let x ∈ Int(U). Then, U is a neighbourhood of x. Hence, there exists Vx ∈ O
such that x ∈ Vx ⊂ U. But U ⊂ B. Thus, Vx ⊂ B. Hence, B is a neighbourhood of x
and x ∈ Int(B). ⊓
⊔
Theorem 1.2. Let (X, O) be a topological space and B ⊂ X. Then, Int(B) is
open.
Proof. If we show that Int(Int(B)) = Int(B), then Theorem 1.1 will imply that
Int(B) ∈ O.
1. We show the inclusion Int(Int(B)) ⊂ Int(B). Let x ∈ Int(Int(B)). Then, Int(B) is
a neighbourhood of x and by definition of neighbourhood x ∈ Int(B).
2. We prove Int(B) ⊂ Int(Int(B)). Let x ∈ Int(B) = {x ∈ X | ∃Vx ∈ O s.t. x ∈ Vx ⊂ B}.
We show that x ∈ Int(Int(B)) = {x ∈ X | ∃Nx ∈ O s.t. x ∈ Nx ⊂ Int(B)}. Since x ∈
Int(B), B is a neighbourhood of x and there exists a Vx ∈ O such that x ∈ Vx ⊂ B.
But Vx ∈ O, therefore Theorem 1.1 implies that Vx = Int(Vx ). Hence, Int(Vx ) is a
neighbourhood of x and there exists a set Nx ∈ O with x ∈ Nx ⊂ Int(Vx ). Moreover,
Lemma 1.3 implies that Int(Vx ) ⊂ Int(B). Hence, we proved the existence of a set
Nx ∈ O such that x ∈ Nx ⊂ Int(B). Thus, x ∈ Int(Int(B)). ⊓
⊔
Corollary 1.1. Let (X, O) be a topological space and B ⊂ X. Then, Int(B) is
the largest open set contained in B.
Proof. Theorem 1.2 ensures that Int(B) is open. Theorem 1.1 implies that Int(Int(B)) =
Int(B). Let x ∈ Int(B). Then, x ∈ Int(Int(B)) and Int(B) is a neighbourhood of x.
Hence, there exists Vx ∈ O such that x ∈ Vx ⊂ Int(B). Since x ∈ Int(B), B is also a
neighbourhood of x and x ∈ B. Thus, x ∈ Vx ⊂ Int(B) ⊂ B. Taking the union over all
x ∈ Int(B) we find
1.1 ABC of topology
5
[
x∈Int(B)
Vx = Int(B) ⊂ B,
that is Int(B) is the largest open set contained in B.
⊓
⊔
The following results about the closure of a set are fundamental.
Lemma 1.4. Let (X, O) be a topological space and A, B ⊂ X. If A ⊂ B, then
A ⊂ B.
/ Since
Proof. Let x ∈ A. Then, for every neighbourhood N of x we have N ∩ A 6= 0.
A ⊂ B, we also have N ∩ B 6= 0/ for every neighbourhood N of x. Hence, x ∈ B. ⊓
⊔
Lemma 1.5. Let (X, O) be a topological space and I an indexing set. Further,
consider the family of closed sets G = {Cλ ⊂ X | X −Cλ ∈ O ∀ λ ∈ I}. Then,
\
Cλ is closed.
λ ∈I
Proof. By induction. If λ = 1, clearly C1 is closed. Let λ = 1, 2. We show that
C1 ∩C2 is closed, i.e. X − (C1 ∩C2 ) is open. Using De Morgan’s law we have
X − (C1 ∩C2 ) = (X −C1 ) ∪ (X −C2 ).
Since X −C1 , X −C2 ∈ O and unions of open sets are open by the second property
of topological space, we conclude that X − (C1 ∩C2 ) ∈ O. Let n > 2 be any natural
number. Suppose that the intersection of n − 1 closed sets of G is closed. We want
to show that
n
\
Cλ is closed.
λ =1
Since
C=
n=1
\
Cλ is closed,
λ =1
and
n
\
λ =1
Cλ = C ∩Cn ,
we conclude that (1.1) is closed because intersection of two closed sets. ⊓
⊔
(1.1)
6
1 Fundamental topological concepts
Lemma 1.6. Let (X, O) be a topological space and A ⊂ X. Then,
A closed ⇐⇒ A = A.
Proof. =⇒ Let A be closed. We have to show the following inclusions
1. A ⊂ A: Instead of proving x ∈ A =⇒ x ∈ A we shall show the contrapositive
x∈
/ A =⇒ x ∈
/ A. Let x ∈
/ A. Then, there exists a neighbourhood N of x such that
N ∩ A = 0.
/ Since x ∈ N, it follows that x ∈
/ A.
2. A ⊂ A: Instead of showing x ∈ A =⇒ x ∈ A we prove the contrapositive x ∈
/ A =⇒
x∈
/ A. Let x ∈
/ A. Then, x ∈ X − A. Since A is closed, X − A is open. Hence, it
is a neighbourhood of x. Therefore, we found a neighbourhood of x such that
(X − A) ∩ A = 0/ and we conclude that x ∈
/ A.
⇐= Let A = A. In order to show that A is closed we will prove that X − A ∈ O. Let
x ∈ X − A. Then, x ∈
/ A. Since A = A, also x ∈
/ A. Hence, there exists a neighbourhood
N of x such that N ∩ A = 0.
/ This implies that N ⊂ X − A. Since N is a neighbourhood
of x there exists Ux ∈ O such that x ∈ Ux ⊂ N. Hence, Ux ⊂ X − A. Taking the union
over all x ∈ X − A we find that
[
x∈X−A
Ux = X − A.
Since a union of open sets is open, it must be X − A ∈ O. ⊓
⊔
By means of the above preliminary result we can prove that the closure of a set H
is the smallest closed set containing H. This follows from the next theorem and the
fact that H ⊂ H.
Theorem 1.3. Let (X, O) be a topological space, H ⊂ X and I an indexing set.
Further, consider the family of closed set F = {Kλ ⊂ X | X − Kλ ∈ O ∀λ ∈ I}
such that H ⊂ Kλ for every λ ∈ I. Define the set
K=
\
Kλ .
λ ∈I
Then, H = K.
Proof. We have to show the following inclusions.
1. H ⊂ K: Since H ⊂ Kλ for every λ ∈ I, then H ⊂ K. Lemma 1.4 implies that
H ⊂ K. Since K is closed by Lemma 1.5, we have K = K and therefore H ⊂ K.
1.1 ABC of topology
7
2. K ⊂ H: Instead of showing x ∈ K =⇒ x ∈ H we prove the contrapositive x ∈
/
H =⇒ x ∈
/ K. Let x ∈
/ H. Then, there exists a neighbourhood N of x such that
N ∩ H = 0.
/ Since N is a neighbourhood, there exists a set Ux ∈ O such that
x ∈ Ux ⊂ N. Moreover, X − Ux is closed and H ⊂ K by definition of K. Hence,
H ⊂ K ⊂ X −Ux . Since x ∈
/ H and x ∈
/ X −Ux we conclude that x ∈
/ K. ⊓
⊔
By means of open sets we defined closed sets, neighbourhoods and closures. However, all these concepts are very closely related and one can give an equivalent definition of topological space in terms of closed sets, neighbourhoods or closures.
Definition 1.3. A topological space is a pair (X, C ) with C a collection of
subsets of X called closed sets such that
1. X, 0/ ∈ C ;
2. arbitrary intersections of closed sets are closed;
3. finite unions of closed sets are closed.
The next definition of topological space is the original one given in 1914 by the
German mathematician Felix Hausdorff in his Grundzüge der Mengenlehre.
Definition 1.4. A topological space is a pair (X, U ) consisting of a set X
and a collection of subsets of X called neighbourhoods and denoted by U =
{Ux }x∈X such that
1. X is a neighbourhood of all of its points and if x ∈ X every neighbourhood
of x contains x;
2. if V ⊂ X contains a neighbourhood of x, then V is also a neighbourhood of
x;
3. the intersection of any two neighbourhoods of x is again a neighbourhood
of x;
4. every neighbourhood of x contains another neighbourhood which is a
neighbourhood of all of its points.
Finally, the most elegant definition of topological space has been introduced by
the Polish mathematician Kazimierz Kuratowski in his doctoral thesis in 1920.
Definition 1.5. A topological space is a pair (X, −) consisting of a set X and
a map − : P(X) −→ P(X) called the closure map such that
1. 0/ = 0;
/
8
1 Fundamental topological concepts
2. A ⊂ A for every A ⊂ X;
3. A = A for every A ⊂ X;
4. A ∪ B = A ∪ B for all A, B ⊂ X.
Example 1.2. A given set may have many different topologies. If a set is given a
different topology, it is viewed as a different topological space. Let X be a nonempty
set. The first two examples show that we can always endow X with at least two
topologies.
1. The discrete topological space: (X, O) with O = P(X).
2. The indiscrete or trivial topological space: (X, O) with O = {X, 0}.
/ The trivial
topology is the smallest topology we can impose on X. We say that it is weaker
than any other topology on X, since any other topology will contain X, 0/ as well
as other subsets of X.
3. The Sierpinski topological space: X = {a, b} with O = {0,
/ X, {a}}. This is the
simplest topological space that is neither discrete nor trivial.
4. the real line R with the usual topology, that is R with a collection of intervals of
the form (a, b) with a < b where we also allow for a = −∞ and b = ∞. We denote
this collection by the symbol Ou .
Example 1.3. The following examples show that the property of being open or
closed is not a dual property since we can construct cases where a set is neither
open nor closed at the same time. However, the same set can become closed or
open when viewed as a subset of an appropriate subspace of the original topological
space.
1. Consider (R, Ou ).
a. [a, b] is closed in (R, Ou ) since
R − [a, b] = (−∞, a) ∪ (b, ∞) ∈ Ou .
b. [a, b) is neither closed nor open in (R, Ou ). Clearly, it is not open because
[a, b) ∈ Ou . Moreover, it is not closed since
R − [a, b) = (−∞, a) ∪ [b, +∞) ∈
/ Ou .
2. Consider (R, Ou ) and (a, b) ⊂ R with a < b. Further, let (c, b) ⊂ R with c < a
equipped with the collection of open sets
Os = {(x1 , x2 ) | x1 < x2 , c < x1 < b, c < x2 < b}.
Then [a, b) is closed in ((c, b), Os ) since
(c, b) − [a, b) = (c, a) ∈ Os .
1.1 ABC of topology
9
3. Consider R with the discrete topology. In this case O = P(R) and it contains
all possible subsets of R such as {a}, {a, b}, {a, b, c, · · ·}, (−∞, a), (−∞, a] and
so on. Then [a, b) is open and closed at the same time. Clearly, it is open since
[a, b) ∈ P(R). Moreover, it is also closed because
R − [a, b) = (−∞, a) ∪ [b, ∞) ∈ P(R).
Example 1.4. We compute the closure of some sets.
1. Consider (R, Ou ) and H ⊂ R with H = {0} ∪ [1, 2). We apply the definition of
closure, i.e. x ∈ H if for every neighbourhood N of x we have N ∩ H 6= 0.
/ A
neighbourhood of x will be any interval of R and the interval itself does not need
to be open. Clearly, points in (0, 1) will belong to the closure of H as well as the
points in the set {0, 1, 2}. Hence, H = {0} ∪ [1, 2].
2. Consider (R, Ou ) and B = {n−1 | n ∈ N}. To compute the closure of B it is again
convenient to grasp back to the definition of a point in the closure of a set. Clearly,
n−1 ∈ B for every n ∈ N since every neighbourhood of n−1 contains at least this
point and therefore its intersection with B will not be empty. As expected B ⊂ B.
Moreover, the sequence of points (xn )n∈N with xn = n−1 converges to 0. Hence,
∀ε > 0 ∃N ∈ N such that 1/n < ε whenever n ≥ N. This means, that whatever
neighbourhood of zero we are going to choose we can always fix N large enough
so that the intersection of this neighbourhood with B will contain at least a point
of B. We conclude that 0 ∈ B and B = {0} ∪ B.
Definition 1.6. Let (X, O) be a topological space. A subset A ⊂ X is dense in
X if
A = X.
Theorem 1.4. The set of the rational numbers Q is dense in R equipped with
the usual topology.
Proof. We show that Q = R. The proof consists mainly in showing that
1. Q ⊂ R: taking into account that Q = Int(Q) ∪ ∂ Q. If we show that Int(Q) = 0/
and ∂ Q ⊂ R, then we are done.
a. Int(Q) = 0:
/ note that Int(Q) = {x ∈ R | x is an interior point of Q} = {x ∈
R | ∃ (a, b) s.t. x ∈ (a, b) ⊂ Q}. Since Q ⊂ R, we have (a, b) ⊂ R and therefore
by the density of the irrational numbers (a, b) contains at least an irrational
number. We conclude that the interior of Q is empty.
10
1 Fundamental topological concepts
b. ∂ Q ⊂ R: show that x ∈ ∂ Q =⇒ x ∈ R. Let x ∈ ∂ Q and I denote the set of all
irrational numbers. Then, Q and R − Q = I are not neighbourhoods of x. This
implies that ∀ U ∈ Ou x ∈ U 6⊂ Q and ∀ V ∈ Ou x ∈ V 6⊂ I. Since U and V are
open intervals, we have U,V ⊂ R and therefore x ∈ R.
2. R ⊂ Q: show that x ∈ R =⇒ x ∈ Q. Let x ∈ R. Since R = Q ∪ I, we have two
cases
a. x ∈ Q: let V be any neighbourhood of x. Then, there exists (a, b) ∈ Ou with
a < b and a, b ∈ R. Since between any two real numbers we can always find a
q ∈ Q such that a < q < b, it follows that q ∈ (a, b) and hence q ∈ V implying
that V ∩ Q 6= 0.
/
b. x ∈ I: similar procedure as above. ⊓
⊔
Problems
1.1. Give an example of a set admitting more than one topology.
1.2. Zariski topology
Let X = R and O = {X, 0}
/ ∪ {U ⊂ R | R − U is finite}. Show that (X, O) is a
topological space.
1.3. Is it true or false that the pair (X, O) with X = {1, 2, 3} and O = {0,
/ X, {1}, {1, 2}}
is a topological space?
1.4. Construct an example of a topology on R such that the interval (0, 1) is not open
in this topology.
1.5. Show that the definitions of topological space in terms of open sets and closed
sets are equivalent.
1.6. Accumulation and isolated points
Let (X, O) be a topological space and H ⊂ X. A point x ∈ X is a limit point of H if
every neighbourhood of x contains at least one point of H other than x, i.e.
∀ V ⊂ X ∃ Nx ∈ O s.t. (Nx ∩ H)\{x} 6= 0/
with Nx ⊂ V . Moreover, a point x ∈ X is an isolated point of H if
∃ Ux ∈ O s.t. U ∩ H = {x}.
Let I be the set of all isolated points of H and Σ be the set of all limit points of H.
/ Is it true or false that if x is a
Show that Σ is closed and H = Σ ∪ I with Σ ∩ I = 0.
limit point of H, then x ∈ H?
1.7. Consider R with the usual topology and H = {n−1 | n ∈ N} ⊂ R. Find the limit
and isolated points of H.
1.1 ABC of topology
11
1.8. Consider R with the usual topology and H ⊂ R with H = {0} ∪ (1, 2). Find the
limit and isolated points of H.
1.9. Let (X, O) be a topological space is it true or false that if H, K ⊂ X and H 6= K,
then H 6= K?
1.10. Show that definitions 1.1, 1.3, 1.4 and 1.5 are all equivalent.
1.11. Consider R equipped with the usual topology and 0/ 6= H ⊂ R. Further, suppose
that H is bounded above. Show that u = sup H ∈ H. Similarly, show that if H is
bounded below, then inf H ∈ H.
1.12. Consider R with the usual topology and a subset H = (a, ∞) ⊂ R. Compute
the closure of (a, b) with a < b as a subset of R. Compute again the closure of the
same subset but this time seen as a subset of H. What do you observe?
1.13. Let (X, O) be a topological space and A ⊂ X. Show that A = A.
1.14. Let (X, O) be a topological space. Show that for any indexing set I (finite or
denumerable) and for every Hi ⊂ X with i ∈ I
\
i∈I
Hi ⊂
\
H i.
i∈I
1.15. Let (X, O) be a topological space. Is it true or false that for every i ∈ N with
i ≥ 2 and Hi ⊂ X
[
[
Hi =
Hi ?
i∈N,i≥2
i∈N,i≥2
1.16. Let (X, O) be a topological space and Hi ⊂ X for every i ∈ In . Show that
[
i∈In
Hi =
[
Hi .
i∈In
1.17. Let (X, O) be a topological space and A ⊂ X.
1. Show that X − A = Int(X − A).
2. Let H ⊂ X. Show that ∂ H = H ∩ (X − H).
3. Consider R with the usual topology and H = (0, 1]. Compute ∂ H by means of
the formula established in the previous point.
1.18. Consider R with the usual topology. Compute the closures of the sets H =
{n/(n + 1) | n ∈ N} and A = (−2, 0) ∪ (0, 3).
1.19. Take R with the usual topology and N ⊂ R. Find the closure of N.
1.20. Consider R2 with the topology of the open discs and H ⊂ R2 given by H =
{(x, y) ∈ R2 | − 1/x < y < 1/x, x > 0}. Find H.
1.21. Is it true or false that the interior and the closure of a subset of a set X depend
on X and the topology on this set? Give examples.
12
1 Fundamental topological concepts
1.2 Metric spaces
Metric spaces offer a non trivial example of topological space. Roughly speaking a
metric space is a set of points where we can compute distances among these points.
Definition 1.7. A metric space is a pair (X, d) consisting of a non empty set
and a distance function d : X × X −→ R called metric such that
1. d(x, y) ≥ 0 ∀ x, y ∈ X and d(x, y) = 0 ⇐⇒ x = y;
2. d(x, y) = d(y, x) ∀ x, y ∈ X (symmetry);
3. d(x, y) ≤ d(x, z) + d(z, y) ∀ x, y, z ∈ X (triangular inequality).
The above properties of a metric are inspired by our everyday experience with
the concept of distance in an Euclidean three-dimensional space. The first property
mimics the fact that if we take two arbitrary points in R3 their distance will be
zero if they coincide and positive if they are distinct. The second property tells us
that the distance between two points x and y remains the same whether we measure
from x to y or vice versa. However, there are also physical theories inspired by
Noncommutative Geometry postulating that at the Planck length (10−35 cm) the
world is noncommutative in the sense that d(x, y) 6= d(y, x). The last axiom in the
definition of metric space captures the fact that the sum of the lengths of two sides of
any triangle is not less than the length of the third side. This property becomes very
useful once we introduce the topology of a metric space. Why did we introduce only
these three properties? Because experience has shown that in many proofs about
continuity of maps between Euclidean spaces only these three properties are used
and not the explicit formula for the Euclidean distance.
Example 1.5. In the list below you can find several examples of famous metric
spaces. Some of them play an important role in Functional Analysis.
1. n-dimensional Euclidan space:
s
n
X = Rn ,
d(x, y) =
∑ (xi − yi )2
i=1
∀ x = (x1 , · · · , xn ), y = (y1 , · · · , yn ) ∈ Rn
2. discrete metric space:
X any nonempty set,
d(x, y) =
1 if x 6= y
0 if x = y
∀ x, y ∈ X
3. taxi-cab metric space:
X = Rn ,
n
d(x, y) = ∑ |xi − yi | ∀ x = (x1 , · · · , xn ), y = (y1 , · · · , yn ) ∈ Rn
i=1
1.2 Metric spaces
4. X = Rn ,
13
d∞ (x, y) = max |xi − yi | ∀ x = (x1 , · · · , xn ), y = (y1 , · · · , yn ) ∈ Rn with
i∈In
In = {1, 2, · · · , n}.
5. X = C with d(z, w) = |z − w| ∀ z, w ∈ C
6. X = B([a, b]) = { f : [a, b] ⊂ R −→ R | f is bounded on [a, b]} with the supremum
metric
d∞ ( f , g) = sup | f (x) − g(x)| ∀ f , g ∈ B([a, b])
x∈[a,b]
7.
