LECTURE NOTES ON GENERAL TOPOLOGY BIT, SPRING ...

LECTURE NOTES ON GENERAL TOPOLOGY

BIT, SPRING 2018

DAVID G.L. WANG

Contents

1. Introduction 3

1.1. Who cares topology? 3

1.2. Geometry v.s. topology 5

1.3. The origin of topology 7

1.4. Topological equivalence 8

1.5. Surfaces 9

1.6. Abstract spaces 9

1.7. The classification theorem and more 10

2. Topological Spaces 12

2.1. Topological structures 12

2.2. Subspace topology 17

2.3. Point position with respect to a set 19

2.4. Bases of a topology 24

2.5. Metrics & the metric topology 27

3. Continuous Maps & Homeomorphisms 37

3.1. Continuous maps 37

3.2. Covers 42

3.3. Homeomorphisms 45

Date: June 30, 2018.1

4. Connectedness 53

4.1. Connected spaces 53

4.2. Path-connectedness 62

5. Separation Axioms 66

5.1. Axioms T0, T1, T2, T3 and T4 66

5.2. Hausdorff spaces 70

5.3. Regular spaces & normal spaces 71

5.4. Countability axioms 75

6. Compactness 78

6.1. Compact spaces 78

6.2. Interaction of compactness with other topological properties 81

7. Product Spaces & Quotient Spaces 86

7.1. Product spaces 86

7.2. Quotient spaces 92

Appendix A. Some elementary inequalities 99

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1. Introduction

1.1. Who cares topology?

Question 1.1. Imagine you have super stretchy and bendy pants. Can you turnthem inside out without taking your feet off the ground?

Answer. Yes.

Question 1.2. You have a picture frame with a loop of string fixed on the back,and two nails. Can you hang it onto the wall using both nails so that pulling outany nail from the wall leads the painting falling down?

Figure 1.1. A solution to Question 1.2. Screenshot from a video ofthe PBS Digital Studios.

Answer. Yes. See Fig. 1.1.

Question 1.3. The necklace splitting problem. Its name and solutions are due tomathematicians Alon & West. See Fig. 1.2.

Answer. The version illustrated in Fig. 1.3 can be solved by using Borsuk-Ulamtheorem: every continuous function from an n-sphere into En maps some pair ofantipodal points to the same point.

Question 1.4. The inscribed square problem: Any simple closed curve inscribes asquare. See Fig. 1.4.

Answer. Open up to 2017. The proposition that any simple closed curve inscribe arectangle can be shown with the aid of Mobius strips; see Fig. 1.5

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Figure 1.2. Illustration of one version of the necklace splittingproblem. Screenshot from a video of 3Blue1Brown.

Figure 1.3. Illustration of one version of the necklace splittingproblem. Screenshot from a video of 3Blue1Brown.

Figure 1.4. A simple, closed, polygonal curve inscribing a square.Stolen from Terrence Tao’s manuscript on arXiv:1611.07441.

5

Figure 1.5. August Ferdinand Mobius (1790–1868) was a Germanmathematician and theoretical astronomer. The right part is aMobius strip. Stolen from Wiki.

Homework 1.1. Make a Mobius strip by yourself.

The Nobel Prize in Physics 2016 was awarded with one half to David J. Thouless,and the other half to F. Duncan M. Haldane and J. Michael Kosterlitz “for theo-retical discoveries of topological phase transitions and topological phasesof matter”; see Fig. 1.6.

Topology, over most of its history, has NOT generally been applied outside ofmathematics (with a few interesting exceptions).

WHY?

• TOO abstract? The ancient mathematicians could not even convince of the subject.

• It is qualitative, not quantitative? People think of science as a quantitativeendeavour.

1.2. Geometry v.s. topology. Below are some views from Robert MacPherson,a plenary addresser at the ICM in Warsaw in 1983.

• Geometry (from ancient Greek): geo=earth, metry=measurement.Topology (from Greek): τ oπoσ=place/position, λoγoσ=study/discourse.

• Topology is “geometry” without measurement.It is qualitative (as opposed to quantitative) “geometry”.

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Figure 1.6. Topology was the key to the Nobel Laureates’ discov-eries, and it explains why electronical conductivity inside thin layerschanges in integer steps. Stolen from Popular Science Background ofthe Nobel Prize in Physics 2016, Page 4(5).

Figure 1.7. Screenshot from a video of Robert MacPherson’s talkin Institute for Advanced Study.

• Geometry: The point M is the midpoint of the straight line segment L connectingA to B.Topology: The point M lies on the curve L connecting A to B.

7

• Geometry calls its objects configurations (circles, triangles, etc.)Topology calls its objects spaces.

1.3. The origin of topology. Here are three stories about the origin of topology.

1.3.1. The seven bridges of Konigsberg. The problem was to devise a walk throughthe city that would cross each of those bridges once and only once; see Fig. 1.8.

Figure 1.8. Map of Konigsberg in Euler’s time showing the actuallayout of the seven bridges, highlighting the river Pregel and thebridges. Stolen from Wiki.

The negative resolution by Leonhard Paul Euler (1707–1783) in 1736 laid thefoundations of graph theory and prefigured the idea of topology. The difficulty Eulerfaced was the development of a suitable technique of analysis, and of subsequenttests that established this assertion with mathematical rigor.

Euler was a Swiss mathematician, physicist, astronomer, logician and engineerwho made important and influential discoveries in many branches of mathematicslike infinitesimal calculus and graph theory, while also making pioneering contribu-tions to several branches such as topology and analytic number theory.

1.3.2. The four colour theorem. The four colour theorem states that given anyseparation of a plane into contiguous regions, producing a figure called a map, nomore than four colours are required to colour the regions of the map so that notwo adjacent regions have the same color.

The four color theorem was proved in 1976 by Kenneth Appel and WolfgangHaken. It was the first major theorem to be proved using a computer.

1.3.3. Euler characteristic. χ = v − e+ f ; see Fig. 1.9.

We use the terminology polyhedron to indicate a surface rather than a solid.

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Figure 1.9. Stamp of the former German Democratic Republichonouring Euler on the 200th anniversary of his death. Across thecentre it shows his polyhedral formula. Stolen from Wiki.

Theorem 1.5 (Euler’s polyhedral formula). Let P be a polyhedron s.t.

• Any two vertices of P can be connected by a chain of edges.

• Any cycle along edges of P which is made up of straight line segments (notnecessarily edges) separates P into 2 pieces.

Then the Euler number or Euler characteristic χ = 2 for P .

• 1750: first appear in a letter from Euler to Christian Goldbach (1690–1764).

• 1860 (around): Mobius gave the idea of explaining topological equivalence bythinking of spaces as being made of rubber. It works for concave P .

• David Eppstein collected 20 proofs of Theorem 1.5.

Homework 1.2. Compute the Euler characteristic of the Platonic solids (allregular and convex polyhedra): the tetrahedron, the cube, the octahedron, thedodecahedron, and the icosahedron; see Fig. 1.10.

1.4. Topological equivalence. = homeomorphism; see Section 1.6.

√: thicken, stretch, bend, twist, . . .;

× : identify, tear, . . ..

See Fig. 1.11.

Theorem 1.6. Topological equivalent polyhedra have the same Euler characteristic.

• The starting point for modern topology.

• Search for unchanged properties of spaces under topological equivalence.

9

Figure 1.10. The Plotonic solids. Stolen from Wiki.

Figure 1.11. Some surfaces which are not equivalent. Stolen fromHaldane’s slides on Dec. 8th, 2016.

• χ = 2 belongs to S2, rather than to particular polyhedra → define χ for S2.

• Theorem 1.6: different calculations, same answer.

1.5. Surfaces. What exactly do we mean by a “space”?

• Homeomorphism → Continuity.

• Geometry → Boundedness.

1.6. Abstract spaces. The axioms for a topological space appearing for the firsttime in 1914 in the work of Felix Hausdorff (1868–1942). Hausdorff, a Germanmathematician, is considered to be one of the founders of modern topology, who

10 D.G.L. WANG

contributed significantly to set theory, descriptive set theory, measure theory,function theory, and functional analysis.

How has modern definition of a topological space been formed?

(1) General enough to allow set of points or functions, and performable construc-tions like the Cartesian products and the identifying. Enough information todefine the continuity of functions between spaces.

(2) Cauchy: distances → continuity.

(3) No distance! Continuity ← neighbourhood ← axiom.

A function f : Em → En is continuous if given any x ∈ Em and any neighbour-hood U of f(x), then f−1(U) is a neighbourhood of x.

Definition 1.7. Let X be a set. Suppose that for any x ∈ X, a collection Nx ∈ 22X

is assigned to x, satisfying the following 4 axioms:

(a) If x ∈ X and U ∈ Nx, then x ∈ U .

(b) If x ∈ X and U1, U2 ∈ Nx, then U1 ∩ U2 ∈ Nx.(c) If x ∈ X, U ∈ Nx, and U ⊂ V , then V ∈ Nx.(d) If x ∈ X and U ∈ Nx, then y ∈ V : V ∈ Ny ∈ Nx.

Then we call every subset U ∈ Nx a neighbourhood of x. The assignment(x,Nx) : x ∈ X of the collection of neighbourhoods satisfying the above fouraxioms to each x ∈ X is called a topology on X. The whole structure, that is,the set X together with the topology, is called a topological space. A functionf : X → Y between topological spaces is continuous if ∀x ∈ X and ∀U ∈ Nf(x),the set f−1(U) ∈ Nx. It is called a homeomorphism if it is a continuousbijection and has a continuous inverse. We call X and Y are homeomorphic ortopologically equivalent spaces if such a function exists.

• The Euclidean space En.

• The subspace topology → surfaces become topological spaces.

• Metric → topology; e.g., d(f, g) = supx |f(x)− g(x)|.• The peculiar cofinite topology: U ⊆ R is a neighbourhood ⇐⇒ |U c| <∞.

• Every singleton in X is a neighbourhood =⇒ every f : X → Y is continuous.

1.7. The classification theorem and more.

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Theorem 1.8 (Classification theorem). Any closed surface is homeomorphic to S2

with either a finite number of handles added, or a finite number of Mobius stripsadded. No two of these surfaces are homeomorphic.

Definition 1.9. The S2 with n handles added is called an orientable surface ofgenus n. Non-orientable surfaces can be defined analogously.

Historical notes. The classification of surfaces was initiated and carried throughin the orientable case by Mobius in a paper which he submitted for consideration forthe Grand Prix de Mathematiques of the Paris Academy of Sciences. He was 71 atthe time. The jury did not consider any of the manuscripts received as being worthyof the prize, and Mobius’ work finally appeared as just another mathematical paper.

