Appunti Topologia Nosol (1)

Notes on Topology

Fabio Tonoli

Department of Decision Sciences - U. Bocconi - Milano

The proofs of the Facts, Answers to Questions and Solutions of theExercises are left as exercise.

First version: October 2012

Current version: September 2015

Printed the 15th September 2015

Contents

II A short introduction in Topology 5

1 Basic elements of General Topology 7

1.1 Topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Interiors, closures and borders . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3 Dense subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.5 Continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.6 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.7 Semicontinuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.8 Comparing topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.9 Weak topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.10 The product topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.11 Pointwise and uniform convergence . . . . . . . . . . . . . . . . . . . . . . 37

Bibliography 42

3

4 CONTENTS

Part II

A short introduction in Topology

5

Chapter 1

Basic elements of General Topology

In this chapter we introduce some basic elements on General Topology which are im-portant and recurrent in the study of Real Analysis. These elements refers to the partof topology related to the study of convergence and approximation in a general setting,and are fundamental for the study of spaces of real functions and in theory of measureand integration. We follow very closely Chapter 2 of [Aliprantis�Border], also in the topicsubdivision, illustrating each topic with a wide number of examples and expliciting quiteall the proofs. The aim is to obtain a very clear undestandanding of the picture, intendedas a rational and complete understanding of all proofs, wishing that in this way the readerwill develop good skills for an abstract self-reasoning in the subject.

We warn that Topology is a much broader subject, involving all the properties of atopological space which are invariant by homemomorphisms. For example, the notion ofconnectedness and the decomposition of a space into connected components, or the Al-gebraic Topology, which deals with the construction and the study of the fundamentalalgebraic groups �n(X; x0). A very important long-lasting open question (one of the 1million dollar Clay problems) on this topic was solved only some years ago by Perelman:a 3-manifold with the fundamental groups of a 3-sphere is homeomorphic to a 3-sphere.

1.1 Topological spaces

De�nition 1.1.1 A topology � on a set X is a collection of subsets (called open sub-sets) of X satisfying:

(o.i) ;; X 2 � ;(o.ii) � is closed under �nite intersections;(o.iii) � is closed under arbitrary unions.

A nonempty set X equipped with a topology � is called a topological space, and is denoted(X; �). The complement of open sets are called closed subsets.

Example 1.1.2 The trivial topology or indiscrete topology on a set X consists of� = fX; ;g. These are also the only closed sets.

7

8 CHAPTER 1. BASIC ELEMENTS OF GENERAL TOPOLOGY

Example 1.1.3 The discrete topology on a set X consists of all subsets of X: � =P(X) = 2X1. Thus every set is both open and closed.

Exercise 1.1.4 The real line R = (�1;+1) has a natural topology, called euclideantopology. It consists of all subsets U such that for each x 2 U there exists some � > 0s.t. (x� �; x+ �) � U . Prove that this is a topology.

Exercise 1.1.5 The extended real line R = [�1;+1] = R [ f�1;+1g has a naturaltopology. It consists of all subsets U such that for each x 2 U :a. If x 2 R, then there exists some � > 0 s.t. (x� �; x+ �) � U ;b. If x = +1, then there exists some y 2 R with (y;+1] � U ;c. If x = �1, then there exists some y 2 R such that [�1; y) � U .

Prove that this is a topology.

Exercise 1.1.6 Consider an in�nite set X and de�ne as closed the set X and the setsformed by a �nite number of elements (�nite sets): the open sets are ; and the sets whosecomplement is a �nite subset of X. This forms a topology, called co�nite topology.Prove that the co�nite topology is a topology.

Facts 1.1.7 Closed sets satisfy the following properties, which are the dual of the prop-erties of the open sets:

(c.i) ;; X are closed;(c.ii) A �nite union of closed sets is closed;(c.iii) An arbitrary intersection of closed sets is closed.

De�nition 1.1.8 A neighborhood of a point x is any set V containing an open set Usuch that x 2 U � V .

Remark 1.1.9 If V is a neighborhood of x, the open set U in the de�nition is an openneighborhood of x.

From the de�nition, the following remark follows immediately.

Remark 1.1.10 A open () A is a neighborhood of each of its points.

De�nition 1.1.11 The collection of all neighborhoods of a point x is called neighbor-hood system of x and is denoted by Nx.

1The notation 2X stands for the set of functions f : X ! f0; 1g. The 1:1 correspondence between thetwo sets maps a subset S � X to the characteristic function �S associated to S, de�ned by �S(x) = 1 ifx 2 S and �S(x) = 0 if x =2 S.

1.1. TOPOLOGICAL SPACES 9

Facts 1.1.12 A neighborhood system satis�es the following properties:1. X 2 Nx (so Nx is never empty)2. U; V 2 Nx =) U \ V 2 Nx

3. V 2 Nx; V � U =) U 2 Nx

Remark 1.1.13 Notice that if V is an open neighbourhood of x and V � U , it is notnecessarily true that U is an open neighbourhood of x. It is the importance of property(3) the reason for which neighborhoods are de�ned to be not necessarily open.

De�nition 1.1.14 A neighborhood of a set F is any set V containing an open setU such that F � U � V . The collection of all neighbourhoods of a set F is calledneighborhood system of F and is denoted by NF .

De�nition 1.1.15 (Separable axioms) A topological space (X; �) is:

T0 (points are topologically distinguishable) if given any two points x; y 2 X withx 6= y there exists a neighbourhood of one not containing the other:

8x; y 2 X : x 6= y 9Ux 2 Nx : y =2 Ux or 9Uy 2 Ny : x =2 Uy;

T1 (points are separated) if given any two points x; y 2 X with x 6= y each of themhas a neighbourhood not containing the other point:

8x; y 2 X : x 6= y 9Ux 2 Nx;9Uy 2 Ny : x =2 Uy; y =2 Ux;

T2 (X is Hausdor¤ or separated) if given any two points x; y 2 X with x 6= y thereexists a pair of neighbourhood, Ux; Uy, one for each given point, which are disjoint:

8x; y 2 X : x 6= y 9Ux 2 Nx;9Uy 2 Ny : Ux \ Uy = ;;

T3 (X is regular) if given a point x 2 X and a closed set F with x =2 F there exists aneighbourhood Ux of x and a neighbourhood UF of F which are disjoint:

8x 2 X;8F : FC 2 � and x =2 F 9Ux 2 Nx;9UF 2 NF : Ux \ UF = ;;

T4 (X is normal) if given two closed sets F1; F2 with F1 \F2 = ; (i.e., disjoint) thereexists a pair of open sets, each containing one given closed set, which are disjoint:

8F1; F2 : FC1 ; FC2 2 � and F1 \ F2 = ; 9UF1 2 NF1 ;9UF2 2 NF2 : UF1 \ UF2 = ;:

Facts 1.1.16 Given a topological space (X; �) the following holds:

1. T1 () each point forms a closed set, i.e., fxg closed 8x 2 X;2. T2 =) T1 =) T0;3. T1 + T3 =) T2;4. T1 + T4 =) T3; T2.


Example 1.1.17 If jXj > 1 the trivial topology is not T0 (hence not T1; T2) because theunique nonempty open set is X but it is T3; T4 because the unique nonempty closed set isX.

Example 1.1.18 The discrete topology is T0; T1; T2; T3; T4, since every set is open.

Exercise 1.1.19 Prove that the co�nite topology is T1 but not T2; T3; T4.

Exercise 1.1.20 Let N = f1; 2; :::g: Prove that the collection � of sets consisting of theempty set and all sets containing 1 is a topology on N. The closed sets are N and all setsnot containing 1. This space is T0 but not T1; T2; T3; T4.

Exercise 1.1.21 Again let N = f1; 2; : : : g and set Un = fn; n + 1; : : : g. Prove that theempty set and all the Un comprise a topology on N. The closed sets are just the initialsegments f1; 2; : : : ; ng and N itself. Notice that every pair of nonempty open sets dointersect, so again it is not T1; T2; T3; T4 but it is T0.

De�nition 1.1.22 A metric on a set X is a function d : X � X ! R such that forevery x; y; z 2 X:

(m.i) (positivity) d(x; y) � 0;(m.ii) (separation of points) d(x; y) = 0 () x = y;

(m.iii) (simmetricity) d(x; y) = d(y; x);(m.iv) (triangle inequality) d(x; z) � d(x; y) + d(y; z):

The pair (X; d) is then called metric space. If (m.ii) is replaced by the weaker condition

(m.ii�) d(x; x) = 0

then d is called semi-metric.

De�nition 1.1.23 8 x 2 X, 8r > 0 de�ne the open ball of radius r (or r-ball) the setBr(x) = fz 2 X : d(x; z) < �g.

Facts 1.1.24 Let (X; d) be a semi-metric space. Then d induces a topology on X in thefollowing way: a set U is open if every point of U has a open �-ball contained in U :

8x 2 U 9� > 0 : B�(x) � U:

De�nition 1.1.25 A topological space (X; �) is metrizable if there exists a metric d onX that generates the topology � .

Example 1.1.26 The discrete metric (or rubber band metric) is de�ned by :

d(x; y) =

�0 if x = y1 if x 6= y :

It generates the discrete topology: every singleton fxg is an open set, being B1=2 (x) = fxgand thus every subset of X is open.

1.1. TOPOLOGICAL SPACES 11

Example 1.1.27 The zero semi-metric, de�ned by d(x; y) = 0, generates the trivial to-pology.

Exercise 1.1.28 A semi-metric d generates a T0 topology () d is a metric. In thatcase, the generated topology is T2.

De�nition 1.1.29 A normed space is a real vector space V endowed of a norm func-tion, a function k�k : V ! R satisfying 8x; y 2 V; 8c 2 R:

(n.i) (positivity and nondegenerate) kxk � 0 with kxk = 0 () x = 0;(n.ii) (homogeneity) kcxk = jcj kxk ;(n.iii) (triangle inequality) kx+ yk � kxk+ kyk :

Facts 1.1.30 A normed space is also a metric space by de�ning d(x; y) = kx� yk.Moreover, in a normed space the set B�(x) is the translated of B�(0) by x, for everypoint x:

B�(x) = fx+ y : y 2 B�(0)g:

Example 1.1.31 The real line R is a normed space, the norm being the absolute valuefunction: the metric d(x; y) = jx � yj de�nes a topology on R, the euclidean topology.Every open interval (a; b) is an open set.

Proposition 1.1.32 Consider R with the euclidean topology. Every open set is a count-able union of disjoint open intervals (where the end points +1 and �1 are allowed).