L p -space:
X = C([a, b]) = { f : [a, b] ⊂ R −→ R | f is continuous on [a, b]},
Z b
1p
∀ p ≥ 1 and ∀ f , g ∈ C([a, b])
d( f , g) =
dx ( f (x) − g(x)) p
a
8. real Hilbert sequence space ℓ2 :
∞
X = {(xn )n∈N | xn ∈ R ∀ n ∈ N and
∞
d(x, y) =
∑ (xn − yn )2
n=1
!1/2
∑ xn2 < ∞},
n=1
∀ x = (xn )n∈N , y = (yn )n∈N ∈ X
Proof. We show that ℓ2 is a metric space. There are two non trivial things to be
shown, namely: the distance function is well-defined and the triangular inequality
holds.
a. We check that d(x, y) < ∞ for every x, y ∈ X. This amounts to show that the
2
series ∑∞
n=1 (xn − yn ) is convergent. Consider the sequence of partial sums
(sN )N∈N with sN = ∑Nn=1 (xn − yn )2 . Then
N
N
N
N
N
N
N
n=1
n=1
n=1
n=1
n=1
n=1
n=1
∑ (xn − yn )2 = ∑ xn2 + ∑ y2n − 2 ∑ xn yn ≤ ∑ xn2 + ∑ y2n + 2 ∑ |xn ||yn |
Applying the Cauchy-Schwartz inequality
s
s
N
N
N
n=1
n=1
n=1
∑ rntn ≤ ∑ rn2 ∑ tn2
with rn = |xn | and tn = |yn | we obtain
N
N
N
n=1
n=1
n=1
s
N
s
N
∑ (xn − yn )2 ≤ ∑ xn2 + ∑ y2n + 2 ∑ xn2 ∑ y2n =
≤
s
∞
∑
n=1
xn2 +
s
n=1
∞
∑
n=1
y2n
!2
n=1
< ∞ since
(1.2)
s
N
s
N
∑ xn2 + ∑ y2n
n=1
∞
∞
n=1
n=1
n=1
∑ xn2 , ∑ y2n < ∞.
!2
,
14
1 Fundamental topological concepts
Finally,
N
∑ (xn − yn )2 < ∞ =⇒ (sN )N∈N convergent.
N→∞
lim
n=1
b. Let x, y, z ∈ X. Then, from
|xn − yn | ≤ |xn − zn | + |zn − yn |
∀n∈N
we have
N
∑ (xn − yn )2 ≤
n=1
≤
≤
N
N
n=1
n=1
N
N
N
∑ (xn − zn )2 + ∑ (zn − yn )2 + 2 ∑ |xn − zn ||zn − yn |,
∑ (xn − zn )
n=1
2
n=1
s
+ ∑ (zn − yn ) + 2
2
n=1
∞
∞
n=1
n=1
N
N
∑ (xn − zn )2 ∑ (zn − yn )2 ,
n=1
s
s
∞
s
n=1
∞
∑ (xn − zn )2 + ∑ (zn − yn )2 + 2 ∑ (xn − zn )2 ∑ (zn − yn )2 ,
n=1
n=1
= d(x, z) + d(z, y).
after application of (1.2) on the r.h.s. of the first line. Taking the limit for
N → ∞ we obtain
N
∑ (xn − yn )2 = d(x, y) ≤ d(x, z) + d(z, y)
N→∞
lim
n=1
and the proof is completed.
⊓
⊔
9. X = Rn with distance function
n
d p (x, y) =
∑ |xi − yi |
i=1
!1
p
p
defined for all x, y ∈ Rn and p ≥ 1 is a metric space. The triangle inequality can
be verified using the Hölder and Minkowsky inequalities.
10. the sequence space ℓ∞ :
∞
ℓ∞ = {(xn )∞
n=1 | xn ∈ R ∀n ∈ N and (xn )n=1 bounded}
with
ds (x, y) = sup |xn − yn |
n∈N
∀x, y ∈ ℓ∞ .
Points in a metric space can be vectors as in Rn , sequences, functions and so on.
Given an underlying set X we can assign different metrics as it is the case for Rn
(see 1., 3. and 4. in the previous example). We already encountered a similar phenomenon when we treated topological spaces where given a set X we can assign to
it different topologies.
1.2 Metric spaces
15
Lemma 1.7. Subsets of metric spaces are again metric spaces.
Proof. Let (X, d) be a metric space and H ⊂ X. We can construct a metric on H
by considering the restriction of d on H, that is the map dH : H × H −→ R. Then,
(H, dH ) is again a metric space since dH (x, y) = d(x, y) for all x, y ∈ H. In other
words, the metric space axioms hold for dH since they hold for d. ⊓
⊔
The metric dH is called the metric on H induced by d.
Example 1.6. We want to show that (C([a, b]), d∞c )) with
d∞c ( f , g) = sup | f (x) − g(x)| ∀ f , g ∈ C([a, b])
x∈[a,b]
is a metric space. Since every continuous function on a closed set is bounded, it
follows that C([a, b]) ⊂ B([a, b]). Since B([a, b]), d∞ ) is a metric space (see 6. in
the previous example) and d∞c ( f , g) = d∞ ( f , g) for all f , g ∈ C([a, b]), it follows
that d∞c is the metric on C([a, b]) induced by d∞ . Hence, Lemma 1.7 implies that
(C([a, b]), d∞c )) is a metric space.
Definition 1.8. Let (X, d) be a metric space. We say that a subset V ⊂ X is
open if for all x ∈ V there exists some εx > 0 such that the εx -ball with centre
x is contained in V , that is
Bεx (x) = {y ∈ X | d(x, y) < εx } ⊂ V.
The use of the notation εx emphasizes the fact that the choice of the radius of the
ball depends on the centre of the ball. There are situations where one should keep in
mind this aspect.
Example 1.7. Take the metric space (R, d) with d(x, y) = |x − y| for all x, y ∈ R. In
this metric space an εx -ball about the point x is an open interval since
Bεx (x) = {y ∈ R | d(x, y) < εx } = {y ∈ R | |x − y| < εx },
= {y ∈ X | − εx < x − y < εx },
= (x − εx , x + εx ).
If we take R2 with the Euclidean distance
q
d(x, y) = (x1 − y1 )2 + (x2 − y2 )2
∀x = (x1 , x2 ), y = (y1 , y2 ) ∈ R2 ,
16
1 Fundamental topological concepts
then an εx -ball about the point x is the interior of a disc of radius εx and centre
x = (x1 , x2 ) since
q
Bεx (x) = {y ∈ R2 | d(x, y) < εx } = {y ∈ R2 | (x1 − y1 )2 + (x2 − y2 )2 < εx },
= {y ∈ R2 | (x1 − y1 )2 + (x2 − y2 )2 < εx2 }.
Finally, if take R3 with the usual Euclidean metric
q
d(x, y) = (x1 − y1 )2 + (x2 − y2 )2 + (x3 − y3 )2 ∀x = (x1 , x2 , x3 ), y = (y1 , y2 , y3 ) ∈ R3 ,
it is straightforward to see that an εx -ball is the interior of a sphere of radius εx and
centre x = (x1 , x2 , x3 ).
The term open ball is inspired by the above cases in the Euclidean space. The next
example warns us not to take the name ”open ball” too litterally.
Example 1.8. Consider the metric space R2 with the metric
d∞ (x, y) = sup {|x1 − y1 |, |x2 − y2 |} ∀x = (x1 , x2 ), y = (y1 , y2 ) ∈ R2 .
The εa -ball of radius εa and centre a = (a1 , a2 ) ∈ R2 is
Bεa (a) = {x ∈ R2 | d∞ (x, a) < εa } = {x ∈ R2 | sup{|x1 − a1 |, |x2 − a2 |} < εa }.
We have two cases. If |x1 − a1 | > |x2 − a2 |, then |x1 − a1 | < εa and hence, x1 ∈
(a1 − εa , a1 + εa ). If |x2 − a2 | > |x1 − a1 |, then |x2 − a2 | < εa and hence, x2 ∈ (a2 −
εa , a2 + εa ). In this case an εa -ball is a square of centre a = (a1 , a2 ) and sides having
length 2εa .
Example 1.9. Let us take R with the Euclidean metric d(x, y) = |x − y| defined for
all x, y ∈ R. Then, the ball of unit radius and centre one is
B1 (1) = {x ∈ R | d(1, x) < 1} = (0, 2).
Consider now the subset H = [0, 1] of R and let dH be the metric induced by d on
H, that is dH (x, y) = d(x, y) for all x, y ∈ H. What is the open ball of unit radius and
centre one in the metric space (H, dH )? First of all,
BH
1 (1) = {x ∈ H | dH (1, x) < 1}.
Clearly, x = 0 ∈
/ BH
1 (1) since dH (1, 0) < 1 would imply 1 < 1 which is impossible.
If 0 < x < 1, then dH (1, x) < 1 implies that |1 − x| < 1 which is clearly satisfied.
Finally, if x = 1, the inequality dH (1, 1) < 1 is trivially true. Hence, we conclude
that BH
1 (1) = (0, 1].
The last three examples tell us that the shape of an open ball depends on the chosen
metric and that an open ball in R with the usual topology can differ dramatically
1.2 Metric spaces
17
from an open ball of a subset of R equipped with some induced metric. Indeed, we
discovered that B1 (1) is open with respect to the topology O(d) and that BH
1 (1)
is open with respect to the topology O(dH ) but for instance BH
(1)
is
not
open
1
when seen as a subset of R endowed with the usual topology. We saw a similar
phenomenon when we treated topological spaces. At this point we might start to
supsect that metric spaces are indeed topological spaces. To prove this important
fact we first need to show the following preliminary results.
Lemma 1.8. Let (X, d) be a metric space and Bεx (x) ⊂ X. Then, for all y ∈
Bεx (x) there exists some δy > 0 such that Bδy (y) ⊂ Bεx (x), i.e. any open ball is
open.
Proof. Take any y ∈ Bεx (x). Let δy = εx − d(x, y) which is clearly positive. Take any
z ∈ Bδy (y) and show that z ∈ Bεx (x), that is d(z, x) < εx . Note that we have
d(z, x) ≤ d(z, y) + d(y, x) < δy + d(x, y) = εx − d(x, y) + d(x, y) = εx
and we are done.
⊓
⊔
Definition 1.9. Let (X, d) be a metric space and N ⊂ X. We say that N is a
neighbourhood of x ∈ X if there exists an open ball Bεx (x) such that Bεx (x) ⊂
N.
Lemma 1.9. Let (X, d) be a metric space and A ⊂ X. Then, A is open if and
only if A can be written as a union of open balls.
Proof. We need to prove both directions.
1. =⇒: let A be open. Then, Definition 1.8 implies that for all x ∈ A there exists
some εx > 0 such that Bεx (x) ⊂ A but then
[
Bεx (x) = A.
(1.3)
x∈A
2. ⇐=: let A be the union of open ball as in (1.3). Take any y ∈ A. Then, y will
belong to some open ball Bεz (z). By Lemma 1.8 we can find a δy > 0 such that
Bδy (y) ⊂ Bεz (z) ⊂
[
z∈A
Bεz (z) = A
18
1 Fundamental topological concepts
and therefore A must be open.
⊓
⊔
Lemma 1.10. Let (X, d) be a metric space. Then, X and 0/ are open.
Proof. By contradiction. Suppose that 0/ is not open. Then, there exists a point x ∈ 0/
such that for all εx > 0 we have Bεx (x) 6⊂ 0.
/ This would imply that that the empty set
is not empty and this is absurde. Suppose that X is not open. Then, there exists some
x ∈ X such that for all εx > 0 we have Bεx (x) 6⊂ X. This would imply that in general
Bεx (x) contains other points than points of X and this is impossible since there are
no such other points. ⊓
⊔
Lemma
1.11. Let (X, d) be a metric space and U1 , · · · ,Un open subsets of X.
T
Then, ni=1 Ui is open.
Proof. There are two cases.
T
1. If ni=1 Ui = 0,
/ then the statement follows trivially from Lemma .
T
T
2. Let ni=1 Ui 6= 0.
/ Then, there exists some x ∈ X such that x ∈ ni=1 Ui . By definition
of intersection and open set we have that
x ∈ U1 open =⇒ ∃ε1 > 0 such that Bε1,x ⊂ U1 ,
..
.
x ∈ Un open =⇒ ∃εn > 0 such that Bεn,x ⊂ Un .
Take εx = min1≤i≤n {εi }. Then, Bεx (x) ⊂ Ui for all i = 1, 2, · · · , n and hence
Bεx (x) ⊂
as required.
n
\
Ui
i=1
⊓
⊔
Lemma 1.12. Let (X, d) be a metric space and I an indexing set. Further,
consider the family of open sets
Then,
S
i∈I Ui
F = {Ui ⊂ X | Ui open ∀i ∈ I}.
is open.
1.2 Metric spaces
19
Proof. There are two cases.
S
1. If i∈I Ui = 0,
/ then the statement follows trivially from Lemma 1.
S
2. Suppose i∈I Ui 6=
0.
/ We need to construct an open ball contained in the union
S
of all Ui . Let x ∈ i∈I Ui . Then, there exists a k ∈ I such that x ∈ Uk . Since Uk is
open, then there exists some εx > 0 such that Bεx (x) ⊂ Uk . So we have
Bεx (x) ⊂ Uk ⊂
as required.
[
Ui
i∈I
⊓
⊔
Theorem 1.5. Every metric space is a topological space.
Proof. Let (X, d) be a metric space and O(d) the collection of all open subsets of X.
Lemma 1.10 implies that X, 0/ are open. Lemma 1.12 implies that arbitrary unions
of open sets are again open while Lemma 1.11 ensures that a finite intersection of
open sets is again open. ⊓
⊔
Different metrics can give rise to the same topology. If this is the case we say that
these metrics are topologically equivalent.
′
Definition 1.10. Let X be a set equipped with metrics d and d . Consider the
′
′
topological spaces (X, O(d)) and (X, O(d )) and U ⊂ X. We say that d an d
are topologically equivalent whenever
′
U ∈ O(d) ⇐⇒ U ∈ O(d ).
Example 1.10. Take X = R2 and the metrics
q
dE (x, y) = (x1 − y1 )2 + (x2 − y2 )2 , d∞ (x, y) = sup{|x1 − y1 |, |x2 − y2 |}.
We want to verify that (R2 , O(dE )) and (R2 , O(d∞ )) are topologically equivalent.
We need to check the above definition. Take any U ∈ O(d∞ )). Then, for all x ∈ U
there exists an open ball B∞
εx (x) ⊂ U. Clearly, U ∈ O(dE ) if for all x ∈ U we have
∞
E
Bρx (x) ⊂ Bεx (x). From a previous example we know that Bε∞x (x) is a square with
side 2εx . Then, BρEx (x) can be taken as the smallest disc inscribed inside the square
B∞
εx (x). Take ρx = εx and we are done. We need now to show that if U ∈ O(dE )
then U ∈ O(d∞ ). Take any U ∈ O(dE ). Then, for any x ∈ U we have BρEx (x) ⊂ U.
E
If we can show that B∞
ρx (x) ⊂ Bεx (x) then we are done. Take the smallest square of
20
1 Fundamental topological concepts
side 2ρx inscribed in the circle of radius
√ εx . Then, Pythagora’s theorem implies that
4ρx2 + 4ρx2 = 4εx2 and hence ρx = εx / 2.
The next theorem is very useful in constructing a new metric once we already have
a metric space.
Theorem 1.6. Let (X, d) be a metric space. Then,
1.
′
d (x, y) =
d(x, y)
1 + d(x, y)
∀ x, y ∈ X is a metric
′
2. d and d are topologically equivalent.
Proof. Left as an exercise.
Definition 1.11. A topological space (X, O) is metrizable if there exists a metric d on X such that O(d) = O.
Theorem 1.7. Let (X, O) be a trivial topological space. If X contains more
than one element then (X, O) is not metrizable.
Proof. Left as an exercise.
This result is very important since it shows that a topological space does not need to
be a metric space and hence the concept of topological space is more general than
the concept of metric space.
Example 1.11. We want to study open balls in a metric space endowed with a discrete metric. Let (X, d) be a discrete metric space. Then, for some point a ∈ X the
open ball of radius 2 and centre a is
B2 (a) = {x ∈ X | d(x, a) < 2} = X
because for x 6= a we have d(x, a) = 1 < 2 whereas for x = a we have d(x, a) = 0 < 2.
If we consider instead the open ball
B1 (a) = {x ∈ X | d(x, a) < 1},
1.3 Back to topological spaces
21
it turns out that that B1 (a) = {a} because for any x 6= a we have d(x, a) = 1 which
cannot be less than one. Hence, the only admissible case is x = a since then d(x, a) =
d(a, a) = 0 < 1. A similar reasoning gives
B1/2 (a) = {x ∈ X | d(x, a) < 1/2} = {a}.
We conclude that for any εx > 0 and x ∈ X we have
X
if εx > 1,
Bεx (x) =
{x} if 0 < εx ≤ 1.
A remarkable aspect concerning discrete metric spaces is that a set consisting of
only one point is open. This is clearly not the case for R equipped with the Euclidean
distance. Also observe that not all metrics in Rn need to be topologically equivalent.
Take for instance R with the discrete metric. Then, {x} is open but it is not open in
R with the usual Euclidean distance function.
Problems
1.22. Prove Theorem 1.6.
1.23. Prove Theorem 1.7.
1.24. Prove that the metrics
s
n
dE (x, y) =
∑ (xi − yi )2 ,
i=1
n
dT (x, y) = ∑ |xi − yi |,
i=1
d∞ (x, y) = sup {|xi − yi |}
1≤n≤n
defined for any x = (x1 , · · · , xn ), y = (y1 , · · · , yn ) ∈ Rn are all equivalent.
1.3 Back to topological spaces
We introduce two new topologies: the subspace topology and the product topology.
Definition 1.12. Let (X, O be a topological space and X0 ⊂ X. The topology
on X0
O|X0 = {U ∩ X0 | U ∈ O}
is called the induced or subspace topology. We call (X0 , O|X0 ) a subspace of
(X, O).
22
1 Fundamental topological concepts
We check that the subspace topology is indeed a topology.
1. Observe that x0 ∈ O|X0 since X0 = X ∩ X0 and X ∈ O. Moreover, 0/ ∈ O|X0 since
0/ − 0/ ∩ X0 and 0/ ∈ O.
2. Take sets Vα = Uα ∩ X0 with α ∈ I. Using the distributivity property of the intersection over the union we have
!
S
[
α ∈I
Vα =
[
α ∈I
(Uα ∩ X0 ) =
[
α ∈I
Uα
∩ X0 ∈ O|X0
since α ∈I Uα ∈ O.
3. Let U1 ∩ X0 ,U2 ∩ X0 ∈ O|X0 . Then,
(U1 ∩ X0 ) ∩ (U2 ∩ X0 ) = U1 ∩ [X0 ∩ (U2 ∩ X0 )] ,
= U1 ∩ [(X0 ∩U2 ) ∩ X0 ] ,
= U1 ∩ [(U2 ∩ X0 ) ∩ X0 ] ,
= U1 ∩ [U2 ∩ (X0 ∩ X0 )] ,
= U1 ∩ (U2 ∩ X0 ) = (U1 ∩U2 ) ∩ X0 ∈ O|X0
since U1 ∩U2 ∈ O. The rest follows by induction.
Definition 1.13. Consider (X, O) and (X0 , O|X0 ). We say that B ⊂ X0 is open
in the subspace topology of X0 if there exists an open subset W in the topology
of X such that B = W ∩ X0 .
Example 1.12. We look at the following two examples of subspace topologies.
1. Consider R2 with the standard topology generated by the Euclidean metric. The
standard topology is the collection of all open discs. Further, consider the subset X0 = R ⊂ R2 . The induced or subspace topology of R is simply the usual
topology of R consisting of all open intervals.
2. The subspace topology of N as a subset of R with the usual topology is the
discrete topology. For instance, let X0 = N and u = (0, π ). Then,
U ∩ X0 = (0, π ) ∩ N = {1, 2, 3}
and we conclude that O|N = P(N).
Definition 1.14. Let (X, OX ) and (Y, OY ) be topological spaces. The product
space
X ×Y = {(x, y) | x ∈ X and y ∈ Y }
1.3 Back to topological spaces
23
can be turned into a topological space by considering the collection OX×Y
of subsets W ⊂ X × Y where W is open in the following sense: for all pairs
(x, y) ∈ W there exists a neighbourhood U of x ∈ X and a neighbourhood V of
y ∈ Y such that U ×V ⊂ W . We call OX×Y the product topology.
We want to verify that (X ×Y, OX×Y ) is a topological space.
1. We verify that X ×Y ∈ OX×Y . Since X ∈ OX , it follows that X is open and hence,
it is a neighbourhood of all of its points. Take in the above definition U = X.