Decide ∼= or 6∼=.

• ∼=: construct a homeomorphism; techniques vary.

• 6∼=: look for topological invariants, e.g., geometric properties, numbers, algebraicsystems.

Examples to show 6∼=.

• E1 6∼= E2: connectedness, h : E1 \ 0 → E2 \ h(0).• Poincare’s construction idea: assign a group to each topological space so that

homeomorphic spaces have isomorphic groups. But group isomorphism does notimply homeomorphism.

Here are some theorems that the fundamental groups help prove.

Theorem 1.10 (Classification of surfaces). No 2 surfaces in Theorem 1.8 haveisomorphic fundamental groups.

Theorem 1.11 (Jordan separation theorem). Any simple closed curve in E1

divides E1 into 2 pieces.

See Fig. 1.12.

Theorem 1.12 (Brouwer fixed-point theorem). Any continuous function from adisc to itself leaves at least one point fixed.

See Fig. 1.13.

Theorem 1.13 (Nielsen-Schreier theorem). A subgroup of a free group is alwaysfree.

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Figure 1.12. Marie Ennemond Camille Jordan (1838–1922) wasa French mathematician, known both for his foundational work ingroup theory and for his influential Cours d’analyse. The Jordancurve (drawn in black) divides the plane into an “inside” region (lightblue) and an “outside” region (pink). Stolen from Wiki.

Figure 1.13. Luitzen Egbertus Jan Brouwer (1881–1966), usuallycited as L. E. J. Brouwer but known to his friends as Bertus, was aDutch mathematician and philosopher, who worked in topology, settheory, measure theory and complex analysis. He was the founder ofthe mathematical philosophy of intuitionism. Stolen from Wiki.

2. Topological Spaces

The definition of topological space fits quite well with our intuitive idea of whata space ought to be. Unfortunately it is not terribly convenient to work with. Wewant an equivalent, more manageable, set of axioms!

2.1. Topological structures. We redefine topological spaces as follows, whichcan be shown to be compatible with that defined in Def. 1.7.

Definition 2.1. Let X be a set and Ω ∈ 2X .

• A (topological) space is a pair (X,Ω), where the collection Ω, called a topologyor topological structure on X, satisfies the axioms

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(i) ∅ ∈ Ω and X ∈ Ω;

(ii) the union of any members of Ω lies in Ω;

(iii) the intersection of any two members of Ω lies in Ω.

• A point in X: p ∈ X.

• An open set in (X,Ω): a member in Ω.

A closed set in (X,Ω): a subset A ⊆ X s.t. Ac ∈ Ω.

A clopen set in (X,Ω): a subset A ⊆ X which is both closed and open.

Remark 1. Being closed is not the negation of being open! A set might be

• open but not closed, or

• closed but not open, or

• clopen, or

• neither closed nor open.

Remark 2. Why do we use the letter O and the letter Ω?

• Open in English

• Ouvert in French

• Otkrytyj in Russian

• Offen in German

• Oppen in Swedish

• Otvoren in Croatian

• Otevreno in Czech

• Open in Dutch

Here are some topological spaces that we will meet frequently in this note.

Space 1. An indiscrete or trivial topological space: (X, ∅, X).

Space 2. A discrete topological space: (X, 2X).

A space is discrete ⇐⇒ every singleton is open.

Space 3. A particular point topology (X,Ω):

Ω = ∅, X ∪ S ⊆ X : p ∈ S,

where p is a particular point in X.

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Figure 2.1. Wac law Franciszek Sierpinski (1882–1969) was a Polishmathematician. He was known for outstanding contributions toset theory (research on the axiom of choice and the continuumhypothesis), number theory, theory of functions and topology. Hepublished over 700 papers and 50 books. Stolen from Wiki.

Space 4. An excluded point topology (X,Ω):

Ω = ∅, X ∪ S ⊆ X : p 6∈ S,where p is a particular point in X.

Space 5. The real line (R, ΩR):

ΩR = unions of open intervalsis the canonical or standard topology on R.

Space 6. The cofinite space (R, ΩT1):

ΩT1 = ∅ ∪ complements of finite subsets of Ris the cofinite topology or finite-complement topology or T1-topology.

Space 7. The arrow (X,Ω):

X = x ∈ R : x ≥ 0 and Ω = ∅, X ∪ (a,∞) : a ≥ 0.

Space 8. The Sierpinski space (X,Ω):

X = a, b and Ω = ∅, a, a, b.

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Homework 2.1. Find a smallest topological space which is neither discrete norindiscrete.

Question 2.2. Does there exist a topology which is both a particular point topologyand an excluded point topology?

Example 2.3. The set 0 ∪ 1/n : n ∈ Z+ is closed in the real line.

“Think geometrically, prove algebraically.” — John Tate

Figure 2.2. John Tate (1925–) is an American mathematician, dis-tinguished for many fundamental contributions in algebraic numbertheory, arithmetic geometry and related areas in algebraic geometry.He is professor emeritus at Harvard Univ. He was awarded the AbelPrize in 2010. Stolen from Wiki.

Homework 2.2. Find a topological space (X,Ω) with a set A ⊂ X satisfying allthe following properties:

a) A is neither open nor closed;

b) A is the union of an infinite number of closed sets; and

c) A is the intersection of an infinite number of open sets.

Answer. The interval [ 0, 1) in R.

Example 2.4. The Cantor ternary setK is the number set created by iterativelydeleting the open middle third from a set of line segments, i.e.,

K =

∑k≥1

ak3k

: ak ∈ 0, 2

=

0.a1a2 · · · : ai 6∈ 1, 4, 7⊂ [0, 1].

See Fig. 2.3. The set K was discovered by Henry John Stephen Smith in 1874,and introduced by Georg Cantor in 1883. It has a number of remarkable and deepproperties. For instance, it is closed in R.

16 D.G.L. WANG

Figure 2.3. The left most is Henry John Stephen Smith (1826–1883), a mathematician remembered for his work in elementarydivisors, quadratic forms, and Smith-Minkowski-Siegel mass formulain number theory. The middle is Georg Cantor (1845–1918), aGerman mathematician who invented set theory. The right mostillustrates the Cantor ternary set. Stolen from Wiki and MathCounterexamples respectively.

Recall that the neighbourhoods of an element has been defined in Def. 1.7, bywhich we defined topology. Now we give an alternative definition of neighbourhoods.

Definition 2.5. Given (X,Ω). Let p ∈ X. A neighbourhood of p is a subsetU ⊆ X s.t.

∃ O∈ Ω s.t. p ∈ O ⊆ U.

Remark 3. In literature the letter U is used to indicate a neighbourhood since it isthe first letter of the German word “umgebung” which means neighbourhood.

Remark 4. We are following the Nicolas Bourbaki group and define the term“neighbourhood” in the above sense. There is another custom that a neighbourhoodof a point p is an open set containing p. Nicolas Bourbaki is the collectivepseudonym under which a group of (mainly French) 20th-century mathematicians,with the aim of reformulating mathematics on an extremely abstract and formalbut self-contained basis, wrote a series of books beginning in 1935. With the goalof grounding all of mathematics on set theory, the group strove for rigour andgenerality. Their work led to the discovery of several concepts and terminologiesstill used, and influenced modern branches of mathematics. While there is noone person named Nicolas Bourbaki, the Bourbaki group, officially known as theAssociation des collaborateurs de Nicolas Bourbaki (Association of Collaborators

of Nicolas Bourbaki), has an office at the Ecole Normale Superieure in Paris.

Question 2.6. A topology can be defined by assigning neighbourhoods or opensets; see Defs. 1.7 and 2.1. Can it be defined by assigning closed sets?

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Answer. Yes. Here is a list of axioms for assigning closed sets:

(i)’ ∅ and X are closed;

(ii)’ the union of any finite number of closed sets is closed;

(iii)’ the intersection of any collection of closed sets is closed.

Remark 5. Given (X,Ω). Since the union of all members in Ω is X, the topology Ωitself carries enough information to clarify a topological space. However, thetopological space (X,Ω) is often denoted simply by X, because different topologicalstructures in the same set X are often considered simultaneously rather seldom.Moreover, to exclaim a set is in general easier than clarifying a topology. As willbe seen in §2.2, subspace topology helps the clarification by calling the knowledgeof common topologies.

2.2. Subspace topology.

Definition 2.7. Given (X,Ω) and A ⊆ X. The subspace topology (or inducedtopology) of A induced from (X,Ω):

ΩA = O ∩ A : O ∈ Ω.The topological subspace induced by A: (A, ΩA).

Question 2.8. Describe the topological structures induced

1) on Z+ by ΩR;

Answer. The discrete topology.

2) on Z+ by the arrow;

Answer. All sets of the form n ∈ N : n ≥ a where a ∈ N.

3) on the two-element set 1, 2 by ΩT1 ;

Answer. The discrete topology.

4) on the two-element set 1, 2 by the arrow topology.

Answer. ∅, 2, 1, 2.

18 D.G.L. WANG

Theorem 2.9. Let (X,Ω) be a topological space and A ⊆ X. The subspace topologyof A induced from (X,Ω) can be defined alternatively in terms of closed sets as

S is closed in A ⇐⇒ S = F ∩ A, where F is a closed set in X.

Theorem 2.10. Let X be a topological space and let A ⊆M ⊆ X.

1) If A is open in X, then A is open in M .

If A is closed in X, then A is closed in M .

2) If A is open in M , and if M is open in X, then A is open in X.

If A is closed in M , and if M is closed in X, then A is closed in X.

Proof. The openness of A in M follows from the formula A = A ∩M . Conversely,if A is open in M , then A = O ∩M , where O is open in X. As a consequence, thisintersection A = O ∩M is open in X as long as M is also open in X. The “closed”version can be shown along the same line with aid of Theorem 2.9.

Remark 6. The condition that M is open/closed in X in Theorem 2.10 is necessary.For instance, the set x ∈ Q : x >

√2 is clopen in Q, but neither closed nor open

in R.

Homework 2.3. Let A ⊆M ⊆ X.

(1) If A is open in M , can we infer that A is open in X?

(2) If A is closed in M , can we infer that A is closed in X?

Answer. No to both.

Theorem 2.10 is about the openness of A in the set inclusion structure A ⊆M ⊆X. Theorem 2.11 concerns the topology of A in that structure.

Theorem 2.11. Let A ⊆ M ⊆ X. Then the topology of A induced from thetopology of X coincides with the topology of A induced from ΩM , where ΩM is thesubspace topology of M induced from the topology of X. In other words, it is safeto say “the subspace topology of A”.

Proof. O ∩ A : O ∈ ΩM = (O′ ∩M) ∩ A : O′ ∈ ΩX = O′ ∩ A : O′ ∈ ΩX.