Example 1.1.33 In Rn the function kxk2 =pPn

i=1 x2i de�nes a norm, called euclidean

norm. The induced metric d2(x; y) =qPn

i=1 (xi � yi)2 is called euclidean distance,

and the topology induced is called euclidean topology.

Exercise 1.1.34 Prove that the euclidean norm is a norm.

Exercise 1.1.35 Prove that in Rn the functions kxk1 =Pn

i=1 jxij and kxk1 = supi=1;:::;n jxijde�nes two norms, called respectively the pedestrian norm and the sup norm. The in-duced metric d1 and d1 are called pedestrian distance and the sup distance.

Exercise 1.1.36 Prove that the pedestrian distance and the sup distance de�nes the sametopology as the euclidean one.

Exercise 1.1.37 Consider R2 with the semi-norm k(x1; x2)k = jx1j (just the �st coordin-ate counts for this semi-norm). Prove that d(x; y) = jy1 � x1j is a semi-distance and thatall the points having the same �rst coordinate have zero distance.


Figure 1.1: Open balls of radius � in the three metrics d1; d2; d1.

Facts 1.1.38 The intersection of a nonempty family A of topologies on a set is again atopology.

Facts 1.1.39 If A is an arbitrary nonempty family of subsets of X, then there exists asmallest (with respect to set inclusion) topology that includes A. This topology �A is calledthe topology generated by A and consists precisely of ;; X and all sets of the form[�V�, where each V� is a �nite intersection of sets of A.

We described the topology in metric spaces by describing the open balls centered atthe various points (if the metric derive from a norm then they are also the translated ofopen balls centered at the origin, but we don�t care of this, it is the concept of a topologicalgroup). This generalizes to the concept of base and subbase of a topology.

De�nition 1.1.40 A base for a topology � is a subfamily B of � such that

8U 2 �; 8x 2 U; 9V 2 B s:t: x 2 V � U:

Example 1.1.41 Consider R with the euclidean topology � . A base of � is the family ofall open intervals

B = f(a; b) : 8a; b 2 R; a � bg:

Facts 1.1.42 1. B is a base of � if 8U 2 � is a union of members of B.2. If B is a family of sets that is closed under �nite intersections and [V 2BV = X, thenthe family of all unions of members of B is a topology for which B is a base.

De�nition 1.1.43 A subfamily S of a topology � is a subbase for � if the collection ofall �nite intersections of members of S is a base for � .

Facts 1.1.44 A family S is a subbase for the topology it generates together with ;; X.

De�nition 1.1.45 A topological space (X; �) is called second countable if it has acountable base.

Facts 1.1.46 1. (X; �) is second countable if and only if � has a countable subbase.2. The euclidean topological space (Rn; k�k2) is second countable.

1.2. INTERIORS, CLOSURES AND BORDERS 13

De�nition 1.1.47 A neighborhood base at x is a collection Bx of neighborhoods of xwith the property:

U 2 Nx =) 9V 2 Bx with V � U:

De�nition 1.1.48 A topological space (X; �) is called �rst countable if every point hasa countable neighborhood base.

Facts 1.1.49 1. Every semi-metric space is �rst countable: the balls of radius 1naround

x form a countable neighborhood base at x.2. Every second countable space is also �rst countable, the converse is false.

De�nition 1.1.50 If Y is a subset of a topological space (X; �), then the collection

�Y = fV \ Y : V 2 �g

is a topology on Y , called the relative topology (induced by � on Y ).

Proof We prove that �Y is a topology on Y . The sets ; = ; \ Y; Y = X \ Y are clearlyin �Y . Given W� 2 �Y then there exist V� 2 � such that W� = V� \ Y . But then[W� = [ (V� \ Y ) = ([V�) \ Y and \W� = \ (V� \ Y ) = (\V�) \ Y . �

When Y � X is equipped with its relative topology, we call Y a topological subspaceof X. An element of �Y is called relatively open in Y , its complement in Y is calledrelatively closed in Y . For example, every subset Y of X is both relatively open andclosed in itself.

Exercise 1.1.51 The relatively closed subsets of Y are the restrictions to Y of the closedsubsets of X.

Facts 1.1.52 For a semimetric topology, the relative topology is derived from the samesemimetric restricted to Y .

Exercise 1.1.53 A countable intersection of open sets needs not be open. Similarly, acountable union of closed sets needs not be closed. Give an example for each case.

1.2 Interiors, closures and borders

Let (X; �) be a topological space, and let A be any subset of X. The topology de�nestwo sets intimately related to A: the interior and the closure of of A.

De�nition 1.2.1 The interior of A, denoted A�, is the largest open set included in A.The closure of A, denoted A, is the smallest closed set including A.


Exercise 1.2.2 1. Prove that A� and A exist.2. Show that there are A with no largest closed sets included in A and that there are Awith no smallest open set including A.

Facts 1.2.3 The following holds:

1. A� is the union of all open subsets of A and A is the intersection of all closed setsincluding A.

2. A � B implies A� � B� and A � B;3. A is open () A = A�, B is closed () B = B;

4. For any set A, A�� = A� and A = A;

5. For any set A, (A�)C = AC and�AC��=�A�C;

6. V 2 Nx () x 2 V �and, similarly, V 2 NF () F � V �.

Exercise 1.2.4 1. If A is open, A � A�, but they are not necessarily equal. Provide an

example where they di¤er.2. For any set A, let A+ = A

�. Then A++ = A+.

3. For any set A, let A� = A�. Then A�� = A�.

De�nition 1.2.5 A point x 2 A is called an interior point of A if there is an open setU such that x 2 U � A.

Remark 1.2.6 A point x 2 A is interior i¤ A 2 Nx.

De�nition 1.2.7 A point x (not necessarily in A) is called a closure point of A if everyneighborhood of x intersects A.


1. A open () A coincides with the set of interior points of A2. A closed () A coincides with the set of closure points of A3. For any sets A;B, (A \B)� = A� \B� and A [B = A [B.

Exercise 1.2.9 Find A;B s.t. (A [B)� 6= A� [B� and C;D s.t. C \D 6= C \D.

De�nition 1.2.10 A point x (not necessarily in A) is an accumulation point (or limitpoint, or cluster point) of A if for each neighborhood V of x we have (V �fxg)\A 6= ;.The set of the accumulation points of A is denoted by A0 and is called the derived set ofA. A set A is called perfect if A = A0.

Remark 1.2.11 In general, it is not true that internal points are accumulation points,that is, A� � A0. For example, take the discrete topology, where every point in a set isinternal and there are no accumulation points.

1.3. DENSE SUBSETS 15

De�nition 1.2.12 A point x (not necessarily in A) is a boundary point of A if eachneighborhood V of x satis�es both V \ A 6= ; and V \ AC 6= ;. The set of the boundarypoints of A is denoted by @A and is called the boundary of A.


1. @A � A, A0 � A;2. A = A� t @A, thus @A = A� A�.3. A = A [ A0;4. @A = @(AC) = A \ AC;5. A closed () @A � A () A0 � A.6. A open () @A � AC.

Example 1.2.14 Consider the set A = [0; 1)[ f2g � R. Then 2 is a closure point of A,but not an accumulation point. The point 1 is both a closure point and a limit point of A.Moreover, A� = (0; 1), A0 = [0; 1], @A = f0; 1; 2g, A = [0; 1] [ f2g.

Exercise 1.2.15 Given A = f 1n: n 2 Ng in R with the euclidean topology, compute A0.

De�nition 1.2.16 A point x 2 A is called an isolated point of A if there exists aneighbourhood V such that V \ A = fxg (this means that fxg is open relatively to A).The set of isolated points of A is denoted by I(A).

Facts 1.2.17 Concerning isolated points, the following holds:

1. I(A) = A� A0;2. @A \ A0 = @A \ (I (A))C;3. A closed =) A = A0 t I(A);4. A = A

� t (@A \ A0) t (@A \ I (A)).

1.3 Dense subsets

De�nition 1.3.1 A subset D of a topological space (X; �) is dense if D = X.

Thus D is dense i¤ every nonempty open subset of X contains a point in D. Inparticular, if D is dense in X and x 2 X, then every neighborhood of x contains a pointin D. This means that any point in X can be approximated arbitrarily well by points inD.

De�nition 1.3.2 A topological space is separable if it includes a countable dense subset.

Lemma 1.3.3 Every second countable space is separable.


Proof Let B = fB1; B2; : : : g be a countable base for the topology, and pick xi 2 Bi foreach i. We claim that D = fx1; x2; : : : g is dense.Indeed every nonempty open subset of X is a union of members of B by Fact (1.1.42),

thus it contains at least one of the points of D. �

The converse is true for metric spaces, but not in general: a metric space is separablei¤ it is second countable (see [Aliprantis�Border], Lemma 3.4).

Example 1.3.4 Let X be an uncountable set and �x x0 2 X. Take the topology consistingof the empty set and all sets containing x0. The set fx0g is dense in X, so X is separable.Furthermore, each set of the form fx0; xg 8x 2 X is open, so there is no countable base.

Example 1.3.5 (Di¢ cult!) Consider the space l1 of all absolutely summable real se-quences, the sequences x = fxng s.t.

Pjxnj <1. Its dual l1 is the space of all bounded

sequences, endowed with the sup norm: an element ' = f'ng 2 l1 de�nes a linearfunction ' : l1 ! R by setting '(x) := hx; 'i =

Pxn'n 2 R.

The weak topology �(l1; l1) on l1 is de�ned as follows: a subbase for the weak topologyis the collection of all sets of the form '�1(U), where ' 2 l1 and U is an open subset ofR. Fixed ' 2 l1 and an open U � R, the corresponding set in the subbase is the set ofall sequences x = fxng 2 l1 such that hx; 'i =

Pxn'n 2 U .2

Now, de�ne D as the countable set of all eventually zero sequences with rationalcomponents. Taken ' 2 l1 and an open U � R, it is possible to de�ne a sequencex = fxng 2 D such that hx; 'i = '(x) =

Pxn'n 2 U (for simplicity, think of U as an

open interval). Therefore any open set in l1 (in the weak topology) contains an elementof D. This proves that D is a dense subset of l1, so (l1; �(l1; l1)) is a separable space.

However the space (l1; �(l1; l1)) is not �rst countable, otherwise it would be �nitedimensional: a vector space is �nite dimensional i¤ the weak topology is �rst countable(see [Aliprantis�Border], Theorem 6.26).