Similarly, since Y ∈ OY , it follows that Y is open and hence, it is a neighbourhood
of all of its points. Take in the above definition V = Y and we are done. Let us
check that 0/ ∈ OX×Y . We do it by contradiction. Suppose that 0/ ∈
/ OX×Y . Then,
0/ is not open and there exists a pair (x, y) ∈ 0/ such that for every neighbourhood
U of x in X and for every neighbourhood V of y in Y we have U ×V ⊂ 0.
/ This is
absurde since 0/ is empty.
2. Consider the following family of subsets of X ×Y , namely
F = {Wα ⊂ X ×Y | Wα ∈ OX×Y ∀α ∈ I}
where I is an arbitrary indexing set. We want to show that
[
α ∈I
Wα ∈ OX×Y
S
that is we need to verify that for all pairs (x, y) ∈ α ∈I Wα there exists a neighbourhood
U of x in XS and a neighbourhood V of y in Y such that U × V ⊂
S
.
Let
(x, y) ∈ α ∈I Wα . Then, there exists a β ∈ I such that (x, y) ∈ Wβ .
W
α ∈I α
Since Wβ ∈ OX×Y , then there exists a neighbourhhod Uβ of x in X and a neighbourhood Vβ of y in Y such that Uβ ×Vβ ⊂ Wβ and hence
Uβ ×Vβ ⊂
[
β ∈I
Uβ ×Vβ ⊂
[
Wβ .
β ∈I
Take U = Uβ and V = Vβ and we are done.
T
3. We show that ni=1 Wi ∈ OX×Y with Wi ∈ OX×Y for all i ∈ In . The proof is by
induction and we will limit us to prove the case W1 ∩ W2 ∈ OX×Y whenever
W1 ,W2 ∈ OX×Y . Note that W1 ∩ W2 will open in the product topology if for all
pairs (x, y) ∈ W1 ∩ W2 there exist neighbourhoods U of x in X and V of y in Y
such that U ×V ⊂ W1 ∩W2 . We need to consider two cases.
a. If W1 ∩W2 = 0,
/ then the result follows trivially since 0/ ∈ OX×Y .
b. Let W1 ∩W2 6= 0.
/ Then, there exists a pair (x, y) ∈ W1 ∩W2 such that (x, y) ∈ W1
and (x, y) ∈ W2 . Since W1 ∈ OX×Y there exists a neighbourhood U1 of x in X
and a neighbourhood V1 of y in Y such that U1 ×V1 ⊂ W1 . Similarly, since W2 ∈
OX×Y there exists a neighbourhood U2 of x in X and a neighbourhood V2 of y
in Y such that U2 × V2 ⊂ W2 . Observe that U1 ∩ U2 is again a neighbourhood
24
1 Fundamental topological concepts
of x in X and V1 ∩ V2 is again a neighbourhood of y in Y . We will verify
it only for U1 ∩ U2 . Since U1 and U2 are neighbourhoods of x, there exist
open sets Nx , Rx ∈ OX such that x ∈ Nx ⊂ U1 and x ∈ Rx ⊂ U2 . But then x ∈
Nx ∩ Rx ⊂ U1 ∩ U2 . Since Nx ∩ Rx ∈ OX and it contains x, U1 ∩ U2 must be
a neighbourhood of x. Going back to the original proof, since the pair (x, y)
belongs to W1 and W2 and (x, y) belongs to U1 ×V1 and U2 ×V2 , we have
(U1 ×V1 ) ∩ (U2 ×V2 ) ⊂ W1 ∩W2 .
Using a result about the Cartesian product of intersections we find that
(U1 ∩U2 ) × (V1 ∩V2 ) ⊂ W1 ∩W2 .
Let U = U1 ∩ U2 and V = V1 ∩ V2 . Then, U and V are neighbourhoods since
intersections of neighbourhoods.
Remark 1.1. At this point some remarks are in order.
1. The neighbourhhods U and V in Definition 1.14 do not need to be open in OX
and OY , respectively.
2. If U ⊂ X and V ⊂ Y are open neighbourhoods, then U × V ⊂ X × Y is an open
box.
3. The collection of all open boxes does not form a topology. This is pretty clear,
since the union of two open boxes does not need to be an open box.
1.4 Bases of topological spaces
We already know that in a metric space any open set is a union of open balls where
in general infinitely many balls can be involved in the union. In a topological space
it is often convenient to have some subcollection of open sets, called a basis, having
the aforementioned property of the open balls in a metric space. We will also see
that the concept of topological basis is useful to prove continuity of a map in an
efficient way.
Definition 1.15. Let (X, O) be a topological space. A collection B of open
sets is called a basis of the topology O if every element in O can be written
as a union of sets from B.
Example 1.13. Let us look closer to some examples of topological bases.
1. Let us verify that every metrizable topological space has a basis. Let (X, O) be
a metrizable topological space. Then, there exists a distance function d such that
1.4 Bases of topological spaces
25
O = O(d) with O(d) = {U ⊂ X | U ∈ O(d)}. The set of all open balls is a basis
for O(d) since
a. Lemma 1.9 implies that every open set can be written as a union of open balls.
b. (X, d) is a metric space and since every metric space is a topological space,
we clearly have X, 0/ ∈ O(d). Since X and 0/ are open, Lemma 1.9 ensures they
can be written as a union of open balls. In particular, for any x ∈ X and ρx > 0
0/ =
[
λ ∈0/
Bρx ,λ (x).
The basis for O(d) can be written as
B = {Bρ (x) ⊂ X | ρ > 0}.
The set of all open balls does not need to form a topology because for instance
in R2 the intersection of two partially overlapping discs is not a disc.
2. The collection of open boxes forms a basis of the product topology.
Proof. Let W ∈ OX×Y . Then, there exists a neighbourhood U of x in X and a
neighbourhood V of y in Y such that U × V ⊂ W but since U and V are neighbourhoods of x and y, respectively, there exist Nx ∈ OX and My ∈ Oy such that
x ∈ Nx ⊂ U and y ∈ My ⊂ V . Clearly, Nx and My are open neighbourhoods and
Nx × My ⊂ U ×V ⊂ W . Take the union over all pairs (x, y) ∈ W . We obtain
[
(x,y)∈W
and this concludes the proof.
Nx × My = W
⊓
⊔
3. The collection of all open balls in Rn with rational centers and rational radius is
a basis for Rn equipped with the Euclidean metric.
Proof. It is sufficient to prove that for all p ∈ Rn and for any open neighbourhood
U of p there exists an open ball
Bρe (e
q) ∈ B = {B p (q) | ρ ∈ Q+ and q ∈ Qn }
q) ⊂ U. Take any p ∈ Rn and let U be some open neighbourhood
such that p ∈ Bρe (e
of p. Since U belongs to the usual topology of the Euclidean space, there exists
an open ball Br (p) ⊂ U. If we show that p ∈ Bρe (e
q) ⊂ Br (p) for some ρe ∈ Q+ ,
then we are done. Clearly, p ∈ Br (p). By the density of the rationals there exists
ρ ∈ Q+ such that 0 < ρ < r. Consider the ball Bρ (p). Clearly, p ∈ Bρ (p) ⊂
Br (p). The density of the rationals also implies that we can find qe ∈ Qn such
q) ⊂ Br (p). First of all, p ∈ Bρ /2 (e
q)
that qe ∈ Bρ /2 (p). We claim that p ∈ Bρ /2 (e
since qe ∈ Bρ /2 (p) implies that d(p, qe) < ρ /2. We need to prove the inclusion.
Let x ∈ Bρ /2 (e
q). Let us verify that x ∈ Br (p). Note that
26
1 Fundamental topological concepts
d(p, x) ≤ d(p, qe) + d(e
q, x) <
and this completes the proof.
ρ ρ
+ =ρ <r
2 2
⊓
⊔
Theorem 1.8. Let (X, O) be a topological space. Then, B is a basis of the
topology O if and only if for all x ∈ X and any open set U containing x there
exists a set B ∈ B such that x ∈ B ⊂ U.
Proof. We need to prove both directions.
1. =⇒: let B be a basis of O. Take any x ∈ X and some open set U containing x.
Since B basis, then U = Ui∈I Bi with Bi ∈ B for all i ∈ I. Hence, there exists k ∈ I
such that x ∈ Bk and Bk ⊂ U.
2. ⇐=: let U be an open set of x, then U = Int(U) and for all x ∈ U there exists
Nx ∈ O such that x ∈ Nx ⊂ U. Then, clearly
U=
[
Nx .
(1.4)
x∈U
Identify B with Nx and in virtue of (1.4) we can conclude that B is a basis.
⊓
⊔
1.5 Continuous maps
Definition 1.16. Let (X, OX ) and (Y, OY ) be topological spaces. Further consider a map f : x −→ Y . We say that f is continuous if the inverse image of an
open set in the topology OY is open in the topology OX , that is
f −1 (U) ∈ OX
∀ U ∈ OY .
We recall that the inverse image of a set is defined as f −1 (U) = {x ∈ X | f (x) ∈ U}.
Keep in mind that the above definition does not make any assertion about direct
images of open sets. We will later show that this definition of continuos map is a
generalization of the usual ε -δε definition of continuity.
Lemma 1.13. Let (X, O) be a topological space. The identity map IdX : X −→
X such that x 7−→ IdX (x) = x is continuous.
1.5 Continuous maps
27
Proof. Take any U ∈ O. We need to show that Id−1
X (U) ∈ O. Note that
Id−1
X (U) = {x ∈ X | IdX (x) ∈ U} = {x ∈ X | x ∈ U} = X ∩U = U
and this completes the proof.
⊓
⊔
In the next result we show that the composition of continuous maps is again continuous.
Lemma 1.14. Let (X, OX ), (Y, OY ) and (Z, OZ ) be topological spaces. Further, consider the continuous maps f : X −→ Y and g : Y −→ Z. Then,
g ◦ f : X −→ Z is continuous.
Proof. We show that (g ◦ f )−1 (W ) ∈ OX for all W ∈ OZ . Note that since g and f
are continuous we have that g−1 (W ) ∈ OY for all W ∈ OZ and f −1 (U) ∈ OX for all
U ∈ OY . Hence, we can choose U = g−1 (W ) and conclude that
f −1 (U) = f −1 (g−1 (W )) = (g ◦ f )−1 (W ) ∈ OX .
This completes the proof.
⊓
⊔
The next result answer the following question: under which condition any map between two topological spaces is continuous?
Lemma 1.15. Any map from a discrete topological space to a topological
space is continuous.
Proof. Consider the topological spaces (X, P(X)) and (Y, OY ). Further, consider
a map f : X −→ Y . Take any U ∈ OY . Then, f −1 (U) ⊂ X. Hence, by definition of
discrete topology f −1 (U) ∈ P(X). ⊓
⊔
The next example shows that a map which is continuous with respect to a certain
topology does not need to be continuous with respect to another topology.
Example 1.14. Consider a map f : R −→ R such that f (x) = x2 for all x ∈ R.
1. Consider the usual topology Ou of the open intervals on R and interval (a, b)
with a < b. Then,

if a < b < 0,

√Ou
√ 0/ ∈
−1
if a < 0 < b,
(− b, b) ∈ O√u
f (a, b) =
√
√
 √
(− b, − a) ∪ (− a, b) ∈ Ou if 0 < a < b.
28
1 Fundamental topological concepts
2. Take R with the topology
O = {R, 0}
/ ∪ {(−∞, x) | x ∈ R}.
Then, f is not continuous since for instance if a > 0 then
√ √
f −1 (−∞, a) = (− a, a) ∈
/ O.
Lemma 1.16. Let (X, O) be a topological space. Then, the constant map f :
X −→ {a} ⊂ X such that f (x) = a for all x ∈ X and some a ∈ X is continuous.
Proof. Equip the set {a} with the subspace topology O|{a} = {U ∩ {a} | U ∈ OX }.
Take any set U ∩ {a} in the subspace topology. Then,
f −1 (U ∩ {a}) = f −1 (U) ∩ f −1 ({a}) = f −1 (U) ∩ X
but
f
−1
(U) = {x ∈ X | f (x) ∈ U} = {x ∈ X | a ∈ U} =
Hence,
f
−1
(U ∩ {a}) =
0/ if U = 0/ or a ∈
/ U,
X
if a ∈ U.
0/ if U = 0/ or a ∈
/ U,
X
if a ∈ U.
and it belongs to O. This concludes the proof.
⊓
⊔
The concept of topological basis is very useful in order to prove continuity of maps
between topological spaces. Suppose that we have topological spaces (X, OX ) and
(Y, OY ). In the case we need to prove that a map f : X −→ Y is continuous, we can
introduce a basis B for the topology OY . If we can show that f −1 (Bi ) ∈ OX for all
Bi ∈ B where i belongs to some indexing set I, then we are done because B basis
S
implies that U = i∈I Bi for all U ∈ OY and
!
f −1 (U) = f −1
[
i∈I
Bi
=
[
i∈I
f −1 (Bi ) ∈ OX .
We develop some equivalent characterizations of continuity.
Lemma 1.17. Let (X, OX ) and (Y, OY ) be topological spaces. Then, the map
f : X −→ Y is continuous if and only if for all U ⊂ Y closed in Y the inverse
image f −1 (U) is closed in X.
1.5 Continuous maps
29
Proof. We need to prove both directions.
1. =⇒: let f be continuous. Then, for all subsets V open in Y the inverse image
f −1 (V ) is open in X. Take any subset V open in Y and define a new subset
U = Y −V which is closed in Y . Then,
f −1 (U) = f −1 (Y −V ) = f −1 (Y ) − f −1 (V ) = X − f −1 (V )
is closed in X since f −1 (V ) is open in X.
2. ⇐=: take any subset U closed in Y . Then, the subset V = Y −U is open in Y and
f −1 (V ) = f −1 (Y −U) = f −1 (Y ) − f −1 (U) = X − f −1 (U)
is closed in X since f −1 (U) is open in X.
This completes the proof.
⊓
⊔
The next criterion is used in some books to define continuity at a point.
Lemma 1.18. (Local criterion of continuity)
Let (X, OX ) and (Y, OY ) be topological spaces. Then, the map f : X −→ Y
is continuous if and only if for all neighbourhoods U ⊂ Y of f (x) in Y the
inverse image f −1 (U) is a neighbourhood of x in X.
Proof. We need to prove both directions.
1. =⇒: let f be continuous. Take any neighbourhood U of f (x). Show that f −1 (U)
is a neighbourhood of x. If U is a neighbourhhod of f (x), then there exists an
open subset N f (x) in Y such that f (x) ∈ N f (x) ⊂ U. Since f is continuous, the
inverse image f −1 (N f (x) ) is open in X. Moreover, x ∈ f −1 (N f (x) ) ⊂ f −1 (U).
Hence, f −1 (U) is a neighbourhood of x.
2. ⇐=: suppose that for any neighbourhood U of f (x) the inverse image f −1 (U)
is a neighbourhood of x. We need to show that for any open subset V in Y the
inverse image f −1 (V ) is open in X. Let V be an open subset of Y .Then, V is a
neighbourhood of all of its points and f −1 (V ) is a neighbourhhod of all of its
points in X. Hence, f −1 (V ) is open in X.
This completes the proof.
⊓
⊔
Continuity can also be characterized in terms of closures as follows. Let (X, OX )
and (Y, OY ) be topological spaces. Then, the map f : X −→ Y is continuous if and
only if f −1 (B) ⊂ f −1 (B) for all B ⊂ Y . We want to show that from the characterization of continuity in terms of neighbourhoods it is possible to derive a definition
of continuity for a map between metric spaces. Moreover, this definition contains as
a special case the usual ε -δε definition of continuity of real functions. This can can
be done in two steps plus a short lemma.
30
1 Fundamental topological concepts
1. Let (X, OX ) and (Y, OY ) be topological spaces. Lemma 1.18 ensures that the map
f : X −→ Y is continuous if and only if for every neighbourhood U of f (x) in Y
the inverse image f −1 (U) is a neighbourhood of x in X.
2. Suppose that the above topological spaces are metrizable. Then, there exist met′
′
′
rics d and d such that OX = O(d) and OY = O(d ). Then, (X, d) and (Y, d ) are
′
metric spaces. Moreover, f will be continuous at x if for any subset U ∈ O(d )
−1
−1
containing f (x) the inverse image f (U) ∈ O(d) with x ∈ f (U). Since U is
′
open there exists an open ball Bε ( f (x)) ⊂ U. On the other hand, f −1 (U) ∈ O(d)
means that there exists an open ball Bδ (x) ⊂ f −1 (U). Hence, we can claim that
′
′
f : X −→ Y ⇐⇒ ∀ Bε ( f (x)) f −1 Bε ( f (x)) ∈ O(d).
′
In order that f −1 Bε ( f (x)) ∈ O(d) there must be an open ball Bδ (x) ∈ O(d)
′
such that Bδ (x) ⊂ f −1 Bε ( f (x)) .
Lemma 1.19. Let f : X −→ Y and N ⊂ X and M ⊂ Y . Further, suppose that
N ⊂ f −1 (M). Then, f (N) ⊂ M.
Proof. Let f (x) ∈ f (N) with f (N) = { f (x) ∈ Y | x ∈ N}. We want to show that
f (x) ∈ M. If f (x) ∈ f (N), then x ∈ N but N ⊂ f −1 (M) and hence, x ∈ f −1 (M) =
{x ∈ X | f (x) ∈ M}. We conclude that f (x) ∈ M. ⊓
⊔
This observations together with Lemma 1.19 suggest the following definition.
′
Definition 1.17. Let (X, d) and (Y, d ) be metric spaces. Then, the map f :
X −→ Y is continuous at x ∈ X if and only if
′
′
∀ Bε ( f (x)) ∃ Bδ (x) such that f (Bδ (x)) ⊂ Bε ( f (x))
′
Note that f (Bδ (x)) ⊂ Bε ( f (x)) means that for all xe ∈ X such that xe ∈ Bδ (x), i.e.
′
′
d(x, xe) < δ we have f (e
x) ∈ Bε ( f (x)) that is d ( f (x), f (e
x)) < ε . This motivates the
following equivalent definition of continuity.
′
Definition 1.18. Let (X, d) and (Y, d ) be metric spaces. We say that f : X −→
Y is continuous at x ∈ X if
′
∀ ε > 0 ∃δε > 0 such that d ( f (e
x, f (x)) < ε whenever d(e
x, x) < δ ∀e
x ∈ X.
1.5 Continuous maps
31
′
If X = Y = R and d = d is the familiar Euclidean distance the previous definition
gives rise to the usual definition of continuity one encounters in real analysis, namely
the real function f : R −→ R is continuous at x0 ∈ R if for all ε > 0 there exists a
δε > 0 such that | f (x) − f (x0 )| < ε whenever |x − x0 | < δε for all x ∈ R.
Theorem 1.9. (Open mapping theorem)
′
Let (X, d) and (Y, d ) be metric spaces. A map f : X −→ Y is continuous if
and only if for every open subset U in Y the inverse image f −1 (U) is open in
X.
Proof. We have to prove both directions.
1. =⇒: this implication is straightforward because of Definition 1.15 and the fact
′
that (X, O(d)) and (Y, O(d )) are topological spaces.
′
2. ⇐=: it follows from Lemma 1.18. Let U ∈ O(d )). Then, U is open in Y and
hence, it is a neighbourhood of all of its points and in particular, a neighbourhood
of f (x) ∈ U. Since f −1 (U) ∈ O(d)), then f −1 (U) is open in X and therefore, it
is a neighbourhood of all of its points. In particular, it is a neighbourhood of x.
⊔
⊓
We can also introduce the concept of convergent sequence.
Definition 1.19. Let (X, d) be a metric space. We say that a sequence (xn )n∈N
of points in X converges to some x ∈ X if
∀ ε > 0 ∃ Nε > 0 such that xn ∈ Bε (x) ∀ n > Nε .
Lemma 1.20. A sequence (xn )n∈N of points in a metric space (X, d) converges
to x ∈ X if and only if
lim d(xn , x) = 0.
n→∞
Proof. Observe that
lim xn = x ⇐⇒ ∀ ε > 0 ∃ Nε > 0 such that xn ∈ Bε (x) ∀ n > Nε ,
n→∞
⇐⇒ ∀ ε > 0 ∃ Nε > 0 such that d(xn , x) < ε ∀ n > Nε ,
⇐⇒ (d(xn , x))n∈N converges to 0
and this completes the proof.
⊓
⊔
32
1 Fundamental topological concepts
Theorem 1.10. (Sequential characterization of continuity)
′
Let (X, d) and (Y, d ) be metric spaces. The map f : X −→ Y is continuous
if and only if for every sequence (xn )n∈N of points in X converging to some
x ∈ X the sequence ( f (xn ))n∈N converges to f (x) ∈ Y .
