Lecture Topology: Part 1 - Huynh Quang Vu

Part 1 of lecture Topology presents the following content: general topology; infinite sets; topological space; continuity; connectedness; convergence; compactness; product of spaces; real functions and spaces of functions; quotient space;...

Lecture notes on Topology

Huỳnh Quang Vũ

Version of August 26, 2022

This is a set of lecture notes for a series of introductory courses in topologyfor undergraduate students at the University of Science, Ho Chi Minh City It iswritten to be delivered by myself, tailored to my students I did not write it withother lecturers or self-study readers in mind.

In writing these notes I intend that more explanations and discussions will becarried out in class I hope by presenting only the essentials these notes will bemore suitable for use in classroom Some details are left for students to fill in or to

be discussed in class

A sign✓in front of a problem notifies the reader that this is a typical problem

or an important one (the result is used later) A sign * indicates a relatively moredifficult problem

Huỳnh Quang VũAddress: Faculty of Mathematics and Computer Science, University of ScienceVietnam National University - Ho Chi Minh City Email: hqvu@hcmus.edu.vn,Web:https://sites.google.com/view/hqvu/

The latest version of this set of notes, together with the source file, are availableat

https://sites.google.com/view/hqvu/teaching

This work is released to Public Domain (CC0) wherever applicable, see

http://creativecommons.org/publicdomain/zero/1.0/, otherwise it is licensed der the Creative Commons Attribution 4.0 International License, see

un-http://creativecommons.org/licenses/by/4.0/

Introduction 1

I General Topology 3 1 Infinite sets 5

2 Topological space 14

3 Continuity 23

4 Connectedness 31

5 Convergence 41

6 Compactness 49

7 Product of spaces 57

8 Real functions and Spaces of functions 65

9 Quotient space 74

Other topics 88

II Algebraic Topology 91 10 Structures on topological spaces 93

11 Classification of compact surfaces 101

12 Homotopy 109

13 The fundamental group 113

14 The fundamental group of the circle 120

15 Van Kampen theorem 125

16 Simplicial homology 131

17 Singular homology 138

18 Homology of cell complexes 147

Other topics 149

III Differential Topology 151 19 Smooth manifolds 153

20 Tangent spaces and derivatives 161

21 Regular values 167

22 Critical points of real functions 172

23 Flows 181

24 Boundary 190

iii

25 Orientation 197

26 Degrees of maps 205

27 Integration of real functions 211

Guide for further reading 215

CONTENTS 1

Introduction

Topology is a mathematical subject that studies shapes The term comes from theGreek words “topos” (place) and “ology” (study) A set becomes a topologicalspace when each element of the set is given a collection of neighborhoods Oper-ations on topological spaces must be continuous, bringing certain neighborhoodsinto neighborhoods There is no notion of distance Topology is a part of geometrythat does not concern distance

Figure 0.1: How to make a closed trip such that every bridge is crossed exactly once?This is the problem “Seven Bridges of Konigsberg”, studied by Leonard Euler in the18th century It does not depend on the sizes of the bridges

Characteristics of topology

Operations on topological objects are more relaxed than in geometry: beside ing around (allowed in geometry), stretching or bending are allowed in topology(not allowed in geometry) For example, in topology circles - big or small, anywhere

mov are same Ellipses and circles are same On the other hand in topology tearing orbreaking are not allowed: circles are still different from lines While topologicaloperations are more flexible they still retain some essential properties of spaces.Contributions of topology

Topology provides basic notions to areas of mathematics where there is a need for

a notion of continuity Topology focuses on some essential properties of spaces Itcan be used in qualitative study It can be useful where metrics or coordinates arenot available, not natural, or not necessary

Initially developed in the late nineteenth century and early twentieth century

to provide basis for abstract mathematical analysis, topology gradually became aninfluential subject, reaching many achievements in the mid and late twentieth cen-tury

Topology often does not stand alone: nowadays there are fields such as gebraic topology, differential topology, geometric topology, combinatorial topol-ogy, quantum topology, Topology often does not solve a problem by itself,but contributes important understanding, settings, and tools Topology featuresprominently in differential geometry, global analysis, algebraic geometry, theoret-ical physics Currently we see topology appearing in contemporary research inapplied fields such as computation and data analysis.

al-Studying topology

Emerging later than major branches of mathematics as geometry, algebra, and ysis, it could be said that topology contains concepts, methods, and ideas which aresignificant advances in mathematical thinking Thus it is very valuable for a stu-dent of mathematics to study topology to develop ability to think in new general,abstract, but also concrete manner

anal-Notions and results on continuity is now used throughout mathematics Theideas of homology is now used widely in geometry and algebra The concept

of manifold has become fundamental in geometry and physics The widespreaduse of topology implies that studying topology can open students to much widerchoices of areas for further works

Part I General Topology

3

of natural numbers and of the set of real numbers.

Example. If the reader likes to, we can go into some more details on maps andCartesian product of two sets

Given a set S, an ordered pairs of two elements a and b of S, written(a, b), can

be defined as the set{a,{a, b}} Thus(b, a) = {b,{b, a}} ̸= (a, b)

The Cartesian product a set A with a set B is the set of all ordered pairs (a, b)such that a is in A and b is in B (so both are in A∪B), written as A×B= {(a, b) |a ∈

A, b∈ B}

A map or a function f from a set A to a set B is a subset of the set A×Bsuchthat for each a ∈ Athere is a unique b ∈ Bsuch that(a, b) ∈ f We usually write

b= f(a)

We should be aware of certain problems in naive set theory

Example (Russell’s paradox) Consider the set S = {x |x /∈ x}(the set of all setswhich are not members of themselves) Then whether S∈ Sor not is undecidable,since answering either yes or no to this question leads to contradiction.1

The example above shows that using the notion of set without definition maylead to trouble In the Von Neumann–Bernays–Godel system a more general notionthan set, calledclass(lớp), is used In this course, we do not distinguish set, class,

the term collection For more one can read [Hal74], [End77, p 6], [Dug66, p 18–20]

Indexed collection

A map f : I→ Awhere I is a set and A is a collection is called anindexed collection,

indexed collection f by(fi)i∈I or{fi}i∈I Notice that it can happen that fi = fj forsome i̸=j

In a village there is a barber; his job is to do hair cut for a villager if and only if the villager does not cut his hair himself Consider the set of all villagers who had their hairs cut by the barber Is the barber himself a member of that set?

