Download Introduction to basic topology and metric spaces

Introduction to basic topology and metric spaces
(with a bit a measure theory)
AKA Introduction to Functional Analysis
Marco Di Francesco
1
A bit of set theory
Although set theory is supposed to be a prerequisite to this course, reviewing here some basic concepts in set
theory will be useful. Moreover, we shall introduce here a basic concept of mathematics such as the Axiom of
Choice and its consequences.
Let A be a set. Let A 3 α 7→ Eα be a map which sends an element α ∈ A into another set Eα . The image
of
that
map is called a family of sets {Eα }α∈A . The set A is called the set of indexes. We use the notation
S
E
α∈A α to denote the set
Tcontaining all elements belonging to at least one of the sets Eα for some α ∈ A
(union), and the notation α∈A Eα to denote the set of all elements belonging to every set Eα for all α ∈ A
(intersection). The notation A ⊆ B between two sets A and B (A subset of B) means that every element in
A belongs to B. The notation A ⊂ B is totally equivalent. The former is used to emphasize the fact that A
may be the whole B. If one wants to point out that A is a subset of B but A cannot be the whole B, one uses
A B. For a given set A, the set of all subsets of A, also called the power set of A, is denoted by P(A).
The complement of a set A in some environment set (a priori known as a global set in which a certain theory
is developed, e. g. the Euclidean space Rd ) is denoted by Ac . The notation A \ B (set difference) stands for
A ∩ (B c ). The notation A∆B (symmetric difference) stands for A∆B = (A \ B) ∪ (B \ A). Given two sets A
and B, we define the Cartesian product A × B as the set of ordered pairs (x, y), with x ∈ A and y ∈ B.
Exercise 1.1. The following are satisfied:
S
S
S
•
E
∩
F
α∈A α
β∈B β =
(α,β)∈A×B (Eα ∩ Fβ ).
•
T
•
S
•
T
T
T
E
∪
F
= (α,β)∈A×B (Eα ∪ Fβ ).
α
β
α∈A
β∈B
α∈A
S
S
Eα ×
β∈B Fβ =
(α,β)∈A×B Eα × Fβ .
T
T
E
×
F
α∈A α
β∈B β =
(α,β)∈A×B Eα × Fβ .
A subset of the cartesian product R ⊆ A × B is said a relation within A and B. A relation R within A
and B is called a function (or a map, or a mapping) if for every x ∈ A there exists a unique y ∈ B such that
(x, y) ∈ R. A function is usually denoted with lower case letters, say f , and the unique element y above is called
f (x). Given a function f , the graph of f is the corresponding relation, with the notation above Gr(f ) = R.
The set {x ∈ A : (x, f (x)) ∈ Gr(f )} is called the domain of f . Let C ⊂ B. The inverse image (or pre-image)
of C via the function f is the set f −1 (C) = {x ∈ A : f (x) ∈ C}.
Exercise 1.2. The following are satisfied:
S
S
−1
• f −1
(Eα ).
α∈A Eα =
α∈A f
T
T
−1
• f −1
(Eα ).
α∈A Eα =
α∈A f
• f −1 (E \ F ) = f −1 (E) \ f −1 (F ).
S
S
• f
α∈A Eα =
α∈A f (Eα ).
T
T
• f
α∈A Eα ⊆
α∈A f (Eα ).
For an arbitrary family of indexes A, one can define infinite Cartesian products
(
)
[
Πα∈A Eα = c : A →
Eα : c(α) ∈ Eα , for all α ∈ A .
α∈A
1
We often write c(α) as cα , and an element c ∈ Πα∈A Eα as c = (cα )α∈A .
Let A be a set. A relation r ∈ A × A is called an equivalence in A if it satisfies the following properties:
• r is reflexive, i. e. (a, a) ∈ r for all a ∈ A.
• r is symmetric, i. e. if (a, b) ∈ r, then (b, a) ∈ r.
• r is transitive, i. e. if (a, b) ∈ r and (b, c) ∈ r, then (a, c) ∈ r.
An equivalence is sometimes denoted with the symbol ∼, i. e. (x, y) ∈ r if and only if x ∼ y. Given a set A
and an equivalence ∼ on that set, for a given a ∈ A we denote by [x]∼ = {x ∈ A : a ∼ x} ⊂ A the equivalence
class containing a. The set of equivalent classes is called quotient set of the equivalence ∼ in A, and is denoted
by A/ ∼.
Example 1.1. Examples of equivalences are (check!)
• Identity between real numbers, i. e. x ∼ y if and only if x = y, x, y ∈ R.
• On R, we set x ∈ y if and only if x − y ∈ Z.
• On R, we set x ∈ y if and only if x − y ∈ Q.
2
A crash course on basic topology
Topology, sometimes referred to as ‘the mathematics of continuity’, or ‘the theory of abstract topological spaces’,
is all of these, but, above all, it is a language, used by mathematicians in practically all branches of science. In
this chapter, we will learn the basic words and expressions of this language as well as its ‘grammar’, i.e. the
most general notions, methods and basic results of topology.
2.1
Topological spaces
The notion of topological space is defined by means of rather simple and abstract axioms. It is very useful as
an ‘umbrella’ concept which allows to use the geometric language and the geometric way of thinking in a broad
variety of vastly different situations. Because of the simplicity and elasticity of this notion, very little can be
said about topological spaces in full generality. And so, as we go along, we will impose additional restrictions on
topological spaces, which will enable us to obtain meaningful but still quite general assertions, useful in many
different situations in the most varied parts of mathematics.
Very often, e. g. when introducing a notion of continuity for a given function between two sets, we use
notions as open interval, or open sphere, to denote ‘small’ sets surrounding given points, in which our function
should pick elements (or to which it should send element), in order to satisfy a certain definition. Notions such
as ‘interval’ and ‘sphere’ obviously rely on a concept of distance, often the so-called Euclidean distance.
Imagine we are now in a ‘space’ in which a notion of distance is not available, but still we would like to have
notions such as continuity (or compactness!). After all, what we need in order to set up a notion of continuous
function is a list of all the neighborhoods of each point in both sets. This is, roughly speaking, what a topological
space is: it is just a set in which we have decided that some of its subset are upgraded to the level of open sets.
2.1.1
Basic definitions and first examples
Definition 2.1. A topological space is a pair (X, τ ), where X is a set and a τ ⊂ P(X) is a family of subsets of
X called the topology of X, whose elements are called open sets, such that
(i) ∅, X ∈ τ (the empty set and the whole set are open sets).
(ii) If {Oα }α∈A ⊂ τ is an arbitrary family of open sets, then
of open sets is open.
S
α∈A
Oα ∈ τ (the union of an arbitrary family
(iii) If {Oj }N
j=1 ⊂ τ , then O1 ∩ . . . ∩ ON ∈ τ (the intersection of finite number of open sets is open).
If x ∈ X, then an open set containing x is called an (open) neighbor of x. The complements of open sets are
called closed sets.
We will often omit the topology τ , and refer to X as a topological space assuming that the topology has
been described.
2
Example 2.1. The Euclidean space Rd acquires the structure of a topological space if its open sets are defined
as in the calculus or elementary real analysis course (i.e a set A ⊂ Rd is open if for every point x ∈ A a certain
open ball centered in x is contained in A, see Definition 6.4 of the preliminary material and the basic properties
of open sets in Rd in the same section).
Example 2.2. If in the set of real numbers R we declare open (besides the empty set and R) all the half-lines
{x ∈ R : x ≥ a}, a ∈ R, then we do not obtain a topological space: the first and third axiom of topological
spaces hold, but the second one does not (e.g. for the collection of all half lines with positive endpoints).
Example 2.3. Let X be a set. The discrete topology on X is τ = P(X). Check that τ is a topology. The
indiscrete topology, or trivial topology on X is τ = {∅, X}. Check that the trivial topology is also a topology.
Exercise 2.1.T Let {Cα }α∈A be an arbitrary family of closed sets in a topological space X. Prove that the
intersection α∈A Cα is still a closed set (Hint: use the definition of closed set and the axioms of a topology).
We now start introducing concepts that are already known in the topology of the Euclidean space Rd , but
we need to set them in the general context of a topological space.
Definition 2.2. The closure A of a set A ⊂ X is the smallest closed set containing A, that is
\
A=
{C : A ⊂ C , C closed set} .
A set A ⊂ X is called dense (or everywhere dense) in X if A = X. A set A ⊂ X is called nowhere dense
if X \ A is dense. A point x ∈ X is called an accumulation point (or a limit point) of a set A ⊂ X if every
neighborhood of x contains infinitely many points of A. A point x ∈ A is called an interior point of A if there
exists a neighborhood of x entirely contained in A. The set of all interior points of A is called the interior of
◦
A, and is denoted by A.
◦
Exercise 2.2. Prove that a set A is open if and only if A = A.
Definition 2.3. A topological space X is said to be separable if there is a countable set S ⊂ X such that
S = X.
Exercise 2.3. Show that R with the usual Euclidean topology is a separable space. Show that R equipped with
the discrete topology (every set is open) is not separable.
Definition 2.4. A point x ∈ X is called a boundary point of a set A ⊂ X if it is neither an interior point of A
nor it is an interior point of Ac = X \ A. The set of boundary points of A is called the boundary of A and is
denoted by ∂A.