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2.3. Point position with respect to a set.

Definition 2.12. Given (X,Ω) and A ⊆ X.

• A limit point of A: a point p ∈ X s.t.

U ∩ (A \ p) 6= ∅, ∀ neighbourhood U of p.

An isolated point of A: a point p ∈ A s.t.

∃ a neighbourhood U of p s.t. U ∩ (A \ p) = ∅.

• A is perfect, if it is closed and has no isolated points.

• The closure of A: the union of A and its limit points, denoted A, i.e.,

A = x ∈ X : N ∩ A 6= ∅, ∀ neighbourhood N of x.

It is alternatively written as ClA when considered as a set operator.

An adherent point of A: a point in A.

• An interior point of A: a point having a neighbourhood in A.

An exterior point of A: a point having a neighbourhood in Ac.

A boundary point of A: a point s.t. each neighbourhood meets both A and Ac.

• The interior of A: A = interior points = ∪O ∈ Ω: O ⊆ A;The exterior of A w.r.t. X: (Ac) = exterior points;The boundary of A: ∂A = boundary points = A \ A.

Remark 7. The symbols Cl(A), Int(A), and Ext(A) are used to denote the closure A,the interior A, and the exterior (Ac) resp., when one emphasizes that they areset operators. The symbol Bd(A) is sometimes recognized as the boundary of A,but used uncommonly since the symbol ∂ well plays the role of a set operator.

Example 2.13. ∀A ⊆ X, we have

A ⊆ A, ExtA ⊆ Ac, and Ext ∅ = X.

In R, we have IntQ = ∅ and ExtQ = ∅. For the Cantor set in Eg. 2.4, we have

K = K, K = ∅, ExtK = I \K, and ∂K = K.

Notation 2.14. The unit interval: the interval [0, 1] in R, denoted by I.

Puzzle 2.15. The Smith-Volterra-Cantor set K ′ is a set of points on the realline R; see Fig. 2.4. It can be obtained by removing certain intervals from I asfollows. After removing the middle 1/4 from I, remove the subintervals of length

20 D.G.L. WANG

1/4n from the middle of each of the remaining intervals. For instance, at the firstand second step the remaining intervals are[

0,3

8

]∪[

5

8, 1

]and

[0,

5

32

]∪[

7

32,3

8

]∪[

5

8,25

32

]∪[

27

32, 1

].

Show that the set K ′ is nowhere dense.

Figure 2.4. The Smith-Volterra-Cantor set. Stolen from Bing image.

Theorem 2.16 (Characterization of the closure and interior). The closure of aset is the smallest closed set containing that set, and the interior of a set is thelargest open set contained in that set.

Proof. The closure of a set is the intersection of all closed sets containing it, andthe interior of a set is the union of all open sets contained in it.

Proposition 2.17. Given a topological space X.

1) The interior and exterior of a set are open, and the boundary is closed.

2) Any set A in X can be decomposed as

A = Lim(A) t Iso(A),

where Lim(A) and Iso(A) are the sets of limits and isolated points of A resp.

3) The whole set X can be decomposed w.r.t. a subset A:

X = (∂A) t (IntA) t (ExtA) = (ClA) t (ExtA),

4) ∂A = A ∩ Ac.5) ∂(A) ⊆ ∂A.

Definition 2.18. Let f be an operator.

• f is an involution: f f = id. is the identity map.

• f is idempotent: f f = f .

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In the operator theory, we denote the set operator of closure, interior, complementresp. by k, i, and c. For instance, the set operator c is an involution.

Theorem 2.19. Given (X,Ω) and A ⊆ X.

1) Both the set operations interior and closure preserve the inclusion, i.e.,

A ⊆ X =⇒ A ⊆ X and A ⊆ X.

2) ci = kc.

3) The operators k, i, and ki = kckc are idempotent.

4) The distributivity of k with ∪, and the distributivity of i with ∩:

A ∪B = A ∪B and (A ∩B) = A ∩B.The operators k does not work well with ∩, neither does i with ∪:

A ∩B ⊆ A ∩B and (A ∪B) ⊇ A ∪B.

Remark 8. It is clear that the exterior does not preserves the inclusion, and that itis not idempotent. From Theorem 2.19, one may see that the set operation closureis “more dual” to the set operation interior, comparing to the set operator exterior.

However, there are some properties that closure and interior do not share.

Theorem 2.20. Let X be a topological space and A ⊆M ⊆ X. Then

ClM(A) = ClX(A) ∩M,

but IntM(A) 6= IntX(A) ∩M in general.

Proof. The formula for closure holds since

ClM(A) =⋂A⊆F

F is closed in M

F =⋂A⊆C

C is closed in X

(C ∩M) = M ∩⋂A⊆C

C is closed in X

C = M ∩ ClX(A),

The other formula holds false for instance X = R2, M = A = R.

Puzzle 2.21 (Kuratowski’s closure-complement problem). How many pairwisedistinct sets can one obtain from of a given subset of a topological space by usingthe set operators k and c?

Answer. 14. An example in the real line: (0, 1) ∪ (1, 2) ∪ 3 ∪([4, 5] ∩Q

).

The answer 14 was first published by Kuratowski in 1922. A subset realising themaximum of 14 is called a 14-set.

22 D.G.L. WANG

Figure 2.5. Kazimierz Kuratowski (1896–1980) was a Polish math-ematician and logician. He was one of the leading representatives ofthe Warsaw School of Mathematics. Stolen from Wiki.

Puzzle 2.22. Recall that we can define a topology either in terms of open sets, orin terms of closed sets, or in terms of neighbourhoods. Can we define a topologywith the aid of the closure operation, or the interior operation?

Answer. Let X be a set. Let Cl∗ be a transformation on the power set 2X s.t.

(i) Cl∗ ∅ = ∅;(ii) A ⊆ Cl∗A;

(iii) distributive with the union operation: Cl∗(A ∪B) = Cl∗A ∪ Cl∗B;

(iv) idempotent: Cl∗Cl∗A = Cl∗A.

Then the set Ω = O ⊆ X : Cl∗(Oc) = Oc is a topology on X. Moreover, the set

Cl∗A is the closure of a set A in the topological space (X,Ω).

Definition 2.23. Given (X,Ω) and A,B ⊆ X.

• A is dense in B: B ⊆ A.

• A is everywhere dense: A = X, i.e., ExtA = ∅.• A is nowhere dense: ExtA is everywhere dense, i.e., Ext ExtA = ∅.

Remark 9. Concerning topics on subset density, an often helpful fact is Theorem 2.16;see the proofs of Theorem 2.25 and Corollaries 2.26 and 2.27.

23

Example 2.24. The whole set is everywhere dense, and the empty set is noteverywhere dense. Continuing Eg. 2.13, the set Q is everywhere dense in R, andthe Cantor set is nowhere dense in I.

Theorem 2.25 (Characterization for everywhere density). A set is everywheredense ⇐⇒ it meets every nonempty open set ⇐⇒ it meets every neighbourhood.

Proof. The second equivalent is clear. We show the first. Let (X,Ω) be a topologicalspace with A ⊆ X.

⇒. Let O ∈ Ω s.t. A ∩ O = ∅. Then A is a subset of the closed set Oc. It followsthat X = A ⊆ Oc. Hence O = ∅.

⇐. Let F ⊂ X be a proper closed set containing A. From premise, we haveA ∩ F c 6= ∅, that is, A 6⊆ F . Therefore, the smallest closed set containing Ais X. By Theorem 2.16, we have A = X. Hence A is everywhere dense.

This completes the proof.

Corollary 2.26. If A is everywhere dense and O is open, then O ⊆ A ∩O.

Proof. If not, then ∃ x ∈ O with a neighbourhood U s.t. U ∩ (A∩O) = ∅. W.l.o.g.,we can suppose that U is open. Then the everywhere dense set A does not meetthe open set U ∩O. By Theorem 2.25, we infer that U ∩O = ∅, contradicting thefact that x ∈ U ∩O.

Corollary 2.27. Let X be a topological space.

(1) X is indiscrete ⇐⇒ only the empty set ∅ is not everywhere dense.

Proof. Let A ⊆ X s.t. A 6∈ ∅, X. If X is indiscrete, then there is onlyone nonempty open set, that is, the whole set X. Certainly A meets X. ByTheorem 2.25, the set A is everywhere dense. Conversely, since A = X 6= A, noA is closed. Hence X is indiscrete.

(2) X is discrete ⇐⇒ only the whole set X is everywhere dense.

Proof. Let A ⊆ X s.t. A 6∈ ∅, X. If X is discrete, then A is closed, andA = A 6= X. Conversely, by Theorem 2.25, the set A does not meet somenonempty open set. Taking A to be the complement of each singleton, we findthat every singleton is open. Hence X is discrete.

(3) A set S is everywhere dense in the arrow ⇐⇒ supS =∞.

24 D.G.L. WANG

Proof. If supS =∞, then (s,∞)∩S 6= ∅, ∀ s ≥ 0. It is clear that [ 0,∞)∩S 6= ∅.Thus S meets every nonempty open set, and is everywhere dense by Theorem 2.25.Conversely, if supS 6= ∞, then ∃ M > 0 s.t. s < M , ∀ s ∈ S. It follows thatS ∩ (M,∞) = ∅. By Theorem 2.25, we infer that S is not everywhere dense.

(4) A set S is everywhere dense in the cofinite space ⇐⇒ S is infinite.

Theorem 2.28. A set A is nowhere dense in a topological space X ⇐⇒each open set in X contains an open set that is contained in Ac.

Proof. By Theorem 2.25, we deduce that A is nowhere dense

⇐⇒ ExtA is everywhere dense

⇐⇒ ExtA meets every neighbourhood

⇐⇒ each neighbourhood contains an exterior point of A

⇐⇒ each neighbourhood contains a neighbourhood that is contained in Ac

⇐⇒ each open set contains an open set that is contained in Ac.

This completes the proof.

Theorem 2.29. The real line is not the union of a countable number of nowheredense sets.

Proof. Assume to the contrary that R = ∪n∈NYn, where Yn is nowhere dense in R.Since Y0 is nowhere dense, by Theorem 2.28, ∃ [a0, b0] ⊆ Y c

0 . Since Y1 is nowheredense, by Theorem 2.28, ∃ [a1, b1] ⊆ [a0, b0] ⊆ Y c

1 . Continuing in this way, oneobtains a nested sequence of closed intervals [an, bn] which is contained in Y c

n . Fromcalculus, we know that the set A = ∩n∈N[an, bn] 6= ∅. But

A ⊆⋂n∈N

Y cn =

(⋃n∈N

Yn

)c= Rc = ∅,

a contradiction.