Example. A sequence of elements in a set A is a collection of elements of A indexed

by the setZ+of positive integer numbers, written as(an)n∈Z+

Relation

When(a, b) ∈Rwe often say that a is related to b and often write a∼Rb

A relation said to be:

(a) reflexive (phản xạ) if∀a∈S,(a, a) ∈R

(b) symmetric (đối xứng) if∀a, b∈ S,(a, b) ∈R⇒ (b, a) ∈R

(c) anti-symmetric (phản đối xứng) if ∀a, b ∈ S,((a, b) ∈R∧ (b, a) ∈R) ⇒ a =b

(d) transitive (bắc cầu) if∀a, b, c∈S,((a, b) ∈R∧ (b, c) ∈R) ⇒ (a, c) ∈R

If R is an equivalence relation on S then anequivalence class(lớp tương đương)represented by a ∈ S is the subset[a] = {b ∈ S | (a, b) ∈ R} Two equivalenceclasses are either coincident or disjoint The set S is partitioned (phân hoạch) intothe disjoint union of its equivalence classes

When(a, b) ∈Rwe often write a≤b When a≤band a̸=bwe write a<b

If any two elements of S are related then the order is called atotal order(thứ tựtoàn phần) and(S,≤)is called atotally ordered set

Example. The setR of all real numbers with the usual order≤is totally ordered

Example. Given a set S thecollection of all subsetsof S is denoted byP (S)or 2S.With the inclusion relation, (2S,⊆) is a partially ordered set, but is not a totallyordered set if S has more than one element

Example. Let(S1,≤1)and(S2,≤2)be two ordered sets The following is an order

on S1×S2: (a1, b1) ≤ (a2, b2)if(a1 < a2)or((a1 = a2) ∧ (b1 ≤ b2)) This is called

In an ordered set, thesmallest element(phần tử nhỏ nhất) is the element that issmaller than all other elements More concisely, if S is an ordered set, the smallest

1 INFINITE SETS 7

element of S is an element a∈ Ssuch that∀b∈S, a≤b The smallest element of S

if exists is unique, denoted by min S

than More concisely, a minimal element of S is an element a ∈ Ssuch that∀b ∈

S, b≤a ⇒b= a There can be more than one minimal element

that is smaller than or equal to any element of the subset More concisely, if A⊂ Sthen a lower bound of A in S is an element a∈ Ssuch that∀b∈ A, a≤b

The definitions of largest element, maximal element, and upper bound are ilar

If a set is not finite we say that it isinfinite

A set is calledcountably infinite(vô hạn đếm được) if it is equivalent toZ+ Aset is calledcountableif it is either finite or countably infinite

Intuitively, a countably infinite set can be “counted” by using the positive gers The elements of such a set can be indexed by the set of all positive integers as

inte-a sequence inte-a1, a2, a3,

Example. The setZ of all integer numbers is countable We can count alternatively

the positive and the negative integers We can work out a formula for the counting,such as

Proof. The statement is equivalent to the statement that any subset ofZ+is able Suppose that A is an infinite subset ofZ+ Let a1 = min A, whose exis-tence is due to the well-ordering propertyof Z+: every non-empty subset of Z+has a smallest element For each n > 1, let an = min A\ {a1, a2, , an− 1} Then

count-an − 1< anand the set B= {an|n∈Z+}is a countably infinite subset of A

Now we show that any element m of A is an anfor some n, and therefore B= A.Let C = {an | an ≥ m}, then C ̸= ∅ since B is infinite Let an0 = min C, then

an0 ≥ m Since an0− 1 < an0 we have an0− 1 ∈/ C, thus an0− 1 < m This implies

m ∈ A\ {a1, a2, , an0− 1} Since an 0 = min(A\ {a1, a2, , an0− 1})we must have

an0 ≤ m Thus an0 =m

Corollary If there is an injective map from a set S toZ+then S is countable.

1.1 Proposition If there is a surjective map fromZ+to a set S then S is countable Proof Suppose that there is a surjective map ϕ : Z+ → S For each s ∈ Sthe set

ϕ−1(s)is non-empty Let ns = min ϕ−1(s) The map s 7→ ns is an injective mapfrom S to a subset ofZ+, therefore S is countable

Z+×Z+→Z+

(m, n) 7→ (1+2+ · · · + ((m+n−1) −1)) +m= (m+n−2)(m+n−1)

We will check that this map is injective Let k = m+n Suppose that (k−2)(2k−1) +

m = (k′−2)(2k′−1) +m′ If k = k′ then the equation certainly leads to m = m′ and

1 INFINITE SETS 9

a contradiction

1.2 Theorem The union of a countable collection of countable sets is a countable set.

Proof. The collection can be indexed as A1, A2, , Ai, (if the collection is finite

we can let Aibe the same set for all i starting from a certain index) The elements ofeach set Ai can be indexed as ai,1, ai,2, , ai,j, (if Ai is finite we can let ai,j be thesame element for all j starting from a certain index) This means there is a surjectivemap from the index setZ+×Z+to the unionS

i ∈ IAiby(i, j) 7→ai,j

Theorem The set Q of all rational numbers is countable.

Proof. One way to prove this result is by writingQ=S

1.3 Theorem The set R of all real numbers is uncountable.

Proof. The proof uses the Cantor diagonal argument We use the property thatevery real number can be expressed in decimal forms Notice that there are realnumbers whose decimal presentations are not unique, such as 1

2 = 0.5000 =0.4999 (See 1.17 for more details on this topic.)

Suppose that set of all real numbers in the interval [0, 1] is countable, and isenumerated as a sequence{ai| ∈Z+} Let us write

a1 =0.a1,1a1,2a1,3

a2 =0.a2,1a2,2a2,3

a3 =0.a3,1a3,2a3,3

Choose a number b = 0.b1b2b3 such that bn ̸= 0, bn ̸= 9, and bn ̸= an,n Then

b̸=anfor all n Thus the number b is not in the above table, a contradiction

Example. Two intervals[a, b]and[c, d]on the real number line are equivalent Thebijection can be given by a linear map x 7→ d − c

b − a(x−a) +c Similarly, any twointervals(a, b)and(c, d)are equivalent

1.4 Example. The interval (−1, 1) is equivalent toR via a map related to the tan

function:

x7→ √ x

1−x2

1.5 Theorem A non-empty set is not equivalent to the set of all of its subsets More

specifically, for any set S̸=∅ there is no surjective map from S to 2S.

Note that there is an injective map from S to 2S, namely a 7→ {a}.

Proof Let ϕ be any map from S to 2S Let X = {a ∈ S |a /∈ ϕ(a)} Suppose thatthere is x∈ Ssuch that ϕ(x) =X Then the truth of the statement x∈ X(whether

it is true or false) is undecidable, a contradiction Therefore there is no x ∈ Ssuch

that ϕ(x) =X, so ϕ is not surjective

This result implies that any set is “strictly smaller” than the set of all of its sets So there can not be a set that is “larger” than any other set There is no “uni-versal set”, “set which contains everything”, or “set of all sets”

sub-The Axiom of choice

Proposition The following statements are equivalent:

on this collection, called a choice function , associating each set in the collection with

an element of that set.

then X has a maximal element.

Zorn lemma is often a convenient form of the Axiom of choice

Intuitively, a choice function “chooses” an element from each set in a given lection of non-empty sets The Axiom of choice allows us to make infinitely manyarbitrary choices.2 This is also often used in constructions of functions, sequences,

col-or nets, see one example at 5.4 One common application is the use of the product

of an infinite family of sets – the Cartesian product, discussed below

pairs of shoes does not need the Axiom of choice (because in a pair of shoes the left shoe is different from the right one so we can define our choice), but usually in a pair of socks the two socks are identical, so choosing one sock from each pair of socks from an infinite collection of pairs of socks needs the Axiom of choice.