Exercise 2.4. For very set A ⊂ X, prove that A = A ∪ ∂A. Consequently, prove that a set C ⊂ X is closed if
and only if C contains its boundary.
◦
Exercise 2.5. Prove that for any set A in a topological space we have ∂A ⊂ ∂A and ∂(A) ⊂ ∂A. Give an
example when all these three sets are different.
Definition 2.5. A sequence {xn }n∈N ⊂ X is said to converge to x ∈ X if for every open O containing x there
exists N ∈ N such that {xn }n≥N ⊂ O. Any such point x is said a limit for the sequence {xn }n .
Example 2.4. In the case of the Euclidean space Rd with the standard topology, the above definitions (of
neighborhood, closure, interior, convergence, accumulation point) coincide with the ones familiar from the
calculus or elementary real analysis course.
Exercise 2.6.
• Let X = R with the discrete topology (all sets are open). Prove that any subset S ⊂ R has neither
accumulation nor boundary points, prove that the closure (as well as the interior) of every set S is the S
itself, prove that the sequence 1/n does not converge to 0.
• Let X = R with the trivial topology (only the empty set and R are open). Prove that every sequence in
R is convergent to any arbitrary point x ∈ R.
The latter example above shows in particular that limits may be not unique in a general topological space.
We will see later on how to achieve the uniqueness of the limit of a sequence by requiring some little more
properties for X.
3
2.1.2
Base of a topology
In practice, it may be awkward to list all the open sets constituting a topology; fortunately, one can often define
the topology by describing a much smaller collection, which in a sense generates the entire topology.
Definition 2.6. A base for a topology τ is a subfamily β ⊂ τ such that every element of τ can be written
as union of elements of β. A topological space with a countable base is called second countable. A base of
neighborhoods of a point x ∈ X is a collection B of open neighborhoods of x such that any neighborhood of x
contains an element of B. If any point of a topological space has a countable base of neighborhoods, then the
space (or the topology) is called first countable.
Example 2.5. The Euclidean space Rd with the standard topology (the usual open and closed sets) has bases
consisting of all open balls, open balls of rational radius, open balls of rational center and radius. The latter is
a countable base.
Example 2.6. The real line (or any uncountable set) in the discrete topology (all sets are open) is an example
of a first countable but not second countable topological space.
Proposition 2.1. Every topological space with a countable base is separable.
Proof. Let {Ok }k∈N ⊂ τ be a countable base. For all k ∈ N choose xk ∈ Ok . The set {xk }k is at most countable.
X.
We claim that such set is also dense. Assume by contradiction that this is not true. This means {xk }k
Since X \ {xk }k is an open set, then there exists an element of the base {Ok }k such that Ok ⊆ O. In particular,
xk 6∈ {xk }k , a contradiction.
Proposition 2.2. A sequence {xn }n∈N converges to x if and only if for all basic open neighborhood N of x
there exists N ∈ N such that {xn }n≥N ⊂ N .
Proof. One implication is trivial. Assume the latter property is satisfied. We want to prove that {xn }n∈N
converges to x. Let O be an open set containing x. By definition there exists a basic neighborhood N ⊂ O
which contains x, and the assumption implies {xn }n≥M ⊂ N ⊂ O for a suitable integer M .
2.1.3
Comparison of topologies
Definition 2.7. On a given set X, a topology σ is said to be stronger (or finer ) than another topology τ if
τ ⊂ σ, and weaker (or coarser ) if σ ⊂ τ .
Clearly, the discrete topology on a given set X is finer than any other topology on X, whereas the trivial
topology is coarser than any other topology. We now address the problem of constructing a topology containing
a given family of subsets C ⊂ P(X) with the least effort, i. e. we want to construct the coarsest topology
containing C.
Proposition 2.3. For a given family C ⊂ P(X), there exists a unique coarsest (or weakest) topology such that
all elements of C are open. We call such topology τ (C), and we refer to it as the topology generated by C.
Proof. Consider the family I consisting of all the finite intersections of elements of C, i. e.


N


[
I= I=
Cj : C1 , . . . , CN ∈ C .


j=1
Now, let τ be the family of arbitrary unions of elements of I, the empty set, and the whole set X. Clearly, τ
is closed upon arbitrary unions, since an arbitrary union of elements of τ is by definition an arbitrary union of
elements of I, and is therefore an element of τ . Moreover, take the intersection of finitely many elements in
τ . By distributing properties of union and intersections, such set is the arbitrary union of the intersection of
finitely many elements of I, and therefore an element of τ . It remains to prove that such topology is finer than
any other topology containing C. Let τ 0 be a topology containing C. From the axioms of a topology, τ 0 contains
all arbitrary unions and finite intersections of element of C. By definition, those sets belong to τ . Clearly, τ is
uniquely defined.
We say that a family C is a subbase for a topology τ if τ = τ (C), or equivalently C is a subbase for τ if C
generates τ as the coarsest topology containing C.
Exercise 2.7. Prove that any weaker topology than a separable topology is also separable.
2.2
Continuity
In this section, we study, in the language of topology, the fundamental notion of continuity.
4
2.2.1
Continuous maps
The topological definition of continuity is simpler and more natural than the , δ definition familiar from the
elementary real analysis course.
Definition 2.8. Let (X, τ ), (Y, σ) be topological spaces. A map f : X → Y is said to be continuous if O ∈ σ
implies f −1 (O) ∈ τ (pre-images of open sets are open). f is an open map if O ∈ τ implies f (O) ∈ σ (images of
open sets are open). f is continuous at a point x ∈ X if for any neighborhood A of f (x) in Y , the pre-image
f −1 (A) contains a neighborhood of x in X.
Exercise 2.8. Prove that a function f is continuous if and only if it is continuous are every point.
In the previous subsection we showed how to generate a topology from an arbitrary family of sets. Here we
shall construct a topology with the least effort in order to make a family of functions continuous.
Proposition 2.4. Let (Y, σ) be a topological space. For any collection F of maps from a set X (without a
topology) to Y there exists a unique weakest topology on X which makes all maps from F continuous. This is
exactly the weakest topology with respect to which pre-images of all open sets in Y under the maps from F are
open. Such topology is called the inverse limit topology of the family F. If F consists of a single map f , this
topology is sometimes called the pullback topology on X under the map f .
Proof. The topology required is constructed as follows. Let C = {f −1 (O) : f ∈ F , O ∈ σ}. We set τ = τ (C), i.
e. τ is the topology generated by the pre-images of all open sets in Y via the maps from F. By definition, such
a topology makes all the maps f ∈ F continuous (check!). Moreover, it is easily checked that every topology
that makes all the maps f ∈ F continuous is finer than τ .
Exercise 2.9. Let p be the orthogonal projection of R2 on one of its sides. Describe the pullback topology of p
on R2 . Will an open (in the usual sense) disk inside K be an open set in this topology?
Exercise 2.10. Let X and Y be two topological spaces with bases B for X and C for Y . Prove that a function
f : X → Y is continuous in x ∈ X if and only if for all basic open neighborhood V of f (x) in Y , there exists a
basic open neighborhood U of x in X such that f (U ) ⊂ V .
2.2.2
Topological equivalence
Definition 2.9. A map f : X → Y between topological spaces is a homeomorphism if it is continuous and
bijective with continuous inverse. If there is a homeomorphism X → Y , then X and Y are said to be
homeomorphic or sometimes topologically equivalent.
A property of a topological space that is the same for any two homeomorphic spaces is said to be a topological
invariant. The relation of being homeomorphic is obviously an equivalence relation (in the technical sense: it
is reflexive, symmetric, and transitive. Thus topological spaces split into equivalence classes, sometimes called
homeomorphy classes.
2.3
2.3.1
Basic constructions
Induced topology
Let (X, τ ) be a topological space. If Y ⊂ X, then Y can be made a topological space in a natural way by taking
the induced topology
τY := {O ∩ Y : O ∈ τ }.
Exercise 2.11. Prove that τY defined above is a topology.
Example 2.7. The topology induced from Rd+1 on the subset
{(x1 , . . . , xd+1 ) :
d+1
X
x2i = 1}
i=1
produces the (standard, or unit) nsphere Sd . For d = 1 it is called the (unit) circle and is sometimes also
denoted by T.
5
2.3.2
Product topology
If (Xα , τα ), α ∈ A are topological spaces and A is any set of indexes, then the product topology on X = Πα∈A Xα
is the topology determined by the base
{Πα∈A Oα : Oα ∈ τα , Oα 6= Xα for only finitely many α} .
Example 2.8. The standard topology in Rd coincides with the product topology on the product of d copies of
the real line R.
2.3.3
Quotient topology
Consider a topological space (X, τ ) and suppose there is an equivalence relation ∼ defined on X. Let π be the
natural projection of X on the set X̂ of equivalence classes, i. e. π : X → X̂ and π(x) = [x]. The identification
space or quotient space X/ ∼:= (X̂, σ) is the topological space obtained by calling a set O ⊂ X̂ open if π −1 (O)
is open, that is, taking on X̂ the finest topology for which π is continuous.