Proof. We need to prove two directions.
1. =⇒: Let f be continuous. Then, for every ε > 0 there exists δε > 0 such that
′
f (Bδ (x)) ⊂ Bε ( f (x)). Take any sequence (xn )n∈N converging to x ∈ X. Then,
for every e
ε there exists Neε > 0 such that xn ∈ Beε (x) for all n > Neε . Since this is
true for all e
ε > 0, it will be in particular true for e
ε = δ . For such a choice there
′
exists Nδ > 0 such that xn ∈ Bδ (x) for all n > Nδ . Since f (Bδ (x)) ⊂ Bε ( f (x)),
′
it follows that f (xn ) ∈ Bε ( f (x)) for all n > Nδ . Hence, ( f (xn ))n∈N converges to
f (x) ∈ Y .
2. ⇐=: by contradiction. Suppose that f is not continuous, that is there exists some
′
ε > 0 such that for all δ > 0 we have f (Bδ (x)) 6⊂ Bε ( f (x)). Since (xn )n∈N converges to x, we have that for every e
ε > 0 there exists some Neε > 0 such that
xn ∈ Beε (x) for all n > Neε . Since the previous statements are true for any e
ε and
for any δ > 0, they remain true when e
ε = δ = 1/n. For such a choice we can
find N1/n > 0 such that xn ∈ B1/n (x) for all n > N1/n . Clearly, it is also true that
′
f B1/n (x) 6⊂ Bε ( f (x)) and we can construct a sequence of points ( f (xn ))n∈N
′
not contained in Bε ( f (x)). Hence, ( f (xn ))n∈N is not convergent. ⊓
⊔
Definition 1.20. Late A and B be sets such that A ⊂ B. The inclusion map is
defined as
i : A ⊂ B −→ B such that i(x) = x ∀ x ∈ A.
The next result turns out to be useful when studying homeomorphisms between
topological spaces.
Lemma 1.21. Let (X, O) be a topological space and H be a subset of X
equipped with the subspace topology O|H . Then, the inclusion map i : H −→
X is continuous.
Proof. Let U ∈ O. Then,
i−1 (U) = {x ∈ H | i(x) ∈ U} = {x ∈ H | x ∈ U} = H ∩U ∈ O|H
1.6 The identification topology
and the proof is completed.
33
⊓
⊔
Consider the topological spaces (X, OX ) and (Y, OY ). I f f : X −→ Y is continuous
and X0 ⊂ X, then the restriction of f on X0 , that is the map f |X0 : X0 −→ Y , is also
continuous. This can be easily seen by introducing the inclusion map i : X0 ⊂ X −→
X which is continuous. Then, f |X0 = f ◦ i is continuous because composition of
continuous maps.
Lemma 1.22. Let (X, OX ), (Y, OY ), and (H, O|H ) with H ⊂ X be topological
spaces. Further, consider a map g : Y −→ H and an inclusion i : H −→ X.
Then, g is continuous if and only if i ◦ g : Y −→ X is continuous.
Proof. We need to prove both directions.
1. =⇒: Since g is continuous and the inclusion map is continuous by Lemma 1.21,
i ◦ g is continuous because composition of continuous maps.
2. ⇐=: let i ◦ g be continuous. We want to show that for any subset V of H open
in the subspace topology the inverse image g−1 (V ) is open in the topology of
Y . Take any V ∈ O|H . Then, V = U ∩ H for some subset U open in X. Hence,
Lemma 1.21 yields
g−1 (V ) = g−1 (U ∩ H) = g−1 (i−1 (U)) = (i ◦ g)−1 (U)
is open in Y because i ◦ g is continuous.
⊓
⊔
Problems
1.25. Let (X, OX ) and (Y, OY ) be topological spaces. Show that the map f : X −→ Y
is continuous if and only if f −1 (B) ⊂ f −1 (B) for all B ⊂ Y .
1.6 The identification topology
This topology also known in the German literature under the name Bildtopologie
plays an important role in constructing the so-called quotient topology for quotient
spaces.
34
1 Fundamental topological concepts
Lemma 1.23. Let X be a set and T = {(Xi , Oi ) | i ∈ I} be a family of
topological spaces. Further, suppose that for every i ∈ I there exists a map
fi : Xi −→ X. Then,
O = {Ω ⊂ X | ∀ i ∈ I fi−1 (Ω ) ∈ Oi }
is a topology on X called the identification topology of the sets Xi under the
action of fi .
Proof. We need to verify the properties of a topological space.
1. First of all 0/ ⊂ X and
fi−1 (0)
/ = {x ∈ Xi | fi (x) ∈ 0}
/ = 0/ ∈ Oi .
Furthermore, X ⊂ X and
fi−1 (X) = {x ∈ Xi | fi (x) ∈ X} = Xi ∈ Oi .
2. Consider an arbitrary of sets in O, say
fi−1
[
Ωj
j∈J
S
!
j∈J Ω j .
=
[
Then,
fi−1 (Ω j )
j∈J
is open in Oi since Ω j ∈ O.
S
3. Consider a finite number of intersections of sets in O, say λ ∈In Ωλ . Then,
!
fi−1
\
λ ∈In
Ωλ
=
\
λ ∈In
fi−1 (Ωλ )
⊔
is open in Oi since Ωλ ∈ O. ⊓
We construct the quotient topology. To this purpose consider a topological space
(X, O) together with an equivalence relation ∼. The set of all equivalence classes
relative to ∼ forms the quotient set X/ ∼. Consider the map π : X −→ X/ ∼. Then,
the quotient topology on X/ ∼ is the identification topology induced by π on X/ ∼,
more precisely
Oπ = {U ⊂ X/ ∼ | π −1 (U) ∈ O}.
Note that the map π is continuous by the next result while Oπ is a topology by
Lemma 1.23.
1.7 Projection maps
35
Lemma 1.24. Let X be a set and T = {(Xi , Oi ) | i ∈ I} be a family of
topological spaces. Further, suppose that for every i ∈ I there exists a map
fi : Xi −→ X. Let O be the identification topology on X induced by the maps
fi ’s. Then,
1. For all i ∈ I the map fi : Xi −→ X is continuous.
′
2. O is the finest topology on X such that all fi ’s are continuous, i.e. if O
is another topology on X such that all fi ’s are continuous, then it must be
′
O ⊂ O.
Proof. We need to verify the following points.
1. It follows directly from the definition of O. Take any i ∈ I and Ω ∈ O. Then,
f −1 (Ω ) ∈ Oi and hence, fi is continuous.
′
2. Let O be another topology on X such that all fi ’s are continuous. We need to
′
′
show that O ⊂ O. Take any Ω ∈ O and any fi continuous. Then, fi−1 (Ω ) ∈ Oi .
Hence, Ω ∈ O. ⊓
⊔
1.7 Projection maps
In general projection maps can be defined also in the absence of topological spaces.
Definition 1.21. Let A1 and A2 be sets. We call projections the maps πi : A1 ×
A2 −→ Ai with i = 1, 2 such that πi (x1 , x2 ) = xi for all = 1, 2 and for all x =
′
′
x(x1 , x2 ) ∈ A1 ×A2 . Let A be an arbitrary set and f : A −→ A1 ×A2 . We define
coordinate maps as follows
′
xi = πi ◦ f : A −→ Ai
∀i = 1, 2.
In the next result we construct the product topology in such a way that the
projection maps are continuous.
Lemma 1.25. Every projection is continuous with respect to the product
topology.
Proof. Let (X, OX ) and (Y, OY ) be topological spaces. Consider the projections πX :
X × Y −→ X and πY : X × Y −→ Y such that (x, y) 7−→ πX (x, y) = x and (x, y) 7−→
36
1 Fundamental topological concepts
πY (x, y) = y, respectively. Equip X × Y with the product topology OX×Y . We only
show that πX is continuous since the proof for πY is similar.We need to show that
for any U ∈ OX the inverse image πX−1 (U) is open in X ×Y . Observe that
πX−1 (U) = {(x, y) ∈ X ×Y | πX (x, y) ∈ U} = {(x, y) ∈ X ×Y | x ∈ U} = U ×Y ∈ OX×Y
since U is a neighbourhood of x and Y is a neighbourhood of y and therefore U ×Y
is an open box. ⊓
⊔
The next example clarifies the reason why the maps π ◦ f are called coordinate maps.
Example 1.15. Consider a curve in R2 , that is a vector-valued function f : [a, b] ⊂
R −→ R2 such that t ∈ [a, b] 7−→ f (t) = (x(t), y(t)) ∈ R2 . The parametric representation of such a curve is given by x = x(t) and y = y(t) with t ∈ [a, b]. For instance,
the curve representing a circle of unit radius will be x(t) = cost and y(t) = sint with
t ∈ [0, 2π ] and in this case
f : [0, 2π ] −→ R2 ,
f (t) = (x(t), y(t)) = (cost, sint).
Note that x = x(t) and y = y(t) are the coordinate maps. We can project an arbitrary point (x(t), y(t)) on the circle down to the x-axis and y-axis by means of the
projections πx and πy . Hence, we have x = πx ◦ f and y = πy ◦ f .
This example suggests that f is continuous on R2 provided that the coordinate maps
x and y are continuous on [a, b]. The next theorem confirms this conjecture.
Theorem 1.11. Consider the topological spaces (X, OX ), (Y, OY ) and
′
′
′
(A , O ). Further, introduce a map f : A −→ X ×Y with X ×Y equipped with
the product topology OX×Y . Then, f is continuous if and only if the coordinate
maps πX ◦ f and πY ◦ f are continuous.
Proof. We need to prove both directions.
1. =⇒: let f be continuous. The projection maps πX and πY are continous by
Lemma 1.25 and the statement follows from the fact that composition of continuous maps is continuous.
′
2. ⇐=: the map f : A −→ X × Y will be continue if for every subset W open in
′
the product topology the inverse image f −1 (W ) is open in the topology of A .
According to a previous remark it is sufficient to test continuity on an arbitrary
element of a basi B of the product topology, since then any element in OX×Y can
be written as a union of sets B. A basis of the product topology is given by the
collection of all open boxes. Hence, we take any U × V ∈ B with U ∈ OX and
′
V ∈ OY and show that f −1 (U ×V ) ∈ O . To this purpose observe that
1.8 First and second-countable topological spaces
37
U ×V = {(x, y) ∈ X ×Y | x ∈ U ⊂ X, y ∈ V ⊂ Y },
= {(x, y) ∈ X ×Y | x ∈ U ⊂ X, y ∈ Y } ∩ {(x, y) ∈ X ×Y | x ∈ X, y ∈ V ⊂ Y },
= (U ×Y ) ∩ (X ×V )
Then,
f −1 (U ×V ) = f −1 ((U ×Y ) ∩ (X ×V )) = f −1 (U ×Y ) ∩ f −1 (X ×V )
(1.5)
Note that πX and πY are continuous and hence
πX−1 (U) = {(x, y) ∈ X ×Y | πx (x, y) ∈ U} = {(x, y) ∈ X ×Y | x ∈ U} = U ×Y
and similarly πY−1 (V ) = X ×V . Hence, from (1.5) we have
f −1 (U ×V ) = f −1 πX−1 (U) ∩ f −1 πY−1 (V ) = (πX ◦ f )−1 (U) ∩ (πY ◦ f )−1 (V )
′
but πX ◦ f : A −→ X is continuous and therefore for all U ∈ OX (πX ◦ f )−1 (U) ∈
′
′
O . Similarly, πY ◦ f : A −→ Y is continuous and therefore for all V ∈ OY (πY ◦
′
′
f )−1 (V ) ∈ O . Hence, we conclude that f −1 (U ×V ) ∈ O . ⊓
⊔
This result allows to prove continuity of functions depending on more than one
variable in a very elegant way. For instance, the function f (x, y) = x2 + xy + y2 will
be continuous since it can be written in terms of sums and products of projections
as follows
f (x, y) = (πX · πX + πX · πY + πY · πY )(x, y)
and projections are continuous maps.
1.8 First and second-countable topological spaces
Definition 1.22. Let (X, O) be a topological space and UO (x) be the collection of all neighbourhoods of x ∈ X with respect to the topology O. We say
that Bx ⊂ UO (x) is neighbourhood basis of x with respect to the topology O
if
∀ U ∈ UO (x) ∃ B ∈ Bx such that B ⊂ U.
Definition 1.23. A topological space (X, O) satisfies
1. the first countability axiom if every x ∈ X has a countable neighbourhood
basis.
38
1 Fundamental topological concepts
2. the second countability axiom if its topology admits a countable basis.
One of the most important properties of first countable spaces is that given a subset
A a point x belongs to the closure of A if and only if there exists a sequence (xn )n∈N
in A converging to x. Moreover, first countable spaces allows a sequential characterization of continuous maps. A very interesting future of second-countable spaces is
that they are all separable, that is they admit a dense subset. For a detailed analysis
of the connections between first and second countable spaces we to [2].
Lemma 1.26. Every metric space is first-countable.
Proof. Let (X, d) be a metric space equipped with the topology O(d). Take any
x ∈ X. We claim that Bx = {B1/n (x) | n ≥ 1 and n ∈ N} is a countable neighbourhood
basis. Let U be a neighbourhood of x. Then, there exists an open ball Br (x) ⊂ U. By
the density of the rationals we can always find a q = 1/n ∈ Q+ such that 0 < q < r.
Hence,
B1/n (x) ⊂ Br (x) ⊂ U.
Clearly, the collection of these balls is a countable set since we cans et up an injection f : Bx −→ N defined as f (B1/n (x)) = n for all n ∈ N. ⊓
⊔
The net result shows that second-countability is a stronger condition than firstcountabilty.
Theorem 1.12. Every second-countable topological space is also firstcountable.
Proof. Let (X, O) be a second-countable topological space. Then, O has a countable
basis B ⊂ O with B = {Uk ∈ O | k ∈ I} where I is an arbitrary indexing set. Take
any x ∈ X and consider the set J = {k ∈ I | x ∈ Uk } ⊂ I. This set is countable and
Bx = {Uk ∈ UO(x) | k ∈ J} is a countable neighbourhood basis of X. ⊓
⊔
Example 1.16. We look at some examples of countable spaces.
1. Rn with the usual topology is second-countable. Although the usual basis of open
balls is not countable, we can consider the set of all rational balls (radius in Q+
and centre in Qn ). This restricted set is countable and forms a basis as we showed
in an earlier section. It is also first countable by the previous lemma.
References
39
2. Metric spaces are first-countable but not necessarily second-countable. Take for
instance (R, d) with the discrete metric. Open balls are
R
if ε > 1,
Bε (x) =
{x} if 0 < ε ≤ 1.
As a metric space (R, d) is first-countable and therefore for any x ∈ X it has a
neighbourhood basis Bx = {{x} | x ∈ R} which is countable because given x Bx
contains only one element. Since for all x ∈ R and for every open neighbourhood
U of x there exists B ∈ Bx such that x ∈ B ⊂ U, we can conclude by a previous
theorem that Bx is also a basis for the topology O(d) but Bx is not countable
since R is not countable. Hence, (R, d) does not satisfy the second-countability
axiom. However, N equipped with the discrete metric is second-countable.
References
1. Tu, Loring W.: An introduction to manifolds, Springer Verlag (2011)
2. Steen, L. A., and Seebach, J. A.: Counterexamples in topology, Dover Publications (1995)
Chapter 2
Homeomorphisms and topological properties
2.1 Homeomorphisms
Definition 2.1. Let (X, OX ) and (Y, OY ) be topological spaces. An homeomorphism between X and Y is a bijective map f : X −→ Y such that f and
f −1 is continuous. In this case we say that the topological spaces involved are
homeomorphic.
Homeomorphisms preserve all those topological properties of X and of its subsets
which can be formulated in terms of open sets. Hence, if f is a homeomorphism we
have
1. A ⊂ X is open in X if and only if f (A) ∈ OY .
Proof. Since f continuous, then for all U ∈ OY the inverse image f −1 (U) ∈ OX .
Let A = f −1 (U). Then, we have f (A) = f ( f −1 (U)) = U which is open in Y . ⊓
⊔
2. A ⊂ X is closed in X if and only if f (A) is closed in Y .
Proof. Observe that by means of the first property
A ⊂ X closed ⇐⇒ X − A ∈ OX ⇐⇒ f (X − A) ∈ OY ,
⇐⇒ f (X) − f (A) ∈ OY ⇐⇒ Y − f (A) ∈ OY ,
⇐⇒ f (A) is closed
and this completes the proof.
⊓
⊔
3. U ⊂ X is a neighbourhood of x ∈ X if and only if f (U) ⊂ Y is a neighbourhood
of f (x) ∈ Y .
Proof. Observe that by means of the first property
41
42
2 Homeomorphisms and topological properties
U ⊂ X neighbourhood of x ⇐⇒ ∃ Nx ∈ OX such that x ∈ Nx ⊂ U,
⇐⇒ ∃ f (Nx ) ∈ OY such that f (x) ∈ f (Nx ) ⊂ f (U),
⇐⇒ f (U) is a neighbourhood of f (x)
and this completes the proof.
⊓
⊔
4. B is a basis of the topology of X if and only if Be = { f (B) ⊂ Y | B ∈ B} is a
basis of the topology of Y .
Proof. The collection B = {Bi ∈ OX | i ∈ I} will be a basis
of the topology OX
S
if and only if for all opens subsets Ω in X we have Ω = j∈I B j with J ⊂ I. By
the first property f (Ω ) ∈ OY but
!
f (Ω ) = f
[
j∈I
Bj
=
[
f (B j )
j∈I
and { f (Bi ) ∈ OY | Bi ∈ B ∀i ∈ I} is a basis of the topology of Y . The other
direction can be obtained by working with the inverse of f . ⊓
⊔
Among all topological properties a topological space may possess we will analyze
in the next sections the following properties
1. connectedness,
2. Hausdorff property,
3. compactness.
These properties allow to classify topological spaces and homeomorphisms play an
important role in that regard because of a topological space X has one of the above
properties, then the target toplogical space Y will have the same property.
Lemma 2.1. Let (X, OX ) and (Y, OY ) be topological spaces. Suppose that B ⊂
Y and let B be equipped with the subspace topology. Consider the inclusion
map i : B ⊂ Y −→ B ⊂ Y . Then, the map g : X −→ B ⊂ Y is continuous if and
only if i ◦ g : X −→ Y is continuous.
Proof. We need to prove two directions.
1. =⇒: the statement follows from the fact that the inclusion map and g are continuous and the composition of continuous maps is continuous as well.
2. ⇐=: let i◦g be continuous. To show that g is continuous we need to verify that for
every subset V open in the subspace topology the inverse image g−1 (V ) is open
in the topology of X. Take any subset V open in the subspace topology. Then, by
definition of subspace topology there exists a subset U open in the topology of Y
such that V = U ∩ B. Moreover,
2.1 Homeomorphisms
43
i−1 (U) = {y ∈ B | i(y) ∈ U} = U ∩ B = V.
Hence,
g−1 (V ) = g−1 (U ∩ B) = g−1 (i−1 (U)) = (i ◦ g)−1 (U) ∈ OX
since i ◦ g is continuous.
⊓
⊔
Theorem 2.1. Let (X, OX ) and (Y, OY ) be topological spaces. Further, suppose that f : X −→ Y is continuous. Then, the graph of f
G f = {(x, f (x)) ∈ X ×Y | x ∈ X, f (x) ∈ Y }
equipped with the subspace topology is homeomorphic to (X, OX ).
Proof. We have to construct an homeomorphism between X and G f . Introduce the
map h : X −→ G f such that x 7−→ h(x) = (x, f (x)) for all x ∈ X. Note that h is
injective since for any x, y ∈ X we have
h(x) = h(y) ⇐⇒ (x, f (x)) = (y, f (y)) ⇐⇒ x = y
and it is also surjective because by construction h(X) = G f . Hence, h is a bijection
and there exists h−1 : G f −→ X such that (x, f (x)) 7−→ x for all x ∈ X. In particular,
we have h−1 = πX ◦ i where πX : X × Y −→ X and i : G f ⊂ X × Y −→ X × Y is
the inclusion map. Since both maps are continuous and composition of continuous
maps is again continuous, we conclude that h−1 is continuous. It remains to show
that h is continuous. To this purpose note that the maps
πX ◦ i ◦ h : X −→ X such that πX (i(h(x))) = πX (i(x, f (x))) = πX (x, f (x)) = x,
πY ◦ i ◦ h : X −→ Y such that πY (i(h(x))) = πY (i(x, f (x))) = πY (x, f (x)) = f (x)
are simply the identity map and the map f , respectively. Both of them are continuous and by Theorem 1.11 we conclude that i ◦ h is continuous. Finally, Lemma 2.1
implies that h is continuous.