1.4 Nets

A sequence in X is a function from N to X, denoted by xn = x(n) 2 X. Because ofthat, we usually think of a sequence as a subset of X indexed by N. A net is a directgeneralization of the notion of a sequence, by taking a more general index set which hasa directed order relation.

De�nition 1.4.1 A preorder � on a set I is a binary relation satisfying:

(o.i) (re�exivity) 8x 2 I x � x;(o.ii) (transitivity) 8x; y; z 2 I x � y; y � z =) x � z.

2In this setting a net of sequences f(xn)�g converges to yn i¤ for every ' 2 l1 it holds ' ((xn)�) !'(yn) in the usual euclidean topology, see the section 1.9 on weak topologies.

1.4. NETS 17

A (partial) order � on a set I is a preorder satisfying

(o.iii) (antisimmetricity) 8x; y 2 I x � y; y � x =) x = y:

A directed order � (or direction) on a set I is a preorder on I satisfying

(o.iv) (upper bound existence) 8x; y 2 I 9z 2 I s.t. z � x; z � y:

A directed set is any set I equipped with a directed order �.

Facts 1.4.2 In a directed set every �nite set has an upper bound.

The master example of a directed set is (N;�), and most (but not all) useful directedorders are partial orders.

Exercise 1.4.3 Prove that the following are directed sets:

1. (0;+1] with the usual order � (assuming +1 � x 8x 2 R).2. (0; 1) with the usual order �.3. A neighborhood system Nx of a point x in a topological space, under the order V �W () V � W .

4. All �nite subsets of a set X under the order V � W () V � W .

If there is no confusion, the direction � is often replaced by �. Notice that Example1.4.3 (3) is the main motivation for introducing nets.

Facts 1.4.4 1. A;B directed sets =) A�B directed set under

(a; b) � (c; d) () a � c; b � d:

2. fDi : i 2 Ig family of directed sets =) D =Yi2IDi directed set under

(ai)i2I � (bi)i2I () ai � bi 8i 2 I

The direction of the cartesian product is called product direction.

De�nition 1.4.5 A net in a set X is a function x : D ! X, where D is a directed set,called the index set of the net. The members � 2 D are called indexes. The functionx(�) is denoted simply by fx�g�2D.

The master examples of nets are sequences.

De�nition 1.4.6 A net fx�g in a topological space (X; �) converges to a limit x 2 Xif it is eventually3 in every neighborhood of x, i.e., if

8V 2 Nx 9�0 2 D s. t. x� 2 V 8� � �0:

This is denoted by writing x� ! x or x��! x or lim�2D x� = x.

3Eventually literally means �from a certain point on�: here it means for all indexes � equal or greaterthan a certain index �0.


Remark 1.4.7 In a metric space x� ! x () d(x�; x)! 0.

The limit of a sequence may be not unique. More precisely, limits are unique exactlyin Hausdor¤ spaces.

Theorem 1.4.8 (Unicity of limit) (X; �) Hausdor¤ () every net converges to atmost one limit.

Proof Suppose that a net fx�g converges to both x; y 2 X, with x 6= y. Then 8U 2Nx;8V 2 Ny the net x� stays eventually in both U and V , so x� 2 U \ V which is thennonempty. This proves that (X; �) is not Hausdor¤.

For the converse, suppose that X is not Hausdor¤. Then 9x; y 2 X with x 6= y ands.t. 8U 2 Nx;8V 2 Ny U \ V 6= ;. Taking an element xU;V 2 U \ V for every pair(U; V ) 2 Nx �Ny we obtain a net which converges to both x and y. �

By the above theorem a space in which a sequence converges to more limits is notHausdor¤. We provide two examples.

Example 1.4.9 Let X be a set with at least two points endowed with the indiscrete to-pology. Then every net converges to every point.

Example 1.4.10 (Line with a doubled origin4) Let X = (R�f0g)[f01; 02g, where 01; 02are two distinct symbols. Consider the family

B = f(a; b) � R� f0ggGf(��; 0) [ f0kg [ (0; �) 8� > 0; k = 1; 2g

and endow X with the topology � generated by B: B is a base of � (by Facts 1.1.42, sinceB is closed under �nite intersection and the union of its members is X), so the open setsare unions of sets in B. X is not Hausdor¤ because every neighbourhood of 01 intersectsevery neighbourhood of 02.Then, for example, the sequence xn = 2�n converges to both 01 and 02.

Recall Facts 1.2.3, in particular D = D [ D0. We want now to characterize thepoints in D. While in metric spaces sequences su¢ ce to describe closure points of sets(and several other properties as well), nets must be used to describe similar properties ingeneral topological spaces.

Exercise 1.4.11 Consider the topology on R in which a set U is open if U = V � C,where V is open in the usual topology and C is countable (open-minus-coutable topology).Then (0; 1) = [0; 1], but no sequence in (0, 1) converges to either 0 or 1. Moreover in thisspace a sequence converges i¤ it is eventually constant.

4The line with the doubled origin serves as an example of a space that is locally homeomorphic to R(every point has an open neighbourhood that is homeomorphic to R) but not Hausdor¤, illustrating thatthe Hausdor¤ requirement in the de�nition of a manifold is not redundant.

1.4. NETS 19

Theorem 1.4.12 The closure D of a set D is the set of limits of nets in D.

Proof Let x 2 D. If V 2 Nx, then V \ D 6= ;, thus 9xV 2 V \ D. This de�nes a netfxV gV 2Nx and xV ! x. (Why?)

Conversely, if a net fx�g in D satis�es x� ! x then, by de�nition of D, x 2 D.(Why?) �

The notion of subnet generalizes the notion of a subsequence.

De�nition 1.4.13 If fx�g and fy�g are nets from directed sets A and B respectively,then fy�g is a subnet of fx�g if there exists a function h : B ! A such that y� = xh(�)8� 2 B satisfying the property:

(sn) 8� 2 A 9�0 2 B such that h(�) � � 8� 2 B with � � �0.

Some authors, instead of property (sn), require that h is order preserving and co�nal:

(sn.i) (h monotone, or order preserving) �0 � � =) h(�0) � h(�);(sn.ii) (h is co�nal) 8� 2 A 9� 2 B such that h(�) � �:

The properties (sn.i) and (sn.ii) imply (sn), but (sn) really includes everything neededfor subnets.

Of course, every subsequence fxnkg of a sequence fxng is a subnet by considering h asthe map h(k) = nk, which is monotone and co�nal. But subnets are much more general:the main feature of subnets is the relaxation on the index set B of the subnet, which isnot a subset of A (as in subsequences) but it only needs to be mapped in A via a maph satisfying (sn). Because of that, the index set B may have larger cardinality than theoriginal one A. For example, under the condition of respecting property (sn), a subnetcan use a same point fx�g multiple times and can permute elements. The next exampleshows how subsequences are pathological w.r.t. a directed set.

Example 1.4.14 (W. Gähler, Grundstrukturen der Analysis I, 1977). A sequence fy�g�2Nis a subnet of a sequence fx�g�2N () the map h : N! N is the composition of a per-mutation with a dilatation, where a dilatation l : N! N is an order preserving map5 anda permutation � : N! N is a 1:1 map.6

Exercise 1.4.15 De�ne the sequence fxng by xn = n2+1. Verify that the net fym;ng(m;n)2N�Nde�ned by ym;n = (m + n)2 + 1 is a subnet of the sequence fxng, by the index functionh : N� N! N; h(m;n) = m+ n (use the product order in N� N).

Exercise 1.4.16 Consider the nets:5Also said isotone map, or monotone map.6Permutations are given by reordering the indexes. Dilatations are given by inserting after an index

some repetitions of it (a �nite number) and by repeating this procedure (eventually in�nitely many times).


� fy�g�2(0;1); y� = 1=� where (0; 1) is directed by �1 � �2 () �1 � �2;� fx�g�2(1;+1); x� = � where (1;+1) is directed by �1 � �2 () �1 � �2:

Then, fy�g is a subnet of fx�g and fx�g is a subnet of fy�g, by the invertible indexfunction h : (0; 1)! (1;+1); h(x) = 1=x.

Subnets are associated with limit points of nets.

De�nition 1.4.17 An element x in a topological space is a limit point (or clusterpoint, or accumulation point) of a net fx�g�2A if 8V 2 Nx, 8� 2 A 9e� � � such thatxe� 2 V . The (possibly empty) set of all limit points of fx�g�2A is denoted Lima2Afx�gor fx�g�2A.

Theorem 1.4.18 x 2 Limfx�g () 9 fy�g subnet of fx�g s.t. y� ! x.

Proof Let x be a limit point of a net fx�g�2A. Then 8 (�; V ) 2 A�Nx (directed by theproduct direction) choose h�;V 2 A s.t. h�;V � � and xh�;V 2 V . De�ne y�;V = xh�;V8 (�; V ) 2 A�Nx: the net fy�;V g is a subnet of fx�g by the index function h (�; V ) = h�;V .(Why?) Further, y�;V ! x. (Why?)

For the converse, assume that a subnet fy�g�2B of fx�g�2A converges to x and leth : B ! A be the index function of the subnet. Fix the choice of �0 2 A and of V 2 Nx:we aim to show that 9 e� 2 A s.t. xe� = xh(e�) = ye� 2 V . Since y� ! x, then 9�0 2 B s.t.y� 2 V 8� � �0. Since fy�g is a subnet of fx�g, then 9�1 2 B s.t. h(�) � �0 8� � �1.Let e� s.t. e� � �0 and e� � �1 and set e� = h(e�): then xe� = xh(e�) = ye� 2 V . �

Theorem 1.4.19 A net converges to a point x i¤ every subnet converges to x.

Proof Let fx�g be a net converging to x: clearly every subnet fy�g satis�es y� ! x.

For the converse, assume that every subnet of fx�g converges to x, and suppose thatx� 9 x. Then 9V 2 Nx s.t. 8� 2 A 9h� � � s.t. xh� =2 V . De�ne fy�g by y� = xh� :fy�g is a subnet of fx�g not converging to x. �

We now describe what happens in R with the euclidean topology: this is exactly aswith sequences.

Facts 1.4.20 (limsup and liminf) 1. Every bounded net fx�g � R has:�a largest limit point, called limit superior and written lim supfx�g;�a smallest limit point, called limit inferior and written lim inffx�g.2. The net finf e��fxe�gg�2A is increasing, the net fsupe��fxe�gg�2A is decreasing.

Moreover, every member of the �rst net is always less than or equal to every member ofthe second net.