Example 2.9. Consider the closed unit interval and the equivalence relation which identifies the endpoints. Other
equivalence classes are single points in the interior. The corresponding quotient space is another representation
of the circle.
Example 2.10. Consider the following equivalence relation in punctured Euclidean space Rd+1 \ {0}:
(x1 , . . . , xd+1 ) ∼ (y1 , . . . , yd+1 )
⇔
yi = λxi for all i = 1, . . . , d + 1
with the same number λ. The corresponding identification space is called the real projective d-space and is
denoted by RP (d).
Exercise 2.12. Describe a homeomorphism between the sphere Sd (Exercise 2.7) and the quotient space of
Example 2.10 in which we require λ > 0.
2.4
Separation properties
Separation properties provide one of the approaches to measuring how fine is a given topology.
2.4.1
T1, Hausdorff, and normal spaces
Here we list, in decreasing order of generality, the most common separation axioms of topological spaces.
Definition 2.10. A topological space (X, τ ) is said to be
• (T1) if any point is a closed set, or equivalently if for any points x1 , x2 ∈ X, there exists a neighborhood
of x1 which doesn’t contain x2 .
• (T2), or Hausdorff space, if any two distinct points possess nonintersecting neighborhoods.
• (T4), or normal space, if it is Hausdorff and any two closed disjoint subsets possess nonintersecting
neighborhoods.
It follows immediately from the definition of induced topology that any of the above separation properties
is inherited by the induced topology on any subset.
Exercise 2.13. Prove that every Hausdorff space is (T1).
Exercise 2.14. Prove that in a (T2) space any sequence has no more than one limit. Show that without the
(T2) condition this is no longer true.
Exercise 2.15. Prove that the product of two (T1) (respectively Hausdorff) spaces is a (T1) (resp. Hausdorff)
space.
2.5
Compactness
The fundamental notion of compactness, familiar from the elementary real analysis course for subsets of the
real line R or of Euclidean space Rd , is defined below in the most general topological situation.
6
2.5.1
Types of compactness
S
A family of open sets {Oα }α∈A ⊂ τ is called an open cover of a topological space X if X = α∈A Oα , and is
a finite open cover if A is finite. If for a given subset B ⊂ A the family {Oα }α∈B is an open over of X, then
{Oα }α∈B is called a subcover of {Oα }α∈A
Definition 2.11. The space (X, τ ) is called
• compact if every open cover of X has a finite subcover,
• sequentially compact if every sequence in X has a convergent subsequence,
• σ-compact if it is the union of countably many compact sets,
• locally compact if every point has an open neighborhood whose closure is compact in the induced topology.
It is known from elementary real analysis that for subsets of a Rd compactness and sequential compactness
are equivalent, see Theorem 6.5 of the introductory material.
Proposition 2.5. Any closed subset of a compact set is compact.
Proof. If K is compact, C ⊂ K is closed, and Γ is an open cover for C, then Γ0 := Γ ∪ {K \ C} is an open cover
for K. Hence, Γ0 contains a finite subcover Γ0 ∪ {K \ C}. Therefore, Γ0 is a finite subcover of Γ for C.
Proposition 2.6. Any compact subset of a Hausdorff space is closed.
Proof. Let X be a Hausdorff space and let C ⊂ X be compact. Fix a point x ∈ X \ C and for each
S y ∈ C (due
to the (T2) assumption) take two neighborhoods Uy 3 y and Vy 3 x such that Uy ∩ Vy = ∅. Then y∈C Uy ⊃ C
TN
is an open cover of C, and it therefore has a finite subcover {Uyi }N
i=1 . Hence Nx :=
i=1 Vyi is a neighborhood
of x disjoint from C. Hence, we have constructed an open neighborhood of x for an arbitrary x ∈ X \ C, which
shows that X \ C is open, and therefore C is closed.
Exercise 2.16. Prove that any compact Hausdorff space is (T4).
Proposition 2.7. Let (X, τ ), (Y, σ) be topological spaces. Let X be compact. Let f : X → Y be a continuous
map. Then f (X) is compact in Y .
Proof. Let {Eα }α∈A be a cover of f (X). All the sets f −1 (Eα ) are open sets because f is continuous. Hence,
{f −1 (Eα )}α∈A is an open cover of X (let x ∈ X, f (x) ∈ f (X), then f (x) ∈ Eα for some α ∈ A, which means
x ∈ f −1 (Eα )). As X is compact, there exists α1 , . . . , αm such that X ⊂ f −1 (Eα1 ) ∪ . . . ∪ f −1 (Eαm ). This
means that Eα1 , . . . , Eαm covers f (X). Indeed, let y ∈ f (X), then there exists x ∈ X with f (x) = y. Let
i0 ∈ {1, . . . , m} with x ∈ f −1 (Ei0 ). This means that y = f (x) ∈ Ei0 . Hence, f (X) is compact.
Corollary 2.8. Let (X, τ ) be a compact topological space, and let f : X → R be a continuous function. Then
f achieves maximum and minimum on X.
Proof. From Proposition 2.7, f (X) is compact, and thus closed and bounded from Theorem 6.4 of the introductory material. Therefore, f (X) has maximum and minimum.
To verify the compactness property, it suffices to do so for basic open covers (i.e. coverings of the whole
space by basic open sets).
Proposition 2.9. Let (X, τ ) be a topological space with a base B. Then the following are equivalent:
• Every open cover has a finite subcover (i.e. X is compact);
• Every basic open cover has a finite subcover.
Proof. The first statement trivially implies the second one. Assume the second statement, and let {Oα }α∈A be
an open cover for X. Since every open setScan be written as union of basic open sets, there exists a family of
basic open sets {Bβ }β∈C such that X ⊂ β∈C Bβ . Moreover, each Oα can be written as the union of some
Bβ ranging in some set of indexes depending of α, β ∈ Cα . Finally, the basic open sets can be chosen such
that every basic open set is contained in some Oα . Due to the assumptions, there exist β1 , . . . , βm ∈ C such
that X ⊂ Bβ1 ∪ . . . ∪ Bβm . By construction of the family Bβ , each Bβj is contained in some Oαj , and hence
Oα1 , . . . , Oαm is a finite subcover of {Oα }α∈A .
7
2.5.2
Alexander’s sub-base theorem
We recall the definition of sub-base.
Definition 2.12. Let (X, τ ) be a topological space. A sub-base for this space is a collection B of subsets of X
such that τ is the weakest topology that makes all the sets B open (i.e. τ is generated by B). Elements of B
are referred to as sub-basic open sets.
Observe for instance that every base is a sub-base. The converse is not true: for instance, the half-open
intervals (−∞, a), (a, +∞) for a ∈ R form a sub-base for the standard topology on R, but not a base. In
contrast to bases, no property is required on a collection B in order for it to be a sub-base; every collection of
sets generates a unique topology with respect to which it is a sub-base.
The precise relationship between sub-bases and bases is given by the following exercise.
Exercise 2.17. Let (X, τ ) be a topological space, and let B be a collection of subsets of X. Then the following
are equivalent:
(i) B is a sub-base for (X, τ ).
(ii) The space B ∗ := {B1 ∩ . . . ∩ Bk : B1 , . . . , Bk ∈ B} of finite intersections of B (including the whole space
X, which corresponds to the case k = 0) is a base for (X, τ ).
Many topological facts involving open sets can often be reduced to verifications on basic or sub-basic open
sets, as the following exercise illustrates.
Exercise 2.18. Let (X, τ ) be a topological space, and B be a sub-base of X, and let B ∗ be a base of X.
• Show that a sequence xn ∈ X converges to a limit x ∈ X if and only if every sub-basic open neighbourhood
of x contains xn for all sufficiently large n.
• Show that a point x ∈ X is adherent to a set E if and only if every basic open neighbourhood of x
intersects E. Give an example to show that the claim fails for sub-basic open sets.
• Show that a point x ∈ X is in the interior of a set U if and only if U contains a basic open neighbourhood
of x. Give an example to show that the claim fails for sub-basic open sets.
• If Y is another topological space, show that a map f : Y → X is continuous if and only if the inverse
image of every sub-basic open set is open.
There is a useful strengthening of Proposition 2.9 in the spirit of the above exercise, namely the Alexander
sub-base theorem:
Theorem 2.10 (Alexander sub-base theorem). Let (X, τ ) be a topological space with a sub-base B. Then the
following are equivalent:
(i) Every open cover has a finite subcover (i.e. X is compact);
(ii) Every sub-basic open cover has a finite subcover.
Proof. The proof will be omitted.
2.5.3
Compactness under products, maps, and bijections
Exercise 2.19. Let {(Xα , τα )}α∈A be an infinite family of topological spaces. Prove that the product topology
on Πα∈A Xα defined in subsection 2.3.2 is generated by the subbase
{πα−1 (Oα ) : Oα ∈ τα , α ∈ A}.
The following result has numerous applications in analysis, PDE, and other mathematical disciplines.
Theorem 2.11 (Tychonoff1 ). The product of any family of compact spaces is compact.
1 The theorem is named after Andrey Nikolayevich Tychonoff, who proved it first in 1930 for powers of the closed unit interval
and in 1935 stated the full theorem along with the remark that its proof was the same as for the special case. The earliest known
published proof is contained in a 1937 paper of Eduard Čech.