1 INFINITE SETS 11

The Axiom of choice is needed for many important results in mathematics,such as the Tikhonov theorem in Topology, the Hahn-Banach theorem and Banach-Alaoglu theorem in Functional analysis, the existence of a Lebesgue unmeasurableset in Real analysis,

There are cases where this axiom could be avoided For example in the proof

of 1.1 we used the well-ordered property ofZ+instead See for instance [End77, p.151] for further material on this subject

Cartesian product

Let (Ai)i∈I be a collection of sets indexed by a set I The Cartesian product(tíchDescartes) ∏i ∈ IAi of this indexed collection is defined to be the collection of allmaps a : I → S

i ∈ IAi such that a(i) ∈ Ai for every i ∈ I The existence of such

a map is a consequence of the Axiom of choice An element a of∏i ∈ IAi is oftendenoted by(ai)i∈I, with ai = a(i) ∈ Ai being the coordinate of index i, in analogywith the finite product case

i Ai) =T

i f−1(Ai).

1.7. Let f be a map Check that:

(a) f(f−1(A)) ⊂A If f is surjective (onto) then equality happens.

(b) f−1(f(A)) ⊃A If f is injective then equality happens.

1.8. Show that if A is infinite and B is countable then A∪B is equivalent to A.

1.9. Give another proof of 1.2 by checking that the mapZ+×Z+→Z+ ,(m, n) 7→2m3nis injective.

1.10. ✓Show that the set of points inRn with rational coordinates is countable.

1.11. Show that if A has n elements then|2A| =2n.

1.12. Show that the set of all functions f : A→ {0, 1}is equivalent to 2 A

1.13. A real number α is called an algebraic number if it is a root of a polynomial with

integer coefficients Show that the set of all algebraic numbers is countable.

A real number which is not algebraic is called transcendental For example it is known

that π and e are transcendental Show that the set of all transcendental numbers is

un-countable.

1.14. A continuum set is a set which is set-equivalent toR Show that a countable union of

continuum sets is a continuum set.

one gets a space consisting of two intervals[0,13] ∪ [23, 1] Continuing, delete the intervals

(19,29) and(79,89) In general on each of the remaining intervals, delete the middle open interval of13the length of that interval The Cantor set is the set of remaining points It can

be described as the set of real numbers ∑∞n=1 3ann , a n = 0, 2 In other words, it is the set of real numbers in[0, 1]which in base 3 could be written without the digit 1.

Show that the total length of the deleted intervals is 1 Is the Cantor set countable?

subset of B and B is set-equivalent to a subset of A then A and B are set-equivalent Suppose that f : A 7→ B and g : B 7→ A are injective maps Let A0 = A and B0 = B For n∈Z, n≥0, let Bn+1= f(A n)and An+1=g(B n).

(a) Show that An+1⊂A n and Bn+1⊂B n

(b) Show that A n+2 is set-equivalent to A n , and A n\A n+1 is set-equivalent to A n+2\

deduce that A is set-equivalent to A1.

For more, see [KF75, p 17], [End77, p 148].

1.17. Show that any real number could be written in base d with any d∈Z, d ≥2 More specifically, any positive real number x can be written as

plane is set-equivalent to a line, or the setC of complex numbers is equivalent to the set R

of real numbers As a corollary,Rn is equivalent toR.

Let us construct a map from[0, 1) × [0, 1)to[0, 1)as follows: the pair of two real bers in decimal forms 0.a1a2 and 0.b1b2 corresponds to the real number 0.a1b1a2b2

num-In view of 1.17, we only allow decimal presentations in which not all digits are 9 starting from a certain digit Check that this map is injective.

1.19. We prove that 2Z+ is set-equivalent toR.

(a) Show that 2Z+ is set-equivalent to the set of all sequences of binary digits Deduce that there is an injective map from[0, 1]to 2Z+.

1 INFINITE SETS 13

(b) Consider a map f : 2Z+ → [0, 2], for each binary sequence a = a 1 a 2 a 3· · ·, if ing from a certain digit all digits are 1 then let f(a) = 1.a 1 a 2 a 3· · · in binary form, otherwise let f(a) =0.a 1 a2a3· · · Show that f is injective.

start-The statement that there does not exist a set that is strictly more thanZ+ but is strictly less thanR (meaning a set S for which there is an injective map from Z+ to S but there is not a bijective map, and there is an injective map from S toR but there is not a bijective map) is

called theContinuum hypothesis.

every non-empty subset A of S has a smallest element, i.e. ∃a ∈ A, ∀b ∈ A, a ≤ b For example with the usual order,N is well-ordered while R is not Notice that a well-ordered

set must be totally ordered Ernst Zermelo proved in 1904, based on the Axiom of choice, that any set can be well-ordered.

Prove the following generalization of the principle of induction Let A be a well-ordered set Let P(a)be a statement whose truth depends on a∈A Suppose that if P(a)is true for all a<b then P(b)is true Then P(a)is true for all a∈ A.

2 Topological space

When we discuss dependence of an object on another object we are often interested

in continuity of the dependence This notion is often understood as that variation

of the dependent object can be controlled by variation in the independent object.The sets that appear in this discussion are called open sets Briefly, a topology is asystem of open sets Topology is a general setting for discussion on continuity.First we recall how continuity arises in metric spaces

Metric space

The notion of open sets and continuity in metric spaces have been very successful

We shall deduce what we consider essential properties of these notions in order togeneralize them

Recall that, briefly, a metric space is a set equipped with a distance betweenevery two points Namely, a metric space is a set X with a map d : X×X7→ R such

that for all x, y, z∈X:

(a) d(x, y) ≥0 (distance is non-negative),

(b) d(x, y) = 0 ⇐⇒ x = y(distance is zero if and only if the two points cide),

coin-(c) d(x, y) =d(y, x)(distance is symmetric),

(d) d(x, y) +d(y, z) ≥d(x, z)(triangular inequality)

A ball is a set of the form B(x, r) = {y∈X|d(y, x) <r}where r∈R, r>0

In the theory of metric spaces, a subset U of X is said to be open if for all x in U

there is ϵ>0 such that B(x, ϵ)is contained in U This is equivalent to saying that anon-empty open set is a union of balls

We see at once that a union of open sets is an open set

Consider the intersection of two open sets A and B in a metric space X Let

x ∈ A∩B Since x ∈ Athere is a ball B(x, r) ⊂ A, and since x ∈ Bthere is a ball

B(x, s) ⊂ B Let t=min{r, s}, then B(x, t) ⊂Aand B(x, t) ⊂B, so B(x, t) ⊂ A∩B.Thus A∩B is open in X By induction, we see that the intersection of any finitecollection of open sets is an open set