3. It holds

lim inf�

fx�g = lim�

�infe��fxe�g

�� lim

�

�supe��fxe�g

�= lim supfx�g

1.5. CONTINUOUS FUNCTIONS 21

4. x� ! x () x = lim inf�fx�g = lim supfx�g.5. Every net fx�g � R has:

�a largest limit point, called limit superior and written lim supfx�g;�a smallest limit point, called limit inferior and written lim inffx�g.These may be �1. The results in points (2)�(4) hold again, and may be proven exacltyin the same way.

1.5 Continuous functions

De�nition 1.5.1 A function f : X ! Y between topological spaces is continuous at apoint x 2 X if f�1(V ) is a neighborhood of x for each neighborhood V of f(x).

A function f : X ! Y between topological spaces is continuous if f�1(U) is open in Xfor each open set U � Y .

Remark 1.5.2 In a metric space, continuity at a point x reduces to the familiar " � �de�nition:

8" > 0 9� = �(") > 0 : f(B�(x)) � B"(f(x)):or, in X � R, the more common one:

8" > 0 9� = �(") > 0 : 8x0 2 X with jx0 � xj < � it holds jf(x0)� f(x)j < �:

Proof If f is continuous at x, then B"(f(x)) is a neighborhood of f(x), so f�1(B"(f(x))is a neighborhood of x and 9� > 0 s.t. B�(x) � f�1(B"(f(x)), which is equivalent to sayf(B�(x)) � B"(f(x)).Conversely, if 8" > 0 9� > 0 s.t. f(B�(x)) � B"(f(x)) and V is a neighborhood of f(x),

then 9" > 0 s.t. B"(f(x)) � V . For such � there exists � > 0 with f(B�(x)) � B"(f(x)),so B�(x) � f�1(B"(f(x)) � f�1(V ) and then f�1(V ) is a neighborhood of x. �

Facts 1.5.3 For a function f : X ! Y between topological spaces the following areequivalent:

1. f is continuous;2. f is continuous at every point;3. C � Y; C closed =) f�1(C) is a closed;4. 8B � Y f�1(B�) � (f�1(B))�;5. 8A � X f(A) � f(A);6. f�1(V ) open for each V in some subbase for the topology on Y.

Continuity is often more easily expressed in terms of convergence of nets.

Theorem 1.5.4 A function f : X ! Y between topological spaces is continuous at x 2 Xi¤ for all nets fx�g � X such that x� ! x then f(x�)! f(x).


Proof ()) Assume that x� ! x: we aim to show that f(x�) ! f(x). For any neigh-borhood of f(x) , say V 2 Nf(x), by continuity of f at x there exists U 2 Nx such thatf(U) � V . Since x� ! x, 9�0 such that 8� � �0 it holds x� 2 U and thus f(x� ) 2 V .Summing up, 8V 2 Nf(x) 9�0 2 A such that f (x� ) 2 V 8� � �0.(() Assume that f is not continuous at x. Then 9V 2 Nf(x) such that 8U 2 Nx it

holds f(U) * V . But this means 8U 2 Nx 9xU 2 U s.t. f(xU) =2 V . Consider the netfxUgU2Nx: clearly xU ! x. Moreover f (xU) 9 f (x), since V 2 Nf(x) and f (xU) =2 V8U 2 Nx. �

Theorem 1.5.5 If f; g : X ! R are continuous real�valued functions on a topologicalspace, then the functions f + g, f � g, minff; gg, maxff; gg, jf j are also continuous. Ifmoreover g(x) 6= 0 8x 2 X, then f=g also is continuous.

Theorem 1.5.6 The composition of continuous functions is continuous.

De�nition 1.5.7 A function f : X ! Y is called a homeomorphism if it is a 1:1continuous function whith continuous inverse f�1. Two topological spaces X and Y arehomeomorphic if there is a homeomorphism between them: in this case we write X ' Y .

From the topological point of view two homeomorphic spaces are identical, and cannotbe distinguished: indeed, any topological property, that is, any property de�ned in termsof the topology, possessed by one space is also possessed by the other.

Exercise 1.5.8 Find a homeomorphism (�1; 1) ' R. Can you extend it to [�1; 1] ' R?

Theorem 1.5.9 (Di¢ cult) Euclidean spaces of di¤erent dimensions are not homeomorphic.

De�nition 1.5.10 A mapping f : X ! Y between two topological spaces is an embed-ding if f : X ! f(X) is a homeomorphism. In this case we can identify X with its imagef(X) � Y .

1.6 Compactness

De�nition 1.6.1 Let K be a subset of a topological space X. A cover of K is a collectionfD�g�2A of subsets D� � X s.t. K � [�2AD�.

De�nition 1.6.2 A subset K of a topological space X is compact if every open cover ofK admits a �nite subcover, that is, every family fU�g�2A of open sets s.t. K � [�2AU�admits a �nite subfamily U�1 ; U�2 ; : : : ; U�n s.t. K � [ni=1U�i.

De�nition 1.6.3 A topological space is called a compact space if it is a compact set.

1.6. COMPACTNESS 23

Example 1.6.4 In the trivial topology every set is compact.

Example 1.6.5 In the discrete topology, a set is compact i¤ it is a �nite set.

Lemma 1.6.6 If K is a compact subset of a Hausdor¤ space, and x =2 K, then 9U; Vopen and disjoint with K � U and x 2 V . In particular, compact subsets of Hausdor¤spaces are closed.

Proof Since X is Hausdor¤, for each choice of a point y 2 K there exist open neighbor-hoods Uy 2 Ny of y and Vy 2 Nx of x which are disjoint. Since K � [y2KUy, there is a�nite subfamily Uy1 ; : : : ; Uyn covering K. Take U = [ni=1Uyi and V = \ni=1Vyi: since forevery i = 1; : : : ; n it holds V � Vyi, it follows that V does not intersect any Uyi and thusit does not intersect U .

To show that K is closed, apply the lemma for each x =2 K: then exists an openneighborhood Vx 2 Nx of x not intersecting K. Now realize that KC = [x=2KVx, so K isclosed. �

Example 1.6.7 Compact subsets of non-Hausdor¤ spaces need not be closed. For ex-ample, let X = fx; yg with the indiscrete topology: any singleton is compact, but X is theonly nonempty closed set.

Theorem 1.6.8 (X; d) metric space, K is compact =) K closed and bounded.

Proof Being X Hausdor¤, by the previous Lemma K is closed. Since X = [n2NBn(0)and K is compact, K is contained in a �nite union of Bn(0). Let M be the maximumindex in this union: then K � BM(0) and it is bounded. �

The converse is true in Rn with the usual topology but it is not true in general.

Theorem 1.6.9 (Heine�Borel) In (Rn; d2) K closed and bounded =) K is compact.

Because of the equivalence that follows in Rn between compact sets and closed andbounded sets, Heine�Borel Theorem is often mistaken for the de�nition of compactness:In a general metric space this theorem is false: for instance, consider an in�nite set withthe discrete metric, where every set is bounded and closed.

Facts 1.6.10 1. Finite sets are compact.

2. Finite unions of compact sets are compact.

3. Closed subsets of compact sets are compact.

4. If K � Y � X, then K is a compact subset of X i¤ K is a compact subset of Y (inthe relative topology).


We now come to the characteriziation of compact sets via nets.

Theorem 1.6.11 For a topological space X, the following are equivalent:

1. X is compact.

2. Every net in X has a convergent subnet.

Proof )) Let fx�g be a net not having limit points: 8x 2 X by de�nition of a limitpoint, it holds 9Ux 2 Nx and 9�x 2 A s.t. x� =2 Ux 8� � �x. But then fUxgx2X is anopen cover of X s.t. any �nite subfamily Ux1 ; Ux2 ; : : : ; Uxn does not contain the points x�for � � �x1 ; �x2 ; : : : ; �xn (such a � exists by property o.iv). In other words, fUxgx2X isan open cover of X not admitting a �nite subcover, thus X is not compact.

() Assume X not compact: there exists an open cover fU�g�2A of X which does nothave a �nite subcover. This means that for any �nite subset S � A then [�2SU� 6= Xand thus 9xS =2 [�2SU�. But then, the set �(A) = fS � A; S �niteg is a directed setw.r.t. the natural inclusion and the terms fxSgS2�(A) form a net which does not have alimit point, which is absurd.

We prove this last statement. By construction of xS, xS =2 [�2SU� 8S 2 �(A). IfxS ! x, taken �o 2 A s.t. x 2 U�o it holds 9S 2 �(A), S � f�og s.t. xS 2 U�0. For suchan S we have xS 2 U�0, U�0 � [�2SU� since f�og � S, and xS =2 [�2SU�. �

Property (2) stated with sequences is not equivalent to compactness in a generic to-pological space, and is referred to as sequentially compactness. This is one of the mainmotivations that brought to the concept of nets, as a �working� generalization of se-quences.

De�nition 1.6.12 A subset A of a topological space is sequentially compact if everysequence in A has a subsequence converging to an element of A.

Exercise 1.6.13 1. In a topological space, if a sequence fxng converges to a point x, thenthe set S = fx; x1; x2; :::g is compact.2. (Nets need not exhibit this property) Let A = (0; 1) \Q, directed by the usual ordering� on R. Consider the space X = [0; 1] and the net fx� = �g�2D. Show that:a. x� ! 1 in X;b. the support set S of fx�g is not compact;c. every y 2 (0; 1) is an accumulation point of S but not a limit point of fx�g�2D.

However, in some cases, one can use sequences instead of nets. These cases are not sorare: all we need is to require that the space X is �rst countable (see De�nition 1.1.48),so, for example, we can do that on all metric spaces, since they are �rst countable.

Facts 1.6.14 Let X be a �rst countable topological space.

1. If A � X, then x 2 A i¤ there is a sequence in A converging to x.2. A function f : X ! Y is continuous i¤ xn ! x implies f(xn)! f(x).

1.6. COMPACTNESS 25

Another (not so useful) characterization of compactness is via the �nite intersectionproperty.

De�nition 1.6.15 A family of sets has the �nite intersection property (FIP) if every�nite subfamily has a nonempty intersection.