8
Proof. Let X = Πα∈A Xα , for a given (possibly infinite) family of topological spaces {(Xα , τα )}α∈A . From
Exercise 2.19 and from Theorem 2.10, it suffices to show that every sub-basic open cover of X has a finite
subcover, where the subbase B is the one in Exercise 2.19. Let {πα−1
(Uβ )}β∈B be a sub-basic cover of X, here
β
B is a set of indexes, αβ ∈ A for all β ∈ B, and Uβ ∈ ταβ for all β ∈ B.
For each α ∈ A, consider the sub-basic open sets πα−1 (Uβ ) that are associated to those β ∈ B with αβ = α. If
the open sets Uβ here cover Xα , then by compactness of Xα , a finite number of the Uβ already suffices to cover
Xα , and so a finite number of the πα−1 (Uβ ) cover X, and the proof would be completed. So we may assume
that the Uβ do not cover Xα , thus for each α ∈ A there exists xα ∈ Xα that avoids all the Uβ with αβ = α (if
there are some). One then sees that the point (xα )α∈A in X avoids all of the πα−1 (Uβ ), a contradiction. The
claim follows.
3
Metric spaces and normed spaces
We recall the concept of vector space, which is supposed to be known from basic linear algebra.
Definition 3.1. A real (complex resp.) vector space (or linear space) is a nonempty set E equipped with two
operations:
(A) Sum of two vectors: E × E 3 (u, v) → u + v ∈ E,
(B) Multiplication by a scalar real (complex resp.) number: R × E 3 (λ, u) → λu ∈ E
such that
1. for all u, v ∈ E, u + v = v + u,
2. for all u, v, w ∈ E, (u + v) + w = u + (v + w),
3. there exists an identity element (also called zero) denoted by 0 ∈ E such that u + 0 = u for all u ∈ E,
4. for all u ∈ E there exists a unique element v ∈ E such that u + v = 0. Such element v is denoted by −u,
5. for all u ∈ E and α, β ∈ R (α, β ∈ C resp.), α(βu) = (αβ)u,
6. 1 · u = u for all u ∈ E,
7. for all α ∈ R (α ∈ C resp.) and u, v ∈ E, α(u + v) = αu + αv,
8. for all α, β ∈ R (α, β ∈ C resp.) and u ∈ E, (α + β)u = αu + βu.
A subset L ⊆ E is called a linear subspace of E if
• for all u, v ∈ L, u + v ∈ L,
• for all α ∈ R (α ∈ C resp.) and u ∈ L, αu ∈ L.
Notation 3.1. Often in this course we shall state results which holds on vector space constructed on either R or
C. For simplicity, we shall use the notation K for a field being either R or C.
3.1
Basic definitions
Definition 3.2. Let X be a set. A distance (or metric) d on X is a function
d : X × X → R,
such that for all x, y, z ∈ X one has
(a) d(x, y) ≥ 0, and d(x, y) = 0 if and only if x = y.
(b) d(x, y) = d(y, x).
(c) d(x, y) ≤ d(x, z) + d(z, y) (triangular inequality).
The pair (X, d) is called a metric space. The elements of a metric space are called points.
Definition 3.3. Let X be a vector space on K. A norm on X is a function k · k, X 3 x → kxk ∈ R such that,
for al x, y ∈ X and for all λ ∈ K,
1. kxk ≥ 0, and kxk = 0 if and only if x = 0,
9
2. kλxk = |λ|kxk,
3. kx + yk ≤ kxk + kyk (triangular inequality).
The number kxk is called the norm of x, and the pair (X, k · k) is called a normed space.
Clearly, metric spaces and normed spaces do have something in common. The first thing that catches our
attention is that a normed space structure needs a vector space structure underneath, whereas a metric space
structure can, in principle, be defined on an set which is not a vector space. On a vector space, every norm
induces a metric, as seen in the next remark.
Remark 3.1. Let (X, k · k) be a normed space. For all x, y ∈ X set d(x, y) = kx − yk. Then, d is a metric
(check!). d is called the metric induced by the norm k · k.
Exercise 3.1. Let X = Rd . The Euclidean norm k · k is a norm. The induced distance is the usual Euclidean
distance.
The main properties of the Euclidean norm are recalled in Subsection 6.1 of the introductory material. The
statement in remark 3.1 shows in particular that every normed space is a metric space. Even if we assume we
are working on a vector space, the opposite is not true, i. e. not all distances can be induced by a norm.
Example 3.1. On a vector space E on K, for x, y ∈ E let
(
0 if x = y
d(x, y) =
.
1 if x 6= y
We claim that d cannot induce a norm. By contradiction, assume there exists a norm k · k on E such that
kx − yk = d(x, y). Now, let x ∈ E \ {0} (0 being the identity element in E), let λ ∈ K \ {0}. Then 1 = d(λx, 0) =
kλxk = |λ|kxk = |λ|d(x, 0) = |λ|, and this is a contradiction unless |λ| = 1.
Let us now start introducing an example which somehow introduces to the core of our course, namely working
with spaces of functions.
Example 3.2. Let A ⊂ R be an open set. Let Cb (A) be the set of all continuous and bounded functions on A.
First of all, Cb (A) is a real vector space with the two operations
• f, g ∈ Cb (A), f + g ∈ Cb (A) defined by (f + g)(x) = f (x) + g(x),
• f ∈ Cb (A), λ ∈ R, λf ∈ Cn (A) defined by (λf )(x) = λf (x).
The identity element of Cb (A) is the zero function f ≡ 0. Now, let us set for any f ∈ Cb (A)
kf k∞ = sup |f (x)|.
x∈A
One can easily prove that k·k∞ is a norm on Cb (A). Clearly kf k∞ ≥ 0 for all f ∈ Cb (A). Moreover, if kf k∞ = 0
then clearly |f (x)| = 0 for all x ∈ A, hence f = 0. For all λ ∈ R, kλf k∞ = supx∈A |λf (x)| = |λ| supx∈A |f (x)| =
|λ|kf k∞ . Finally, for all f, g ∈ Cb (A),
kf + gk∞ = sup |f (x) + g(x)| ≤ sup (|f (x)| + |g(x)|) ≤ sup |f (x)| + sup |g(x)| = kf k∞ + kgk∞ .
x∈A
x∈A
x∈A
x∈A
Hence, k · k∞ is a norm, and (Cb (A), k · k∞ ) is a normed space.
3.2
Topology in metric spaces
Let (X, d) be a metric space. Many of the definitions in this section will be generalisations of analogous definition
on the Euclidean space Rd , see subsection 6.1 of the introductory material. We begin with the definition of a
ball in a metric space.
Definition 3.4. Let x0 ∈ X and r > 0. The open ball (or sphere, or spherical neighborhood) with center x0
and radius r is the set
Br (x0 ) := {y ∈ X : d(x, y) < r}.
We are now ready to define the topology of a metric space.
Definition 3.5 (Open sets in a metric space). Let A ⊂ X. A point x0 ∈ A is said to be an interior point of A
if there exists a ball with center x0 entirely contained in A. The set of interior points of A is called the interior
◦
◦
of A and is denoted by A. A is an open set of X if A = A.
10
Remark 3.2. We could equivalently define the topology τ on a metric space (X, d) as the one with base B made
up by all the open balls of X. This procedure allows to define a topology automatically from a metric, and is
therefore called the topology induced by the metric. Sometimes the opposite procedures holds. More precisely,
assume (X, τ ) is a given topological space. Assume that all open sets O ∈ τ can be characterised as those
sets O for which every element x ∈ O has an open ball contained in O, where open balls are defined on the
basis of a given metric d as in definition 3.4. Then we say that the topological space (X, τ ) is metriced by
the distance d. Life for a mathematician would be much easier if all topologies would be metriced by some
distance. Unfortunately this is not the case, as we shall see during the course when introducing the notion of
weak topology.
Remark 3.3. All the concepts and definitions defined in subsection 2.1.1 apply to metric spaces. We recall them
quickly. The closed sets of X are characterised as those sets the complement
of which is open. For a given set
T
A, the closure of A is the smallest closed set containing A, i. e. A = {C : C ⊃ A, C is closed}. A set A ⊂ X
is dense in X if A = X. A set N is a neighbourhood of a point x ∈ X if N 3 x and N is open. A point x ∈ X
is an accumulation point for a set A is every neighborhood of x contains infinitely many points of A. A point
x is said in the interior of a set A if there exists an open neighborhood of x contained in A. A metric space is
separable if there is a countable set S ⊂ X such that S = X. A point x ∈ X is a boundary point of a set A ⊂ X
if it is neither an interior point of A nor an interior point of X \ A.
Similarly, we can define the fundamental concept of converging sequence in a metric space.
Definition 3.6. A sequence {xn }n∈N ⊂ X is said to converge to x ∈ X if for every open neighborhood O of x
there exists N ∈ N such that {xn }n≥N ⊂ O.
Note that in view of Proposition 2.2 the above concept of convergence is equivalent to the following one:
{xn }n∈N ⊂ X is said to converge to x ∈ X if for all r > 0 there exists N ∈ N such that {xn }n≥N ⊂ Br (x).