However, an infinite intersection of open sets, such asT ∞

n = 1B(x,n1) = {x}, maynot be an open set

Let(X, dX)and(Y, dY)be metric spaces Recall that a map f :(X, dX) → (Y, dY)

is continuous at x∈ Xif and only if

∀ϵ>0,∃δ >0,∀y∈ X, dX(y, x) <δ =⇒ dY(f(y), f(x)) <ϵ,

2 TOPOLOGICAL SPACE 15equivalently,

∀ϵ>0,∃δ>0,∀y∈ BX(x, δ), f(y) ∈BY(f(x), ϵ),

or even shorter,

∀ϵ>0,∃δ>0, f(BX(x, δ)) ⊂BY(f(x), ϵ),

That means given any ball B(f(x), ϵ)centered at f(x)there is a ball B(x, δ)centered

at x such that f brings B(x, δ)into B(f(x), ϵ) Without mentioning balls, a map f

is continuous at x if f(y)is arbitrarily close to f(x)provided that y is sufficientlyclose to x

Going a bit further, writing Uxto indicate that x ∈U, we can phrase the notion

of continuity equivalently as:

∀Uf(x)open in Y,∃Vxopen in X, f(V) ⊂U

In this form continuity only needs a notion of open sets

an open set containing x Note that a neighborhood does not need to be open3.The notion of continuity can be phrased in terms of neighborhoods as: for anygiven neighborhood Uof f(x)there is a neighborhood V of x so that f(V) ⊂ U.Thus continuity means the values of the map can be controlled and assured to bewithin any given neighborhood provided the inputs are only taken within a certainneighborbood

Topology

We see that in metric space continuity can be expressed through a notion of borhoods of points In metric space, the notion of neighborhood is expressed usingdistance To generalize without the present of distance, we directly specify whichsets are open sets We shall choose some properties of open sets in metric spaces

neigh-to retain There are many approaches for various purposes Since the early 20thcentury the following setting has shown to be sufficiently general and effective formany areas of mathematics

Taking properties of open sets in metric spaces, we propose:

Definition. Atopology on a set X is a collection τ of subsets of X satisfying the

following conditions:

(a) The sets∅ and X are elements of τ In symbols: ∅∈τ, X ∈τ

(b) A union of elements of τ is an element of τ In symbols: F ⊂ τ =⇒

S

O ∈ FO∈τ; or∀F⊂ τ,S

O ∈ FO∈τ

but Munkres [Mun00] requires a neighborhood to be open.

(c) A finite intersection of elements of τ is an element of τ In symbols: ((F ⊂

τ) ∧ (|F| <∞)) =⇒ T

O ∈ FO∈τ; or∀(F⊂τ,|F| <∞),T

O ∈ FO∈ τ

The elements of τ are called the open sets of X in the topology τ.

In short, a topology on a set X is a collection of subsets of X which includes∅and X and is “closed” under unions and finite intersections

A set X together with a topology τ is called a topological space, denoted by(X, τ)

or X alone if we do not need to specify the topology An element of X is often called

Example. On any set X there is the trivial topology (tôpô hiển nhiên) {∅, X}.There is also thediscrete topology(tôpô rời rạc) whereas any subset of X is open.Thus on a set there can be many topologies

Example. Let X= {1, 2, 3} The collection τ = {∅,{1, 2},{2, 3},{2},{1, 2, 3}}is atopology on X

Notice that the statement “intersection of any finitely many open sets is open”

is equivalent to the statement “intersection of any two open sets is open”, by tion

induc-Example (topology generated by metric) A metric space is canonically a ical space with the topology generated by the metric, i.e non-empty open sets areunions of balls When we speak about topology on a metric space we mean this

Example (normed space) Recall that a normed space (không gian định chuẩn) isbriefly a vector space equipped with lengths of vectors Namely, a normed spaceover the field of real numbers is a set X with a structure of a vector space overthe real numbers and a real function X → R, x 7→ ∥x∥, called a norm (chuẩn),satisfying:

(a) ∥x∥ ≥0 and∥x∥ =0⇐⇒ x=0 (length is non-negative),

(b) ∥cx∥ = |c| ∥x∥for c∈R (length is proportionate to vector),

i = 1x2i1/2 The topology generated

by this norm is called theEuclidean topology(tôpô Euclid) ofRn

2 TOPOLOGICAL SPACE 17

On a set X we can compare topologies by using the order relation by set

inclu-sion Namely, let τ1 and τ2 be two topologies on X, if τ1 ⊂ τ2 we say that τ2 is

smaller) than τ2

Example. On any set the trivial topology is the coarsest topology and the discretetopology is the finest one

A complement of an open set is called aclosed set

Proposition In a topological space X:

(a) ∅ and X are closed.

(b) A finite union of closed sets is closed.

(c) An intersection of closed sets is closed.

Proof. This is simply a property of operations on sets We deduce the property fromthe set identities

[

U

(X\U) =X\ \

UU

!,and

\

U

(X\U) =X\ [

UU

!

Remark. Notice that being closed and being open are not exclusive (in contrast touses of the terms outside of topology): a subset of space, such as∅ and the wholespace itself, can be both closed and open in that space, or can be neither closed noropen We used to work in Euclidean spaces, where except the empty set and thewhole space, being open and being closed are exlusive, see

Interior – Closure – Boundary

Let X be a topological space and let A be a subset of X A point x in X is said to be:(a) aninterior point(điểm trong) of A in X if there is an open set of X containing

xthat is contained in A,

(b) acontact point(điểm dính) (or point of closure) of A in X if any open set of

Xcontaining x contains a point of A,

(c) alimit point(điểm tụ) (or cluster point, or accumulation point) of A in X ifany open set of X containing x contains a point of A other than x,

(d) aboundary point(điểm biên) of A in X if every open set of X containing xcontains a point of A and a point of the complement of A In other words, aboundary point of A is a contact point of both A and the complement of A.With these notions we define:

(a) The set of all interior points of A is called theinterior(phần trong) of A in X,denoted by ˚Aor int(A),

(b) The set of all contact points of A in X is called theclosure(bao đóng) of A in

met-Example. On the Euclidean lineR, consider the subset A= [0, 1) ∪ {2} Its interior

is intA= (0, 1), the closure is clA= [0, 1] ∪ {2}, the boundary is ∂A= {0, 1, 2}, theset of all limit points is[0, 1]

Bases of a topology

Definition. Given a topology, a collection of open sets is abasis (cơ sở) for thattopology if every non-empty open set is a union of members of that collection

More concisely, let τ be a topology of X, then a collection B⊂ τis called a basis

for τ if for any∅ ̸=V∈ τthere is C⊂ Bsuch that V =S

Example. In a metric space the collection of all balls is a basis for the topologygenerated by the metric

Example. The Euclidean plane has a basis consisting of all open disks Since anydisk contains a square with the same center, an open set is also a union of opensquares, thus the Euclidean plane also has a basis consisting of all open squares In

a similar manner one can introduce other shapes as bases for this topology, see fig.2.1, and 2.16

So notice that a topology can have many bases

Definition. A collection S⊂τis called asubbasis (tiền cơ sở) for the topology τ if the collection of all finite intersections of members of S is a basis for τ.