Theorem 1.6.16 For a topological space X the following are equivalent:

1. X is compact

2. Every family of closed subsets of X with the FIP has a nonempty intersection.

Proof )) Let fC�g�2A be a family of closed subsets having the FIP. If \� 2 AC� = ;,then [�2ACc� = X and fCC� g�2A is an open cover of X. Being X compact, 9�1; �2; : : : ; �ns.t. X = [ni=1Cc�i, but then (\

ni=1C�i)

C = [ni=1Cc�i = X, i.e., \ni=1C�i = ;. () LetfU�g�2A be an open cover of X. Since [�2AU� = X, \�2AUC� = ; and thus the FIP mustfail. But this means 9�1; �2; : : : ; �n s.t. \ni=1C�i = ;, i.e., [ni=1U� = X. �

We now study the correlation apsects between continuous maps and compact sets.

Theorem 1.6.17 Every continuous function between topological spaces carries compactsets to compact sets.

Proof Let f : X ! Y be the continuous function and K the compact subset of X.

Consider an open cover of f(K), say fU�g. Clearly ff�1(U�)g is an open cover ofK and thus 9�1; �2; : : : ; �n s.t. K � [ni=1f�1(U�i). Applying f again, we have f(K) �f ([ni=1f�1(U�i)) = [ni=1f (f�1(U�i)) = [ni=1U�i. �

As a corollary of this result and Theorem 1.6.8, we obtain the Weierstrass theorem.

Theorem 1.6.18 (Weierstrass) A continuous real-valued function de�ned on a compactspace achieves its maximum and minimum values.

Proof If f : K ! R is a continuous function, then f(K) is a compact set, and thus itis closed and bounded. Since it is a bounded set, 9N = inf f(K), M = sup f(K), sinceit is closed the sup and the inf are points in f(K), hence 9x1; x2 s.t. N = f(x1) andM = f(x2), i.e., N = minx2K f(x) and M = maxx2K f(x). �

Facts 1.6.19 A 1:1 continuous function from a compact space onto a Hausdor¤ space isa homeomorphism.


1.7 Semicontinuous functions

Suppose X is a topological space, R = [�1;+1] with the usual topology, and f : X ! Ris a function from X into the extended real numbers R. We want to study

De�nition 1.7.1 A function f : X ! R on a topological space X is:

(lsc) lower semicontinuous (LSC) if the strictly�upper level sets are open:

L>a := f�1(a;+1) = fx 2 X : f(x) > ag open 8a 2 R;

(usc) upper semicontinuous (USC) if the strictly�lower level sets are open

L<a := f�1(�1; a) = fx 2 X : f(x) < ag closed 8a 2 R:

Facts 1.7.2 The following holds for a function f : X ! R:1. f LSC() L�c = fx 2 X : f(x) � cg closed 8c 2 R:2. f USC() L�c = fx 2 X : f(x) � cg closed 8c 2 R.3. f continuous () f USC and LSC

As with continuity, we may de�ne a notion of semi-continuity at a point xo 2 X.

De�nition 1.7.3 Given a function f : X ! R on a topological space X and a pointxo 2 X, the function f is:

(lsc.p) lower semicontinuous at x0 if 8" > 0 9U 2 Nx0 s.t. f(x) � f(x0)� " 8x 2 U ;(usc.p) upper semicontinuous at x0 if 8" > 0 9U 2 Nx0 s.t. f(x) � f(x0) + " 8x 2 U:

Facts 1.7.4 Given a function f : X ! R on a topological space X, then:1. f LSC() f LSC at each point xo 2 X.2. f USC() f USC at each point xo 2 X.

Proof Assume f : X ! R lower semicontinuous and choose xo 2 X and " > 0. Theset U = f�1(f(x0)� ";+1) is open and contains x0. Thus U 2 Nx0 and clearly f(x) �f(x0)� " 8x 2 U .Conversely, assume that f : X ! R is lower semicontinuous at each xo 2 X and

choose a 2 R. Consider the set A = f�1(a;+1) and a point x0 2 U . Since f is lowersemicontinuous at xo there exists an open neighborhood Ux0 of x0 s.t. f(x) � f(x0) � "8x 2 Ux0. But then A = [x02AUx0 is an open set. �

Theorem 1.7.5 Given a function f : X ! R on a topological space X and a pointx0 2 X, then:1. f is lower semicontinuous at x0 i¤ x� ! x =) lim inf

�f(x�) � f(x0);

2. f is upper semicontinuous at x0 i¤ x� ! x =) lim sup�

f(x�) � f(x0):

3. When X is �rst countable, nets can be replaced by sequences

1.8. COMPARING TOPOLOGIES 27

Proof For the de�nition of lim inf or lim sup see Facts 1.4.20. We establish the lowersemicontinuous case, being the upper semicontinuous case analogous.

Assume f lower semicontinuous at x0 and x� ! x. Choose " > 0 and let U 2 Nx0 asin the de�nition. Since x� ! x, 9�0 s.t. x� 2 U 8� > �0. But then f(x�) � f(x0) � "8� > �0, that is inf

��0f(x�) � f(x0)� ". Recall that lim inf

�f(x�) = sup

�o

inf��0

f(x�): then

we have proved that lim inf�

f(x�) � f(x0)� " 8" > 0, that is lim inf�

f(x�) � f(x0).

For the converse, suppose that f is not lower semicontinuous at x0. Then 9" > 0 s.t.8U 2 Nx0 9xU 2 U with f(xU) < f(x0)�". As a consequence, the net fxUgU2Nx0 satis�esxU ! x (by construction) and lim inf

Uf(xU) < f(x0). Indeed 8U0 9U 2 Nx0, U � U0 and

for such U we have f(xU) < f(x0) � ": this implies infU�U0

f(xU) < f(x0) � " 8U0 2 Nx0,

and thuslim inf

Uf(xU) = lim

U0infU�U0

f(xU) � f(x0)� " < f(x0):

We do not prove the third statement (see also Fact 1.6.14). �

The following result generalizesWeierstrass�Theorem (Theorem 1.6.18) on the extremevalues of continuous functions.

Theorem 1.7.6 (Weierstrass) A real-valued lower semicontinuous function on a com-pact space attains a minimum value, and the nonempty set of minimizers is compact.

Similarly, an upper semicontinuous function on a compact set attains a maximumvalue, and the nonempty set of maximizers is compact.

Proof Let X be a compact space and let f : X ! R be lower semicontinuous. PutA = f(X), and for each c 2 A, consider the lower level set L�c = fx 2 X : f(x) � cg:since f is LSC, 8c 2 A the set L�c is closed and nonempty. Furthermore, the familyfL�c : c 2 Ag has the �nite intersection property: \ni=1L�ci = L�minfc1;c2;:::;cng, which isnotempty. But then, since X is compact, \c2AL�c is compact and nonempty. Note thatany element x0 2 \c2AL�c is a point of minimum: f(x0) � c 8c 2 A = f(X). �

1.8 Comparing topologies

De�nition 1.8.1 The family of all topologies on X is partially ordered by set inclusion.If � 0 � � , that is, if every � 0-open set is also � -open, then we say that � 0 is weaker orcoarser than � and that � is stronger or �ner than � 0.

Facts 1.8.2 For two topologies � and � 0 on a set X the following are equivalent:

1. � 0 is weaker than � , that is, � 0 � � .2. Every � 0-closed set is also � -closed.

3. The identity mapping id : x 7! x, from (X; �) to (X; � 0), is continuous.


4. Every � -convergent net is also � 0-convergent to the same point.

5. The � -closure of any subset is included in its � 0-closure.

6. The � 0-interior of any subset is included in its � -interior.

Facts 1.8.3 If � 0 is weaker than � , then each of the following holds:

1. Every � -compact set is also � 0-compact.

2. Every � -dense set is also � 0-dense.

3. Every � 0-continuous function on X is also � -continuous.

When we have a choice of what topology to put on a set, there is the following roughtradeo¤. The �ner the topology, the more open sets there are, so that more functions fromthat set are continuous. On the other hand, there are also more insidious open covers ofa set, so there tend to be fewer compact sets and there are less continuous functions tothat set.

1.9 Weak topologies

There are two classes of topologies that are are of wide interest: the class of topologiesgenerated by a metric and the one of weak topologies.

De�nition 1.9.1 Let X be a nonempty set, let f(Yi; �i)gi2I be a family of topologicalspaces and fi : X ! Yi be a family of function. The weak topology on X generated bythe family of functions ffigi2I is the weakest topology on X that makes all the functionsfi continuous:

� =\

�0: fi continuous 8i2I

�0:

Facts 1.9.2 1. A subbase for the weak topology is:

Sw0 = ff�1i (V ) : 8i 2 I; 8V 2 �ig:

Hence, a subset U � X is open in the weak topology i¤ U = [�W�, where each W� is a�nite intersection of sets of the form '�1(V ) with ' 2 ffigi2I and V 2 �i.2. If Si is a subbase for �i, then a subbase for the weak topology is:

Sw = ff�1i (V ) : 8i 2 I; 8V 2 Sig:

3. If Si is a subbase for �i, then a base for the weak topology is:

Bw = fn\

ik=1

f�1ik (Vik) : 8n 2 N; 8ik 2 I; 8Vik 2 Sik ; for all k = 1; : : : ; ng:

1.9. WEAK TOPOLOGIES 29

Lemma 1.9.3 (Weak convergence) In the weak topology, x�w�! x i¤ fi(x�)

�i�!fi(x) 8i 2 I.

Proof One side is obvious: if x� ! x then, since each fi is weak continuous, fi(x�) !fi(x) for all i 2 I.Conversely, assume that fi(x�)

�i�! fi(x) 8i 2 I and let V = \nik=1f�1ik(Vik) be a

neighborhood of x in Bw. Since fi(x�)�i�! fi(x) 8i 2 I, then 8k = 1; : : : ; n 9�k s.t.

8� � �k it holds x� 2 f�1ik (Vik). But then pick �0 = maxf�1; : : : ; �kg: 8� � �0 x� 2 Vik8k = 1; : : : ; n and thus x� 2 V . This means exactly x� ! x. �

An apparently particular case is the weak topology generated by a family of realfunctions F on X. Despite the appearence, this case is very important and, a posteriori(see Theorem 1.9.11), not so particular! From the rest of this section we assume to havea family of real functions F on a set X and we will study the weak topology generatedby that family F on X.

De�nition 1.9.4 Let F be a family of real functions on X. The weak topology generatedby F is denoted by �(X;F).