Example 2.1. Consider f (x) = tan x. Since f is continuous on the interval (−π /2, π /2),
the previous theorem implies that the graph of f will be homeomorphic to the interval (−π /2, π /2).
44
2 Homeomorphisms and topological properties
2.2 Connectedness
Intuitively a set is connected if it cannot be written as a disjoint union of open
subsets. We start with a more abstract definition of connectedness and later we show
that it is equivalent to the intuitive idea of connected set.
Definition 2.2. A topological space (X, O) is connected if only 0/ and X are at
the sime time open and closed.
The next result gives an equivalent formulation of connectedness.
Lemma 2.2. A topological space (X, O) is connected if and only if X cannot
be written as a disjoint union of non-empty open subsets.
Proof. We prove the contrapositive, that is
∃ A1 , A2 ∈ O such that A1 ∩ A2 = 0,
/ A1 ∪ A2 = X ⇐⇒ (X, O) not connected
1. =⇒: let A1 , A2 ∈ Obe such that A1 ∩ A2 = 0/ and A1 ∪ A2 = X. Then, A1 and A2
are also closed since X − A1 = A2 and X − A2 = A1 are both closed in X. Hence, 0/
and X are not the only sets at the same time both open and closed. Hence, (X, O)
cannot be connected.
2. ⇐=: let (X, O) be not connected. Then, X and 0/ are not the only sets at the same
time open and closed. Hence, there must exist at least one non-empty set subset
of X, say A1 , that is open and closed at the same time but then A2 = X −A1 will be
open and closed as well. Hence, we found A1 , A2 ∈ O such that they are disjoint
and the union is X. ⊓
⊔
Since in many cases the connectedness of complicated topological spaces derives
from the connectedness of the intervals of the real line, we will show that any interval of the real line is connected. This will imply that the intervals of the real line
equipped with the usual topology are connected subspaces.
Definition 2.3. Let (X, O) be a topological space and A ⊂ X be equipped with
the subspace topology O|A . We say that (A, O|A ) is connected if the only
subsets of A that are simultaneously open and closed in A are A and the 0.
/
2.2 Connectedness
45
Example 2.2. Consider the real line equipped with the usual topology and the subset
A = [0, 1]∪(2, 3) equipped with the subspace topology. A is disconnected since [0, 1]
is open and closed in the subspace topology. First of all, note that even though [0, 1]
is closed in R with the usual topology, it is open in the subspace topology of A since
it can be written as the intersection of an open interval of the real line and A as
follows
1 3
∩ A.
[0, 1] = − ,
2 2
Moreover, [0, 1] is also closed in the subspace topology of A because
3
A − [0, 1] = (2, 3) =
,4 ∩A
2
is open in the subspace topology.
We want to show that the real line with the usual topology is connected and that
open intervals of R equipped with the subspace topology are connected as well.This
together with the fact that homeomorphisms preserve connectedness will allow us
to conclude that every open interval of the real line is connected as well.
Lemma 2.3. Let (X, O) be not connected. Then, there exists a continuous surjection f : X −→ {−1, 1}.
Proof. Let (X, O) be connected. Then, there exists A1 , A2 ⊂ X open in X and
nonempty such that A1 ∩A2 = 0/ and A1 ∪A2 = X. Construct a map f : X −→ {−1, 1}
as follows
−1 ∀ x ∈ A1 ,
f (x) =
+1 ∀ x ∈ A2 .
Then, f is surjective by construction since f (X) = {−1, 1}. Let us equip {−1, 1}
with the discrete topology OD = {0,
/ {−1}, {1}, {−1, 1}}. Then, f is continuous
since
f −1 ({−1, 1}) = X,
f −1 (0)
/ = 0,
/
f −1 ({−1}) = A1 ,
f −1 ({1}) = A2 ,
that is the inverse image of any element in OD is open in the topology of X.
⊓
⊔
The contrapositive of this lemma that is ”if there does not exist a continuous surjection f : X −→ {−1, 1}, then (X, O) is not connected” is particularly useful to prove
connectedness of topological spaces.
Theorem 2.2. The real line equipped with the usual topology (R, Ou ) is connected.
46
2 Homeomorphisms and topological properties
Proof. By contradiction. Suppose that (R, Ou ) is not connected. Then, there exist
A1 , A2 ⊂ R open in R and both nonempty such that A1 ∪ A2 = R and A1 ∩ A2 = 0.
/
Since A1 , A2 6= 0/ there exists a ∈ A1 and b ∈ A2 . Suppose without loss of generality
that a < b and consider the closed interval [a, b] ⊂ R. We construct a sequence of
nasted closed intervals with endpoints in A1 and A2 , respectively. To this purpose
consider the midpoint a+b
2 ∈ [a, b]. There are two cases
a+b a+b
1. 2 ∈ A1 and then 2 , b ⊂ [a, b] with a+b
2 ∈ A1 and b ∈ A2 ,
a+b a+b
2. 2 ∈ A2 and then a, 2 ⊂ [a, b] with a ∈ A1 and a+b
2 ∈ A2 .
In any case we end up with a closed interval having endpoints in A1 and A2 , respectively, the interval is contained
in
[a,b] and
has length (b − a)/2. Considering the
a+b
midpoint of the intervals a+b
,
b
or
a,
⊂ [a, b] and proceeding as before we
2
2
a+b ⊂ [a, b] having
can construct new closed intervals contained in a+b
2 , b or a, 2
the property that their endpoints belong to A1 and A2 , respectively. These intervals have length (b − a)/4. Proceeding inductively we end up with a sequence of
n
nasted
intervals
where the n-th interval has length (b − a)/2 if we consider
a+b closed
a+b
⊂ [a, b] to be the first term in such a sequence. Since these in2 , b or a, 2
tervals have decreasing length they will intersect at some point x0 ∈ R. On the other
side R = A1 ∪ A2 and we have two cases
1. if x0 ∈ A1 , since A1 ∈ Ou there exists an open ball Bεx0 (x0 ) ⊂ A1 where in the
present case Bεx0 (x0 ) = (x0 − εx0 , x0 + εx0 ). Moreover, we can always find a closed
interval contained in (x0 − εx0 , x0 + εx0 ) by requiring that
b−a
< 2εx0
2n
that is
n > log2
b−a
2εx0
but such an interval would have an endpoint in A2 and at the same time it would
be contained in A1 . This is a contradiction and hence, x0 ∈
/ A1 .
2. If x0 ∈ A2 we can proceed as above and end up with a contradiction so that
x0 ∈ A2 .
Therefore, such a point x0 does not exist meaning that a and b cannot have existed.
Therefore, either A1 is empty or A2 is empty. ⊓
⊔
In order to show that any open interval with the subspace topology is homeomorphic
to the real line equipped with the usual topology and thus it is connected, we need
some auxiliary results.
Lemma 2.4. Let (X, OX ) and (Y, OY ) be topological spaces. Suppose that
(X, OX ) is connected and f : X −→ Y is continuous and surjective. Then,
(Y, OY ) is also connected.
2.2 Connectedness
47
Proof. We have to show that Y and 0/ are the only sets to be open and closed at the
same time. In other words, we have to show that any two subsets A, B ⊂ Y such that
A ∪ B = Y and A ∩ B = 0/ where A and B are open and closed at the same time must
coincide with Y and 0.
/ The continuity of f and a previous lemma about equivalent
characterizations of continuity yield that f −1 (A) and f −1 (B) are both closed and
open in X. Since (X, OX ) is connected, we must have
f −1 (A) = X,
f −1 (B) = 0/
or
f −1 (A) = 0,
/
f −1 (B) = X.
Since f is surjective, it must be A = Y , B = 0/ or A = 0,
/ B = Y. ⊓
⊔
What happens if we drop the requirement that f is surjective? Then, f (Y ) ⊂ Y and
we can still show that f (Y ) equipped with the subspace topology is connected but
nothing can be said about Y .
Theorem 2.3. Let (X, OX ) and (Y, OY ) be homeomorphic. Then, (X, OX ) is
connected if and only if (Y, OY ).
Proof. Since (X, OX ) and (Y, OY ) are homeomorphic, there exists a homeomorphism f : X −→ Y and f −1 is also a homeomorphism. The direction =⇒ follows
from Lemma 2.4 because f homeomorhpism ensures that f is continuous and surjective. The other direction follows again from the same lemma with f substituted
by f −1 . ⊓
⊔
Theorem 2.4. Any interval (a, b) with a < b and equipped with the subspace
topology is connected.
Proof. Without loss of generality consider the interval (−π /2, π /2) since we can
always use a continuous and bijective map g : (a, b) −→ (−π /2, π /2) such that
g(x) =
π (2x − a − b)
2(b − a)
to switch from one interval to the other. Clearly, g is a homeomorphism. On the
other hand, (R, Ou ) is connected and f (x) = tan x with x ∈ (−π /2, π /2) is a homeomorphism. The required homeomorphism is f ◦ g : (a, b) −→ R since composition
of homeomorphisms is a homeomorphism. ⊓
⊔
Lemma 2.5. The interval [0, 1] equipped with the subspace topology is connected.
48
2 Homeomorphisms and topological properties
Proof. By contradiction. Suppose that [0, 1] is not connected. Then, [0, 1] and 0/
are not the onlys sets to be open and closed at the same time. Hence, there exists
nonempty disjoint subsets A and B such that A ∪ B = [0, 1] and A ∩ B = 0.
/ Let a =
sup{A} and b = sup{B}. Since A and B closed, a theorem in analysis implies that
a ∈ A and b ∈ B. If we show that a = b, then A ∩ B 6= 0/ and we have a contradiction.
We claim that a = b = 1. To this purpose consider the set X coinciding with A or B
and such that 1 ∈
/ X. Let α = sup{X}. Since 1 ∈
/ X, α must be an interior point of
[0, 1] and therefore, it must be within either X or [0, 1] − X. If α is within X, since X
is open there exists εα > 0 such that (α − εα , α + εα ) ⊂ X. If α is within [0, 1] − X,
′
′
′
since this set is open, there exists εα > 0 such that (α − εα , α + εα ) ⊂ [0, 1] − X. In
both cases, α cannot be a supremum since we can find upper bounds α + (εα /2)
′
and α + (εα /2). Hence, it must be sup{X} = 1 but sup{X} = sup{A} = sup{B} and
this completes the proof. ⊓
⊔
2.3 Path-connectedness
In addition to connectedness we also have a stronger concept called path-connectedness.
As the name already suggests we could say that a subset A of R2 or R3 is pathconnected if we can take any two points in A and join them with a continuous curve
completely contained in A.
Definition 2.4. Let (X, O) be a topological space and [0, 1] ⊂ R with R is
equipped with the usual topology. We say that (X, O) is path-connected if for
all a, b ∈ X there exists a path, that is a continuous map f : [0, 1] −→ X such
that f (0) = a and f (1) = b and f (t) ∈ X for any t ∈ (0, 1).
Lemma 2.6. If a topological space is path-connected, it is also connected.
Proof. By contradiction. Suppose that (X, O) is path-connected but not connected.
Since it is not connected, there exists A ⊂ X with A 6= open and closed at the same
time. Let B = X − A. Clearly, B 6= 0/ and B is also open and closed. Since A and
B are not empty, there exist a ∈ A and b ∈ B with a 6= b because A ∩ B = 0.
/ On
the other hand, (X, O) is path-connected. Hence, there exists a continuous map f :
[0, 1] −→ X such that f (0) = a and f (1) = b. Since 0 ∈ f −1 (A) but 1 ∈
/ f −1 (A)
−1
−1
−1
−1
−1
because 1 inf (B) and f (A ∩ B) = f (A) ∩ f (B), we have f (A) ⊂ [0, 1].
Since A is open and closed and f continuous, it follows that f −1 (A) is open and
closed in [0, 1]. Moreover, f −1 (A) 6= 0/ and f −1 (A) ⊂ [0, 1]. Hence, we can conclude
that [0, 1] is not connected and we have a contradiction. ⊓
⊔
2.3 Path-connectedness
49
The converse of the above result is not in general true, i.e. a connected topological
space does not need to be path-connected. In order to fully appreciate the next example of a topological space which is connected but not path-connected we need
some preliminary results.
Lemma 2.7. Let (X, OX ) and (Y, OY ) be topological spaces with (X, OX ) connected. Further, suppose that f : X −→ Y is continuous. Then, the graph of f
denoted by G f is connected.
Proof. Theorem 2.1 implies that G f with the subspace topology is homeomorphic
to (X, OX ) but then Theorem 2.4 forces G f to be connected. ⊓
⊔
We construct now an example showing that connectedness does not imply pathconnectedness.
Example 2.3. Consider R2 with the topology of the open discs induced by the Euclidean metric and the function f : (0, ∞) −→ R defined through
1
f (x) = sin .
x
Since f continuous on (0, ∞) which is connected because it is homeomorphic to the
interval (0, 1), where the homeomorphism is given by the map g : (0, 1) −→ (0, ∞)
with g(x) = x/(1 − x), Lemma 2.7 implies that the graph of f
1
∈ R2 | x ∈ (0, ∞)
Gf =
x, sin
x
is connected. We construct the closure of the graph of f and prove that it is connected but not connected. We do this in steps.
1. G f = {0} × [−1, 1] ∪ G f .
Proof. We know that G f = ∂ G f ∪ Int(G f ). Since f is continuous, Theorem 2.1
implies that ((0, ∞), O|(0,∞) and G f equipped with the subspace topology are
homeomorphic. This in turn implies that there exists a homeomorphism ϕ :
(0, ∞) −→ G f . Hence, ϕ is surjective and thus ϕ (0, ∞) = G f . Moreover, (0, ∞)
is open in the subspace topology and the first part of the remark following Definition311d implies that ϕ (0, ∞) is open in the corresponding subspace topology.
Hence, G f is open and therefore Int(G f ) = G f . Hence, we have G f = ∂ G f ∪ G f .
To get a guess for ∂ G f we look at the graph of f which oscillates and the oscillations become stronger and stronger as we approach 0 from the right showing
the tendency of the graph to pile up on the vertical segment [−1, 1] of the y-axis.
This suggests that the boundary points of G f should be of the form (0, y) with
y ∈ [−1, 1]. It is sufficient to prove (0, y) with y ∈ [−1, 1] belongs to ∂ G f . We
50
2 Homeomorphisms and topological properties
verify that every point p = (0, y p ) ∈ {0} × [−1, 1] also belongs to G f . This will
be the case if every neighbourhood Up of p meets G f , i.e. Up ∩ G f 6= 0.
/ Since U
neighbourhood of p there exists an open ball Bε p (x) ∈ Ou such that Bε p (x) ⊂ U. If
we show that Bε p (x) ∩ G f 6= 0,
/ then we are done. In particular, we will construct
a point q ∈ R2 such that q ∈ Bε p (x) and q ∈ G f . To this purpose, note that
sin
and
sin
1
= 1 for
x
1
= −1 for
x
xn =
2
π (1 + 4n)
xen =
2
π (3 + 4n)
with n ∈ N. Consider the closed interval
2
2
∆n =
.
,
π (3 + 4n) π (1 + 4n)
Since the funxtion f is continuous on ∆n and by the intermediate value theorem
[−1, 1] is the image of ∆n under the function f then there exists x0 ∈ ∆n such that
sin
1
= yp.
x0
Consider the point
1
q = x0 , sin
x0
= (x0 , y p ) ∈ G f .
We show we can always find a ball with centre p containing q. Note that
q
2
1
1
1
.
d(p, q) = (0 − x0 )2 + (y p − y p )2 = x0 ≤
=
<
1
π (1 + 4n) 2nπ 1 + 4n
2nπ
Let
1
<ε
2nπ
and we are done because given any ε < 1 we can choose n such that the above
inequality is satisfied. ⊓
⊔
2. We have to show that G f is connected. This part follows directly from this lemma
Lemma 2.8. Let (X, O) be a topological space and (A, O|A ) connected
with A ⊂ X. Then, (A, O|A ) is also connected.
Proof. (A, O|A ) will be connected if A and 0/ are the only sets to be open and
closed at the same time. To prove uniqueness suppose that A = U ∪ V with U ∩
2.3 Path-connectedness
51
V = 0.
/ Observe that since A ⊂ A we have
A = A ∩ A = A ∩ (U ∪V ) = (A ∩U) ∪ (A ∩V ) = A1 ∪ A2
and by construction
A1 ∩ A2 = (A ∩U) ∩ (A ∩V ) = A ∩ (U ∩V ) = A ∩ 0/ = 0.
/
Since (A, O|A ) connected it must be
A ∩U = 0,
/
A ∩V = A or
A ∩U = A,
A ∩V = 0.
/
Hence, we conclude that U = 0/ and V = A or U = A and V = 0.
/
3. We show that G f is not path-connected.
⊓
⊔
Proof. By contradiction. Suppose that for all a, b ∈ G f there exists a path represented by the map ψ : [0, 1] −→ G f such that ψ (0) = a ∈ G f and ψ (1) = b ∈ G f .
Consider in particular a ∈ G f amd b = (0, 1) ∈ {0} × [−1, 1]. Then, there exists
a path
γ : [0, 1] −→ G f
such that
γ (0) = a ∈ G f ,
γ (1) = (0, 1).
Since γ continuous, the inverse image of a closed set under γ is closed. Take the
closed disc in G f characterized by
1
d(γ (t), (0, 1)) ≤ .
2
We want to show that it is possible to find a point in such a disc such that its
distance from the centre is greater than 1/2. The set B1/2 (γ (t)) ∩ G f will be
mapped by γ −1 to the closed interval [1 − δ1/2 , 1] ⊂ [0, 1]. Consider the set γ ([1 −
δ1/2 , 1]). It is connected since [1− δ1/2 , 1] is connected and continuous maps send
connected sets into connected sets. Consider the restriction of γ on [1 − δ1/2 , 1],
that is
γ |[1−δ1/2 ,1] −→ A ⊂ G f , γ |[1−δ1/2 ,1] (t) = (x, y).
Let
γ |[1−δ1/2 ,1] (1 − δ1/2 ) = (x0 , y0 ).
Introduce the projection πX : R2 −→ R such that (x, y) 7−→ x. Then,
πX ◦ γ |[1−δ1/2 ,1] : [1 − δ1/2 , 1] −→ R
is continuous because composition of continuous maps. Observe that
a. (πX ◦ γ )([1 − δ1/2 , 1]) is a connected subset of R
b. (πX ◦ γ )([1 − δ1/2 , 1]) 6= 0/ since
(πX ◦ γ )(1) = πX (γ (1)) = πX (0, 1) = 0
52
2 Homeomorphisms and topological properties
and
1
(πX ◦ γ )(1 − δ1/2 ) = πX (γ (1 − δ1/2 )) = πX x0 , sin
= x0 .
x0
This implies that
(πX ◦ γ )([1 − δ1/2 , 1]) = [0, x0 ].
Hence,
1
.
∀ x1 ∈ [0, x0 ] ∃ t ∈ [1 − δ1/2 , 1] such that γ (t) = x1 , sin
x1
For n large enough we can always construct a point xe1 ∈ (0, x0 ) such that
xe1 =
but then
1
2nπ − π2
1
π
= −1
= sin 2nπ −
2
xe1
sin
for some e
t ∈ [1 − δ1/2 , 1]. Hence, by construction the point
γ (e
t) =
1
, −1
2nπ − π2
belongs to B1/2 (γ (t)) ∩ G f and
dE
s
1
1
, −1 , (0, 1) =
2 + 4 > 2
2nπ − π2
2nπ − π2
This completes the proof.
⊓
⊔
Finally, remember that the interior of a connected set does not need to be connected.
Problems
2.1. Let (X, OX ) and (Y, OY ) be topological spaces. Show that
1. If (X, OX ) is path-connected and f : X −→ Y continuous, then (Y, OY ) is pathconnected.
2. If {Ai ⊂ X | i ∈ I} is a family of subsets of X such that the family of topological
T
S
spaces T = {(Ai , O|Ai ) | i ∈ I} is (path-) connected and i∈I Ai 6= 0,
/ then i∈I Ai
is (path-) connected.
2.4 The Hausdorff property
53
3. T = {(A1 , O|A1 , · · · , (Am , O|Am } is a family of (path-) connected topological
spaces if and only if A = A1 × · · · × Am equipped with the product topology is
(path-) connected.