2 TOPOLOGICAL SPACE 19

Figure 2.1: Different bases for the Euclidean topology of the plane

Clearly a basis for a topology is also a subbasis for that topology Briefly, given atopology, a subbasis is a subset of the topology that can generate the entire topology

by finite intersections and then by unions

Example. On X = {1, 2, 3}the topology τ= {∅,{1, 2},{2, 3},{2},{1, 2, 3}}has asubbasis{{1, 2},{2, 3}}

2.2 Example. The collection of all open rays, that are, sets of the forms (a,∞)and(−∞, a), is a subbasis for the Euclidean topology ofR.

Generating topologies

Suppose that we have a set and we want a topology such that certain subsets of thatset are open sets, how do find a topology for that purpose?

Theorem Let S be a collection of subsets of X whose union equal X, if this is not the case

then add X to S4 The collection τ consisting of ∅ and all unions of finite intersections of

members of S is the smallest topology on X containing S, called the topology generated

Proof It is clear that any topology of X containing S must contain τ We only need

to check that τ is a topology Notice that X belongs to τ because of our assumption that the union of S equal X We check that τ is closed under unions Let σ ⊂ τ,considerS

!

U ∈( S

A σF A )U,

empty collections.

for a reason we want certain sets to be open, we can always build a topology forthat purpose.

Example. Let X= {1, 2, 3, 4} The set{{1},{2, 3},{3, 4}}generates the topology{∅,{1},{3},{1, 3},{2, 3},{3, 4},{1, 2, 3},{1, 3, 4},{2, 3, 4},{1, 2, 3, 4}}.

A basis for this topology is{{1},{3},{2, 3},{3, 4}}

Example (ordering topology) Let(X,≤)be a totally ordered set The collection ofsubsets of the forms{β∈ X|β<α}and{β∈ X|β> α}generates a topology on

X, called theordering topology

Example. The Euclidean topology on R is the ordering topology with respect to

the usual order of real numbers (This is just a different way to state 2.2.)

Problems

whose complements are finite Check that this is indeed a topology.

of X whose complements are countable Check that this is indeed a topology Compare the countable complement topology to the finite complement topology.

2.5. Let X be a set and p∈X Show that the collection consisting of ∅ and all subsets of X containing p is a topology on X This topology is called theparticular point topologyon

X, denoted by PPX p Describe the closed sets in this space.

2.6. Show that

(a) The interior of A in X is the largest open subset of X that is contained in A A subset

is open if and only if all of its points are interior points.

2 TOPOLOGICAL SPACE 21

(b) The closure of A in X is the smallest closed subset of X containing A A subset is closed if and only if it contains all of its contact points.

2.7. Let X be a topological space and A⊂X.

(a) Show that A is the disjoint union of ˚ Aand ∂A.

(b) Show that X is the disjoint union of ˚ A, ∂A, and X\A.

2.8. In a metric space X, a point x ∈ X is a limit point of the subset A of X if and only

if there is a sequence in A\ {x}converging to x (This is not true in general topological spaces, see 5.4 )

2.9. Find the closures, interiors and the boundaries of the interval [0, 1) under the clidean, discrete and trivial topologies ofR.

Eu-2.10. Show that

(a) In a normed space, the boundary of the ball B(x, r)is the sphere{y| ∥x−y∥ =r}, and so the ball B′(x, r) = {y| ∥x−y∥ ≤r}is the closure of B(x, r).

(b) In a metric space, the boundary of the ball B(x, r)is a subset of the sphere{y|d(x, y) =

r} Is the ball B′(x, r) = {y|d(x, y) ≤r}the closure of B(x, r)?

2.11. Let O n = {k∈Z+|k≥ n Check that{∅} ∪ {O n|n ∈ Z+}is a topology onZ+ Find the closure of the set{5} Find the closure of the set of all even positive integers.

2.12. Show that any open set in the Euclidean line is a countable union of open intervals.

2.13. In the real number line with the Euclidean topology, is the Cantor set (see 1.15) closed

or open, or neither? Find the boundary and the interior of the Cantor set.

2.14. ✓ Show that the intersection of a collection of topologies on a set X is a topology

on X If S is a subset of X, then the intersection of all topologies of X containing S is the smallest topology that contains S Show that this is exactly the topology generated by S.

2.15. ✓A collection B of open sets is a basis if for each point x and each open set O taining x there is a U in B such that U contains x and U is contained in O.

con-2.16. ✓Show that two bases generate the same topology if and only if each member of one basis is a union of members of the other basis, equivalently, if and only if whenever a point

x belongs to an element U of one basis then there is an element V of the other basis such that x∈V⊂U.

2.17. Let B be a collection of subsets of X Then B∪ {X}is a basis for a topology on X if and only if the intersection of two members of B is either empty or is a union of some members

of B (In several textbooks to avoid adding the element X to B it is required that the union

of all members of B is X.)

2.18. In a metric space the set of all balls with rational radii is a basis for the topology The set of all balls with radii21m , m≥1 is another basis.

center has rational coordinates forms a basis for the Euclidean topology ofRn

2.20. Let d1and d2be two metrics on X If there are α>0, β>0 such that αd1≤d2≤ βd1

then the two metrics are said to be equivalent Show that two equivalent metrics generate same topologies.

2.21. Let(X, d)be a metric space.

(a) Let d1(x, y) = min{d(x, y), 1} Show that d1 is a metric on X generating the same topology as that generated by d Is d 1 equivalent to d?

(b) Let d2(x, y) = 1+d(x,y)d(x,y) Show that d2is a metric on X generating the same topology

as the topology generated by d Is d2equivalent to d? Is d2equivalent to d1?

Eu-clidean norm, and let∥·∥be any norm.

(a) Check that the map x7→ ∥x∥from(Rn ,∥·∥2)to(R,∥·∥2)is continuous.

(b) Let Sn be the unit sphere under the Euclidean norm Show that the restriction of the map above to Sn has a maximum value β and a minimum value α Hence α ≤x

∥x∥2 ≤βfor all x̸=0.

(c) Deduce that any two norms inRn generate equivalent metrics, hence all norms inRn

generate the Euclidean topology.

2.23. Is the Euclidean topology onR2 the same as the ordering topology onR2 with respect

to the dictionary order? If it is not the same, can the two be compared?

2.24. The collection of all intervals of the form[a, b)generates a topology onR (the

Sorgen-frey’s line) Compare this topology with the Euclidean topology.

2.25. On the set of all integer numbersZ, consider arithmetic progressions

Sa,b =a+bZ,

where a∈Z and b∈Z+

(a) Show that these sets form a basis for a topology onZ.