Lemma 1.9.5 A subbase of the weak topology �(X;F) is the family of the sets

U(f; x; ") = fy 2 X : jf(y)� f(x)j < "g; 8f 2 F ; 8x 2 X; 8" > 0:

Proof Recall Facts 1.9.2 and notice that here the index set I is the given family F ofreal functions. Then use as subbase Sf of the standard topology on the codomain R of afunction f 2 F the open balls with "-radius centered at the point f(x), for any choice of" > 0, x 2 X:

Sf = fV (f; x; ") = B"(f(x)) : 8" > 0;8x 2 Xg:Finally, note that the preimage of the set V (f; x; ") via the function f is precisely the setU(f; x; "). �

De�nition 1.9.6 A family F of real functions on X separates points if

8x; y 2 X : x 6= y 9f 2 F : f(x) 6= f(y):

Lemma 1.9.7 The weak topology �(X;F) is Hausdor¤ i¤ F separates points.

Proof Assume that F separates points: taken x; y 2 X with x 6= y, then 9f 2 F s.t.f(x) 6= f(y) and (being R Hausdor¤) there exist opens U 2 Nf(x); U

0 2 Nf(y) withU \ U 0 = ;. But then f�1(U), f�1(U 0) are two disjoint open sets containing x and yrespectively.

Conversely, suppose that F does not separate points, that is, 9x; y 2 X with x 6= ysuch that 8f 2 F it holds f(x) = f(y). But then, 8f 2 F it holds y 2 U(f; x; "): sincethese sets form a subbase for �(X;F) (see Lemma 1.9.5), then every open set containingx contains also y and thus �(X;F) is not Hausdor¤. �


Lemma 1.9.8 (Relative weak topology) Let F be a family of real functions on Xand Y � X. Then

�(X;F)jY = �(Y;FjY ):

Proof By Lemma 1.9.5, a subbase for the restricted weak topology �(X;F)jY is

S = fU(f; x; ") \ Y : 8f 2 F ; x 2 X; " > 0g:

De�neUY (f; x; ")g = fy 2 Y : jf(y)� f(x)j < "g 8f 2 F ; x 2 Y; " > 0:

By the same Lemma, a subbase for the weak topology �(Y;F) is:

S 0 = fUY (f; x; ") : 8f 2 F ; x 2 Y; " > 0g:

Clearly, for all x 2 Y it holds U(f; x; ") \ Y = UY (f; x; "). If instead x =2 Y , choosefor each z 2 U(f; x; ") \ Y a su¢ ciently small "z > 0 s.t. U(f; z; "z) � U(f; x; "): then

U(f; x; ") \ Y =[

z2U(f;x;")\Y

U(f; z; "z) \ Y =[

z2U(f;x;")\Y

UY (f; z; "z):

and therefore the sets in S can be constructed as unions of sets in S 0. Since converselythe sets in S 0 are particular sets of S, it follows that �(X;F)jY = �(Y;F). �

Usually the choice of the family F is one of the following vector spaces:

1. RX = ff : X ! Rg, all real-valued functions on X;2. C(X) = ff : X ! R, f continuousg, all continuous real-valued functions on X.3. Cb(X) = ff : X ! R, f continuous and boundedg, all continuous and boundedreal-valued functions on X;

4. Cc(X) = ff : X ! R, f continuous, supp f compactg, all continuous real-valuedfunctions on X with compact support, where the support of a function is theclosure of the set of points where f is not zero-valued:

supp f = fx 2 X : f(x) 6= 0g:

Lemma 1.9.9 Let X be any topological space. Then �(X;C(X)) = �(X;Cb(X)).

Proof Clearly, since Cb(X) � C(X), the sets fU(f; x; ") : 8f 2 Cb(X); x 2 X; " > 0gare contained in the sets fU(f; x; ") : 8f 2 C(X); x 2 X; " > 0g and thus �(X;Cb(X)) ��(X;C(X)):

Conversely, consider a subbasic open set U(f; x; ") where f 2 C(X); x 2 X; " > 0.De�ne the function g : X ! R by

g(z) =

8<:f(x)� " if f(z) � f(x)� "f(z) if f(x)� " < f(z) � f(x) + "f(x) + " if f(z) > f(x) + "

:

1.9. WEAK TOPOLOGIES 31

Then g 2 Cb(X)and U(g; x; ") = U(f; x; "). Thus �(X;C(X)) � �(X;Cb(X)): �

We now show how weak topologies generated by a family of real-valued functions arenot so �particular�.

De�nition 1.9.10 (Separable axioms) A topological space (X; �) is completely reg-ular if given a point x 2 X and a closed set F with x =2 F there exists a continuousfunction f : X ! R s.t. f(x) = 0 and f(y) = 1 8y 2 F , shortly:

8x 2 X;8F : FC 2 � and x =2 F 9f continuous s.t. f(x) = 0 and f jF = 1:

It is immedate to show that completely regular spaces are also regular, by using thefunction f to provide the required open sets (see De�nition 1.1.15). We can use weaktopologies to characterize completely regular spaces.

Theorem 1.9.11 A topological space (X; �) is completely regular i¤ � = �(X;C(X)).7

Proof Clearly, for any topological space (X; �) we have �(X;C(X)) � � .Assume that (X; �) is completely regular and choose U 2 � . For any point x 2 U pick

fx 2 C(X) satisfying fx(x) = 0 and fxjUc = 1. Then x 2 U(fx; x; 1) = fy 2 X : jfx(y)j <1g � U and so U =

Sx2UU(fx; x; 1) 2 �(X;C(X)).

Suppose now that � = �(X;C(X)). Let F be closed and x =2 F . Since x 2 FC 2�(X;C(X)), by Lemma 1.9.5 there exists an open neighborhood U � FC of x of the form:

U =N\i=1

U(fi; x; "i) =N\i=1

fy 2 X : jfi(y)� fi(x)j < "ig,

where each fi 2 C(X).For each i = 1; : : : ; N de�ne efi = 1

"ifi. Then efi 2 C(X) and

U =

N\i=1

U(efi; x; 1) = N\i=1

fy 2 X : jefi(y)� efi(x)j < 1g.For each i = 1; : : : ; N take gi (y) = minfjefi(y)� efi(x)j; 1g and g (y) = maxfgi (y)g. Thengi (y) 2 C(X), g (y) 2 C(X) and

g (x) = 0, g (y) = 1 8y 2 UC ,

since if y 2 UC there exists an index i s.t. jefi(y)� efi(x)j � 1, so gi (y) = 1 and g (y) = 1.Summing up, g is continuous, g(x) = 0 and gjF = 1, since F � UC . Thus X is completelyregular. �

7Recall that �(X;C(X)) = �(X;Cb(X)) by Lemma 1.9.9.


Corollary 1.9.12 The completely regular spaces are precisely those whose topology is theweak topology generated by a family of real functions.

Proof If (X; �) is completely regular, then by Theorem 1.9.11 � = �(X;C(X)).

Conversely, suppose � = �(X;F) for a given family F of real functions. We wantto show that (X; �) is completely regular by applying again Theorem 1.9.11. Since � =�(X;F), it follows that F � C(X), so � = �(X;F) � �(X;C(X)). On the other hand,for any topological space (X; �) we have �(X;C(X)) � � . Thus � = �(X;C(X)) and wecan conclude that (X; �) is completely regular. �

The next easy corollary of Theorem 1.9.11 and Lemma 1.9.3 characterizes convergencein completely regular spaces.

Corollary 1.9.13 If (X; �) is completely regular, x� ! x i¤ f(x�)! f(x) 8f 2 C(X).8

1.10 The product topology

De�nition 1.10.1 (Product space) Let f(Xi; �i)gi2I be a family of topological spaces.The cartesian product is the set X de�ned as

X =Yi2IXi = f(xi)i2I : xi 2 Xi 8i 2 Ig:

For each j 2 I the projection pj : X ! Xj is de�ned by pj ((xi)i2I) = xj. The producttopology is the weak topology on X generated by the family of projections � = fpi : i 2 Igand is denoted by �(X; �). The product space is the pair (X; �(X; �)). A functionf : X ! Y is called jointly continuous if it is continuous with respect to the producttopology.

Facts 1.10.2 1. A subbase for the product topology consists of all subsets of the formp�1j (Vj) =

Qi2I Vi where Vj 2 �j and Vi = Xi 8i 2 I, i 6= j.

2. A base for the product topology consists of all subsets of the formQi2I Vi where Vi 2 �i

and Vi = Xi for all but �nitely many i 2 I.3. A net f(x�i )i2Ig in X converges to a point (xi) and we write (x�i ) �! (xi) i¤ x�i

�i�! xi8i 2 I.4. The euclidean metric on Rn induces the product topology on Rn =

Qni=1R.

Proof 1. A subbase for �(X; �) is S = fp�1j (Vj) : 8j 2 I; 8Vj 2 �jg: just realize that(xi) 2 p�1j (Vj) () xj 2 Vj (the other xi�s are free in Xi). Thus p�1j (Vj) =

Qi2I Vi where

Vj 2 �j and Vi = Xi 8i 2 I, i 6= j.8Again, by Lemma 1.9.9 C(X) can be replaced by Cb(X).

1.10. THE PRODUCT TOPOLOGY 33

2. A base for �(X; �) is formed by �nite intersections of members of S: this intersectionis then of the form

Qi2I Vi where Vi 2 �i and Vi = Xi for all but �nitely many i 2 I.

3. Suppose (x�i ) �! (xi) in �(X; �). We consider its j-th projection 8j: pj is con-tinuous thus pj(x�i )! pj(xi), that is, x�j ! xj.Conversely, suppose x�i ! xi 8i 2 I. For any neighborhood U containing (xi) there existsa basic open set V satisfying (xi) 2 V � U : this open set is of the form

Qi2I Vi where

Vi 2 �i and Vi = Xi for all but �nitely many i 2 I. Let i1; i2; : : : ; in be these indices: 8j =i1; i2; : : : ; in 9�0j such that x�j 2 Vj 8� � �0j . But then de�ne �0 = maxf�01; �02; : : : ; �0ng:it holds x�j 2 Vj 8� � �0 and so (x�i ) 2 V � U 8� � �0, which means (x�i ) �! (xi) in�(X; �).

4. Let � be the euclidean topology on Rn. If V is a standard open set in R, thenp�1j (V ) is the rectangle having the j-th side equal to V and all the other sides equal toR: this is clearly an open set according to � , and thus �(X; �) � � .Conversely, let U 2 � be an euclidean open set and x 2 U a point in it. Then there existsan open rectangle Vx =

Qni=1 V

xi where V

xi = (ai; bi) 8i = 1; : : : ; n satisfying x 2 V � U .