2.4 The Hausdorff property
The Hausdorff property is our second example of a topological property. As we
will soon see this property ensures that every convergent sequence has a unique
limit. Topological spaces with the Hausdorff property are particularly important in
differential geometry.
Definition 2.5. We say that a topological space (X, O) has the Hausdorff
property if and only if for all x, y ∈ X with x 6= y there exists neighbourhoods
Ux and Uy of x and y, respectively, such that Ux ∩Uy = 0.
/
Lemma 2.9. Every metrizable topological space has the Hausdorff property.
Proof. Let (X, O) be a metrizable topological space. Then, there exists a metric d
such that O coincides with the topology O(d) induced by the metric d. Take the
open balls as a basis of the induced topology. Let x, y ∈ X with x 6= y and d(x, y) = r
for some r > 0. We can always construct neighbourhoods Ux = Br/2 (x) and Uy =
Br/2 (y) such that Ux ∩Uy = 0.
/ ⊓
⊔
The above result tells us that every metric space has the Hausdorff property. Can we
find some examples of topological spaces that do not have such a property? Yes and
we give some examples here below.
Example 2.4. Some topological spaces that do not possess the Hausdorff property
are
1. Consider X = {a, b} equipped with the trivial topology O = {0,
/ X}. Although
a 6= b we cannot find any neighbourhood of a and b, respectively, since there is
only one set at disposal and this is X itself.
2. A less trivial example is represented by R equipped with the Zariski topology
OZ = {U ⊂ R | R −U finite} ∪ {0,
/ R}.
This topology is used in algebraic topology.
54
2 Homeomorphisms and topological properties
Example 2.5. Every set equipped with the discrete topology is a topological space
with the Hausdorff property. Let us consider (X, O) with O = P(X). For any x, y ∈
X with x 6= y we can take as neighbourhoods Ux = {x} and and Uy = {y} in P(X).
Clearly, Ux ∩Uy = 0.
/
One of the advantage of working with a topological space having the Hausdorff
property is that convergent sequences have a unique limit.
Definition 2.6. Let (X, O) be a topological space and (xn )n∈N a sequence of
points in X. We say that the sequence converges to a ∈ X if
∀ Un neighbourhood of a ∃ N ∈ N such that xn ∈ Un ∀ n ≥ N.
Example 2.6. In a trivial topological space every sequence is convergent and its limit
is not unique since it converges to all points in X. To see that, take a set X and the
trivial topology. Let (xn )n∈N be a sequence in X and a ∈ X. The only neighbourhood
of a we can consider is Ua = X and we have xn ∈ X for all n ≥ 1. Take N = 1 and
conclude that xn → a as n → ∞. Hence, the sequence is convergent. Since we started
with an arbitrary a ∈ X, the conclusion will hold for any a ∈ X.
Lemma 2.10. In every topological space with the Hausdorff property the limit
of a convergent sequence is unique.
Proof. Let (X, O) have the Hausdorff property. Further, consider a convergent sequence (xn )n∈N in X. Let a ∈ X be its limit. Further suppose that b 6= a is also a limit
of the same sequence. Then, for every neighbourhhod Ub of b there exists N ∈ N
such that xn ∈ Ub for all n ≥ N. Take any neighbourhood Ub . Since (X, O) is Hausdorff and b 6= a there exists a neighbourhood Va of a such that Va ∩Ub = 0/ but then
pnly a finite number of terms in the sequence can be found in Va which contradict
the initial assumption that the sequence converges to a. ⊓
⊔
The uniqueness of the limit of convergent sequences in metric spaces is a trivial
implication of Lemma 2.9 and Lemma 2.10. In the next result we collect some
useful facts about the Hausdorff property.
Lemma 2.11. The following statements are true
2.4 The Hausdorff property
55
1. Let (X, O) be a topological space with the Hausdorff property and H ⊂
X be equipped with the subspace topology. Then, (H, O|H ) has also the
Hausdorff property.
2. The topological product of two topological spaces with the Hausdorff property is Hausdorff.
3. Let (X, OX ) and (Y, OY ) be topological spaces. Suppose that f : X −→ Y
is an injective and continuous map. If (Y, OY ) has the Hausdorff property,
then (X, OX ) has also the Hausdorff property.
4. The Hausdorff property is a topological property, that is let (X, OX ) and
(Y, OY ) be homeomorphic topological spaces. Then, (X, OX ) has the Hausdorff property if and only if (Y, OY ) has the Hausdorff property.
Proof. The proofs of the above statements are as follows.
1. Take any x, y ∈ H such that x 6= y. Since (X, O) is Hausdorff, we can find neighbourhoods Ux and Uy of x and y, respectively, such that their intersection is empty.
Recall that the subspace topology O|H is defined as O|H = {U ∩ H | U ∈ O}.
Note that Ux and Uy do not need to be open in X. Since Ux and Uy are neighbourhoods, there exist Nx , My ∈ O such that x ∈ Nx ⊂ Ux and y ∈ My ⊂ Uy . Conex = Nx ∩ H and
sider the following sets in the subspace topology of H, that is U
ex and y ∈ Vey . Finally, observe that
Vey = My ∩ H. By construction x ∈ U
ex ∩ Vey = (Nx ∩ H) ∩ (My ∩ H) = (Nx ∩ My ) ∩ H.
U
ex ∩ Vey = 0/ ∩ H = 0.
Since Ux ∩Vy = 0,
/ we also have Nx ∩ My = 0/ and therefore U
/
This means that H with the susbpace topology has the Hausdorff property. ⊓
⊔
2. Let (X, OX ) and (Y, OY ) be topological spaces with the Hausdorff property. Consider the product space (X ×Y, OX×Y ). Let (x1 , y1 ), (x2 , y2 ) ∈ X ×Y with x1 6= x2
and y1 6= y2 . Since (X, OX ) and (Y, OY ) are Hausdorff, there exist neighbourhoods
Ux1 , Ux2 and Vy1 , Vy2 such that Ux1 ∩Ux2 = 0/ and Vy1 ∩Vy2 = 0.
/ Moreover, there exists Nx1 , Nx2 ∈ OX such that x1 ∈ Nx1 ⊂ Ux1 and x2 ∈ Nx2 ⊂ Ux2 and My1 , My2 ∈ OY
such that y1 ∈ My1 ⊂ Vy1 and y2 ∈ My2 ⊂ Vy2 . Recall that open boxes form a bae(x ,y ) = Nx × My and
sis for the product topology. Let us consider the sets U
1
1
1 1
e(x ,y ) = Nx × My which are clearly neighbourhoods of the points (x1 , y1 ) and
U
2
2
2 2
(x2 , y2 ), respectively. Then,
e(x ,y ) ∩ U
e(x ,y ) = (Nx × My ) ∩ (Nx × My ),
U
1
1
2
2
1 1
2 2
= (Nx1 ∩ Nx2 ) × (My1 ∩ My2 ) = 0/ × 0/ = 0/
and this concludes the proof. ⊓
⊔
3. We need to show that for all x1 , x2 ∈ X with x1 6= x2 there exist neighbourhoods
ex and Vex such that U
ex ∩ Vex = 0.
U
/ Take any two points x1 6= x2 . Since f is in1
2
1
2
jective, we have f (x1 ) 6= f (x2 ). Moreover, (Y, OY ) is Hausdorff and therefore we
56
2 Homeomorphisms and topological properties
can find neighbourhoods U f (x1 ) and V f (x2 ) such that U f (x1 ) ∩V f (x2 ) = 0.
/ Since f is
continuous, the inverse images f −1 (U f (x1 ) ) and f −1 (V f (x2 ) ) are neighbourhoods
of x1 and x2 , respectively. Moreover,
f −1 (U f (x1 ) ) ∩ f −1 (V f (x2 ) ) = f −1 (U f (x1 ) ∩V f (x2 ) ) = f −1 (0)
/ = 0.
/
ex = f −1 (U f (x ) ) and Vex = f −1 (V f (x ) ) and we are done.
Choose U
1
2
1
2
4. We need to prove both directions.
⊓
⊔
a. =⇒: let (X, OX ) be Hausdorff. Since (X, OX ) and (Y, OY ) are homeomorphic,
there exists a homeomorphism h : X −→ Y . Since h−1 : Y −→ X is continuous
and injective, we can apply 3. to conclude that (Y, OY ) is Hausdorff.
b. ⇐=: let (Y, OY ) be Hausdorff. Apply 3. to h : X −→ Y which is injective and
continuous. ⊓
⊔
The quotient space of a Hausdorff topological space does not need to be Hausdorff.
Problems
2.2. Show that R equipped with the Zariski topology
OZ = {U ⊂ R | R −U finite} ∪ {0,
/ R}.
is a topological space. Verify that it has the Hausdorff property.
2.5 Compactness
This topological property plays an important role for example in real analysis where
one can show that if a function is continuous on a compact interval then it mus be
bounded there.
Definition 2.7. Let (X, O) be a topological space. A cover for X is a collection
S
of open sets, say U = {Uλ ⊂ X | Uλ ∈ O ∀λ ∈ I} such that X ⊆ λ ∈I Uλ .
A finite
subcover or refinement is a finite subcollection V ⊂ U such that
S
X ⊆ ni=1 Uλi . Finally, we say that a topological space is compact if every
open cover has a finite subcover.
Clearly, a topological space is not compact if we can find an open cover without a finite subcover. In a compact topological space if some property holds
locally, that is for all x ∈ X there exists Ux ∈ O having the property P, then
2.5 Compactness
57
the same property P must be true globally, i.e. for the whole space X. This
is not surprising because X compact implies that X can be written as a finite
union of open sets where each open set has the property P and therefore, the
property P will be transmitted to X.
Example 2.7. Let (X, O) be compact and consider the real line equipped with the
usual topology. Suppose that f : X −→ R is locally bounded, that is for every U ⊂ X
with U ∈ O there exists M ≥ 0 such that | f (x)| ≤ M for all x ∈ U. Then, f is bounded
on X. This can be verified as follows. Since (X, O) is compact, there exists a finite
S
subcover {U1 , · · · ,Un } with Ui ∈ O for all i = 1, · · · , n and X ⊆ ni=1 Ui . On the other
hand, f is bounded and therefore, for all Ui we can find Mi ≥ 0 such that | f (x)| ≤ Mi
for all x ∈ Ui . Take M = max1≤i≤n {Mi }. Then, | f (x)| ≤ M for all x ∈ X and we
conclude that f is bounded on X.
Example 2.8. We present some examples of compact and noncompact topological
spaces.
1. (0, 1) equipped with the subspace topology is not compact.
Proof. By contradiction. Suppose the given interval is compact. Take the open
cover
1
, 1 ⊂ (0, 1) | n ∈ N .
U =
n
Note that
[
n≥2
1
, 1 = (0, 1).
n
Take a finite subcover
Then,
N
[
i=1
1
1
,1 ,···,
,1
.
n1
nN
1
1
,1 =
, 1 ⊂ (0, 1)
ni
M
with M = max{n1 , · · · , nN }.
2. The real line equipped with the ussual topology is not compact.
⊓
⊔
Proof. By contradiction. Consider the open cover U = {(n − 1, n + 1) ⊂ R | n ∈
S
N} and note that Un∈N (n − 1, n + 1) = R since the open intervals form a basis
for R. Consider the finite subcover {(n1 − 1, n1 + 1), · · · , (nN − 1, nN + 1)}. Then,
N
[
(ni − 1, ni + 1) = (M − 1, M + 1) ⊂ R,
i=1
M = max{n1 , · · · , nN }
58
2 Homeomorphisms and topological properties
and clearly, points of R such that x ≤ M − 1 or x ≥ M + 1 will not be covered.
⊓
⊔
3. A discrete topological space is not compact. Take for instance R equipped with
the discrete topology P(R). Since (n − 1, n + 1) ∈ P(R) we can reason as in 2.
and conclude that the real line equipped with the discrete topology is not compact.
4. Every finite set with an arbitrary topology is compact. This follows from the fact
that every open cover consists of a finite number of open sets since X finite and
therefore any subcover must be finite as well.
5. Every trivial topological space is compact. This is pretty clear since there is only
one open cover and it contains only X. This open cover is also the only one finite
subcover.
6. If X is finite, then (X, P(X)) is compact. This follows from 4..
7. [0, 1] ⊂ R with the subspace topology is compact.
Proof. By contradiction. Suppose [0, 1] is not compact. Then, there exists an open
cover U without finite subcover. Decompose [0, 1] into [0, 1/2] and [1/2, 1].
Then, U ∩ [0, 1/2] and U ∩ [1/2, 1] are open covers for the intervals [0, 1/2]
and [1/2, 1], respectively. At least one of these open covers does not have a finite
subcover because if both of them would have a finite subcover, then the union of
these subcovers would be a finite subcover for [0, 1]. Let I1 = [a1 , b1 ] be one of
the intervals [0, 1/2] and [1/2, 1] having a finite subcover. There are two cases
a. if I1 = [0, 1/2], we consider the intervals [0, 1/4] and [1/4, 1/2].
b. if I1 = [1/2, 1], we consider the intervals [1/2, 3/4] and [3/4, 1].
Applying the same reasoning we can show that in the first case one of the intervals [0, 1/4] and [1/4, 1/2] cannot have a finite subcover. Similarly for the second
case. Hence, in any case we can construct a closed interval I2 ⊂ I1 . Proceeding in
this way we can construct a sequence of closed nasted intervals, say
· · · In ⊂ In−1 ⊂ · · · ⊂ I2 ⊂ I1 ,
where none of them has a finite subcover and the legth of In is 1/2n . Since the
length of these intervals decreases and they are all closed we must have
∞
\
n=1
In = {x}
with x ∈ [0, 1].
Since [0, 1] has an open cover, there exists U ∈ U such that x ∈ U. Since U
is open in the subspace topology, there exists Bεx (x) ∩ [0, 1] such that Bεx (x) ∩
[0, 1] ⊂ U with Bεx (x) = (x − εx , x + εx ). Requiring that 1/2n < 2εx , that is n >
log2 21εx , we can always construct an interval In ⊂ Bεx (x) ∩ [0, 1] ⊂ U and U is a
finite subcover for In which is impossible by the way we constructed In . Hence,
[0, 1] mus be compact. ⊓
⊔
2.5 Compactness
59
Similarly, one can prove that any interval of the form [a, b] with a < b must be
compact in the subspace topology.
Lemma 2.12. Continuous images of compact topological spaces are compact.
Proof. Let (X, OX ) be compact. Suppose that f : X −→ f (X) is continuous. We
need to show that f (X) with the subspace topology is compact. Consider an open
cover for f (X) such as
U = {Ui ⊂ f (X) | Ui ∈ O| f (X) ∀ i ∈ I} such that
f (X) ⊆
[
Ui∈I Ui .
Since f continuous, f −1 (Ui ) ∈ OX for all i ∈ I. Moreover, since f is surjective
!
[
X = f −1 ( f (X)) ⊆ f −1
Ui
=
[
f −1 (Ui ).
i∈I
i∈I
Hence, { f −1 (Ui ) | i ∈ I} is an open cover for X but X is compact and therefore,
there exists a finite subcover { f −1 (U1 ), · · · , f −1 (UN )}. Thus,
X⊆
N
[
f −1 (Ui )
i=1
if and only if
f (X) ⊆ f
N
[
i=1
f
−1
(Ui )
!
=
N
[
f ( f −1 (Ui )) =
i=1
N
[
Ui .
i=1
Hence, {U1 , · · · ,UN } is a finite subcover for f (X). ⊓
⊔
Lemma 2.13. Suppose (X, OX ) and (Y, OY ) are homeomorphic. Then,
(X, OX ) is compact if and only if (Y, OY ) is compact.
Proof. =⇒: since (X, OX ) and (Y, OY ) are homeomorphic, there is a homeomorphism f : X −→ Y . Let (X, OX ) be compact. Since f homeomorphism, f is surjective and f (X) = Y . Moreover, f is continuous. Hence, Lemma 2.12 implies that
(Y, OY ) is compact. The other direction follows again from Lemma 2.12 applied to
f −1 . ⊓
⊔
We want to find a characterization for compactness in the framework of metric
spaces. To this purpose we need some auxiliary results.
60
2 Homeomorphisms and topological properties
Definition 2.8. Let (X, d) be a metric space. We say that H ⊂ X is bounded if
there exists an open ball containing H.
Lemma 2.14. Let (X, d) be a metric space and H ⊂ X. If (H, O|H ) is compact, then it is also bounded.
Proof. Since (H, O|H ) is compact, there exists a finite subcover V = {Ui ⊂ H | Ui ∈
S
O|H ∀ i = 1, · · · , n} such that H ⊆ ni=1 Ui . On the other side, Ui O|H and it can be
represented as Ui = Bri (xi ) ∩ H with xi ∈ X for all i = 1, · · · , n because the open balls
form a basis for the topology induced by the metric. Hence,
H⊆
n
[
Bri (xi ).
i=1
S
Let R = max1≤i≤n {r1 + d(x, xi ) + ri }. Then, ni=1 Bri (xi ) ⊆ BR (x1 ). Indeed, if x ∈
i=1 Bri (xi ), then x will belong to at least one open ball, say Br j (xi ). We have x ∈
BR (x1 ) since
Sn
d(x, x1 ) ≤ d(x, x j ) + d(x j , x1 ) < r j + d(x j , x1 ) < r j + d(x j , x1 ) + r1 ≤ R.
Thus H ⊆ BR (x1 ) and H is bounded.
⊓
⊔
Lemma 2.15. Let (X, O) be Hausdorff and (H, O|H ) compact with H ⊂ X.
Then, H is closed.
Proof. We need to show that X − H is open in X, that is for all x ∈ X − H there exists
Ux ∈ O such that x ∈ Ux ⊂ X −H. Take any x ∈ X −H. Since (X, O) is Hausdorff, we
(y)
(y)
can always find disjoint neighbourhoods U1 of x and U2 of y ∈ H. The superscript
y emphasizes that the choice of the neighbourhoods U1 and U2 depends on the point
y we consider in H. Repeating the procedure for every y ∈ H we can construct a
family of neighbourhoods
(y)
(y)
{U2 ⊂ X | U2 neighbourhood of y ∀ y ∈ H}
(y)
whose neighbourhoods are disjoint from teh corresponding neighbourhood U1 of x
and such that they cover H. To get an open cover for H we observe that by definition
2.5 Compactness
61
(y)
(y)
e ∈ O such
of neighbourhood each neighbourhood U2 will have an open subset U
2
(y)
(y)
e
that y ∈ U2 ⊂ U2 . Consider the open cover for H given by
(y)
(y)
e ∩H | U
e ∈ O, y ∈ H}.
U = {U
2
2
Since (H, O|H ) is compact, there exists a finite subcover
e (y) ∩ H | yi ∈ H ∀ i = 1, · · · , n} such that
V = {U
2
H⊆
n
[
i=1
e (yi ) .
U
2
To this finite subcover there will correspond a finite number of neighbourhoods
(y )
(y )
U2 1 , · · · ,U2 n covering H. From the initial construction to these neighbourhoods
(y )
(y )
there will correspond a finite number of neighbourhoods of x, say U1 1 , · · · ,U1 n
(yi )
such that U1
have
(y j )
∩ U2
= 0.
/ From the Hausdorff definition of topological space we
Ux =
n
\
(yi )
U1
i=1
(y )
(y )
is again a neighbourhood of x which is disjoint to all neighbourhoods U2 1 , · · · ,U2 n .
S
(y )
(y )
Since H ⊆ ni=1 U2 i and Ux ∩ U2 i = 0/ for all i = 1, · · · , n, we conclude that
e ∈ O withx ∈ U
e ⊂ Ux
Ux ⊂ X − H. Since Ux is a neighbourhood of x there exists U
e
and hence U ⊂ X − H showing that X − H ∈ O. ⊓
⊔
Theorem 2.5. Every compact subspace of a metric space is closed and
bounded.
Proof. Let (X, d) be a metric space and H ⊂ X be equipped with the subspace topology. Then, Lemma 2.14 ensures that H is bounded. Since every metric metric space
is Hausdorff, Lemma 2.16 implies that H is closed. ⊓
⊔
Finally, the results obtained so far allow us to generalize a famous result in analysis
concerning boundedness of continuous functions on a compact interval and prove
the Heine-Borel theorem.
Theorem 2.6. Every real-valued function with compact domain is bounded.