(b) Show that with this topology each set Sa,bis closed.

a topology we specify a collection of sets to be the collection of open sets Therefore

we can propose continuity to be generalized to topological spaces:

Definition. Let X and Y be topological spaces We say a map f : X→Yis

Vof X containing x such that f(V)is contained in U Writing Ux to indicate that

x ∈U, we can write the definition in symbolic form as:

if we look at a metric space as a topological space with the topology generated

by the metric then continuity in the metric space is the same as continuity in thetopological space Thereforewe inherit all results concerning continuity when the

Example. WhenR is equipped with the Euclidean topology then the sine function

sin :R→R is topologically continuous, since we know the sine function is

contin-uous under the Euclidean metric, and the Euclidean metric generates the Euclideantopology

Example. Let X and Y be topological spaces

(a) The identity function, idX: X →X, x7→ x, is continuous

(b) The constant function, with a given a∈Y, x7→a, is continuous

(c) If Y has the trivial topology then any map f : X →Yis continuous

(d) If X has the discrete topology then any map f : X→Yis continuous

Theorem A map is continuous if and only if the inverse image of an open set is an open

set.

Proof. (⇒) Suppose that f : X → Y is continuous Let U be an open set in Y Let

x ∈ f−1(U) Since f is continuous at x and U is an open neighborhood of f(x),there is an open set Vxcontaining x such that Vxis contained in f−1(U) Therefore

f−1(U) =S

x ∈ f − 1 ( U )Vxis open

(⇐) Suppose that the inverse image of any open set is an open set Let x ∈ X.Let U be an open neighborhood of f(x) Then V= f−1(U)is an open set containing

x, and f(V)is contained in U Therefore f is continuous at x

Sometimes it is convenient to use the following property:

Proposition Suppose that f : X→Y and S is a subbasis for the topology of Y Then f is

continuous if and only if the inverse image of any element of S is an open set in X.

Example. A map from a topological space to a metric space is continuous if andonly if the inverse image of any open ball is an open set

A map f : X → R where R has the Euclidean topology is continuous if and

only if for any a∈R the sets f− 1((−∞, a))and the set f−1((a,∞))are open.The following property is familiar to us from metric spaces:

3.1 Proposition A map is continuous if and only if the inverse image of a closed set is a

closed set.

Topology generated by maps

In many circumstances we have situations where we have maps first and we want

to build topologies so that these maps become continuous

In one situation, let(X, τX)be a topological space, Y be a set, and f : X →Ybe

a map, we want to find a topology on Y such that f is continuous The requirement

for such a topology τY is that if U ∈ τY then f−1(U) ∈ τ The trivial topology

on Y is the coarsest topology satisfying that requirement The collection {U ⊂

Y| f−1(U) ∈τ }is the finest topology satisfying that requirement

In another situation, let X be a set,(Y, τY)be a topological space, and f : X→Y

be a map, we want to find a topology on X such that f is continuous The

require-ment for such a topology τX is that if U ∈ τY then f−1(U) ∈ τ The discretetopology on X is the finest topology satisfying that requirement The collection

τ = {f−1(U) |U ∈ τY}is the coarsest topology satisfying that requirement Wecan observe further that if the collection SYgenerates τYthen τXis generated by thecollection{f−1(U) |U∈SY}

Example. We often encounter a situation where we have a real-valued function on aset X, that is a map f : X→R, for example a grading or a ranking, where R has the

usual order and metric Suppose we want the map f to be continuous, that is, wewant the change of values of f is arbitrarily small provided we stay within a certainneighborhood Then the coarsest topology on X for that purpose is generated bythe collection of all subsets of the form f−1((a, b))where(a, b)is any interval ofR.

Subspace

Let(X, τ)be a topological space and let Y be a subset of X We want to define atopology on Y that can be naturally considered as being “inherited” from X Thus

Definition. Let Y be a subset of the topological space X Thesubspace topologyon

Y, also called therelative topology(tôpô tương đối) with respect to X, is defined

to be the collection of restrictions of the open sets of X to Y, that is, the collection{O∩Y|O ∈ τ} With this topology we say that Y is asubspace(không gian con)

is open in the subspace[0, 1], but is not open inR. When we say that a set is open,

Proposition Let X be a topological space and let Y⊂ X The subspace topology on Y is

the coarsest topology on Y such that the inclusion map i : Y ,→ X, x 7→ x is continuous.

In other words, the subspace topology on Y is the topology generated by the inclusion map from Y to X.

Proof. If O is a subset of X then i−1(O) =O∩Y Thus the topology generated by i

is{O∩Y|O∈ τ }, exactly the subspace topology of Y

Example (subspaces of a metric space) The notion of topological subspaces iscompatible with the earlier notion of metric subspaces Let(X, d)be a metric spaceand let Y ⊂ X Then, as we know, Y is a metric space with the metric inheritedfrom X, namely dY =dX|Y×Y A ball in Y is a set of the form, for a∈Y, r>0:

BY(a, r) = {y∈Y|d(y, a) <r} = {x∈ X|d(x, a) <r} ∩Y =BX(a, r) ∩Y.Any open set A in Y is the union of a collection of balls in Y, i.e

Example. For n ∈ Z+ define the sphereSn to be the subspace of the EuclideanspaceRn + 1consisting of all points of distance 1 to the origin:

Sn= {(x1, x2, , xn+ 1) ∈Rn + 1|x21+x22+ · · · +x2n+1 =1}

Homeomorphism

A map from one topological space to another is said to be ahomeomorphism(phépđồng phôi) if it is a bijection, is continuous and its inverse map is also continuous.Two spaces are said to behomeomorphic(đồng phôi) if there is a homeomorphismfrom one to the other This is a basic relation among topological spaces

It is easy to see that a homeomorphism brings an open set onto an open set, aclosed set onto a closed set, see 3.7

Proposition A homeomorphism between two spaces induces a bijection between the two

topologies.

Proof. A homeomorphism f :(X, τX) → (Y, τY)induces a map

˜f : τX → τY

O 7→ f(O),which is a bijection

Roughly speaking, in the field of Topology, when two spaces are phic they are considered the same For example a “topological sphere” means atopological space which is homeomorphic to a sphere

homeomor-3.2 Example. Any two balls in a normed space are homeomorphic via translations(tịnh tiến) and dilations (scaling) (co dãn, vị tự)

3.3 Example. Any ball in a normed space is homeomorphic to the whole space

We can consider a map from the unit ball B(0, 1)to the whole space in the form

x 7→ f(x) = φ(∥x∥) x

∥ ∥, where φ is a bijection from(0, 1)onto(0,∞) Such a map

φ has appeared in 1.4: we can take φ(t) = √ t

1 − t 2, or φ(t) = 1−tt For example, take

y= f(x) = x

1 −∥ x , then x = f−1(y) = y

1 +∥ ∥

topological space Y is a homeomorphism from X to a subspace of Y, i.e it is a map

f : X → Ysuch that the restriction ˜f : X → f(X)is a homeomorphism If there is

an imbedding from X to Y then we say that X can beembeddedin Y

Example. With the subspace topology the inclusion map is an embedding

Example. The Euclidean lineR can be embedded in the Euclidean plane R2 as aline in the plane

3 CONTINUITY 27

Example. Suppose that f : R → R is continuous under the Euclidean topology.