But such a set Vx is a basic open set for �(X; �) (it is the intersection of the n rectangleswhich have the j-th side equal to V xj and all the other sides equal to R for j = 1; : : : ; n).Since U = [x2UVx, it holds U 2 �(X; �). �

Having de�ned products, we can now study an important topological object relatedto a function f : X ! Y between topological spaces: the graph (and the epigraph andhypograph).

De�nition 1.10.3 The graph of a function f : X ! Y is the set

Gr f = f(x; y) 2 X � Y : y = f(x)g:

For a function f : X ! R the epigraph is the set

epi f = f(x; c) 2 X � R : c � f(x)g � X � R

and the hypograph is the set

hyp f = f(x; c) 2 X � R : c � f(x)g � X � R:

Theorem 1.10.4 A continuous function into a Hausdor¤ space has closed graph.

Proof Assume that f : X ! Y is continuous and Y is Hausdor¤. The claim is that Gr fis a closed subset of X � Y . Suppose (x�; y�) ! (x; y) and that y� = f(x�) (this means(x�; y�) 2 Gr f). Since f is continuous, y� = f(x�) ! f(x). Since also y� ! y and Y isHausdor¤ (so the limit is unique) then y = f(x), so (x; y) 2 Gr f . �

As consequence, examples of continuous function with non-closed graph require a non-Hausdor¤ codomain (for instance, Y not a metric space).


Theorem 1.10.5 (Closed Graph Theorem) A function from a topological space intoa compact Hausdor¤ space is continuous i¤ its graph is closed.

Proof We already proved one direction. For the converse, assume that f : X ! Y , Y iscompact Hausdor¤ and Gr f is closed. Assume by contradiction that there exists a netfx�g�2A such that x� ! x and f(x�)9 f(x).

Since f(x�)9 f(x) it holds

9V 2 Nf(x) s.t. 8�0 9� � �0 s.t. f(x�) =2 V:

Consider the transition function h : B = A! A de�ned by h(�0) = � where the �0 and� are as above: then the corresponding subnet fx0� = xh(�)g�2B satis�es f(x0�) =2 V 8�(verify that h satisfy the property (sn) of the de�nition of a subnet).

Since Y is compact, the net ff(x0�)g has a convergent subnet, say ff(x00 )g 2� s.t.f(x00 ) ! y. Thus (x00 ; f(x

00 )) ! (x; y) and, since Gr f is closed, it follows that (x; y) 2

Gr f , which means y = f(x). Therefore f(x00 )! f(x), absurd since f(x00 ) =2 V 8 . �

Example 1.10.6 Let f : R ! R be de�ned by f(x) = 1=x if x 6= 0 and f(0) = 0. Itsgraph is closed (it is a union of three closed sets: the two branches of the hyperbola andthe origin) and f is not continuous.

Example 1.10.7 Consider as f the identity map in X = Y = [0; 1], but equip X withthe euclidean topology and Y with the discrete topology (richer). Then the graph is closedbut f is not continuous.

The closure of the epigraph/hypograph is instead equivalent to lower/upper semicon-tinuity.

Theorem 1.10.8 For a function f : X ! R the following holds:

1. f LSC () epi f closed.

2. f USC () hyp f closed.

Proof It is an immediate consequence of Theorem 1.7.5 and the de�nition of the producttopology. �

We now come to one of the most important compactness results in mathematics.

Theorem 1.10.9 (Tychono¤Product Theorem) The product of a family of topolo-gical spaces is compact in the product topology i¤ each factor is compact.

1.10. THE PRODUCT TOPOLOGY 35

We put a few comments before the proof of Tychono¤�s Theorem. Tychono¤originallyproved that an arbitrary product of compact intervals is compact, the general form is dueto µCech.

Moreover, one direction of the theorem is a trivial application of Theorem 1.6.17: ifX is compact then Xi = pi(X) is also compact since pi is a continuous function.

The core of the theorem is the converse part. There are di¤erent historical proofsof this direction, all involve some form of the axiom of choice. We provide two proofs:the �rst is the �Bourbaki�ultra�lter proof, given by Cartan, but rewritten without usingultra�lters and using families of closed sets having the Finite Intersection Property (FIP).The second is an alternative simple proof due to Cherno¤ (see [Cherno¤]), which usesonly nets. Both proofs use the Zorn�s Lemma to show the existance of a maximal set.

Lemma 1.10.10 (Zorn�s Lemma) Let P be a nonempty partially ordered set, such thatfor every totally ordered subset L, there exists an upper bound9 for L. Then P has amaximal element.

Cartan�s proof of Tychono¤�s Theorem. Let fXi : i 2 Ig be a family of topologicalspaces and X =

Qi2I Xi. We assume that each Xi is compact and prove that X is

compact by Theorem 1.6.16, that is, by proving that every family of closed subsets of Xhaving the �nite intersection property (FIP) has a nonempty intersection.

Let A be a family of closed subsets of X having the FIP. Forget that the sets in A areclosed and apply Zorn�s Lemma: A is contained into some maximal set B of subsets of Xhaving the FIP. Now 8i 2 I it is easy to see that the family of closed sets fpi(B) : B 2 Bghas the FIP and therefore, since Xi compact, by Theorem 1.6.16 9bi 2 pi(B) 8B 2 B.We go back to X and de�ne b = (bi)i2I .

We claim that to showTA2AA 6= ; it su¢ ces to check that B contains all the neigh-

borhoods of b. Indeed, if this is true, then all neighborhoods of b intersect each B 2 B.In particular, they intersect each A 2 A: hence 8A 2 A it holds b 2 A = A and thusb 2

TA2AA.

To show the claim, note that each neighborhood of b contains a basic open V =Qi2I Vi

where Vi 2 �i and Vi = Xi for all but �nitely many i 2 I and that it is enough to verify V 2B. Let i1; : : : ; in be the indices such that Vik 6= Xik and observe that V = \nk=1p�1ik (Vik).It is then enough to recall that bi 2 pik(B) 8B 2 B and so Vik \ pik(B) 6= ; 8B 2 B: thisimplies p�1ik (Vik) \B 6= ; 8B 2 B, 8k = 1; : : : ; n and therefore V \B 6= ; 8B 2 B.But then V 2 B. Indeed, B is a maximal family characterized by the FIP, so any �nite

intersectionTni=1Bi of elements in B is again in B: for any �nite family fBn+1; : : : ; Bmg �

B then (Tni=1Bi) \

�Tmi=n+1Bi

�=Ti=1;:::;mBi 6= ;. But then, since V \ B 6= ; 8B 2 B,

V intersect every �nite intersection of elements in B, and so V 2 B. �

Cherno¤�s proof of Tychono¤�s Theorem. Let fXi : i 2 Ig be a family of topologicalspaces andX =

Qi2I Xi. We assume that eachXi is compact and prove thatX is compact

9Let X be a partially ordered set and S � X. An upper bound for S is an element K 2 X s.t. x � K8x 2 S.


by Theorem 1.6.11, that is, by proving that every net has a convergent subnet, that is,that every net has a limit point.

First remark that the statement holds for the product of �nitely many factors: choosea �rst subnet whose projection converges in X1, a second subnet of the �rst subnet whoseprojection converges in X2, and so on: in a �nite number of steps you get a subnet whosecomponents all converge, that is, a convergent subnet. We now go back to the generalcase X =

Qi2I Xi with I in�nite set.

Let fx� : � 2 Ag be a net in X. By a partially de�ned member x of X we meanan element x 2 XJ =

Qi2J Xi for some J � I; note that to J is associated a projection

pJ : X ! XJ . We say that x 2 XJ is a partial limit point of fx�g if x is a limit point ofthe net fpJ(x�)g. Thus a partial limit point x de�ned on all X is a limit point of fx�g:its existence is precisely our aim.

Let P be the set of all partial cluster points of the net fx�g. Note that P is non-empty(see the case of a product of �nitely many factors). Note also that P is partially orderedby inclusion: g1 � g2 i¤ g1 2 XJ1, g2 2 XJ2, J1 � J2 and g1; g2 have the same commoncomponents (those with indexes J1).

Moreover, any totally ordered subset L = fx� : � 2 �g of P has an upper bound.Indeed, since each pair of x��s agree on the common components, we may de�ne J = [�2��and as upper bound the element x 2 J whose �-components are the same as x�. To showthat x 2 P we need to show that it is a partial limit point. But every basic open of x isof the form V = \nk=1p�1ik (Vik) (see Fact 1.10.2) and therefore 9� 2 � s.t. fi1; : : : ; ikg � �(we already proved the case of �nitely many components). Since x� is a limit point, thereexists a subnet fx�g of fx�g s.t. p�(x�) ! x�, thus p�(x�) 2 p�(V ) for � big enough,which means x� 2 V for � big enough. So x is a partial limit point.

Thus, by Zorn�s Lemma, P has a maximal element x 2 XJ : we assert that x is de�nedon the whole X = XI , that is J = I. If this is not the case, then 9k 2 I � J . Sincex 2 P, there exists a subnet fx�g whose projection in XJ converges to x in XJ . SinceXk is compact, there exists a subnet fx g of fx�g whose projection in Xk converges. Butthen fx g is a subnet of fx�g whose projection to XJ[fpgconverges to a point y 2 P andthus x < y, absurd. �

An important consequence in measure theory and probability is the following.

Corollary 1.10.11 A countable product of compact and metrizable spaces is compact andmetrizable.

An analogous resut for uncountable products does not hold in general: Tychono¤Theorem ensures that an uncountable product of compact sets is still compact but anuncountable product of of metrizable spaces is not necessarily metrizable.

Example 1.10.12 Are you able to provide an example?

1.11. POINTWISE AND UNIFORM CONVERGENCE 37

We conclude this section by citing a consequence of Tycono¤ Theorem which is usedto prove results on compact-valued correspondences (thus here it is not of common use).The result is somehow expected. However the proof, even if straightforward, is long andnot so instructive (see [Aliprantis�Border]).

Lemma 1.10.13 Let fXigi2I be a family of topological spaces, and 8i 2 I let Ki be acompact subset of Xi. If G is an open subset of

Qi2I Xi then there exists a basic open setQ

i2I Vi s.t. Yi2IKi �

Yi2IVi � G:

1.11 Pointwise and uniform convergence

De�nition 1.11.1 Let X be any set (not necessarily a topological space). A net ff�g ofreal-valued functions on X converges pointwise to a real-valued function f if f�(x)!f(x) 8x 2 X, that is:

8" > 0 8x 2 X 9�0 = �0("; x) s.t. jf�(x)� f(x)j < " 8� � �0:

De�nition 1.11.2 Let X be any set (not necessarily a topological space). A net ff�g ofreal-valued functions on X converges uniformly to a real-valued function f if

8" > 0 9�0 = �0(") s.t. jf�(x)� f(x)j < " 8x 2 X 8� � �0:

Remark 1.11.3 Uniform convergence implies pointwise convergence.