Proof. Let (X, O) be compact and R be equipped with the usual topology. Consider
the continuous function f : X −→ R. Then, Lemma 2.12 implies that f (X) ⊂ R with
the subspace topology is compact. Since R with the usual topology is a metric space,
62
2 Homeomorphisms and topological properties
Theorem 2.12 implies that f (X) is closed. Since it is closed, we have sup{ f (X)}
and inf{ f (X)} belong to f (X). Hence, there exists xm , xM ∈ X such that f (xm ) =
inf{ f (X)} and f (xM ) = sup{ f (X)} thus implying that f (x) ≤ f (xM ) for all x ∈ X.
⊔
⊓
Theorem 2.7. (Heine-Borel)
Consider R with the usual topology. Then, every subset K ⊂ R is compact if
and only if K is closed and bounded.
Proof. The direction =⇒ follows from Theorem 2.5 since the real line equipped
with the usual topology is a metric space. Concerning the other direction, let K ⊂ R
be closed and bounded. Since it is bounded there exists an open interval (−a, a) with
a > 0 such that K ⊂ (−a, a). Moreover, (−a, a) ⊂ [−a, a]. Hence, K ⊂ [−a, a] but
[−a, a] is compact and any finite subcover of [−a, a] will be also a finite subcover
for K. Hence, K is compact. ⊓
⊔
This theorem can be generalized to compact subsets of Rn . The part of the proof
treating the implication =⇒ does not change. Concerning the part ⇐=, since K is
bounded there exists an open ball BR (a) ∈ O(dE )such that K ⊂ BR (a) ⊆ BR (a) is
closed. Moreover, BR (a) is bounded because we can take an open ball BR+ε (a) with
ε > 0 since then BR (a) ⊂ BR+ε (a). Then, BR (a) is compact and any finite subcover
of BR (a) will also be a finite subcover for K.
Theorem 2.8. Let (X, OX ) and (Y, OY ) be topological spaces. Then, the product space (X × Y, OX×Y ) is compact if and only if (X, OX ) and (Y, OY ) are
compact.
Proof. We need to prove both directions.
1. =⇒: introduce projections πX : X × Y −→ X and πY : X × Y −→ Y . Since they
are continuous, Lemma 2.12 implies that (X, OX ) and (Y, OY ) are compact.
2. ⇐=: let (X, OX ) and (Y, OY ) be compact. Take an arbitrary open cover for X ×Y ,
that is
U = {Wλ ⊂ X ×Y | Wλ ∈ OX×Y }
with X ×Y ⊆
[
Wλ .
λ ∈I
Construct a finite subcover. Take any (x, y) ∈ X × Y . Since U is an open cover,
λ such that (x, y) ∈ Weλ . On the other side, the
there must exists some λ ∈ I, say e
choice of such λ depends on the choice of the point (x, y). We emphasize these
fact by introducing the notation e
λ = λ(x,y) and we will have (x, y) ∈ Wλ(x,y) . Since
2.5 Compactness
63
Wλ(x,y) ∈ OX×Y , there exists an open box U(x,y) × V(x,y) with U(x,y) ,V(x,y) ∈ OX×Y
such that (x, y) ∈ U(x,y) ×V(x,y) ⊂ Wλ(x,y) . Keep x fixed. Then,
f= {V(x,y) ⊂ Y | y ∈ Y }
U
is an open cover of Y . Since (Y, OY ) is compact, there exists a finite subcover
Ve = {V(x,y1 (x)) , · · · ,V(x,yn (x)) } with y1 (x), · · · , yn (x) ∈ Y.
Remember that these points depend on the original choice of x. Let
Ux =
n
\
U(x,yi (x) .
i=1
Since (X, OX ) is compact, {Ux1 , · · · ,Uxm } will cover X with x1 , · · · , xm ∈ X.
Hence, we constructed m vertical stripes covering X ×Y . Moreover, we covered
X ×Y with a finite number of open boxes Uxi ×V(xi ,y j (xi )) , that is
X ×Y ⊆
m [
n
[
i=1 j=1
Uxi ×V(xi ,y j (xi )) .
Each of these open boxes will be contained in some Wλ (xi ,y j (x)) that is
Uxi ×V(xi ,y j (xi )) ⊂ Wλ (xi ,y j (x)) .
This implies that
X ×Y ⊆
m [
n
[
i=1 j=1
Uxi ×V(xi ,y j (xi )) ⊆
m [
n
[
Wλ (xi ,y j (x)) .
i=1 j=1
This implies that
{Wλ (xi ,y j (x)) | 1 ≤ i ≤ m, 1 ≤ j ≤ n}
is a finite subcover.
⊓
⊔
The above result can be extended to a finite product of compact topological spaces
by induction and it has been generalized to the product of infinitely many compact
spaces by Tychonoff (1930). Sometimes in proving that two topological spaces are
homeomorphic we face the situation where we are able to construct a bijective and
continuous map f but it results difficult to verify that f −1 is continuous. The next
result is a solution to this problem when certain conditions on the domain and range
of f are satisfied.
64
2 Homeomorphisms and topological properties
Theorem 2.9. Let (X, OX ) be compact and (Y, OY ) be Hausdorff. Suppose
that f : X −→ Y is bijective and continuous. Then, f −1 is continuous and
hence, f is a homeomorphism.
Proof. Let g = f −1 : Y −→ X. Show that for all A ⊂ X closed the inverse image
g−1 (A) is closed in Y . Take any A ⊂ X closed in X. Since X is compact, Lemma 2.16
implies that A is compact. Moreover, g−1 (A) = f (A). Since f continuous, f (A) is
compact in Y by Lemma 2.12. Now f (A) is a compact subspace of a Hausdorff
space and it must be closed. Hence, g−1 (A) is closed in Y . ⊓
⊔
Lemma 2.16. Let (X, O) be compact and H ⊂ X closed. Then, H equipped
with the subspace topology is compact.
Proof. Since (X, O) is compact, it has an open cover U . With this open cover we
can cosntruct an open cover for H by considering U ∩H. Since H is closed, X −H ∈
O and therefore we can construct a new open cover for X given by U ∩H and X −H.
Since (X, O) is compact, there exists a finite subcover of X
{U1 ∩ H, · · · ,Un ∩ H} ∪ {X − H}
where {U1 ∩ H, · · · ,Un ∩ H} is a finite subcover of H.
⊓
⊔
Chapter 3
Topological vector spaces
3.1 Topological vector spaces
The main goal of this chapter is to present a series of examples of topological spaces
which play a fundamental role in functional analysis, namely the topological vector
spaces.
Definition 3.1. Let K be a real or complex field. A vector space V on K which
is also a topological space is said to be a topological vector space if
1. the addition map + : V ×V −→ V such that (u, v) 7−→ u + v for all u, v ∈ V
is continuous.
2. the scalar multiplication · : K × V −→ V such that (λ , u) 7−→ λ u for all
λ ∈ K and for all u ∈ V is continuous.
The first example of topological vector space we are going to study is represented
by normed spaces.
Definition 3.2. Let V be a vector space on K. A norm on V is a map k · k :
V −→ R such that
1. kuk ≥ 0 for all u ∈ V and kuk = 0 if and only if u is the zero vector.
2. kλ uk = |λ |kuk for all λ ∈ K and for all u ∈ V .
3. ku + vk ≤ kuk + kvk for all u, v ∈ V .
The pair (V, k · k) is called a normed space.
65
66
3 Topological vector spaces
Lemma 3.1. Let (V, k · k) be a normed space. Then,
|kuk − kvk| ≤ ku − vk.
Proof. Observe that
kuk = ku − v + vk ≤ ku − vk + kvk
and hence kuk − kvk ≤ ku − vk. Moreover,
kvk = kv − u + uk ≤ kv − uk + kuk = ku − vk + kuk
and hence −ku − vk ≤ kuk − kvk. This completes the proof.
⊓
⊔
Lemma 3.2. Every norm on a vector space induces a metric.
Proof. Let (V, k · k) be a normed space. We want to show that d(u, v) = ku − vk
defined for all u, v ∈ V generates a metric d : V ×V −→ R. We only need to check
the axioms of a metric space.
1. Note that a vector space is closed under addition. Hence, there exists w = u − v
and then
d(u, v) = ku − vk = kwk ≥ 0.
Clearly, kwk = 0 if and only if w = 0 and hence if and only if u = v.
2. Take any u and v in V . Then,
d(u, v) = ku − vk = kv − uk = d(v, u).
3. Take any u, v, w ∈ V . Then,
d(u, v) = ku − vk = ku − w + w − vk ≤ ku − wk + kw − vk = d(u, w) + d(w, v).
This completes the proof.
⊓
⊔
The metric d introduced in Lemma 3.2 induces a topology O(d) on V , thus making
(V, O(d)) a topological space. It is not difficult to check that Definition 3.1 applies
and therefore (V, k · k) is a topological vector space. Furthermore, the norm is a
continuous map. Last but not least, since every normed space is a metric space, it
will also have the Hausdorff property.
3.2 Finite-dimensional normed spaces
67
Problems
3.1. Consider a normed vector space (V, k·k) with norm defined as d(u, v) = ku−vk
for all u, v ∈ V . Let O(d) be the topology induced on V . Show that (V, k · k) is a
topological vector space. Furthermore, using Lemma 3.1 show that the norm is a
continuous map.
3.2 Finite-dimensional normed spaces
Example 3.1. We present some examples of finite-dimensional vector spaces.
1. Let V = Rn with n ≥ 1 and consider the norm
s
n
kuk =
∑ u2i
i=1
∀ u = (u1 , · · · , un ) ∈ Rn .
k · k is called the canonical Euclidan norm on Rn . The metric
d(u, v) = ku − vk ∀ u, v ∈ Rn
is called the standard metric on Rn .
2. Let V = Rn and consider the supremum norm
kuk = max{|u1 |, · · · , |un |} ∀ u = (u1 , · · · , un ) ∈ Rn .
The main goal of this section is to show that all norms on Rn induce the same
topology. To prove this we need some preliminary results.
′
Definition 3.3. Let V be a vector space over K. Two norms k · k and k · k on
V are equivalent if there exists k1 , k2 > 0 such that
′
k1 kuk ≤ kuk ≤ k2 kuk
∀ u ∈ V.
The above definition introduces an equivalence relation on the set of all norms.
′
Two equivalent norms on V induce the same topology on V because if k · k and k · k
′
are such norms and d and d the associated metrics, then
′
Br/k2 (u, d) ⊂ Br (u, d ) ⊂ Br/k1 (u, d)
for all u ∈ V and r ∈ R with r > 0. We finally prove that all norms on Rn are equivalent. This together with the result in Problem3.4 implies that all induced topologies
68
3 Topological vector spaces
on Rn are equivalent, so that in principle we can say that there is only one induced
topology on Rn .
′
′
Theorem 3.1. Let k · k be an arbitrary norm on Rn . Then, k · k and k · k∞ are
equivalent. In particular, all norms in Rn induce the same topology.
′
Proof. Show that there exists k1 , k2 > 0 such that k1 kuk ≤ kuk ≤ k2 kuk∞ for all
u = (u1 , · · · , un ) ∈ Rn. Consider the hypersphere Sn = {u ∈ Rn | kuk∞ = 1} and the
′
′
continuous map k · k : S −→ R. Since Sn is a compact subset of Rn and k · k is
S
S
continuous, then the image of Sn under the given map is a compact subset of R by
Lemma 2.13. Heine-Borel theorem requires that this compact subset of R must be a
closed and bounded interval, say [k1 , k2 ]. Hence, for all u ∈ Sn there exists k1 , k2 > 0
′
such that k1 ≤ kuk ≤ k2 . Moreover, for all u ∈ Rn \{0} we have
′
u kuk
≤ k2 kuk∞
kuk =
kuk∞ = kuk∞
kuk∞ ′
and
′
1 ′
′
′
1
kuk∞
kuk =
kuk
′ kuk = u kuk ≤
kuk k
kuk
1
∞
′
′
and hence, k1 kuk∞ ≤ kuk . Combining both inequalities obtained above we find that
′
k1 kuk∞ ≤ kuk ≤ kuk∞ and since this inequality also holds for u = 0, we conclude
that these norms are equivalent on Rn . ⊓
⊔
Problems
3.2. Show that V = Rn equipped with the supremum norm
kuk = max{|u1 |, · · · , |un |} ∀ u = (u1 , · · · , un ) ∈ Rn .
is a normed space.
3.3. Show that Definition 3.3 introduces an equivalence relation on the set of all
norms.
3.4. Show that two equivalent norms on V induce the same topology on V , that is if
′
′
k · k and k · k are such norms and d and d the associated metrics, prove that
′
Br/k2 (u, d) ⊂ Br (u, d ) ⊂ Br/k1 (u, d)
for all u ∈ V and r ∈ R with r > 0.
3.3 Continuous maps between normed spaces
69
3.3 Continuous maps between normed spaces
The goal of this section is to show that linear maps between normed spaces are also
continuous and vice-versa. Hence, on a normed space linearity is a synonimous of
continuity and this is a wonderful result emerging from the marriage between linear
algebra (introduced by means of vector spaces) and topology (introduced through
the distance induced by the norm and its continuity).
Definition 3.4. Let (V, k · kV ) and (W, k · kW ) be normed spaces with V,W vector spaces over K. A map ϕ : V −→ W is said to be linear if
ϕ (λ u + µ v) = λ ϕ (u) + µϕ (v) ∀ λ , µ ∈ K and ∀ u, v ∈ V.
Theorem 3.2. Let (V, k · kV ) and (W, k · kW ) be normed spaces and ϕ : V −→
W as in the definition above. Then, the following statements are equivalent
1. ϕ is continuous on V .
2. ϕ is continuous at 0V .
3. there exists M > 0 such that kϕ (u)kW ≤ MkukV for all u ∈ V .
Proof. We show that 1. =⇒ 2. =⇒ 3. =⇒ 1..
1. 1. =⇒ 2.: trivial since ϕ is continuous on V , it follows that ϕ is continuous at
every u ∈ V and in particular at u = 0V .
2. 2. =⇒ 3.: let O(dV ) and O(dW ) be the topologies induced by the metrics dV and
dW generated by the norms k · kV and k · kW , respectively. Note that the linearity
of ϕ implies that the zero vector of V is mapped to the zero vector of W
ϕ (0V ) = ϕ (0 · 0V ) = 0ϕ (0V ) = 0W .
Since ϕ is continuous, we have that for every open neighbourhood of 0W the
inverse image is an open neighbourhood of 0V . Hence, without loss of generality
′
let B = {w ∈ W | kwkW < 1} be the ball of radius one and centre 0W . Then,
′
ϕ −1 (B ) is an open neighbourhood 0V . Since it is an open set we can always find
a ball of radius r > 0 and centre 0V such that
′
′
Br (0V ) ⊂ ϕ −1 (B ) =⇒ ϕ (Br (0V )) ⊂ B
Let v ∈ V \{0}V . Note that
70
3 Topological vector spaces
rv r
r
2kvkV = 2kvkV kvkV = 2 .
V
This implies that
rv
∈ Br (0V )
2kvkV
and hence,
ϕ
rv
2kvkV
We can conclude that
=
′
r
ϕ (v) ∈ B .
2kvkV
r
< 1
ϕ
(v)
2kvkV
W
and it follows that
2
kϕ (v)kW < kvkV .
r
If v = 0V note that
2
kϕ (0V )kW = k0W kW = 0 = 0 · k0kV .
r
Hence, we can conclude that
2
kϕ (v)kW < kvkV
r
∀ v ∈ V.
3. 3. =⇒ 1.: we show that ϕ is continuous at any v0 ∈ V , that is for every ε > 0
there exists δε > 0 such that dW (ϕ (v), ϕ (v0 )) < ε whenever dV (v, v0 ) < δε . The
linearity of ϕ together with 3. implies that
′
dW (ϕ (v), ϕ (v0 )) = kϕ (v) − ϕ (v0 )kW = kϕ (v − v)kW ≤ M kv − v0 kV
′
′
= M dV (v, v0 ) < M δε .
′
Take δε = ε /M for any ε > 0 and we are done.
′
Corollary 3.1. Let k · k be any norm on Rn . Further, consider the normed
space (V, k · kV ) and a linear map ψ : Rn −→ V . Then, ψ is continuous on
′
(Rn , k · k ).
Proof. Left as an exercise.
3.3 Continuous maps between normed spaces
Problems
3.5. Prove Corollary 3.1.
71
Appendix A
The Continuum Hypothesis
The continuum hypothesis is a hypothesis, introduced by the German mathematician
Georg Cantor in 1877, concerning sizes of infinite sets. More precisely, he stated that
it is not possible to find a set having cardinality strictly between that of the integers
and that of the real numbers. In the year 1900 the German mathematician David
Hilbert included the proof of the truth or falsehood of the continuum hypothesis
in his famous twenty-three problems. Only later the Austrian mathematician Kurt
Gödel in 1940 [1] and the American mathematician Paul Cohen in 1963 [2] were
able to prove that the hypothesis can neither be disproved nor be proved using the
axioms of Zermelo - Fraenkel set theory. In other words, this hypothesis is undecidable. The name of the hypothesis originates from the term the continuum for the
real numbers. We will mainly follow the presentation given in [3].
A.1 Cardinality
The concept of cardinality has been developed to compare the sizes of two sets.
Consider for instance the sets S = {x ∈ R | x2 = 4} and T = {1, 2, 3} having two and
three elements, respectively. Intuitively, we would say that the size of T is greater
than the size of S because T has more elements than S. Even though this idea works
well for finite sets we need to formalize it also for infinite sets like the set of the
natural numbers N and the set of all real numbers R.
Definition A.1. Two sets S and T have the same size, (or are equivalent or equinumerous) and we write S ∼ T if there exists a bijection f : S −→ T .
Definition A.2. A set S is finite if S = 0/ or there exists a natural number n such that
S ∼ In = {1, 2, · · · , n}. In is called the indexing set. We say that S is infinite if it is
not finite. Further, S is denumerable if S ∼ N. Finally, a set is countable if it is finite
or denumerable. A set is uncountable if it is not countable.
Example A.1. Let S and T be two sets having both n elements. We want to show
that these sets are equinumerous, that is we have to construct a bijection h : S −→ T .
73
74
A The Continuum Hypothesis
Since S has n elements, then S ∼ In and we have a bijection f : In −→ S. Similarly, we
can construct a bijection g : In −→ T . Since f −1 is also a bijection and composition
of bijective maps is bijective, we can set up a bijection h : S −→ T such that h =
f −1 ◦ g.
The property of two sets being equinumerous is fundamental for the introduction
of the concept of cardinal number. To this purpose, consider a countable family
of sets F = {Ui | i ∈ I} with indexing set I = In or I = N. The property of being
equinumerous is an equivalence relation on F . This can be easily verified as follows
1. reflexivity: a natural bijection is represented by the identity map f : Ui −→ Ui
and hence Ui ∼ Ui .
2. symmetry: if Ui ∼ U j with i 6= j then there exists a bijection f : Ui −→ U j . On the
other hand f −1 : U j −→ Ui is also a bijection and we can conclude that U j ∼ Ui .
3. transitivity: if Ui ∼ U j and U j ∼ Uk with i 6= j 6= k, it follows that Ui ∼ Uk because
composition of bijective maps is again a bijection.
Therefore, ∼ induces a partition of F into equivalence classes [Ui ] = {Uk ∈
F | Uk ∼ Ui }.
Definition A.3. A cardinal number or cardinality of a set is defined to be the size
of the sets belonging to an equivalence class determined by the relation ∼.
Clearly, two sets will have the same cardinal numbers if they are equinumerous.
Example A.2. The cardinal number of the set In is n. If S ∼ In , than S belongs to
the equivalence class having In as a representative and therefore we say that the
cardinality of S is n. The empty set 0/ has cardinality 0.
Definition A.4. If a cardinal number is not finite, it is called transfinite. The cardinal
number of a denumerable set is denoted by ℵ0 .
It is in general not true that every infinite set has cardinal number ℵ0 because an infinite set does not need to be denumerable. This suggests that we could have different
infinite sets having different sizes of infinity.
A.2 Countable sets
In the world of denumerable sets it could be very dangerous to try to apply some
common sense gained from the experience of working with finite sets. Consider,
for instance, the set N and the set O of all odd natural numbers. Is the size of O
smaller than the size of N? The answer is no since we can always set up a bijection
f : N −→ O such that f (n) = 2n − 1. Hence, O and N are equinumerous and as a
consequence they have the same size.
Exercise A.1. We want to show that the set of the integer numbers Z is denumerable
and it has the same cardinality of N. We need to set up a bijection f : N −→ Z. To
A.2 Countable sets
75
do that we observe that N is given by the union of the set E of all even numbers and
the set O of all odd numbers. The required bijection is given by
n
if n ∈ E,
f (n) = 2 n−1
− 2 if n ∈ O.