Exercise 3.2. Prove that every metric space is a Hausdorff space.
3.3
Continuity in metric spaces
Let (X, d) and (Y, d1 ) be metric spaces. The following definition is equivalent to the usual definition of continuity
for topological spaces in view of exercise 2.10.
Definition 3.7. Let A ⊂ X and x0 ∈ A. A function f : A → Y is said to be continuous in x0 if for all > 0
there exists δ > 0 such that f (Bδ (x)) ⊂ B (f (x)).
Exercise 3.3. For a fixed x0 ∈ X, prove that the distance function X 3 x 7→ d(x, x0 ) ∈ R is a continuous
function.
Exercise 3.4. A function f : X → Y is continuous if and only if, for all sequences xn → x in X, we have
f (xn ) → f (x).
We now introduce the following two concepts, already known for real functions, both stronger than continuity.
Definition 3.8. Let A ⊂ X. A function f : A → Y is said to be uniformly continuous in A if, for all > 0
there exists δ > 0 such that, for all x, y ∈ A with d(x, y) < δ, one has d1 (f (x), f (y)) < . f is called a Lipschitz
continuous function with constant k > 0 if
d1 (f (x), f (y)) ≤ kd(x, y)
for all x, y ∈ A.
The notions of bounded set and bounded function are also introduced.
Definition 3.9. A set A ⊆ X is called bounded if it is contained in an open ball. A function f : X → Y is
called bounded if its image f (X) is a bounded set in Y .
Now, the notion of continuity of a map between metric spaces gets even more interesting if the two metric
spaces are normed spaces. Let (E, k · kE ) and (F, k · kF ) be two normed spaces on R. We recall that E and F
are linear spaces, therefore the notion of linear operator between E and F makes sense as in any other context
in linear algebra. In particular, we say that a map T : E → F is a linear operator if it is linear as a map, i. e.
if T (x + y) = T (x) + T (y) for all x, y ∈ E and if T (λx) = λT (x) for all λ ∈ R and x ∈ E. We recall that the
same concept can be formulated on linear spaces on the field C.
Let E be a vector space over R. We recall that a functional on E is a function defined on E, or on some
subspace of E, with values in R.
The notion of linearity of a map between linear spaces expresses the compatibility of the two linear structures
via the map T . Since the spaces we are considering are normed spaces, besides their linear structure they have
a topology induced by their norm, which is compatible with the linear structure. Hence, one may ask whether
or not a linear map is e. g. continuous.
11
Exercise 3.5. Let (X, k · k) be a finite dimensional normed space on R. Let (e1 , ·, eN ) be a basis for the vector
space X. This means that for every vector x ∈ X there exists a unique set of N numbers x1 , . . . , xN in R such
that x = x1 e1 + . . . + xN eN . For all x ∈ X let us set T (x) = (x1 , . . . , xN ) ∈ RN , hence T : X → RN . Prove
that the map T is an isomorphism2 between X and RN . Moreover, prove that T is a continuous mapping with
a continuous inverse between X and RN where RN is considered equipped with the Euclidean topology (i. e.
X and RN are topologically equivalent). Hint: for the last part of the proof, use the exercise 3.4.
Exercise 3.6 (Linear maps on finite dimensional normed spaces are continuous). Let T : X1 → X2 be a linear
map between two normed spaces (X1 , k · k1 ) and (X2 , k · k2 ). Prove that T is continuous (Hint: use the canonical
isomorphism in the Exercise 3.5 for both spaces. This reduces the exercise to prove that a linear mapping
between two Euclidean spaces RN and RM is continuous!).
An interesting question arises at this stage. In Chapter 2 we have seen that several distinct topologies can
be defined in the same set, even on a finite dimensional set (even on a finite set!). Now, is this still true if we
additionally require the topology to be induced by a norm? As far as finite dimensional spaces are concerned,
the answer is provided in the following theorem.
Theorem 3.1 (All finite dimensional norms are equivalent). Let (E, k · k) be a finite dimensional normed space.
Then, any other norm on E induces the same topology as (E, k · k).
Proof. Let E be d dimensional. Then, there is a canonical isomorphism between E and Rd : let {e1 , . . . , ed } be
a basis for E, which means that every vector x ∈ E can be uniquely decomposed as x = x1 e1 + . . . + xn en . We
then construct T : E → Rd as T (x) = (x1 , . . . , xd ). From the Exercise 3.5, we know that T is a linear invertible
mapping between E and Rd .
Pd
Now, for a given x ∈ E with x = x1 e1 + . . . + xn en , let us set kxk1 = i=1 |xi |. We have, from the triangular
inequality,
d
d
d
X
X
X
kxk = k
xi ei k ≤
|xi |kei k ≤ c1
|xi | = c1 kxk1 ,
i=1
i=1
i=1
where c1 := maxi=1,...,d kei k. We now wish to prove that there exists a constant C such that kxk1 ≤ Ckxk.
Assume by contradiction that such C does not exist. Then for all k ∈ N there exists a vector xk such that
kxk k1 > kkxk k. This implies in particular kxk k1 > 0, so that we can set uk := xk /kxk k1 . Then, kuk k =
kxk k/kxk k1 < 1/k and kuk k1 = 1. Now, the sequence T (uk ) ∈ Rd is uniformly bounded (this can be seen by
contradiction: a subsequence of T (uk ) diverging to +∞ in the Euclidean norm would contradict the fact that T is
continuous from the Exercise 3.5), hence the closure of the set {T (uk )}k∈N is closed and bounded. From Theorem
6.4 of the introductory material, such set is compact. Therefore, there exists a subsequence {T (ukh )}h∈N which
converges to some vector in Rd , and since T is invertible and linear this implies the existence of a ξ ∈ E such
that ukh → ξ. Now, the triangular inequality implies kξk ≤ kξ − ukh k + kukh k ≤ kξ − ukh k + k1h , and we can
send h → +∞ to obtain kξk = 0, and hence ξ = 0. On the other hand,
|kξk1 − kukh k1 | ≤ kξ − ukh k1 =
d
X
|T (ξ)i − T (ukh )i |,
i=1
and the right hand side goes to zero as h → +∞ because T is linear (and continuous). Since kukh k1 = 1, this
implies kξk1 = 1, but that contradicts ξ = 0. Hence, there exist two constants c1 , c2 > 0 such that
c1 kxk1 ≤ kxk ≤ c2 kxk1 .
This shows (as a simple exercise) that (X, k · k) and (X, k · k1 ) have the same topology (use basic open sets to
prove that). Since the norm k · k was arbitrary, and the property of inducing the same topology is an equivalence
between distances (check!), the above inequality holds for any arbitrary pair of norms.
Remark 3.4. An important property of metric spaces is that the topology of a metric space is entirely determined
by its convergent sequences, something which does not hold in general for a topological space. To see this, let
(E, d1 ) and (E, d2 ) be two metric spaces on the same set E. Assume the two spaces have the same converging
sequence. Then we claim the two distances d1 and d2 induce the same topology. Let Br1 (x) be an open ball of
radius r and center x in the d1 distance. If we prove that Br1 (x) is open in the d2 topology, then the assertion
will be proven. Assume by contradiction that Br1 (x) is not open in the d2 topology. Hence, there exists a point
y ∈ Br1 (x) for which no basic open neighborhoods in d2 is entirely contained in Br1 (x). Equivalently, for all
s > 0, the d2 -open ball Bs2 (y) is not contained in Br1 (x). Therefore, there exists some z ∈ Bs2 (y) which is not
in Br1 (x). Choosing s = 1/n, we thus construct a sequence zn which converges to y in the d2 topology. On the
other hand, since zn 6∈ Br1 (x) for all n ∈ N, and since Br1 (x) is an open set containing y, this clearly means
that zn does not converge to y in d1 , which contradicts the fact that the two spaces have the same convergent
sequences.
2 An
isomorphism between two vector spaces A and B is a linear bijection T : A → B
12
3.4
Completeness and Banach spaces
Let s(X, d) be a metric space.
Definition 3.10. A sequence {xn }n∈N is called a Cauchy sequence if for all > 0 there exists N ∈ N such
that d(xn , xm ) < for all n, m ≥ N . The metric space (X, d) is called a complete metric space if every Cauchy
sequence is convergent (i. e. it has a limit x ∈ X).
We immediately recognise the definition of the Cauchy criterion in Theorem 6.1 of the introductory material
for sequences in Rd . Theorem 6.1 of the introductory material actually shows that the metric space Rd equipped
with the standard Euclidean norm is a complete metric space.
Exercise 3.7. Let (X, d) be a complete metric space let C ⊂ X be a closed subset of X. Prove that (C, d) is a
complete metric space.
Clearly, the definition of Cauchy sequence holds in normed spaces as well: {xn } is a Cauchy sequence in
(X, k · k) if for all > 0 there exists N ∈ N such that kxn − xm k < for all n, m ≥ N .
Definition 3.11. A complete normed space is called a Banach space.
d
Example 3.3. The normed space
(Q , k · k)1 where
n k · k is the standard Euclidean norm is not a Banach space.