2.4. Bases of a topology.

Definition 2.30. Given (X,Ω).

• A base for Ω: a collection β of open sets s.t.

any open set of X is the union of some members of β.

• Ω is generated by β: β is a base for a topology Ω; denoted Ω = 〈β〉.

25

Remark 10. In contrast to a basis of a vector space in linear algebra, a base need notto be maximal, e.g., any open set can be safely added to a base. Moreover,a topological space may have disjoint bases of distinct sizes, e.g., the standardtopology ΩR of the real line has a base of all open intervals with rational ends, andanother base of all open intervals with irrational ends.

Question 2.31. Here are some questions about bases.

(1) Describe a smallest base for an indiscrete space.

Proof. The whole set.

(2) Describe a smallest base for a discrete space. Is it unique?

Answer. The collection of all singletons. Yes.

(3) Describe a smallest base for the arrow. Is it unique?

Answer. (r,∞) : r ∈ Q+ ∪ [0,∞). No.

Here is a criterion for a collection to be a base of a given topology.

Theorem 2.32. Given (X,Ω) and β ⊆ Ω. Then β is a base for Ω

⇐⇒ for any pair (x,O) where x ∈ O ∈ Ω, ∃ B ∈ β s.t. x ∈ B ⊆ O.

Proof. Let O ∈ Ω. ⇒: Since β is base, ∃ Γ ⊆ β s.t. O = ∪B∈ΓB.

⇐: ∀x ∈ O, let Bx ∈ β such that x ∈ Bx ⊆ O. Then O = ∪x∈OBx.

Puzzle 2.33. Any base of the real line is reducible. In other words, for any base βof the real line R, ∃ B ∈ β s.t. β \ B is a base of R.

Proof. Assume that R has an irreducible base β. Let A ∈ β. By Theorem 2.32,there exists a pair (xA, OA) where xA ∈ OA ⊆ ΩR s.t. the set A is the uniqueM ∈ β s.t. xA ∈ M ⊆ OA. Note that the open set A is the union of some openintervals. Let IA be the maximal open interval in A which contains xA.

We shall show that A = IA. In fact, since β is a base, the open set IA can begenerated by β. If A 6= IA, then IA ⊂ A, and IA can be generated by β \ A. Itfollows the existence of B ∈ β \ A s.t. xA ∈ B ⊆ IA. Hence, both A,B ∈ β wellplay the role of M , contradicting the uniqueness of M . This proves IA = A.

Let A = (u, v). Suppose that u ∈ R. Since for each n ≥ 1, the open intervalJn = (u + (v − u)/n, v) can be generated by β \ A, the interval A = ∪n≥1Jn

26 D.G.L. WANG

can be generated by β \ A. This proves that the base β can be reduced byremoving the member A. Along the same lines we can prove the reducibility ifv ∈ R. Otherwise A = (−∞,∞). Since the interval (−n, n) can be generated byβ \ A for each n ∈ Z+, so can the interval A. Hence β is reducible.

Next is a criterion for a collection to be a base of a certain topology.

Theorem 2.34. Given X and β ⊆ 2X . Then

β is a base for a certain topology on X

⇐⇒

X is the union of all members of β,

the intersection of any two members of β is the union of some sets in β.

Proof. ⇒: Since any member of β is open for the topology generated by β, theintersection of any two members is open and hence can be generated by thebase β. ⇐: Define Ω ⊆ 2X to be the collection of unions of members of β. Byusing distributive law for the unions and intersections, one may verify that Ω is atopology on X.

In contrast to Euclid’s classical proof of the infinitude of primes recorded inthe Elements, Hillel (Harry) Furstenberg published a topological proof in 1955 (H.Furstenberg, On the infinitude of primes, Amer. Math. Monthly 62 (5) 353), whenhe was an undergraduate student at Yeshiva Univ.; see Fig. 2.6.

Figure 2.6. Hillel (Harry) Furstenberg (1935 — ) is an American-Israeli mathematician, a member of the Israel Academy of Sciencesand Humanities and U.S. National Academy of Sciences and a laure-ate of the Wolf Prize in Mathematics. Stolen from Wiki.

Space 9. The evenly spaced integer topology (Z,Ω): Ω is generated by allarithmetic progressions.

27

Furstenberg’s proof of the infinitude of primes. In the evenly spaced integer topol-ogy, each set P (a, d) = a+ nd : n ∈ Z is clopen. Note that

Z \ −1, 1 =⋃

p is prime

P (0, d).

If the number of primes is finite, then the set −1, 1 is clopen. However, everyopen set is infinite from definition, a contradiction.

2.5. Metrics & the metric topology.

Definition 2.35. Let X be a set.

• A metric on X: a function d : X ×X → R s.t.

(a) subadditivity or triangle inequality: d(x, y) ≤ d(y, z) + d(z, x);

(b) symmetry: d(x, y) = d(y, x); and

(c) being positive-definite: d(x, y) ≥ 0, and d(x, y) = 0 ⇐⇒ x = y.

• A metric space: the pair (X, d).

Remark 11. In economics, subadditivity is an essential property of some particularcost functions. It implies that production from only one firm is socially lessexpensive (in terms of average costs) than production of a fraction of the originalquantity by an equal number of firms. – from wiki.

We distinguish the notation of a topological space (X,ΩX) and the notation ofa metric space (X, d) by statement. As will be seen in Definition 2.48, a metricinduces a topological structure.

Example 2.36. Let d1 and d2 be metrics on X. Then ρ+ = d1 + d2 is a metric,and so is ρmax = maxd1, d2.

Proof. The symmetry and positive semi-definity for each of the functions ρ+

and ρmax are clear. The subadditivity is shown below.

ρ+(x, z) + ρ+(y, z) =(d1(x, z) + d2(x, z)

)+(d1(y, z) + d2(y, z)

)=(d1(x, z) + d1(y, z)

)+(d2(x, z) + d2(y, z)

)≥ d1(x, y) + d2(x, y) = ρ+(x, y);

ρmax(x, z) + ρmax(y, z) = maxd1(x, z), d2(x, z)+ maxd1(y, z), d2(y, z)≥ maxd1(x, z) + d1(y, z), d2(x, z) + d2(y, z)≥ maxd1(x, y), d2(x, y) = ρmax(x, y).

28 D.G.L. WANG

The properties of being definite can be shown as follows.

ρ+(x, y) = 0 ⇐⇒ d1(x, y) = 0 and d2(x, y) = 0 ⇐⇒ x = y.

ρmax(x, y) = 0 ⇐⇒ d1(x, y) = 0 and d2(x, y) = 0 ⇐⇒ x = y.

This completes the proof.

Example 2.37. The map d∆(A,B) = S(A) + S(B)− 2S(A ∩B), where S(C) isthe area of a polygon C, defines the area metric on the set of all bounded planepolygons. In fact, the number d∆(A,B) is the area of the symmetric differenceA∆B. Thus the symmetry and being positive-definite are clear. The subadditivityholds because A∆C ⊆ (A∆B) ∪ (B∆C) for any bounded plane polygons A,B,C.

Homework 2.4. Let d be a metric on a set X, and f a function defined on R≥0.Then the function g(x, y) = f(d(x, y)) is a metric if

i) f(0) = 0;

ii) f is increasing; and

iii) f is subadditive, i.e., f(x+ y) ≤ f(x) + f(y) for any x, y ∈ R.

Proof. We check the 3 conditions for a metric.

(1) Subadditivity.

f(d(x, y)) ≤ f(d(x, z) + d(y, z)) (by the increasing property of f)

≤ f(d(x, z)) + f(d(y, z)) (by the subadditivity of f).

(2) Symmetry. f(d(x, y)) = f(d(y, x)) by the symmetry of d.

(3) Being positive-definite. By the increasing property, we have

f(d(x, y)) ≥ f(0) = 0, and

f(d(x, y)) = 0 ⇐⇒ f(d(x, y)) = f(0) ⇐⇒ d(x, y) = 0 ⇐⇒ x = y.

This completes the proof.

Homework 2.5. For any metric d, the function mind(x, y), 1 is a metric.

Proof. We check that the function ρ(x, y) = mind(x, y), 1 satisfy the 3 conditionsfor a metric.

29

(1) Subadditivity. Since ρ(x, z) = mind(x, z), 1 ≤ 1, by the non-negativity of themetric d, the desired inequality holds if d(x, y) ≥ 1 or d(y, z) ≥ 1. Consequently,we may suppose that d(x, y) < 1 and d(y, z) < 1. Then, by the subadditivityof d, we infer that

ρ(x, y) + ρ(y, z) = d(x, y) + d(y, z) ≥ d(x, z) ≥ ρ(x, z).

(2) Symmetry: Directly by the symmetry of d.

(3) Being positive-definite: we have ρ(x, y) ≥ 0 since d(x, y) ≥ 0, and

ρ(x, y) = 0 ⇐⇒ d(x, y) = 0 ⇐⇒ x = y.

This completes the proof.

Homework 2.4 and Homework 2.5 give two ways of self-production of metrics.

In linear algebra, functional analysis and related areas, we use the concept ofnorm, which has a close relationship with metrics.

Definition 2.38. Let V be a vector space over R.

• A seminorm on V : a function ‖ · ‖ : V → R satisfying

(1) subadditivity: ‖u+ v‖ ≤ ‖u‖+ ‖v‖, ∀u, v ∈ V ; and

(2) absolutely scalability: ‖λv‖ = |λ|· ‖v‖, ∀λ ∈ R, v ∈ V .

• Seminorm is positive semi-definite: ‖v‖ ≥ 0, ∀ v ∈ V .

(2) ⇒ ‖0‖ = 0 and ‖−v‖ = ‖v‖. Taking u = −v in (1) ⇒ ‖v‖ ≥ 0.

• The trivial seminorm: ‖v‖ = 0 ∀ v ∈ V .

• A norm on V : a function ‖ · ‖ satisfying (1),(2), and

(3) being definite: ‖v‖ = 0⇒ v = 0.

• A normed space: a vector space equipped with a norm.

• The metric induced by the norm: d(x, y) = ‖x− y‖.

Notation 2.39. We denote H = x : x ≥ 1 ∪ +∞.Definition 2.40. Let x = (x1, . . . , xn) ∈ Cn and p ∈ H.

• The p-norm of x:

(2.1) ‖x‖p =

( n∑i=1

|xi|p)1/p

.

The subadditivity is called Minkowski’s inequality; see Theorem A.6.

30 D.G.L. WANG

• The grid/Manhattan/taxicab norm or snake distance: the 1-norm. SeeFig. 2.7.

Figure 2.7. The Manhattan distance. Stolen from Bing image.

• The Euclidean norm: the 2-norm.