This implies that the cardinality of the integers is the same as the cardinality of the
natural numbers, that is
|Z| = |N| = ℵ0 .
We want now to look at some counting processes for finite and denumerable sets.
We consider the following two cases
1. Let S = {s1 , s2 , · · · , sn } be a finite set with cardinality n. Then, S ∼ In and there is
a bijection f : In −→ S. We can use f to count off the elements of S as follows
f (1) = s1 ,
f (2) = s2 , · · · , f (n) = sn .
2. Suppose that the set T = {t1 ,t2 , · · ·} is denumerable. Then, there is a bijection
h : N −→ T that can be used to count the elements of T as follows
g(1) = t1 ,
g(2) = t2 , · · · , g(n) = tn , · · ·
.
For this reason finite and denumerable sets are called countable sets. Moreover, it is
not difficult to prove that the finite union of finite sets is finite and every subset of a
finite set is finite.
Theorem A.1. A finite union of finite sets is finite.
Proof. Let A1 , · · · , An be finite sets and let m = max{|A1 |, · · · , |An |}. Then,
n
[
n
n
Ak ≤ ∑ |Ak | ≤ ∑ m = mn.
k=1 k=1
k=1
Note that in the first inequality the equality will hold whenever the sets A1 , · · · , An
are all mutually disjoint. ⊓
⊔
Clearly, arbitrary unions of finite sets do not need to be finite as the following counterexample shows. Consider the finite set Ai = {i} with i ∈ N. Then,
[
Ai = N
i∈N
which is not finite.
Exercise A.2. We want to show that every subset of a finite set is finite. Let B ⊂ A
with A a finite set. We have two cases
1. if A is empty, the only subset of the empty set is the empty set which is trivially
finite.
76
A The Continuum Hypothesis
2. Let A be nonempty and finite with cardinality n ∈ N. Then, A ∼ In and both
belong to the same equivalence class. Let us choose In to be the representative of
this equivalence class. This allows us to prove the statement for any subset of In
without loss of generality. We proceed now by induction
a. if n = 1, then I1 = {1} and the only subsets of I1 are the empty set and I1 itself
which are both finite.
b. Let k > 1 with k < n and assume that every subset of Ik is finite. Consider
Ik+1 and B ⊂ Ik+1 . We have two cases. If k + 1 ∈
/ B then B ⊂ Ik and it must
be finite. If k + 1 ∈ B, let D = B\{k + 1}. Then, D ⊂ Ik must be finite but
B = D ∪ {k + 1} is a finite union of finite sets and therefore it must be finite.
In order to prove the next result we need the so called Well-Ordering Property
(WOP) which can be stated as follows: if S is a nonempty set of N we can always
find an element m ∈ S such that m ≤ k for all k ∈ S.
Theorem A.2. If S is a countable set and T ⊆ S, then T is countable.
Proof. If T = S, the statement is trivial. Let T ⊂ S. We have two cases.
1. If T is finite, then Def. A.2 implies that T is countable.
2. Let T be infinite. Then, by Exercise A.2 we conclude that S is infinite as well.
Since S is infinite and countable, it is denumerable. Let S = {s1 , s2 , · · ·}. Then,
there exists a bijection f : N −→ S such that f (n) = sn for all n ∈ N. Consider the
set A = {n ∈ N | sn ∈ T }. This set cannot be empty because T is not empty since
T ⊂ S and T is infinite. Since A ⊂ N, by the WOP A has a least element n1 ∈ N
to which corresponds an element sn1 ∈ T . Applying the WOP to A\{n1 } ⊂ N
yields that A\{n1 } has a least element n2 ∈ N to which corresponds an element
sn2 ∈ T . If we apply the WOP recursively, then the set A\{n1 , n2 , · · · , nk } has a
least element nk+1 ∈ N to which corresponds an element sn+1 ∈ T . Moreover, we
also have
n1 < n2 < · · · < nk+1 < · · · .
(A.1)
Introduce the map g : N −→ N such that g(k) = nk for all k ∈ N. We verify that g
is a bijection.
a. Take any i, k ∈ N with i < k. Then, g(i) = ni and g(k) = nk . By (A.1) we have
ni < nk and hence g(i) 6= g(k), that is g is injective.
b. We check surjectivity that is we verify that for any nm ∈ N there exists a k ∈ N
such that g(k) = nm with m ∈ N. Take m = k and we are done.
We consider now the map f ◦g which is a bijection since composition of bijective
maps. Observe that the domain of definition of this map is N and its range is T
because for any k ∈ N we have
( f ◦ g)(k) = f (g(k)) = f (nk ) = sk ∈ T.
Hence, f ◦ g : N −→ T is the required bijection making T denumerable.
⊓
⊔
This theorem plays a key role in the derivation of countability criteria of a set.
A.2 Countable sets
77
Theorem A.3. Let S 6= 0/ and J be a set such that J = In if S is finite and otherwise
J = N if S is denumerable. Then, the following statements are equivalent
1. S is countable;
2. there exists an injection f : S −→ J;
3. there exists a surjection g : J −→ S.
Proof. We break down the proof in three parts.
1. 1. =⇒ 2.: let S be countable. Then, S is finite or denumerable and hence there
exists a bijection h : J −→ S such that J = In for some n ∈ N if S is finite and
J = N if S is denumerable. Clearly, h−1 : S −→ J is also a bijection. Let f = h−1
and we are done.
2. 2. =⇒ 3.: let f : S −→ J be an injection. Then, f : S −→ f (S) ⊆ J is a bijection
and therefore f −1 : f (S) −→ S is also a bijection. We use f −1 to construct a
surjective map g : J −→ S as follows
−1
f (n) if n ∈ f (S),
g(n) =
p
for some p ∈ S if n ∈ J\ f (S).
We prove that g is surjective by verifying that g(J) = S. From J = f (S) ∪ (J −
f (S)) we have
g(J) = g( f (S) ∪ (J − f (S))) = g( f (S)) ∪ g(J − f (S)) = f −1 ( f (S)) ∪ {p, q, · · ·}
with {p, q, · · ·} ⊆ S. Hence, we conclude that g(J) = S ∪ {p, q, · · ·} = S.
3. 3. =⇒ 1.: suppose that there exists a surjection g : J −→ S. We want to show
that S is countable. The idea behind the proof is to use the map g to construct
a bijection between S and some countable set. Consider the map h : S −→ N
defined through the requirement that h(s) is the smallest natural number n such
that g(n) = s. Then, h is injective and in particular the map h : S −→ h(S) ⊆ J is
a bijection. Since N is countable and h(S) ⊆ N, then Theorem A.2 implies that
h(S) is countable. On the other hand, h : S −→ h(S) is a bijection and therefore
h(S) and S are equinumerous. Hence, S is countable. ⊓
⊔
The next result is an immediate consequence of the above theorem.
Corollary A.1. Let S and T be sets. Then,
1. S is countable if f : S −→ T injective and T countable;
2. T is countable if g : S −→ T surjective and S countable.
Proof. We have to prove two parts.
1. Since T is countable, by the previous theorem there is an injection h : T −→ J.
Moreover, the map f is injective and hence h ◦ f : S −→ J is injective as well and
again Theorem A.3 ensures that S is countable.
2. Since S is countable, Theorem A.3 implies that there is a surjection h : J −→ S.
Moreover, the map g is surjective and hence g ◦ h is surjective as well. Finally,
Theorem A.3 implies that T is countable. ⊓
⊔
78
A The Continuum Hypothesis
By means of Theorem A.3 it is possible to show that the countable union of countable sets is countable and that the Cartesian product of two countable sets is again
countable.
Theorem A.4. The countable union of countable sets is countable.
Proof. It is sufficient to show that the union of two countable sets is again countable,
since the rest can be proved by induction. Let S and T be two countable sets. We
have two cases.
1. If S = 0/ = T or S = 0,
/ T 6= 0/ or S 6= 0,
/ T = 0/ the statement is trivial.
2. Let S, T 6= 0.
/ Then, since S and T countable there exist surjections f : J −→ S
and g : J −→ T . If we are able to construct a surjection h : J −→ S ∪ T , then
Theorem A.3 will make the rest. Such a surjective map can be constructed as
follows
n+1 f 2 if n odd,
h(n) =
if n even
g n2
and this completes the proof.
⊓
⊔
Theorem A.5. The Cartesian product of two nonempty countable sets is countable.
Proof. Let S and T be nonempty and countable. Then, Theorem A.3 implies that
there exist injections f : S −→ J and g : T −→ J. If we can construct an injection
h : S × T −→ J, then Theorem A.3 will make the rest. Inspired by the fact that every
natural number can be written as a prduct of primes and the representation is unique
up to the order of the factors (for instance 12 = 22 · 3 = 3 · 22 ) we define h as follows
h(s,t) = 2 f (s) 3g(t)
for all s ∈ S and for all t ∈ T . The map h is injective since for any s, u ∈ S and for
any t, v ∈ T we have
h(s,t) = h(u, v) ⇐⇒ 2 f (s) 3g(t) = 2 f (u) 3g(v) ⇐⇒ 2 f (s)− f (u) = 3g(v)−g(t) ,
⇐⇒ f (s) = f (u), g(v) = g(t).
Since f and g are injective, it follows that s = u and v = t.
⊓
⊔
Examples of countable sets are N, Z and Q.
Exercise A.3. Show that Q is countable. We consider the representation Q = Q+ ∪
{0} ∪ Q− . Since {0} is finite and hence countable, if we can prove that Q± are
countable, then Q will be countable because union of countable sets where
n m
o
Q± = ± | m, n ∈ N and m, n relatively prime .
n
To prove that Q+ is countable we set up an injection f : Q+ −→ N and Q+ will be
countable by Theorem A.3. Consider the map
A.2 Countable sets
79
f
m
n
= 2m 3n .
For its injectivity see the proof of Theorem A.5. To show that Q− is countable we
will prove that Q− and Q+ are equinumerous. This will force Q+ to be countable.
This will be the case if we can construct a bijection g : Q+ −→ Q− . Take for instance
m
m
m
=−
∀ ∈ Q+
g
n
n
n
and we are done.
The next theorem provides us with an important example of an uncountable set. This
has been shown by the German mathematician Cantor.
Theorem A.6. R is uncountable.
Proof. We first show that (0, 1) and R are equinumerous by introducing a bijection
f : (0, 1) −→ R defined through f (x) = tan π (x − 1/2). In this way it is sufficient
to prove that (0, 1) is uncountable. We will do it by contradiction and make use of
Cantor’s diagonal argument. Suppose that (0, 1) is countable. Then, there exists a
bijection g : N −→ (0, 1) and the real numbers in (0, 1) can be represented by the
list {x1 , x2 , · · ·}. These numbers will have decimal representations
x1 = 0.a11 a12 a13 a14 · · · ,
x2 = 0.a21 a22 a23 a24 · · · ,
x3 = 0.a31 a32 a33 a34 · · · ,
..
.
Canto’s diagonal argument shows that we can construct a real number which is not
contained in the list above and therefore this number cannot be in correspondence
with any natural number. Let us construct a real number ω = 0.a1 a2 a3 · · · as follows:
a1 must be different from a11 so that ω cannot be x1 , a2 must be different from a22
so that ω cannot be x2 , a3 must be different from a33 so that ω cannot be x3 and so
on. In this way, ω does not coincide with any number xi for all i ∈ N and therefore,
it cannot be put in correspondence with any natural number. ⊓
⊔
Another example of an uncountable set is provided by the set of all irrational numbers.
Exercise A.4. We want to show that the set of all irrational numbers is uncountable. Let I denote such a set. Then, R = I ∪ Q. By contradiction suppose that I is
countable.Then, since Q is countable and the countable union of countable sets is
countable we conclude that the set of all real numbers is countable and hence, we
reach a contradiction.
80
A The Continuum Hypothesis
A.3 The Continuum Hypothesis
We already know that two sets S and T have the same size if they are equinumerous.
This is equivalent to say that they have the same cardinal numbers and we write
|S| = |T | where | · | denotes the cardinal number of a set.
Definition A.5. We say that S is smaller than T and we write |S| ≤ |T |, if there exists
an injection but no surjection f : S −→ T .
Note that if S would be larger than T we would have at least two distinct elements
of S mapped to the same element of T and the map f could not be injective. The
next theorem describes some basic properties of cardinal numbers.
Theorem A.7. Let S, T and U be sets. Then,
1.
2.
3.
4.
5.
|S| ≤ |T | if S ⊆ T ;
|S| ≤ |S|;
|S| ≤ |U| if |S| ≤ |T | and |T | ≤ U|;
|Im | ≤ |In | if m ≤ n with m, n ∈ N;
|S| < ℵ0 if S is finite.
Proof. We prove only properties 3. and 5. the others being trivial.
1. 3.: since |S| < |T | and |T | < |U|, there exist injections f : S −→ T and g : T −→ U
and h = g◦ f : S −→ U is also an injection because composition of injective maps.
Hence, |S| < |U|.
2. 5.: let S be finite. Then, S and In are equinumerous for some n ∈ N but In ⊂ N
and hence |In | < |N| = ℵ0 .
This completes the proof.
⊓
⊔
The cardinal number of R is denoted by c where c stays for continuum. Since Q ⊂ R
and |Q| = |N| the first part of the above theorem implies that ℵ0 < c. Hence, we
can have unequal transfinite cardinals. Can we have other transfinite cardinals? To
answer this question the concept of power set is needed.
Definition A.6. Let S be a set. The power set of S denoted by P(S) or 2S is the set
of all subsets of S.
Example A.3. Consider the set S = {a, b, c}. The corresponding power set is
P(S) = {0,
/ S, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}}.
The cardinality of S is |S| = 3 e that of the power set is |P(S)| = 23 = 8. Note that
|S| < |P(S)|.
The above example suggests the following result.
Theorem A.8. For any set S we have |S| < |P(S)|.
A.3 The Continuum Hypothesis
81
Proof. We need to show that there exists an injection f : S −→ P(S) but no
surjection g : S −→ P(S). Le us consider a map f : S −→ P(S) defined by
f (x) = {x} ∈ P(S). Take any x, y ∈ S. Then, f is clearly an injection since f (x) =
f (y) ⇐⇒ {x} = {y} ⇐⇒ x = y. We prove that there is no surjection between S
and P(S) by contradiction. Suppose that there exists a surjection g : S −→ P(S).
Then, g(S) = P(S) and for any x ∈ S g(x) is a subset of P(S). In general, x does
not belong to g(x). Use this observation to construct the set
B = {x ∈ S | x ∈
/ g(x)} ⊆ S.
Then, B ⊆ S implies that B ∈ P(S) and g surjective implies that there exists an a ∈ S
such that g(a) = B. We have two cases.
1. If a ∈ B, then the way we defined B implies that a ∈
/ g(a) but g(a) = B and hence
a∈
/ B.
2. If a ∈
/ B, then the way we defined B implies that a ∈ g(a) but g(a) = B and hence
a ∈ B.
In any case we have a contradiction and such a surjection cannot exist.
⊓
⊔
Theorem A.9. |P(N)| = c.
Theorem A.10. Every infinite set has a denumerable subset.
Let us take S = N and apply Theorem A.8 recursively. We obtain an infinite sequence
of transfinite cardinals, namely
ℵ0 < c < |P(P(N))| < |P(P(P(N)))| < · · · .
What is the first cardinal greater than ℵ0 ? In other words, is it possible to find a
transfinite cardinal number , say λ , such that ℵ0 < λ < c? This is equivalent to ask
if there exists any set having size between N and R. Such a set has not yet been
discovered and the Cantor was the first to conjecture that such set does not exist.
This conjecture is known under the name Continuum Hypothesis. This conjecture
has been included in 1900 as the first of Hilbert’s 23 unsolved problems and till
now nobody knows if this conjecture is true or not. There is the possibility that this
question is unanswerable since
1. the assumption of the Continuum Hypthesis does not contradict any of the usual
axioms of set theory (Gödel, 1938).
2. It has been shown that the denial of the Continuum Hypothesis does not lead to
any contradiction (Cohen, 1963).
In other words, the Continuum Hypothesis is undecidable, i.e. it can’t be proved nor
disproved, on the basis of the currently accepted axioms for set theory.
Appendix B
Quick review of real analysis
We give a short review of those concepts of real analysis the reader should be familiar with. The exposition follows mainly [4].
B.1 Real numbers
Definition B.1. Let 0/ 6= S ⊂ R. An upper bound of S is a number L ∈ R such that
x ≤ L for every x ∈ S. A lower bound of S is a number ℓ ∈ R such that x ≥ ℓ for
every x ∈ S. If the first condition is satisfied we say that S is bounded from above.
We say that S is bounded from below if the second condition applies.
Upper and lower bounds of a set S do not need to belong to S. Further, a set can have
many upper and/or lower bounds.
Definition B.2. Let 0/ 6= S ⊂ R and S bounded from above. The least upper bound
of S is a number u ∈ R such that u is an upper bound of S and L ≥ u for every upper
bound L of S. If S is bounded from below, the greatest lower bound of S is a number
m ∈ R such that m is a lower bound of S and ℓ ≤ m for every lower bound ℓ of S.
The least upper bound and the greatest lower bound of a set S do not need to belong
to S. Further, they are unique when they exist. By sup (S) we denote the supremum
of S, i.e. the least upper bound of S whereas inf (S) called the infimum of S is the
greatest lower bound of S.
Theorem B.1. Any nonempty subset S of R that is bounded from above has a least
upper bound (similarly for greatest lower bound).
Finally, for every x, y ∈ R
|x + y| ≤ |x| + |y|,
|x − y| ≥ ||x| − |y||.
83
84
B Quick review of real analysis
B.2 Real sequences
Definition B.3. An infinite sequence of real numbers is a map s : N −→ R such that
s(n) = sn for every n ∈ N.
You can think to a sequence as of an infinite string (s1 , s2 , · · ·). This in turn motivates
the notation s = (sn )n∈N . A sequence s is infinite but its image set s(N) does not need
to be infinite.
Definition B.4. A real sequence s = (sn )n∈N converges to ℓ ∈ R if for every ε > 0
there exists an N ∈ N such that |sn − ℓ| < ε whenever n > N.
In general, sequences might or might not converge. However, if a sequence converges its limit is unique. For simple sequences we can often guess the limit and
prove convergence by applying Def. B.4. There are situations when it may be hard
or impossible to guess the limit. The next two theorems combined with the concepts
of bounded monotone sequence and Cauchy sequence allow to prove convergence
without using a known value of the limit.
Definition B.5. A real sequence (sn )n∈N is monotone increasing (decreasing) if
sn+1 ≥ sn (sn+1 ≤ sn ) for every n ∈ N. It is monotone if it has either of these properties.
Theorem B.2. Every bounded monotone sequence converges.
Definition B.6. A real sequence (sn )n∈N is a Cauchy sequence if fore every ε > 0
there exists an N ∈ N such that |sn − sm | < ε for every n, m > N.
The following result is also known as the Cauchy’s convergence criterion.
Theorem B.3. A sequence (sn )n∈N of real numbers converges if and only if it is a
Cauchy sequence.
B.3 Limits of functions
Definition B.7. Let f : D ⊆ R −→ R. We say that f tends to ℓ ∈ R as x tends to
a ∈ R and we write
lim f (x) = ℓ
x→a
if for every ε > 0 there exists a δ > 0 such that for all x ∈ D, | f (x) − ℓ| < ε whenever
0 < |x − a| < δ .
In general, f (a) 6= ℓ. Moreover, the point a does not need to belong to the domain
of definition of f . What is important is that D contains numbers x arbitrarily close
to a so that the distance |x − a| can be made arbitrarily small. The next result is
fundamental to give a sequential characterization of continuity.
References
85
References
1. Gödel, K.: The Consistency of the Continuum-Hypothesis, Princeton University Press (1940)
2. Cohen, P. J.:The Independence of the Continuum Hypothesis, Proceedings of the National
Academy of Sciences of the United States of America 50 1143-48 (1963)
3. Lay, S.: Analysis with an Introduction to Proof, Prentice Hall (2004)
4. Sutherland, W. A.: Introduction to metrical and topological spaces, Clarendon Press Oxford
(1986)
86
B Quick review of real analysis
References
1. Hijab, O.: Introduction to Calculus and Classical Analysis, Springer Verlag, New York, Heidelberg (1997)
2. Sutherland, W. A.: INTRODUCTION TO METRIC AND TOPOLOGICAL SPACES,
Clarendon Press, Oxford (1986)