1 n
, which clearly converges to (e, . . . , e) as n → +∞. {xn } is a
Indeed, choose xn = 1 + n , . . . , 1 + n
Cauchy sequence, but the limit is not in Qd , therefore the space is not complete.
Definition 3.12. The set of all bounded functions f : (X, d) → (Y, d1 ) equipped with the distance
d∞ (f, g) = sup{d1 (f (x), g(x)) : x ∈ X}
is a metric space (check!) and is denoted by B(X, Y ). Moreover, if Y is normed with norm k · kY , then B(X, Y )
is also normed with the norm
kf k∞ = sup{kf (x)kY : x ∈ X}.
The set of continuous and bounded functions f : X → Y equipped with the d∞ distance is a metric space, and
is denoted by Cb (X, Y ). If Y is normed, Cb (X, Y ) is also normed with the k · k∞ norm. The d∞ topology is
often referred to as the uniform convergence topology.
Theorem 3.2. If Y is complete, then so are B(X, Y ) and Cb (X, Y ).
Proof. Let fn be a Cauchy sequence in B(X, Y ). For all > 0 there exists N such that d∞ (fm , fn ) < for
all n, m > N . Therefore, by definition of d∞ , for all x ∈ X we have d1 (fn (x), fm (x)) < , which implies that
fn (x) is a Cauchy sequence in Y . Since Y is complete, fn (x) converges to some point in Y that we call f (x) (it
depends on the point x, so it is a function of x). We now prove that fn converges to f in B(X, Y ). Since the
distance function d(z, ·) is continuous (see exercise 3.3), we can let m → +∞ in d1 (fn (x), fm (x)) < with n
fixed and get d1 (fn (x), f (x)) < . The latter is valid for all x ∈ X and for all n ≥ N . Hence, d∞ (fn , f ) ≤ for
all n ≥ N , which is equivalent to say that fn converges to f in the topology of (B(X, Y ), d∞ ). Hence B(X, Y )
is complete.
We now turn to the completeness of Cb (X, Y ). It suffices to prove that Cb (X, Y ) is a closed subset of
B(X, Y ). Let fn ∈ Cb (X, Y ) converge to some f ∈ B(X, Y ). We need to prove that f is continuous. Let
x0 ∈ X, and let > 0. There exists N such that d∞ (fn , f ) < for all n ≥ N . Since fN is continuous, there
exists δ > 0 such that d(x, x0 ) < δ implies d1 (fN (x), fN (x0 )) < . Hence, if d(x, x0 ) < δ we have
d1 (f (x), f (x0 )) ≤ d1 (f (x), fN (x)) + d1 (fN (x), fN (x0 )) + d1 (fN (x0 ), f (x0 )) ≤ 3,
i. e. f is continuous.
The spaces B(X, Y ) and Cb (X, Y ) are the first examples of function spaces in this course. As soon as the
topology of the metric space X is complex enough to resemble e. g. that of the real line R and Y is not a trivial
space, then the space B(X, Y ) is infinite dimensional.
Example 3.4. Let X = Y = R, both equipped with the Euclidean space. Then B(R, R) is infinite dimensional.
To see this, assume by contradiction that dim(B(R, R)) = M ∈ N. We prove that there exist M + 1 linearly
independent elements in B(R, R). Consider M + 1 disjoint closed intervals I1 , . . . , IM +1 in the real line R, and
set
(
1 if x ∈ Ii
fi (x) =
.
0 if otherwise
PM +1
Let α1 , . . . , αM +1 ∈ R. Assuming i=1 αi fi (x) = 0 for all x ∈ R clearly implies that αi = 0 for all i =
1, . . . , M + 1. Therefore the fi are linearly independent, and that contradicts the dimension of B(R, R) being
equal to M . Since M was arbitrarily chosen, this proves that the dimension of the space B(R, R) is infinite.
13
The following result is of paramount importance in the theory of differential equations.
Theorem 3.3 (Banach fixed point theorem). Let (X, d) a complete metric space. Let f : X → X be a
contraction on X, i. e. there exists α ∈ (0, 1) such that
d(f (x), f (y)) ≤ αd(x, y),
for all x, y ∈ X.
Then, there exists a unique x ∈ X such that f (x) = x, i. e. f has a unique fixed point.
Proof. Let us fix x0 ∈ X arbitrarily, and define a recursive sequence by iterating f several times: x1 := f (x0 ),
x2 = f (x1 ), . . ., xn+1 = f (xn ). Clearly,
d(xn+1 , xn ) ≤ αd(xn , xn−1 ) ≤ α2 d(xn−1 , xn−1 ) ≤ . . . ≤ αn d(x1 , x0 ).
Moreover, by triangular inequality, let m, n ∈ N with m > n, we have
d(xn , xm ) ≤ d(xn , xn+1 ) + d(xn+1 , xn+2 ) + . . . + d(xm−1 , xm )
≤ d(x1 , x0 )
m−1
X
αn − αm
.
1−α
αk = d(x1 , x0 )
k=1
Since the latter term above tends to zero as n, m → +∞, this shows that {xn }n∈N is a Cauchy sequence. Since
X is complete, there exists ξ ∈ X such that xn → ξ. Since f is continuous (easy exercise, consequence of f being
a contraction!), we can let n → +∞ in the identity xn+1 = f (xn ) and get ξ = f (ξ), which proves that ξ is a fixed
point. The uniqueness is proven as follows. Let ξ, η be two fixed points. Then d(ξ, η) = d(f (ξ), f (η)) ≤ αf (ξ, η),
and since α ∈ (0, 1) we get d(ξ, η) = 0, which gives ξ = η.
Exercise 3.8 (Differential equations via Banach fixed point). Assume we have the Cauchy problem on the real
line
(
y 0 = f (y)
(1)
y(0) = y0
for some f ∈ C 1 (R) and y0 ∈ R. Consider the same problem in its mild formulation
t
Z
y(t) = y0 +
f (y(τ ))dτ.
0
In order to prove that the Cauchy problem (1) has a unique local solution y ∈ C 1 (I0 ; R) (where I0 is some open
neighborhood of 0), prove that the mild formulation above admits a unique solution. To do so, consider the
normed space BM := {g ∈ Cb ([−M, M ]; R)} and its closed ball BM,R {g ∈ BM : kg − y0 k∞ ≤ R} for some
R > 0. Define the map T : BM,R → BM,R
Z
T (g)(x) = y0 +
t
f (g(τ ))dτ,
0
and prove that for a suitable choice of M and R the above map is
• well-defined (i. e. the image is included in the (Banach) space BM,R ),
• a contraction.
Use Banach’s fixed point theorem to prove the assertion.
4
Peano-Jordan measure theory
5
Measuring elementary sets
Definition 5.1 (Interval, boxes, elementary sets). An interval is a subset of R of the form [a, b] = {x ∈ R :
a ≤ x ≤ b}, [a, b) = {x ∈ R : a ≤ x < b}, (a, b] = {x ∈ R : a < x ≤ b}, (a, b) = {x ∈ R : a < x < b}, where
a ≤ b are real numbers. We define the length |I| of an interval I = [a, b], [a, b), (a, b], (a, b) to be |I| = b − a.
A box in Rd is a Cartesian product B := I1 × . . . × Id of d intervals I1 , . . . , Id (not necessarily of the same
length), thus for instance an interval is a one-dimensional box. The volume |Bj | of such a box B is defined as
|B| := |I1 | × . . . × |Id |. An elementary set is any subset of Rd which is the union of a finite number of boxes.
14
Exercise 5.1. . Show that if E, F ⊂ Rd are elementary sets, then the union E ∪ F , the intersection E ∩ F , and
the set theoretic difference E \ F := {x ∈ E : x 6∈ F }, and the symmetric difference E∆F := (E \ F ) ∪ (F \ E)
are also elementary. If x ∈ Rd , show that the translate E + x := {y + x : y ∈ E} is also an elementary set.
We now give each elementary set a measure.
Lemma 5.1 (Measure of an elementary set). . Let E ⊂ Rd be an elementary set.
(i) E can be expressed as the finite union of disjoint boxes.
(ii) If E is partitioned as the finite union B1 × . . . × Bk of disjoint boxes, then the quantity m(E) := |B1 | +
. . . + |Bk | is independent of the partition. In other words, given any other partition B10 , . . . , Bk0 0 of E, one
has |B1 | + . . . + |Bk | = |B10 | + . . . + |Bk0 0 |.
We refer to m(E) as the elementary measure of E.
Proof. We first prove (i) in the one-dimensional case d = 1. Given any finite collection of intervals I1 , . . . , Ik ,
one can place the 2k endpoints of these intervals in increasing order (discarding repetitions). Looking at the
open intervals between these endpoints, together with the endpoints themselves (viewed as intervals of length
zero), we see that there exists a finite collection of disjoint intervals J1 , . . . , Jk0 such that each of the I1 , . . . , Ik
are a union of some sub-collection of the J1 , . . . , Jk0 . This already gives (i) when d = 1. To prove the higher
dimensional case, we express E as the union B1 , . . . , Bk of boxes Bi = Ii1 × . . . × Iid . For each j = 1, . . . , d,
we use the one-dimensional argument to express I1,j , . . . , Ik,j as the union of sub-collections of a collection
J1,j , . . . , Jkj0 ,j of disjoint intervals. Taking Cartesian products, we can express the B1 , . . . , Bk as finite unions of
boxes Ji1 ,1 × . . . × Jid ,d , where 1 ≤ ij ≤ kj0 for all 1 ≤ j ≤ d. Such boxes are all disjoint, and the claim follows.