• The infinity/maximum norm: ‖x‖∞ = maxi|xi|.In fact, if M = maxi|xi|, then

‖x‖∞ = limp→∞

( n∑i=1

|xi|p)1/p

= M · limp→∞

( n∑i=1

|xi|p

Mp

)1/p

= M · limp→∞|1 ≤ i ≤ n : |xi| = M|1/p = M.

Example 2.41. Let p ∈ H. Consider the metric

(2.2) ρ(p)(x, y) =

( n∑i=1

|xi − yi|p)1/p

induced by the p-norm.

(1) ρ(1)(x, y) =∑n

i=1 |xi − yi|.

(2) ρ(2)(x, y) =√∑n

i=1(xi − yi)2 is the Euclidean metric.

(3) The Chebyshev/chessboard distance or L∞-metric:

ρ(∞)(x, y) = maxi|xi − yi|.

The Chebyshev distance is used in warehouse logistics, as it effectively measuresthe time an overhead crane takes to move an object. In fact, the crane can moveon the x and y axes at the same time but at the same speed along each axis;see Fig. 2.8.

31

Figure 2.8. Pafnuty Lvovich Chebyshev (1821–1894) was a Rus-sian mathematician. His name can be alternatively transliteratedas Chebychev, Chebysheff, Chebychov, Chebyshov; or Tchebychev,Tchebycheff (French transcriptions); or Tschebyschev, Tschebyschef,Tschebyscheff (German transcriptions). The right part is a photo foran overhead crane in Qingdao Beihai Shipbuilding Heavy Industryin China. Stolen from Wiki and Bing image.

Definition 2.42 (generalization). Let x = (x1, . . . , xn) ∈ Cn and p ∈ H.

• The `p-space: the set of infinite complex-valued sequences (x1, x2, . . .) withconvergent `p-norm

‖(x1, x2, . . .)‖ =

(∑i

|xi|p)1/p

.

• The Lp-space: the set of functions f on X ⊆ R with convergent Lp-norm

‖f‖p,X =

(∫X

|f(x)|p dx)1/p

.

Definition 2.43. Let (X, d) be a metric space. Let r ≥ 0.

• The open ball of radius r centerd at x ∈ X:

B(x, r) = y ∈ X : d(x, y) < r.

• The disk of radius r centerd at x ∈ X:

D(x, r) = y ∈ X : d(x, y) ≤ r.

• The sphere of radius r centerd at x ∈ X:

S(x, r) = y ∈ X : d(x, y) = r.

32 D.G.L. WANG

Example 2.44. The names in Definition 2.43 come from the Euclidean space.

• Dn: the unit disk in Rn, i.e., the disk of radius 1 centered at 0.

D1 is the interval [−1, 1], D2 is a plane disk, and D3 a ball.

• Sn: the unit sphere in Rn+1, i.e., the sphere of radius 1 centered at 0.

S0 is the set −1, 1, S1 is a circle, and S2 a sphere.

However, they may rather surprising in other metric spaces.

• The unit ball in R2 equipped with ρ(1) is a closed square of side length√

2.

• The unit ball in R2 equipped with ρ(∞) is a closed square of side length 2.

Definition 2.45. Let (X, d) be a metric space with A,B ⊆ X and x ∈ X.

• The diameter of A: diam(A) = supx,y∈A d(x, y).

• A is bounded: A is contained in a ball, i.e., ∃ r s.t. d(x, y) < r, ∀ x, y ∈ A.

• The distance between x and A: d(x,A) = infa∈A d(x, a).

• The Hausdorff distance between A and B:

dH(A,B) = max

supa∈A

d(a,B), supb∈B

d(A, b)

.

See Fig. 2.9.

Figure 2.9. The Hausdorff distance. Stolen from Wiki.

Remark 12. In computer graphics the Hausdorff distance is used to measure thedifference between two different representations of the same 3D object.

Theorem 2.46. Let A be a closed set. Then d(A, b) = 0 ⇐⇒ b ∈ A.

33

Proof. ⇐: Clear. ⇒: Since d(A, b) = 0, each ball centered at b has a nonemptyintersection with A. Hence b 6∈ Ac. Thus b ∈ Cl(A) = A.

Theorem 2.47. For every metric space, the Hausdorff distance is a metric on theset of its closed bounded subsets.

Proof. We should verify the three conditions for a metric. Let (X, d) be a metricspace. Let A,B,C be closed bounded subsets of X.

(1) Symmetry and positivity. Clear from definition.

(2) Being definite. By Theorem 2.46, we can infer that

dH(A,B) = 0 ⇐⇒ supa∈A

d(a,B) = supb∈B

d(A, b) = 0

⇐⇒ d(a,B) = d(A, b) = 0 for all a ∈ A and b ∈ B⇐⇒ a ∈ B and b ∈ A for all a ∈ A and b ∈ B⇐⇒ A ⊆ B and B ⊆ A

⇐⇒ A = B.

(3) Subadditivity. Let r(A,B) = supa∈A d(a,B). Then

dH(A,B) = max(r(A,B), r(B,A)

).

To show the subadditivity of dH , it suffices to show subadditivity of the function r.By the subadditivity of d, for all a ∈ A and all b ∈ B, we have

d(a, C) ≤ d(a, b) + d(b, C) ≤ d(a, b) + r(B,C).

It follows from the arbitrariness of b, we infer that

r(A,C) ≤ d(a, C) ≤ d(a,B) + r(B,C) ≤ r(A,B) + r(B,C).

This completes the proof.

Homework 2.6. Find a metric space and two balls in it s.t. the smaller ballcontains the bigger ball properly. What is the minimal number of points in such aspace? Show that the largest radius in such a space is at most twice the smallerradius.

Homework 2.7. Show that the segment with endpoints a, b ∈ Rn can be describedas

x ∈ Rn : ρ(a, x) + ρ(b, x) = ρ(a, b)under the Euclidean metric. How does the above set look if ρ = ρ(1) or ρ = ρ(∞)?

34 D.G.L. WANG

Definition 2.48. Let X be a set.

• Given a metric d on X. The metric topology of (X, d), or the topology of Xgenerated by d:

ΩX,d = O ⊆ X : ∀x ∈ O, ∃ ε, B(x, ε) ⊆ O.In other words, the collection of open balls in (X, d) is a base of Ωd.

• Given a topology Ω on X. X is metrizable: Ω is generated by a certain metric.

Homework 2.8. Let X be a set. Define the function d0 : X ×X → R≥0 by

(2.3) d0(x, y) =

0, if x = y,

1, if x 6= y.

(1) Show that d0 is a metric.

(2) Let x ∈ X. Find D(x, 1), D(x, 1/2), and S(x, 1/2) in this metric space.

(3) What is the metric topology of (X, d0)?

Theorem 2.49. We have the following.

(1) Any metrizable indiscrete space is a singleton.

Proof. Let (X, ΩX) be an indiscrete space. Then ΩX = ∅, X from definition.Suppose that |X| ≥ 2. Then X contains two distinct points, say, x and y. IfX is metrizable, say, ΩX is generated by some metric d. Then d(x, y) > 0 fromdefinition, and the set S = B(x, d(x, y)/2) is open. But x ∈ S and y 6∈ S, weinfer that S 6∈ ΩX , a contradiction! This completes the proof.

(2) A finite space is metrizable if and only if it is discrete.

Proof. Let (X, ΩX) be a finite space. If X is discrete, then ΩX = 2X fromdefinition. Under the metric d0 defined by Eq. (2.3), every singleton x ∈ Xforms an open set since x = B(x, 1/2) ∈ ΩX . Hence every subset of X is openunder the topology generated by d0. Conversely, suppose that ΩX is generatedby a metric d. Let

mx = mind(x, y) : y ∈ X \ x.Since X is finite, we have mx > 0. Then every singleton x = B(x, mx/2) isopen. Thus ΩX is the discrete topology. This completes the proof.

Proposition 2.50. Let (X, d) be a metric space. Let x ∈ X and A,B ⊆ X.

35

(1) d(x,A) = 0 ⇐⇒ x ∈ A.

(2) dH(A,B) is finite if both A and B are bounded.

(3) dH(A,B) = 0 ⇐⇒ A = B.

(4) For every metric space, the Hausdorff distance is a metric on the set of its closedbounded subsets.

Table 2.1. Openness & closeness of balls and spheres.

open ball disk sphereOpen YES MAYBE MAYBEClosed MAYBE YES YES

Theorem 2.51. Two metrics d1 and d2 on X are equivalent if they generate thesame topology, denoted d1 ∼ d2. If ∃ c, C > 0 s.t. c·d1(x, y) ≤ d2(x, y) ≤ C ·d1(x, y),∀x, y ∈ X, then d1 ∼ d2.

Proof. Write ΩX,di = Ωi for i = 1, 2. The desired equivalence says Ω1 = Ω2.

First, we show that Ω2 ⊆ Ω1. Since any open set O2 ∈ Ω2 is the union of someopen balls w.r.t. the metric d2, it suffices to recast a typical ball of the form

B(x, r) = z ∈ X : d2(x, z) < ras a member in Ω1, i.e., the union of open balls w.r.t. the metric d1. Define

U(x, r) =⋃

z∈B(x,r)

y ∈ X : d1(y, z) < L

where

L =r − d2(x, z)

2C.

Then U(x, r) ∈ Ω1. We shall show that B(x, r) = U(x, r). On one hand, ∀ z ∈B(x, r), since d1(z, z) = 0 < L, we have z ∈ U(x, r). On the other hand, ∀ y ∈U(x, r), ∃ z ∈ B(x, r) s.t. d1(y, z) < L. Then

d2(x, y) ≤ d2(x, z) + d2(y, z) ≤ d2(x, z) + C· d1(y, z) ≤ d2(x, z) + C·L < r.

Thus y ∈ B(x, r). This proves B(x, r) = U(x, r) and Ω2 ⊆ Ω1.

From premise, we have d1(x, y) ≤ c−1· d2(x, y). Along the same lines, one mayshow that Ω1 ⊆ Ω2, which implies Ω1 = Ω2 immediately.

Homework 2.9. Let H = x ∈ R : x ≥ 1 ∪ +∞. The metrics ρ(p) defined inEq. (2.2) for p ∈ H are equivalent.

36 D.G.L. WANG

Hint. By Theorem 2.51.

Homework 2.10. Let d be a metric on a set X. Then the function d/(1 + d) is ametric, and it is equivalent to d.

37

3. Continuous Maps & Homeomorphisms

3.1. Continuous maps.

Definition 3.1. Given (X,ΩX) and (Y,ΩY ). Let f : X → Y be a map.

• f is open: open 7→ open, i.e., f(ΩX) ⊆ ΩY .