ThenR can be embedded into the plane as the graph of f

Figure 3.4: The Euclidean real number line can be embedded to the Euclidean plane

as the graph of the function y=x3

Example. The sphere Snminus one point is homeomorphic to the Euclidean space

Rnvia thestereographic projection(phép chiếu nổi), see Fig 3.5

N

x

y

x n+1

Figure 3.5: The stereographic projection

Specifically, on the sphere Sn minus the North Pole N = (0, 0, , 0, 1), eachpoint x = (x1, x2, , xn+ 1) corresponds to the intersection y = (y1, y2, , yn, 0)between the straight line from N to x with the hyperplane xn + 1 =0 An intersectionequation between the line and the hyperplane is:

(1−t)N+tx=y,

We can also say that the n-dimensional Euclidean space can be embedded ontothe n-sphere minus one point.

Problems

3.6. ✓If f : X→Y and g : Y→Z are continuous then g◦f is continuous.

3.7. Define anopen mapto be a map such that the image of an open set is an open set.

homeomorphism is an open map and is also a closed map.

3.8. Show that a continuous bijection is a homeomorphism if and only if it is an open map.

3.9. Show that(X, PPX p)and(X, PPX q)(see 2.5) are homeomorphic.

3.10. ✓Let X be a set and(Y, τ)be a topological space Let fi: X→Y, i∈ I be a collection

of maps Find the coarsest topology on X such that all maps fi, i∈ I are continuous.

3.11. Suppose that X is a normed space Prove that the topology generated by the norm

is exactly the coarsest topology on X such that the norm and the translations (maps of the form x7→x+a) are continuous.

3.12. ✓Show that a subset of a subspace Y of X is closed in Y if and only if it is a restriction

of a closed set in X to Y.

3.13. The set{x∈Q| −√2<x<√

2}is both closed and open inQ under the Euclidean

topology ofR.

3.14.  Suppose that X is a topological space and Z⊂Y ⊂X Then the relative topology

of Z with respect to Y is the same as the relative topology of Z with respect to X.

3.15. ✓Let X and Y be topological spaces and let f : X→Y.

(a) If Z is a subspace of X, denote by f|Z the restriction of f to Z Show that if f is continuous then f|Zis continuous.

(b) Let Z be a space containing Y as a subspace Consider f as a function from X to Z, that is, let ˜f : X → Z, ˜f(x) = f(x) Show that f is continuous if and only if ˜f is continuous.

3 CONTINUITY 29

are both closed in X Suppose f : X→Y, and f|Aand f|Bare both continuous Then f is continuous.

Another way to phrase the above statement is the following Let g : A→Y and h : B→

Y be continuous and g(x) =h(x)on A∩B Define

homeomor-3.20. In the Euclidean plane the upper half-plane{(x, y) ∈R2|y> 0}is homeomorphic

to the whole plane.

3.21. Is it true that any two balls in a metric space are homeomorphic?

3.22. Show that any two finite-dimensional normed spaces of the same dimensions over the same fields are homeomorphic.

3.23. Let∥ · ∥1and ∥ · ∥2be any two norms onRn Consider the map x 7→ ∥x∥1∥x∥x

2 for

x̸=0 and sending 0 to 0.

(a) Prove that this map is a homeomorphism from(Rn ,∥ · ∥1)to(Rn ,∥ · ∥2).

(b) Deduce that balls with respect to different norms onRn are homeomorphic.

(c) In particular deduce that inRn any Euclidean ball is homeomorphic to an n-dimensional rectangle.

3.24. ✓If f : X → Y is a homeomorphism and Z ⊂ X then X\Z and Y\ f(Z) are homeomorphic.

3.25. On the Euclidean planeR2 , show that:

(a) R2\ {(0, 0)}andR2\ {(1, 1)}are homeomorphic.

(b) R2\ {(0, 0),(1, 1)}andR2\ {(1, 0),(0, 1)}are homeomorphic.

3.26. On the Euclidean plane, show that any two (non-degenerate) triangles are morphic.

homemo-3.27. Show thatN and Z are homeomorphic under the Euclidean topology More generally,

prove that any two set-equivalent discrete spaces are homeomorphic.

3.28. Among the spacesZ, Q, R, each with the Euclidean topology, which one is

homeomor-(a) Show that the sphere S 2 is homogeneous.

(b) Show that being homogeneous is atopological property, meaning that if two spaces are homeomorphic and one space is homogeneous then the other space is also ho- mogeneous.

3.33. On the setR with the finite complement topology, any polynomial is continuous, but

a trigonometric function such as the sine function is not continuous.

3.34. * Anisometry(phép đẳng cấu metric, phép đẳng cấu hình học, hay phép đẳng cự) from a metric space X to a metric space Y is a surjective map f : X → Y that preserves distance, that is d(f(x), f(y)) = d(x, y)for all x, y ∈ X If there exists such an isometry then X is said to beisometricto Y.

(a) Show that an isometry is a homeomorphism.

(b) Show that being isometric is an equivalence relation among metric spaces.

(c) Show that (R2 ,∥ · ∥∞)and (R2 ,∥ · ∥1)are isometric, but they are not isometric to

(R2 ,∥ · ∥2), although the three spaces are homeomorphic, where∥(x, y)∥1 = |x| +

|y| ∥(x, y)∥2= (x2+y2)1/2 ,∥(x, y)∥∞=max{|x| |y|}(see 2.22) (For generalization

to higher dimensions one may use the Mazur-Ulam theorem, see [P Lax, Functional Analysis, John Wiley and Sons, 2002, p 49].)

4 CONNECTEDNESS 31

Intuitively we say a space is connected to indicate that it has “one piece”, whereas

a space is disconnected when it has “several pieces”

Precisely, a space is said to be not connected, ordisconnected, if it is the union

of certain two non-empty disjoint open subsets In symbols,(X, τ)is disconnected

if there exist∅̸=U∈ τ,∅̸=V∈τsuch that X=U∪V

A space is said to beconnected(liên thông) if it is not disconnected, equivalently,

it is not the union of any two non-empty disjoint open subsets In symbols,(X, τ)

is connected if whenever there exist U∈ τ, V ∈ τsuch that X =U∪Vthen either

U=∅ or V =∅

Example. A space containing only one point is connected

When we say that a subset of a topological space is connected we mean that thesubset under the subspace topology is a connected space

Example. The Euclidean real number line minus a point is not connected

The Euclidean plane minus a line is not connected

The Euclidean plane minus a circle is not connected

We can immediately deduce a convenient characterization of connectedness:

Proposition A topological space is connected if and only if the only subsets which are both

closed and open are the empty set and the space itself.

So connectedness means nonexistence of non-empty proper subsets which areboth closed and open In a connected space, aside from the empty set and the spaceitself, the terms “open” and the term “closed” are mutually exclusive

The following result is used often to check that a space is connected:

4.1 Proposition If a collection of connected subspaces of a space has non-empty

intersec-tion then its union is connected.