To prove (ii) we use a discretisation argument. Observe (exercise!) that for any interval I, the length of I
can be recovered by the limiting formula
1
1
] I∩ Z ,
(2)
|I| = lim
N →+∞ N
N
where N1 Z := {n/N : n ∈ Z} and ]A denotes the cardinality of a finite set A. Taking Cartesian products, we
see that
1
1 d
|B| = lim
]
B
∩
Z
,
N →+∞ N d
Nd
for any box B, and in particular that
1
1 d
|B1 | + . . . + |Bk | = lim
] E ∩ dZ .
N →+∞ N d
N
Denoting the right-hand side above as m(E), we obtain the claim (ii).
Remark 5.1. One might be tempted to define the measure m(E) of an arbitrary set E ⊂ Rd by the formula
1
1 d
m(E) = lim
] E ∩ dZ ,
(3)
N →+∞ N d
N
since this worked well for elementary sets. However, this definition is not particularly satisfactory for a number
of reasons. Firstly, one can construct examples
the limit does not exist (As an example, think of
S+∞ in which
k
the set of dyadic rational
numbers
E
=
{a/(2
)
:
a
= 0, . . . , 2k } in the interval [0, 1]. The sequence
k=1
1
d
mN = ] E ∩ N d Z has two subsequences with different limits, namely (m2n )n∈N with limit 1 and (m3n )n∈N
with limit 0). Even when the limit does exist, this concept does not obey reasonable properties such as
translation invariance. For instance, if d = 1 and E := Q ∩ [0, 1] √
:= {x ∈ Q : √
0 ≤ x ≤ 1}, then this definition
would give E a measure of 1, but would give the translate E + 2 := {x + 2 : x ∈ E} a measure of zero.
Nevertheless, the formula (3) will be valid for all Peano–Jordan measurable sets (see below). It also makes
precise an important intuition, namely that the continuous concept of measure can be viewed as a limit of the
discrete concept of (normalised) cardinality.
We now collect the basic properties of the elementary measure.
Lemma 5.2.
(i) For all elementary sets E ⊂ Rd , m(E) ≥ 0.
(ii) Given E1 , . . . , Ek ⊂ Rd disjoint elementary sets, we have the finite additivity
m(E1 ∪ . . . ∪ Ek ) = m(E1 ) + . . . + m(Ek ).
15
(iii) m(∅) = 0.
(iv) For all boxes B ⊂ Rd , m(B) = |B|.
(v) Let E, F ⊂ Rd be elementary sets with E ⊆ F . Then m(E) ≤ m(F ).
(vi) Given E1 , . . . , Ek ⊂ Rd elementary sets (not necessarily disjoint), we have the finite sub-additivity
m(E1 ∪ . . . ∪ Ek ) ≤ m(E1 ) + . . . + m(Ek ).
(vii) For all x ∈ Rd and all elementary sets E ⊂ Rd , we have the translation invariance
m(x + E) = m(E).
Proof. (i), (iii), and (iv) are trivial.
The proof of (ii) is trivial by construction in the case of two elementary sets E1 , E2 . The case of N sets
follows by induction.
(v) follows from (i) and (ii) (exercise).
The proof of (vi) in the case of two sets E1 , E2 is easily proven by writing E1 ∪ E2 as the union of two
disjoint sets E1 ∪ E2 = E1 \ E2 ∪ E2 and using (ii) and (v) to get
m(E1 ∪ E2 ) = m(E1 \ E2 ) + m(E2 ) ≤ m(E1 ) + m(E2 ).
The case of N sets follows by induction.
(vii) is immediate from the definition of elementary measure.
6
A review on the Peano–Jordan measure theory
Peano-Jordan measure is typically considered a subject to be developed in elementary calculus units. For the
sake of selfcontainedness, we shall recall here the main concepts of Peano-Jordan measure.
With the content of the previous section, we now have a satisfactory notion of measure for elementary sets.
But of course, the elementary sets are a very restrictive class of sets, far too small for most applications. For
instance, a solid triangle or disk in the plane will not be elementary, or even a rotated box. On the other hand,
as essentially observed long ago by Archimedes, such sets E can be approximated from within and without by
elementary sets A ⊂ E ⊂ B, and the inscribing elementary set A and the circumscribing elementary set B can
be used to give lower and upper bounds on the putative measure of E. As one makes the approximating sets
A, B increasingly fine, one can hope that these two bounds eventually match. This gives rise to the following
definitions.
Definition 6.1 (Peano–Jordan measure). Let E ⊂ Rd be a bounded set.
• The Peano–Jordan inner measure m∗,(J) (E) of E is defined as
m∗,(J) (E) :=
sup
m(A).
A⊆E, A elementary
• The Peano–Jordan outer measure m∗,(J) (E) of E is defined as
m∗,(J) (E) :=
inf
B⊇E, B elementary
m(B).
• If m∗,(J) (E) = m∗,(J) (E) we say that E is Peano–Jordan measurable, and call
m(E) := m∗,(J) (E) = m∗,(J) (E)
the Peano–Jordan measure of E.
By convention, we do not consider unbounded sets to be Jordan measurable (they will be deemed to have
infinite Jordan outer measure).
Jordan measurable sets are those sets which are ‘almost elementary’ with respect to Jordan outer measure.
More precisely, we have
Exercise 6.1 (Characterisation of Peano–Jordan measurability). Let E ⊂ Rd be bounded. Show that the
following are equivalent:
16
(1) E is Peano–Jordan measurable.
(2) For every > 0, there exist elementary sets A, B such that A ⊂ E ⊂ B such that m(B \ A) ≤ .
(3) For every > 0, there exists an elementary set A such that m∗,(J) (A∆E) ≤ .
As one corollary of this exercise, we see that every elementary set E is Peano–Jordan measurable, and that
Peano–Jordan measure and elementary measure coincide for such sets; this justifies the use of m(E) to denote
both.
Jordan measurability also inherits many of the properties of elementary measure.
Lemma 6.1. Let E, F ⊂ Rd be Peano–Jordan measurable sets.
1. (Boolean closure) Show that E ∪ F , E \ F , E ∩ F , and E∆F are Peano–Jordan measurable.
2. (Non-negativity) m(E) ≥ 0.
3. (Finite additivity) If E, F are disjoint, then m(E ∪ F ) = m(E) + m(F ).
4. (Monotonicity) If E ⊂ F , then m(E) ≤ m(F ).
5. (Finite subadditivity) m(E ∪ F ) ≤ m(E) + m(F ).
6. (Translation invariance) For any x ∈ Rd , E + x is Peano–Jordan measurable, and m(E + x) = m(E).
Proof. Step 1. We use the characterisation of Peano–Jordan measurable sets in the exercise 6.1. Let E and
F are measurable and disjoint. For > 0, let AE , AF , BE , BF be elementary sets such that AE ⊂ E ⊂ BE ,
AF ⊂ F ⊂ BF , m(BE \ AE ) ≤ , and m(BF \ AF ) ≤ . Since AE and AF are disjoint, then we can use the
subadditivity of elementary set and get
m((BE ∪ BF ) \ (AE ∪ AF )) = m(BE \ AE ) + m(BF \ AF ) ≤ 2,
and since BE ∪ BF and AE ∪ AF are elementary sets this proves that E ∪ F is measurable. Now, let E, F ⊂ Rd
be measurable and let AE ⊂ E ⊂ BE , AF ⊂ F ⊂ BF as above. By monotonicity of the measure of elementary
set, since
(BE ∩ BF ) \ (AE ∩ AF ) ⊆ BE ∩ BF ∩ (AcE ∪ AcF ) ⊂ (BE ∩ \AE ) ∪ (BF \ AF )
we have
m((BE ∩ BF ) \ (AE ∩ AF )) ≤ m((BE \ AE ) ∪ (BF \ AF )),
and by subadditivity we get
m((BE ∩ BF ) \ (AE ∩ AF )) ≤ 2.
Since the elementary sets AE ∩ AF and BE ∩ BF satisfy AE ∩ AF ⊂ E ∩ F ⊂ BE ∩ BF , we get that E ∩ F is
measurable. For a given measurable set E ⊂ Rd , the complement E c is clearly measurable due. Indeed, since
E is measurable, for a given > 0 there exists an elementary set A with m(A∆E) ≤ . On the other hand we
also have m(A∆E c ) = m(A∆E) ≤ , and this proves the assertion. For two measurable sets E, F ⊂ Rd , we
have E \ F = E ∩ F c and this is also a measurable set. Moreover, given two measurable sets E, F ⊂ Rd we
write E ∪ F = (E \ F ) ∪ (E ∩ F ) and this is a disjoint union of measurable sets, and hence it is measurable.
This proves the Boolean closure.