• f is closed: closed 7→ closed.

• f is continuous: the preimage of any open set is open.

• f is continuous at a point x ∈ X: the set f−1(V ) is a neighbourhood of xfor every neighbourhood V of f(x).

Example 3.2. Here are some continuous maps.

(1) The identity map. It is also open and closed.

(2) Any constant map. It is also open and closed.

(3) Any inclusion.

(4) Any projection.

Question 3.3. Consider (X,ΩR) and (Y,ΩY ), where X = Y = [0, 2]. Is thefollowing functions f : X → Y continuous?

1) If ΩY = ΩR, and

f(x) =

x, if x ∈ [0, 1);

3− x, if x ∈ [1, 2].

Answer. No.

2) If ΩY is induced from the arrow, and

f(x) =

x, if x ∈ [0, 1];

x+ 1, if x ∈ [1, 2].

Answer. Yes!!

Question 3.4. Given (X,ΩX) and (Y,ΩY ).

1) If ΩX is discrete, then which maps X → Y are continuous?

Answer. All maps.

38 D.G.L. WANG

2) If ΩX is discrete, then which maps Y → X are continuous?

Answer. A map Y → X is continuous ⇐⇒ ΩY is discrete.

3) If ΩX is indiscrete, then which maps X → Y are continuous?

Answer. A map f : X → Y is continuous ⇐⇒ f is constant.

4) If ΩX is indiscrete, then which maps Y → X are continuous?

Answer. All maps.

Theorem 3.5 (Characterizations of map continuity). Given topological spaces(X,ΩX) and (Y,ΩY ). Let f : X → Y be a map. Then the followings are equivalent.

1) The preimage of any open set under f is open.

2) The preimage of any closed set under f is closed.

3) The preimage of any member of a base under f is open.

4) The preimage of any neighbourhood under f is a neighbourhood, i.e., f iscontinuous at each point.

5) f(A) ⊆ f(A), ∀A ⊆ X.

6) f−1(B) ⊆ f−1(B), ∀B ⊆ Y .

Proof. Straightforward by definition: each of Items 2) to 4) is equivalent to Item 1).

Item 4) ⇐⇒ Item 5).

⇒. Let x ∈ A. Let V be a neighbourhood of f(x). By Item 4), the set f−1(V )is a neighbourhood in X. Since f−1(V ) is a neighbourhood of x, we havef−1(M) ∩ A 6= ∅, and

V ∩ f(A) = f(f−1(V ) ∩ A

)6= ∅.

⇐. Let x ∈ X. Let V be a neighbourhood of f(x). Assume that Item 4) is false.Then f−1(V ) is not a neighbourhood of x. Define A = X \ f−1(V ). Then x ∈ A.

It follows that f(x) ∈ f(A) ⊆ f(A). Thus f(A) meets every neighbourhood off(x). In particular, we have f(A) ∩ V 6= ∅, contradicting the definition of A.

Item 6) ⇐⇒ Item 2).

⇒. Let B be a closed subset of Y . Then B = B, and f−1(B) ⊆ f−1(B) = f−1(B).

It follows that f−1(B) = f−1(B) is closed.

39

⇐. Since B ⊆ B, we have f−1(B) ⊆ f−1(B). Since B is closed, we infer that f−1(B)

is closed. Since f−1(B) is the smallest closed set containing f−1(B), we derive

that f−1(B) ⊆ f−1(B).

This completes the proof.

Theorem 3.6. Continuous surjections preserve the property of being everywheredense.

Proof. Let f : X → Y be a surjective continuous map between topological spaces.Let A be an everywhere dense set of X. Since f is surjective, we have

f(A) = f(X) = Y.

Since f is continuous, by Theorem 3.5 Item 5), we have f(A) ⊆ f(A). Therefore,

we obtain f(A) = Y , i.e., f(A) is everywhere dense.

Theorem 3.7. Let f1, . . . , fl, g1, . . . , gm, h1, . . . , hn : X → R be continuous func-tions. Let

A =

x ∈ X :

f1(x) ≤ 0, . . . , fl(x) ≤ 0,

g1(x) = 0, . . . , gm(x) = 0,

h1(x) ≥ 0, . . . , hn(x) ≥ 0

and

B =

x ∈ X :

f1(x) < 0, . . . , fl(x) < 0,

h1(x) > 0, . . . , hn(x) > 0.

Then A is closed and B is open.

Proof. The set A is the intersection of the preimages of the closed sets (−∞, 0 ]or 0 or [ 0,∞) under continuous functions, and hence closed. The set B is theintersection of the preimages of the open sets (−∞, 0) or (0,∞) under continuousfunctions, and hence closed.

Corollary 3.8. The kernel x ∈ X : f(x) = 0 of any continuous f : X → R isclosed.

Corollary 3.9. The set x ∈ R : f(x) = x of fixed points of any continuousf : R→ R is closed.

Lemma 3.10. Let f : X → R be a continuous function. Then ∀ p ∈ X, ∃ aneighbourhood U of p s.t. f is bounded in U . If 0 6∈ f(X), then ∃ a neighbourhood Vof p in which

40 D.G.L. WANG

(i) the absolute value |f(x)| has a positive lower bound in V , and

(ii) f(x)f(p) > 0.

Proof. Let p ∈ X and ε > 0. Since f is continuous, ∀ ε > 0, ∃ a neighbourhood Uof p s.t. |f(x)− f(p)| < ε. Thus |f(x)| ≤ |f(p)|+ ε, i.e., f is bounded in U .

If 0 6∈ f(X), then f(p) 6= 0, one may require the above ε < |f(p)/2|. Then|f(x)| ≥ |f(p) − ε| > ε. Moreover, the inequality |f(x) − f(p)| < ε implies thatf(x)f(p) > 0.

Theorem 3.11. Let f, g : X → R be continuous functions. Then the functionsX → R defined by the following formulas are continuous:

x 7→ f(x)± g(x),

x 7→ f(x)g(x),

x 7→ f(x)/g(x), if 0 6∈ g(X),

x 7→ |f(x)|,x 7→ minf(x), g(x),x 7→ maxf(x), g(x).

Proof. We show them individually. Let p ∈ X and ε > 0.

(1) The continuity of f − g can be obtained by considering the continuous functionsf and −g. Let h(x) = f(x) + g(x). For w ∈ f, g, since w is continuous, ∃ aneighbourhood Uw of p s.t. |w(x)− w(p)| < ε/2, ∀ x ∈ Uw. Hence

|h(x)− h(p)| ≤ |f(x)− f(p)|+ |g(x)− g(p)| < ε, ∀ x ∈ U,

where U = Uf ∩ Ug is a neighbourhood of p. In other words, ∀ open interval Jof R, h−1(J) is a neighbourhood in X. This proves the continuity of h(x).

(2) Let h(x) = f(x)g(x). Since f is continuous, by Lemma 3.10, ∃ a neighbour-hood Uf of p s.t.

|f(x)| < 1 and |g(p)| · |f(x)− f(p)| < ε/2.

Since g is continuous, ∃ a neighbourhood Ug of p s.t.

|g(x)− g(p)| < ε/2.

Therefore, one may infer that

|h(x)− h(p)| ≤ |f(x)| · |g(x)− g(p)|+ |g(p)| · |f(x)− f(p)| < ε, ∀ x ∈ U,

where U = Uf ∩ Ug is a neighbourhood of p. This proves the continuity of h(x).

41

(3) By using the continuity of the product of continuous functions, it suffices toshow that the function h(x) = 1/g(x) is continuous. In fact, by Lemma 3.10, ∃` > 0 and a neighbourhood U1 of p s.t. |g(x)g(p)| ≥ `. By the continuity of g, ∃a neighbourhood U2 of p s.t. |g(x)− g(p)| < ε`. Therefore, we can deduce that∣∣∣ 1

g(x)− 1

g(p)

∣∣∣ =∣∣∣g(x)− g(p)

g(x)g(p)

∣∣∣ < ε, ∀ x ∈ U = U1 ∩ U2.

(4) Let h(x) = |f(x)|. By Lemma 3.10, ∃ a neighbourhood U1 of p s.t. f(x)f(p) > 0.Since f is continuous, ∃ a neighbourhood U2 of p s.t. |f(x)−f(p)| < ε. Therefore,we infer that

|h(x)− h(p)| = |f(x)− f(p)| < ε, ∀x ∈ U1 ∩ U2.

(5) See Homework 3.1. Let h(x) = minf(x), g(x).(6) The continuity of the function maxf, g = −min−f, −g can be obtained by

the continuity of the minimum of continuous functions.

This completes the proof.

Homework 3.1. Let f, g : X → R be continuous functions. Prove that thefunctions x 7→ minf(x), g(x) is continuous.

Proof. Use the continuousness of sum and the continuousness of absolute value ofcontinuous functions, and the fact that

minf(x), g(x) =f(x) + g(x)

2− |f(x)− g(x)|

2.

Theorem 3.12. A function

f : X → Rn

x 7→(f1(x), . . . , fn(x)

)is continuous ⇐⇒ so are all the functions f1, f2, . . . , fn.

Hint. The topology on Rn is generated by the infinity metric

ρ(∞)(x, y) = max|x1 − y1|, . . . , |xn − yn|.

Observe that ρ(∞)(f(x), f(p)) < ε ⇐⇒ |fi(x)− fi(p)| < ε for each i.

42 D.G.L. WANG

Homework 3.2. If a function f : R→ R is continuous, then so is the graph map

Γf : R→ R2

x 7→(x, f(x)

).

Theorem 3.13. For any subset A of a metric space (X, d), the function

dA : X → Rx 7→ d(x,A)

is continuous.

Hint. |dA(x)− dA(y)| =∣∣∣infa∈A d(x, a)− infa∈A d(y, a)

∣∣∣ ≤ d(x, y).

The definition of continuity usually studied in Calculus, when applicable, isequivalent to the definition stated in terms of topological structures.

Theorem 3.14. Let f : (X, dX)→ (Y, dY ) be a map between metric spaces. Thenthe followings are equivalent.

(1) f is continuous at a point a ∈ X.

(2) Each ball centered at f(a) contains the image of a ball centered at a.

(3) ∀ ε > 0, ∃ δ > 0, ∀ x ∈ X, the inequality dX(x, a) < δ implies

dY (f(x), f(a)) < ε.

3.2. Covers. What conditions on a cover of a topological space X ensure that thecontinuity of a map f : X → Y follows from the continuity of its restrictions tomembers of that cover? See Theorem 3.16.

Definition 3.15. Let X be a set.

• A cover or covering of X: a collection Γ of sets s.t. X ⊆ ∪A∈ΓA.

In this case, we say that Γ covers X.