That pointwise convergence implies uniform convergence is false, as shown in the nextexample.

Example 1.11.4 A famous and important example is given by the sequence of powerfunctions in [0; 1], that is, fn : [0; 1] ! [0; 1] de�ned by fn(x) = xn. This sequenceconverges pointwise but not uniformly to the function f de�ned by f(x) = 0 8x 6= 1 andf(1) = 1 (see Figure 1.2).

Example 1.11.5 The sequence of functions de�ned by fn : [0; 1] ! [0; 1], fn(x) = xn=nconverges uniformly to the zero function. Indeed jfn(x) � 0j = xn

n� 1

n! 0, so 8" > 0

take �0 = [1" ] + 1: then 8n � �0 it is jfn(x)� 0j � 11"+1= "

"+1< " (see Figure 1.2).

Example 1.11.6 (Sequences cannot describe a topology not having a countable base, asa product topology over uncountable sets) Consider the space of functions from [0; 1] to[0; 1], regarded as [0; 1][0;1] with its product topology (pointwise converence). Let F denotethe family of indicator functions of �nite subsets of [0; 1]:

�A(x) =

�1 if x 2 A0 if x =2 A 8A �nite, A � [0; 1].


0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 0

0.2

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Figure 1.2: Graphs of f(x) = xn and f(x) = xn=n in [0; 1].

The identically 1 function is not the pointwise limit of any sequence in F : if An is asequence in F , then A = [1n=1An is countable, so 9x =2 A and �An(x) = 0 8n.However, there is a net in F that converges pointwise to 1: the net f�A : A �nite,

A � [0; 1]g directed by set inclusion. Indeed for any x 2 [0; 1] it holds �A(x) = 1 8A s.t.A � fxg.Thus the identically 1 function is in F but is not the limit of any sequence in F .

Theorem 1.11.7 The uniform limit of a net of continuous real functions is continuous.

Proof Let ff�g be a net of continuous real functions on a topological space X thatconverges uniformly to a function f 2 RX . The thesis is to show that if x� ! x thenf(x�)! f(x). First observe that for any x� ! x we can write

jf(x�)� f(x)j � jf(x�)� f�(x�)j+ jf�(x�)� f�(x)j+ jf�(x)� f(x)j:

Now �x a sequence x� ! x and " > 0. Since ff�g converges uniformly to f , 9�0 =�0 (") s.t. 8� � �0 it holds jf�(z) � f(z)j < "=3 8z 2 X. Fix any � � �0 to get inparticular that jf(x�)� f�(x�)j < "=3 and jf�(x)� f(x)j < "=3, so

jf(x�)� f(x)j ��

3+ jf�(x�)� f�(x)j+

�

3:

For the addendum in the middle we use the continuity of the function f�: since x� ! x,9�0 s.t. jf�(x�)� f�(x)j < "=3 8� � �0.Summing up, we have proven that

8x� ! x 8" > 0 9�0 s.t. jf(x�)� f(x)j < " 8� � �0;

or, equivalently, that if x� ! x then f(x�)! f(x). �

Here is a simple su¢ cient condition for a net to converge uniformly.


Theorem 1.11.8 (Dini�s Theorem) If a net of continuous real functions on a compactspace converges monotonically to a continuous function pointwise, then the net convergesuniformly.

Proof Let ff�g be a net of continuous functions on the compact space X satisfyingf�(x) " f(x) for each x 2 X.

Let " > 0. For each x 2 X pick an index �x such that 0 � f(x)� f�x(x) < ". By thecontinuity of f � f�x there is an open neighborhood Vx of x such that f(y)� f�x(y) < "8y 2 Vx. Note that since f�(x) " f(x) for each x 2 X then f(y)� f�(y) � 0 8y 2 X 8�and f(y)� f�(y) < f(y)� f�x(y) 8� � �x 8y 2 X, thus

8" > 0 8x 2 X 9�x 9Vx 2 Nx : 0 � f(y)� f�(y) < " 8� � �x 8y 2 Vx:

Since X is compact and X = [x2XVx, there are x1; x2; : : : ; xn 2 X s.t. X = [ni=1Vxi.De�ne �0 = maxf�x1 ; �x2 ; : : : ; �xng: then

8" > 0 9�0 : 0 � f(y)� f�(y) < " 8� � �0 8y 2 X;

which shows the uniform convergence of ff�g.

If instead f�(x) # f(x) for each x 2 X, then for each x 2 X pick an index �x suchthat 0 � f�x(x)� f(x) < " and repeat the argument. �

We now study some natural topologies on sets of real-valued functions on a given set.

Facts 1.11.9 1. A function f : X ! R can be regarded as an element in the productRX =

Qx2X R by identifying:

f = (f(x))x2X :

2. The projections of RX are called evaluation functionals: px(f) = f(x) consists inevaluating f at the point x, 8x 2 X. Thus the product topology of RX is the weak topologygenerated by the evaluation functionals.

3. Let ff�g be a net of real-valued functions on X. Then:

ff�g converges pointwise () ff�g converges in the product topology

4. Consider RX with the product topology and a subset F . Then:

F compact () F closed and pointwise bounded:

Here are some more advanced topics, concerning duality.


De�nition 1.11.10 A real-valued function f on set X is called a functional. Let F bea family of functionals on X. Then we have a function, called evaluation pairing, fromX �F to R denoted h; i and de�ned by:

h; i : X �F ! R(x; f) 7! hx; fi = f(x) 8x 2 X;8f 2 F :

According to the evaluation pairing h; i, the elements of X can be regarded as evaluationfunctionals on F by setting x = hx;�i, that is x(f) = hx; fi = f(x) 8f 2 F , and theelements f 2 F can be regarded as evaluation functionals on X because f = h�; fi, thatis f(x) = hx; fi 8x 2 X.

Proposition 1.11.11 The evaluation pairing h; i embeds (F ; �(F ; X)) into�RX ; �(X; �)

�.

Moreover, if the family F of real-valued functions on X separates points, then the evalu-ation pairing h; i embeds (X; �(X;F)) into (RF ; �(RF ; �)).

Proof (First embedding) It is clear that F � RX for each family F of functionals onX. Thus the statement means to prove the following claim:

Claim The weak topology �(F ; X) induced on F by the family of functionals on a set Xis equal to the relative topology on F induced by the product topology �(X; �) of RX .Before proving the claim, let us discuss about convergence in those topologies. Recall

�rst that the weak topology �(F ; X) is generated by the evaluation functionals x =hx;�i, which are de�ned by x(f) = f(x) 8f 2 F . Lemma 1.9.3 states that a netff�g � F converges in �(F ; X) i¤ x(f�) ! x(f) 8x 2 X: this means by de�nition thatf�(x) ! f(x) 8x 2 X, that is, pointwise convergence. Recall now that the producttopology of RX =

Qx2X R is the weak topology generated by the projections px. These

projections are again the evaluation functionals at points x 2 X: it follows immediatelythat convergence in the product topology means again pointwise convergence. Finally,the relative topology on F � RX is still generated by evaluation functionals at pointsx 2 X and thus convergence still means pointwise convergence: the only di¤erence is thatin the relative topology only the functionals in F are considered, instead of the wholefunctionals in RX . Therefore the two topologies induces the same notion of convergence.Concerning the open sets of the two topologies, Facts 1.9.2 and 1.10.2 give explicitly a

base for each of the two topologies: if � denotes the standard topology on R, then a basefor both topologies is given by considering all sets of the form

Qx2X Vx where Vx 2 f�1x (�)

for some fx 2 F and Vx = X for all but �nitely many x 2 X. It follows that the inclusionF � X is an homeomorphism according to the topologies �(F ; X) on F and �(X; �) onX.

(Second embedding) If F separates the points in X, the mapping which associatesto a point x the evaluation hx;�i is one-to-one. Again, recall that �(X;F) is generatedby the evaluation functionals f = h�; fi and that in the product topology �(RF ; �) theprojections are again the evaluation functionals f = h�; fi. It follows, exactly as above,that both topologies are the topologies of pointwise convergence on the set F . �

Lemma 1.11.12 A subset of F is weakly compact, that is, compact in weak topology,i¤ it is pointwise bounded and contains the pointwise limits of its nets.


Proof First note that, since (F ; �(F ; X)) ��RX ; �(X; �)

�is an embedding, we may

think to (F ; �(F ; X)) as a subset of RX with the relative topology on F induced byproduct topology on RX . Thus the theorem reduces to prove that

Claim In RX with the product topology, a subset K � F is relatively compact i¤ itis pointwise bounded and contains the pointwise limits of its nets.

Always considering�RX ; �(X; �)

�as the ambient space, property (4) of Facts 1.6.10

states that K � F is relatively compact i¤ K is compact in RX . At this point, thestatemtent is a consequence of Theorem 1.10.9: K is compact in RX i¤ Kx = px(K) iscompact in R = Rx 8x 2 X, exactly the two conditions in the RHS of the statement. �

Bibliography

[Aliprantis�Border] Aliprantis, C.D. and Border, K.C. (2005), In�nite DimensionalAnalysis, A Hitchhiker�s Guide, Third Edition , Springer-Verlag, New York.

[Cherno¤] Cherno¤, P.R., A Simple Proof of Tychono¤�s Theorem ViaNets, The American Mathematical Monthly, Vol. 99, No. 10 (Dec.,1992), pp. 932-934

[Berberian] Berberian, S.K. (1965), Measure and Integration, The Mac-millan Company, New York.

[Chow] Chow, Y.S., e H. Teicher (1988), Probability Theory, Springer-Verlag, New York.

[Kolmogorov] Kolmogorov, A.N., e S.V. Fomin (1975), Introductory RealAnalysis, Dover, New York. (Translation of Elementy TeoriiFunktsij i Funktsional�nogo Analiza, Nauka, Mosca, 1968.)

[Royden] Royden, H.L. (1988), Real Analysis, Prentice Hall, EnglewoodCli¤s.

[Rudin1] Rudin, W. (1976), Principles of Mathematical Analysis,McGraw-Hill, New York.

[Rudin2] Rudin, W. (1987), Real and Complex Analysis, McGraw-Hill,New York.

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