Step 2. The nonnegativity is an immediate consequence of the definition. For the finite additivity, let E, F
be measurable with E ∩ F = ∅. We know from Step 1 that E ∪ F is measurable. Now, for a given > 0 let
AE , AF , BE , BF be elementary sets as in Step 1. Since AE ∩ AF = ∅, we have
m(E) − m(F ) − m(E ∪ F ) ≤ m(BE ) + m(BF ) − m(AE ∪ AF ) =≤ m(BE ) + m(BF ) − m(AE ) − m(AF ) ≤ 2,
and
m(E ∪ F ) − m(E) − m(F ) ≤ m(BE ) + m(BF ) − m(AE ) − m(AF ) ≤ 2,
which means |m(E ∪ F ) − (m(E) + m(F ))| ≤ 2. This proves that m(E ∪ F ) = m(E) + m(F ).
Step 3. The monotonicity property is trivial. Indeed, if E, F are measurable with E ⊂ F , then the infimum
in the definition of m∗,(J) (E) is computed on a larger set of elementary sets than m∗,(J) (F ), and therefore
m∗,(J) (E) ≤ m∗,(J) (F ). The subadditivity follows from the additivity property and the monotonicity:
m(E ∪ F ) = m(E \ F ) + m(E ∩ F ) ≤ m(E) + m(F ).
The translation invariance is trivial and is left as an exercise.
17
Exercise 6.2 (Regions under graphs are Jordan measurable). Let B be a closed box in Rd , and let f : B → R
be a continuous function.
• Show that the graph {(x, f (x)) : x ∈ B} ⊂ Rd+1 is Peano–Jordan measurable in Rd+1 with Peano–Jordan
measure zero. (Hint: on a compact subset of Rd , continuous functions are uniformly continuous, see
Theorem 6.10 of the introductory material.)
• Show that the set {(x, t) : x ∈ B; 0 ≤ t ≤ f (x)} ⊂ Rd+1 is Peano–Jordan measurable.
Exercise 6.3.
• Show that a solid bounded triangle in R2 is Peano-Jordan measurable.
• Show that all open Euclidean balls Br (x) in Rd are Peano–Jordan measurable.
Exercise 6.4. Define a Peano–Jordan null set to be a Peano–Jordan measurable set of measure zero. Show that
any subset of a Peano–Jordan null set is a Peano–Jordan null set (Hint: use Exercise 6.1).
Exercise 6.5. Show that (3) holds for all Peano–Jordan measurable sets E ⊂ Rd .
Solution. From Exercise 6.1 we know that for all > 0 there exist two elementary sets A and B such that
A ⊂ E ⊂ B with elementary measure m(B \ A) ≤ . We clearly have
1 d
1
1 d
1
] A ∩ d Z ≤ lim
] E ∩ dZ
m(A) = lim
N →+∞ N d
N →+∞ N d
N
N
1
1
≤ lim
] B ∩ d Zd = m(B) = m(A) + m(B \ A) ≤ m(A) + N →+∞ N d
N
Hence.
m∗,(J) (E) ≥ m(A) ≥
and
∗,(J)
m
1 d
1
]
E
∩
Z
− ,
N →+∞ N d
Nd
lim
1
1 d
(E) ≤ m(B) ≤ m(A) + ≤ lim
] E ∩ d Z + .
N →+∞ N d
N
Since > 0 is arbitrary, the last two inequalities prove that
1
1 d
m(E) = lim
]
E
∩
Z
.
N →+∞ N d
Nd
Lemma 6.2. Let E ⊂ Rd be a bounded set.
1. E and the closure E have the same Peano–Jordan outer measure.
◦
2. E and the interior E have the same Peano–Jordan inner measure.
3. E is Peano–Jordan measurable if and only if the topological boundary ∂E of E has Peano–Jordan outer
measure zero.
Proof. In order to prove the first statement, let B ⊃ E be an elementary set, and let B be its closure. Since
m(B) = m(B), and since B ⊃ E, we clearly have m∗,(J) (E) = m∗,(J) (E). The proof of the second statement is
◦
◦
◦
◦
very similar: for all elementary sets A ⊂ E, A is still an elementary set with m(A) = m(A), and A ⊂ E, which
proves m∗,(J) (E) = m∗,(J) (E). Now,
◦
m∗,(J) (∂E) = m∗,(J) (E \ E).
Now 1 and 2 imply
◦
◦
◦
m∗,(J) (E) = m∗,(J) (E) ≤ m∗,(J) (E) ≤ m∗,(J) (E) + m∗,(J) (∂E) = m∗,(J) (E) = m∗,(J) (E),
and we have used the additivity of the outer measure m∗,(J) . If E is Peano–Jordan measurable, then the first
and last term in the above inequality coincide, and this implies as a byproduct that m∗,(J) (∂E) = 0. On the
other hand, if m∗,(J) (∂E) = 0 then obviously m∗,(J) (∂E) = 0. Hence,
◦
m∗,(J) (E) = m∗,(J) (E) = m∗,(J) (E),
and
◦
◦
m∗,(J) (E) = m∗,(J) (E) = m∗,(J) (E) + m∗,(J) (∂E) = m∗,(J) (E).
18
◦
By contradiction, if m∗,(J) (E) > m∗,(J) (E), this means that m∗,(J) (E) > m∗,(J) (E).
On the other hand, if m∗,(J) (∂E) = 0, for an arbitrary > 0 there
Sm exist an elementary set C ⊃ ∂E with
m(C) < . By definition of elementary set, B can be written as C = i=1 Qri (yi ), with the open cubes Qri (yi )
defined by the formula Qr (y) := Πdj=1 ((yi )j − ri , (yi )j + ri ), for some y1 , . . . , ym ∈ ∂E and some r1 , . . . , rm > 0.
The constants ri can be chosen in such a way that
C=
m
[
Qri (yi ).
i=1
Now, there are two cases. Either m∗,(J) (E) = 0, or m∗,(J) (E) > 0. In the former case, due to the statement
◦
1, we have m∗,(J) (E) = 0, and m∗,(J) (∂E) = 0 implies m∗,(J) (E) = 0, and therefore E is measurable with
◦
m(E) = 0. In the case m∗,(J) (E) > 0, for each x ∈ E there is an open cube Qrx (x) ⊂ E for some rx > 0.
Hence, the family
◦
{Qrx (x) : x ∈ E} ∪ {Qr1 (y1 ), . . . , Brm (ym )}
‘covers’ E. Hence, from the result in theorem 6.5 of the introductory material there exists a finite number of
◦
points x1 , . . . , xk ∈ E such that
k
m
[
[
E⊂
Qrxj (xj ) ∪
Qr1 (y1 ).
j=1
Hence, since
Sk
j=1
Qrxj (xj ) ∪
Sm
i=1
i=1
Qr1 (y1 ) is an elementary set, we have

m∗,(J) (∂E) ≤ m 
k
[

Qrxj (xj ) + m(C).
j=1
On the other hand, since
Sk
j=1
◦
Qrxj (xj ) ⊂ E, we have
m∗,(J) (∂E) ≥ m(
k
[
Qrxj (xj )),
j=1
and the last two inequalities together with m(C) < imply that E is Peano–Jordan measurable because of the
exercise 6.1.
Exercise 6.6. Show that both the subsets of R2
• [0, 1]2 \ Q2
• [0, 1]2 ∩ Q2
have 0 as inner measure and 1 as outer measure. In particular, none of the two sets is Peano–Jordan measurable.
Informally, any set with a lot of ‘holes’, or a very ‘fractal’ boundary, is unlikely to be Jordan measurable.
In order to measure such sets we will need to develop Lebesgue measure, which is done in the next section.
Exercise 6.7 (Connection with the Riemann integral). For simplicity, we shall consider only the one-dimensional
case. We recall that the basic definitions and properties of the Riemann integral are sketched in the preliminary
material. A first property that can be easily checked as an exercise is the following 3
• If E is an elementary subset of [a, b], then the indicator function 1E : [a; b] → R (defined by setting
Rb
1E (x) := 1 when x ∈ E and 1E (x) := 0 otherwise) is piecewise constant, and p.c. a 1E dx = m(E).
The next step is to prove that indicator functions of Peano-Jordan measurable sets are Riemann integrable.
• If E is a Peano-Jordan measurable of [a, b], then the indicator function 1E : [a; b] → R (defined by setting
Rb
1E (x) := 1 when x ∈ E and 1E (x) := 0 otherwise) is Riemann integral, and p.c. a 1E dx = m(E).
The above property establishes the first important link between the Peano–Jordan measure and the Riemann
integral. The second one is the following.
3 Recall
the definition of piecewise constant integral in the preliminary material.
19
• (Area interpretation of the Riemann integral). Let [a, b] be an interval, and let f : [a, b] → R be a bounded
function. Show that f is Riemann integrable if and only if the sets E+ := {(x, t) : x ∈ [a, b], 0 ≤ t ≤ f (x)}
and E− := {(x, t) : x ∈ [a, b], f (x) ≤ t ≤ 0} are both Jordan measurable in R2 , in which case one has
Z
b
f (x)dx = m(E+ ) − m(E− ).
a
(Hint: First establish this in the case when f is non-negative.)
Exercise 6.8. Extend the definition of the Riemann and Darboux integrals stated in the introductory material
to higher dimensions, in such a way that analogues of all the previous results hold.
20