• A subcover or subcovering of a cover Γ: a subfamily of Γ which covers X.

• A refinement of a cover Γ: a cover of X whose every member is contained in amember of Γ.

• A partition of X: a cover Γ of X s.t. X = tA∈ΓA.

To partition X: to construct a partition of X.

43

• Suppose that X = ∪A∈ΓA is a topological space. We say that the cover Γ is

– fundamental: the following equivalence holds

a set O ⊆ X is open ⇐⇒ O ∩ A is open in A, ∀A ∈ Γ.

– open: every member in Γ is open.

– closed: every member in Γ is closed.

– locally finite: every point in X has a neighbourhood meeting only a finitenumber of members in Γ.

Theorem 3.16. Let f : X → Y be a map between topological spaces. Let Γ be afundamental cover of X s.t. f |A is continuous ∀A ∈ Γ, then f is continuous.

Proof. Let O be an open set in Y . Since f |A is continuous, we have

f−1(O) ∩ A = (f |A)−1(O)

is open in A, ∀A ∈ Γ. Since Γ is a fundamental cover, we find f−1(O) is open in X.Hence f is continuous.

Remark 13. Theorem 3.16 provides a way of proving the continuity of a map inthe spirit of divide and conquer.

Theorem 3.17 (Characterizations of fundamental covers). Let X be a set with acover Γ. Then

Γ is fundamental

⇐⇒ a set O ⊆ X is open, provided O ∩ A is open in A, ∀ A ∈ Γ

⇐⇒ a set F ⊆ X is closed, provided F ∩ A is closed in A, ∀ A ∈ Γ.

Proof. For the first equivalence, it suffices to check from right to left. In fact, whenO is open in X, by definition of the subspace topology, the intersection O ∩ A iscertainly an open set of A.

Let us show the second equivalence. ⇒: ∀ A ∈ Γ, the set (X\F )∩A = A\(F∩A)is open. Thus X \ F is open, i.e., F is closed. The converse statement can beshown similarly.

Theorem 3.18 (Criteria for the fundamentality). A cover Γ is fundamental

1) if it has a fundamental refinement;

44 D.G.L. WANG

Proof. Let O ⊆ X. Suppose that O ∩ A is open in A, ∀A ∈ Γ. Let Γ′ be afundamental refinement of Γ. Let B ∈ Γ′. Then ∃ AB ∈ Γ s.t. B ⊆ AB. ByTheorem 2.11, we can regard the topology of AB ∩B as the one induced fromAB. It follows that O ∩ B = O ∩ AB ∩ B is open in AB ∩ B = B. By thearbitrariness of B, and since Γ′ is fundamental, we infer that O is open in X.This proves that Γ is fundamental.

2) if it is open or finite closed;

Proof. Let O ⊆ X s.t. O ∩ A is open in A, ∀A ∈ Γ. By Theorem 2.10, O ∩ Ais open in X, and so is the set ∪A∈Γ(O ∩ A) = O. By Theorem 3.17, Γ isfundamental. In the same vein, every finite closed cover is fundamental.

3) if it is locally finite and closed;

Proof. Each point p ∈ X has an open neighbourhood Op meeting only a finitenumber of members of Γ. Then the cover

Σ = A ∩Op : A ∈ Γ

of Op is finite and closed, and thus fundamental. Let O ⊆ X be a set s.t. O ∩Ais open in A, ∀A ∈ Γ. We shall show that O is open in X.

In fact, let p ∈ X and A ∈ Γ. Consider the topology of A ∩Op as the oneinduced from the topology of A, we infer that

(O ∩Op) ∩ (A ∩Op) = (O ∩ A) ∩Op

is open in A∩Op. By the arbitrariness of A, and since the cover Σ is fundamental,we infer that the set O∩Op is open in Op, and thus open in X by Theorem 2.10.Hence

O = O ∩X = O ∩⋃p∈X

Op =⋃p∈X

(O ∩Op)

is open in X. This completes the proof.

4) if ∃ a fundamental cover Σ s.t. the cover ΣS = A ∩ S : A ∈ Γ of S isfundamental ∀S ∈ Σ;

Proof. Let O ⊆ X s.t. O∩A is open in A, ∀A ∈ Γ. Let S ∈ Σ. Then O∩A∩Sis open in A∩S. Since the cover ΣS is fundamental, we infer that O ∩S is openin S. Since the cover Σ is fundamental, we derive that O is open in X. Thisproves that Γ is fundamental.

5) if every point has a neighbourhood U s.t. the cover ΓU = A ∩ U : A ∈ Γ of Uis fundamental. In other words, the cover fundamentality is a local property.

45

Proof. ∀p ∈ X, let Up be a neighbourhood of p s.t. the cover ΓUp is fundamental.We claim that the cover Σ = Up : p ∈ X of X is fundamental. In fact, supposethat O ⊆ X s.t. O ∩ Up is open in Up, ∀ p ∈ X. Since Up is a neighbourhoodof p, ∃ open set Op s.t. p ∈ Op ⊆ Up. Therefore, O ∩Op = O ∩ Up ∩Op is openin Op = Up ∩ Op, and thus open in X. Hence the set O = ∪p∈X(O ∩ Op) isopen in X. This proves the claim. Now, by using Item 4), we obtain that Γ isfundamental.

Remark 14. The above characterisations will be used in the proof of Theorem 4.37.

3.3. Homeomorphisms.

Definition 3.19. Given topological spaces X and Y .

• An invertible map f : X → Y is a homeomorphism: both f and f−1 arecontinuous.

• X is homeomorphic to Y : ∃ a homeomorphism between them.

Remark 15. In this note the homeomorphism relation is denoted by ∼=. Thisnotation is not commonly accepted. You may see any sign closed to but distinctfrom the equal sign =, such as ∼, ', ≈ from other resources.

Theorem 3.20. Being homeomorphic is an equivalent relation.

Proof. The identity map of any topological space is a homeomorphism. Bothcomposition and inverse preserve homeomorphisms.

Theorem 3.21. Let h : (X,ΩX)→ (Y,ΩY ) be a homeomorphism. Let x ∈ A ⊆ X.

(1) A ∈ ΩX ⇐⇒ h(A) ∈ ΩY .

Proof. ⇐: A = h−1(h(A)

)is the preimage of the open set h(A) under the

continuous map h. ⇒: Define g = h−1. h(A) = g−1(A) is the preimage of theopen set A under the continuous map g−1 = h.

(2) A is closed ⇐⇒ h(A) is closed.

(3) h is commutative with the set operations Cl and Int.

(4) A is a neighbourhood of x ⇐⇒ h(A) is a neighbourhood of the point h(x).

46 D.G.L. WANG

Remark 16. Homeomorphic spaces are completely identical from the topologicalpoint of view: a homeomorphism X → Y establishes a one-to-one correspondencebetween “all phenomena” in X and Y that can be expressed in terms of topologicalstructures. This justifies our assertion in §1 that the topological spaces (X,ΩX)and (Y,ΩY ) should be thought of as the same topological space.

When the notion of topological space had not yet been developed, mathemati-cians studied only subspaces of Euclidean spaces, their continuous maps, andhomeomorphisms. Christian Felix Klein, in his famous Erlangen Program, classifiedvarious geometries that had emerged up to that time, like Euclidean, Lobachevsky,affine, and projective geometries, and defined topology as a part of geometry thatdeals with properties preserved by homeomorphisms. Try to say some differencesbetween topology and geometry, and then check §1.2.

Figure 3.1. Christian Felix Klein (1849–1925) was a German math-ematician. His 1872 Erlangen Program, classifying geometries bytheir underlying symmetry groups, was a highly influential synthesisof much of the mathematics of the day. Stolen from Wiki.

Theorem 3.22 (A characterization for homeomorphisms). A map f : X → Yis a homeomorphism ⇐⇒ f is bijective and determines a bijection between thetopologies of X and Y .

Hint. Using Theorem 3.21.

Theorem 3.23 is an immediate corollary. We write it as a theorem for it helpsjudge a homeomorphism.

Theorem 3.23. A continuous bijection is a homeomorphism ⇐⇒ it is closed.

Example 3.24. Most of the following can be deduced by using Theorem 3.22.

1) Every bijection on an indiscrete space onto itself is a homeomorphism.

47

2) Every bijection on a discrete space onto itself is a homeomorphism.

3) Every bijection on the cofinite space onto itself is a homeomorphism.

4) A discrete space and a non-one-point indiscrete space are not homeomorphic.

5) Spaces containing different numbers of points are not homeomorphic.

6) The spaces Z, Q, R, RT1 , and the arrow are pairwise non-homeomorphic.

Proof. Here are properties that distinguish each of the spaces from the remainingones: Z is discrete, Q is countable, each proper closed subset of RT1 is finite,and any two nonempty open sets in the arrow have nonempty intersection.

Example 3.25. We have the following examples.

(1) The function f : [ 0, 1)→ C defined by f(x) = e2πi is continuous but not open.

Proof. The image of the open set [ 0, 1/2) is not open.

(2) The function h : R→ (0, 1) defined by h(x) = ex/(1 + ex) is a homeomorphism.

(3) Every nondegenerate affine transformation of Rn is a homeomorphism.

Proof. Recall that an affine transformation f : Rn → Rn is given by the formulay = f(x) = Ax+ b, where A is a matrix and b a vector. f is nondegenerate if Ais invertible, whence

x = A−1(y − b) = A−1(y)− A−1(b),

which means that f is a bijection and f−1 is also a nondegenerate affine transfor-mation. Finally, f and f−1 are continuous, because they are given in coordinatesby linear formulas. This completes the proof.

(4) ∀R > 0, the inversion

τ : Rn \ 0 → Rn \ 0

x 7→ R· x|x|2

is a homeomorphism.

Hint. τ is an involution. Use Theorems 3.11 and 3.12 to show the continuity.

(5) Let H = z ∈ C : =z > 0 be the upper half plane. Let a, b, c, d ∈ R s.t. ad > bc.Then the function

f : H → H

z 7→ az + b

cz + d

48 D.G.L. WANG

is a homeomorphism.

Hint. Since =(f(x+ yi)

)= (ad− bc)y/|cz + d|2, we have f(H) ⊆ H. Find the

inverse and use Theorems 3.11 and 3.12 to show the continuity.

(6) The graph of a continuous real-valued function defined on an interval is homeo-morphic to the interval.

Hint. The vertical projection determines a homeomorphism.

Question 3.26. Let a < b. Prove that

(1) [0, 1] ∼= [a, b];

(2) [0, 1) ∼= [a, b) ∼= (0, 1] ∼= (a, b];

(3) (0, 1) ∼= (a, b);

(4) (−1, 1) ∼= R;

(5) [0, 1) ∼= [0,∞) and (0, 1) ∼= (0,∞).