Proof. Consider a topological space and let F be a collection of connected subspaceswhose intersection is non-empty Let A be the union of the collection, A= S

D ∈ FD.Suppose that C is a subset of A that is both open and closed in A If C ̸= ∅ thenthere is D∈ Fsuch that C∩D̸=∅ Then C∩Dis a subset of D and is both openand closed in D (we are using 3.14 here) Since D is connected and C∩D ̸= ∅,

we must have C∩D = D This implies C contains the intersection of F Therefore

C∩D ̸= ∅ for all D ∈ F The argument above shows that C contains all D in F,that is, C= A We conclude that A is connected

Proposition (continuous image of connected space is connected) If f : X→Y is

continuous and X is connected then f(X)is connected.

Proof. Suppose that U and V are non-empty disjoint open subset of f(X)and U∪

V = f(X) Since f : X → f(X)is continuous (see 3.15), f−1(U)and f−1(V)areopen in X, are non-empty and disjoint, and f−1(U) ∪f−1(V) =X This contradictsthe connectedness of X

The following result says that a locally constant map on a connected space must

be constant, is a demonstration of the use of connectedness assumption:

Proposition Let X be connected, and let f : X → Y be locally constant , that is, each point of X has a neighborhood on which f is constant Then f is constant on X.

Proof. Fix x0 ∈ X and let c = f(x0) Let U = {x ∈ X | f(x) = c}, then U is notempty We shall check that U is both open and closed in X, therefore U = Xsince

Xis connected, and so f =con X

Given x1∈ Xthen since f is locally constant at x1there is an open neighborhood

Vof x1such that f|V = f(x1) If x1 ∈Uthen f(x1) =c, therefore f|V =c, implying

V ⊂ U This shows that U is open in X Similarly if x1 ∈ X\U then f(x1) ̸= ctherefore for all x ∈ V we have f(x) ̸= c, implying V ⊂ X\U So X\Uis alsoopen, therefore U is closed in X

Connected sets in the Euclidean real number line

Theorem A subspace of the Euclidean real number line is connected if and only if it is an

Let us prove thatR is connected Suppose that R contains a non-empty, proper,

open and closed subset C Let x /∈ Cand let D=C∩ (−∞, x) =C∩ (−∞, x] Then

Dis both open and closed inR, and is bounded from above.

If D ̸= ∅, consider s = sup D Since D is closed and s is a contact point of D,

s ∈ D Since D is open s must belong to an open interval contained in D But thenthere are points in D which are bigger than s, a contradiction

If D =∅ we let E=C∩ (x,∞), consider t=inf E and proceed similarly

In the reverse direction, suppose that a subset A of R is connected Suppose

that x, y ∈ A and x < y If x < z < y we must have z ∈ A, otherwise the set{a ∈ A| a < z} = {a ∈ A |a ≤ z}will be both closed and open in A Thus Acontains the whole interval[x, y]

Let a = inf A if A is bounded from below and a = −∞ otherwise Similarlylet b = sup A if A is bounded from above and b = ∞ otherwise Suppose that Acontains more than one element, so that a < b There are sequences(an)n∈Z+ and(bn)n∈Z+ of elements in A such that a< an <bn<b, and an →awhile bn →b By

4 CONNECTEDNESS 33the above argument,[an, bn] ⊂ Afor all n So(a, b) ⊂S ∞

n = 1[an, bn] ⊂ A ⊂ [a, b] Itfollows that A is either(a, b)or[a, b)or(a, b]or[a, b]

4.2 Example (Euclidean spaces are connected) Since the EuclideanRnis the union

of all lines passing through the origin, it is connected Therefore a non-emptyproper subset ofRncannot be both open and closed, in other words, being closedand being open are exclusive

4.3 Proposition (Intermediate value theorem) If X is a connected space and f : X→

R is continuous, where R has the Euclidean topology, then the image f(X)is an interval.

A consequence is the following familiar theorem in Calculus: Let f :[a, b] →R

be continuous under the Euclidean topology, if f(a)and f(b)have opposite signsthen the equation f(x) =0 has a solution

Below is a simple application, a form of Intermediate value theorem (4.3):

Theorem (Borsuk–Ulam theorem) For any continuous real function on the sphere Sn

there must be antipodal (i.e opposite) points where the values of the function are same.5

Proof. Let f : Sn → R be continuous Let g(x) = f(x) − f(−x) Then g is tinuous and g(−x) = −g(x) Since Snis connected (see 4.11), the range g(Sn)is aconnected subset of the EuclideanR, and so it is an interval, containing the inter-

con-val between g(x)and g(−x) = −g(x) Therefore 0 is in the range of g, so there is

x0∈Snsuch that f(x0) = f(−x0)

Connected component

Let us define a relation on a topological space X whereas two points are related

if both belong to a connected subspace of X (we then say that the two points areconnected) Then this relation is an equivalence relation, by 4.1

Proposition Under the above equivalence relation the equivalence class containing a point

x is equal to the union of all connected subspaces containing x, thus it is the largest

con-nected subspace containing x.

Proof. Consider the equivalence class [x]represented by a point x By definition,

y ∈ [x] if and only if there is a connected set Oy containing both x and y Since

Oy ⊂ [x], we have[x] =S

y ∈[ x ]Oy By 4.1,[a]is connected

Under the above equivalence relation, the equivalence classes are called the

components

Example. Consider the space Z with the Euclidean topology inherited from R.

This topology is actually the discrete topology onZ, therefore a subset containing

same!

only one point is open, and any subset containing more than one point is not nected Hence the subsets containing only one point are the connected components

con-ofZ.

Example. Consider the spaceQ with the Euclidean topology inherited from R If a

subset A ofQ contains two elements a and b, a<b, then there is an irrational ber c between a and b, and A= ((−∞, c) ∩A) ∩ ((c,∞) ∩A), a union of two opendisjoint non-empty subsets, thus A is not connected Hence the subsets containingonly one point are the connected components ofQ.

num-4.4 Proposition A map defined on a space with finitely many connected components is

continuous if and only if the map is continuous on each component.

From this result when study a topological space with finitely many connectedcomponents we often only need to work on each connected component, or we mayassume the space itself is connected

Theorem If two spaces are homeomorphic then there is a bijection between the collections

of connected components of the two spaces.

Proof. Let f : X →Ybe a homeomorphism Set

˜f :{[x] |x∈ X} → {[y] |y∈Y}[x] 7→ [f(x)]

We check that ˜f is a well-defined map Suppose that y∈ [x] Since f is continuous,the image f([x])is connected Thus f(y)and f(x)belong to a connected subspace

f([x]), so[f(y)] = [f(x)]

The map ˜f has an inverse map

gf−1:{[y] |y∈Y} → {[x] |x ∈X}

[y] 7→ [f−1(y)].Therefore we have a bijection between the two sets

From this result we say that connectedness is a topological property We alsosay that the number of connected components is a topological invariant If twospaces have different numbers of connected components then they must be differ-ent (that is, not homeomorphic)

Example (line not homeomorphic to plane) Suppose that R and R2 under theEuclidean topologies are homeomorphic via a homeomorphism f Delete any point

xfromR By 3.24 the subspaces R\ {x}andR2\ {f(x)}are homeomorphic But

R\ {x}is not connected whileR2\ {f(x)}is connected (see 4.25), a contradiction.ThusR and R2are not homeomorphic