Proof. Use x 7→ a+ (b− a)x and x 7→ tan(πx/2).

Question 3.27. Let N = (0, 1) ∈ S1 be the North Pole of the unit circle. Provethat S1 \ N ∼= R.

Proof. Define f : (1/4, 5/4) → S1 \ N by f(x) = (cos(2πx), sin(2πx)). Analternative proof which can be generalized is as follows. Define g : S1 \ N → Rby g(x, y) = x/(1− y). Its inverse is g−1(x) = (2x/(x2 + 1), (x2− 1)/(x2 + 1)).

Homework 3.3. Let p ∈ Sn. The punctured sphere Sn \ p is homeomorphic toRn.

Figure 3.2. Illustration for the stereographic projection. Stolenfrom Google image.

49

Proof. Let p = (0, 0, . . . , 1) ∈ Rn+1 be the north pole of Sn. Define two mapsf : Sn \ p → Rn and g : Rn → Sn \ p by

f(x1, . . . , xn+1) =

(x1

1− xn+1

, . . . ,xn

1− xn+1

), and

g(x1, . . . , xn) =

(2x1

|x|2 + 1, . . . ,

2xn|x|2 + 1

,|x|2 − 1

|x|2 + 1

),

where |x|2 =∑n

i=1 x2i . By Theorem 3.12, the maps f and g are continuous.

We claim that the map f is a homeomorphism. In fact, it is routine to checkthat f g is the identity map on Rn, and that g f is the identity map on Sn \ p.This completes the proof.

The map f : Sn \ p → Rn in the above proof is called the stereographicprojection. It is a particular mapping (function) that projects a sphere onto aplane; see Fig. 3.2. It was known to Hipparchus (approx. BC 190–BC 120), andoriginally as the planisphere projection. The term planisphere is still used to referto celestial charts; see Fig. 3.3.

Figure 3.3. A celestial map from the 17th century, by the Dutchcartographer Frederik de Wit.

50 D.G.L. WANG

Example 3.28. D2 ∼= I2.

Proof. Let K = (x, y) ∈ R2 : x, y ∈ [−1, 1]. We have the homeomorphisms

I2 → K

(x, y) 7→ (2x− 1, 2y − 1)

and

K → D2

(x, y) 7→ max|x|, |y|√x2 + y2

· (x, y).

Geometrically, this means that each segment joining the origin with a point on thecontour of the square is linearly mapped to the part of the segment that lies withinthe circle.

Example 3.29. The following plane domains are homeomorphic.

(1) The whole plane R2;

(2) open square I2;

(3) open strip (x, y) ∈ R2 : x ∈ (0, 1);(4) open upper half-plane;

(5) open half-strip (x, y) ∈ R2 : x > 0, y ∈ (0, 1);(6) open rectangle (x, y) ∈ R2 : a < x < b, c < y < d;(7) open quadrant: (x, y) ∈ R2 : x, y > 0;(8) the plane without the ray (x, 0) ∈ R2 : x ≥ 0;(9) open half unit ball;

(10) open sector (x, y) ∈ R2 : x2 + y2 < 1, x > y > 0;(11) unit open ball;

(12) open angle: (x, y) ∈ R2 : x > y > 0.

Proof. Use polar coordinates for the last two.

Example 3.30. Every simple (i.e., without self-intersections) closed polygon in R2

is homeomorphic to the circle S1. Every non-closed simple finite polyline in R2 ishomeomorphic to the unit interval I.

51

Hint. Map each edge of the polygon homeomorphically to a suitable arc of S1.Map each edge of the polyline homeomorphically to a suitable part of I.

One of the classical problems in topology is the homeomorphism problem: tofind out whether or not two given topological spaces are homeomorphic. In order toprove that two spaces are homeomorphic, it suffices to present a homeomorphismbetween them. This is essentially what one usually does in this case. However, toprove that two spaces are not homeomorphic, it is usually impossible to review allthe maps. We may look for a property or a characteristic shared by homeomorphicspaces s.t. one of the spaces has it, while the other one does not.

Definition 3.31. Topological properties or topological invariants: proper-ties and characteristics that are shared by homeomorphic spaces.

Obvious examples are the cardinality of the set of points and the set of opensets. Less obvious properties are the main object of the forthcoming sections.

Example 3.32 (Not proved at this stage). Here are some pairs of spaces whichare not homeomorphic.

(1) Rm 6∼= Rn if m 6= n.

(2) Dp 6∼= Dq if p 6= q.

(3) Sp 6∼= Sq if p 6= q.

(4) The letters A and P are not homeomorphic.

(5) The punctured plane R2 \ (0, 0) is not homeomorphic to the plane with ahole (x, y) ∈ R2 : x2 + y2 ≥ 1.

Definition 3.33. A continuous map f : X → Y is a (topological) embedding ifthe submap f : X → f(X) is a homeomorphism. Denoted f : X → Y .

Theorem 3.34. Any inclusion is an embedding. The composition of embeddingsis an embedding.

Question 3.35. Find topological spaces X and Y s.t. X can be embedded in Y ,and Y can be embedded in X, but X 6∼= Y .

Answer. Take X = I and Y = R.

Question 3.36. Is Q embeddable in Z?

52 D.G.L. WANG

Answer. No. The topology on Z is discrete but that on Q is not discrete.

Homework 3.4. Is a discrete space embeddable in an indiscrete space? Howabout vice versa?

Definition 3.37. Given metric spaces (X, dX) and (Y, dY ), with f : X → Y .

• f is an isometric embedding if dY (f(a), f(b)) = dX(a, b) ∀ a, b ∈ X.

• f is an isometry if it is a bijective isometric embedding.

Lemma 3.38. ∀ metric spaces (X, dX) and (Y, dY ), ∃ a metric space Z s.t. Xand Y can be isometrically embedded in Z.

Proof. Consider the set W = X t Y of disjoint union of X and Y . Fix x0 ∈ X andy0 ∈ Y . For x, y ∈ W , we define a function

f(x, y) =

dX(x, y), if x, y ∈ X;

dY (x, y), if x, y ∈ Y ;

dX(x, x0) + dY (y, y0) + 1, if x ∈ X and y ∈ Y .We shall show that f is a metric on W . The conditions of being positive-definiteand symmetric are clear. It remains to show the triangle inequality for f w.r.t.three elements in W . If the three elements are all in X, or all in Y , then thesubadditivity of X or Y itself works. Otherwise, we can suppose that x1, x2 ∈ Xand y ∈ Y , and it suffices to show that

f(x1, x2) + f(x1, y) ≥ f(x2, y) and f(x1, y) + f(x2, y) ≥ f(x1, x2).

that is,

dX(x1, x2) + dX(x1, x0) ≥ dX(x2, x0) and

dX(x1, x0) + dX(x2, x0) + 2dY (y, y0) + 2 ≥ dX(x1, x2),

both of which hold for the triangle inequality w.r.t. the elements x0, x1, x2 in X.

Definition 3.39. Isometrically embedding two metric space in a single one, wecan consider the Hausdorff distance between their images. The infimum of suchHausdorff distances over all pairs of isometric embeddings of metric spaces X andY in metric spaces is the Gromov-Hausdorff distance between X and Y .

Puzzle 3.40. Is the Gromov-Hausdorff distance a metric on the set of metricspaces?

53

Figure 3.4. Mikhail Leonidovich Gromov (also Mikhael Gromov,Michael Gromov or Mischa Gromov) (1943–present), is a French-Russian mathematician known for work in geometry, analysis andgroup theory. He is a permanent member of IHES (English: Insti-tute of Advanced Scientific Studies, a French institute supportingadvanced research in mathematics and theoretical physics) in Parisand a professor of Mathematics at New York Univ. Gromov haswon the Abel Prize in 2009 “for his revolutionary contributions togeometry”. Stolen from Wiki.

4. Connectedness

4.1. Connected spaces. Being connected means, intuitively, being all in onepiece.

Definition 4.1. Given (X,Ω).

• X is connected: X has only two clopen subsets, i.e., ∅ and X.

• X is disconnected: X is not connected.

• A (connected) component of X: a maximal nonempty connected subsetof X.

• X is totally disconnected: X is disconnected, and every point of X is acomponent.

Whenever the connectedness of a specific topological space is considered, pleaseask yourself “how far is the truth away from my intuition?”

Example 4.2. Examples for connected spaces.

(1) The empty set ∅.(2) Any singleton space.

54 D.G.L. WANG

(3) Any indiscrete space.

(4) The cofinite space.

(5) The arrow.

(6) Any two-element non-discrete space.

Example 4.3. Examples for totally disconnected spaces.

(1) Any discrete topological space X with |X| ≥ 2.

Proof. Let A ⊆ X s.t. |A| ≥ 2. Let S be a singleton subset of A. ThenA = S ∪ Sc is the union of nonempty open sets of A, and thus disconnected.Thus any connected component of X is a singleton.

(2) Q.

Proof. Let A ⊆ Q s.t. |A| ≥ 2. Let x, z ∈ A and x < z. Then ∃ y ∈ R \Q s.t.x < y < z. It follows that

(4.1) A =(A ∩ (−∞, y)

)∪(A ∩ (y, ∞)

)is the union of nonempty open sets, and thus disconnected. Hence any connectedcomponent of Q has at most one element.

(3) R \Q.

Homework 4.1. Give an example of an uncountable closed totally disconnectedsubset of the real line.

Answer. The Cantor set.

Theorem 4.4. Let X be a topological space. Then

X is disconnected

⇐⇒ ∃ nonempty open sets A and B s.t. X = A tB⇐⇒ ∃ nonempty closed sets A and B s.t. X = A tB⇐⇒ ∃ nonempty clopen sets A and B s.t. X = A tB⇐⇒ ∃ a continuous surjection X → S0 = −1, 1.

Proof. If X is disconnected, then X has a nonempty proper clopen subset, say, A.Let B = Ac. Then B is nonempty and open, and X = A tB.

55

If X = A tB for some nonempty open A and B, then A = Bc and B = Ac areclosed. Hence both A and B are clopen. In this case, we can define f : X → S0 by

f(x) =

−1, if x ∈ A;

1, if x ∈ B.Since A and B are nonempty, the function f is surjective. Since the preimage ofevery subset of S0 is open, we infer that f is continuous.

Suppose that ∃ a continuous surjection f : X → S0. Since the singleton 1 is aclopen subset of S0, the preimgae A = f−1(1) is clopen. Since f is surjective, wehave A 6= ∅ and A 6= X. This proves that X is disconnected.

Remark 17. As will be seen, Theorem 4.4 is powerful in proving the connectednessof a space. For convenience, we call the last sufficient and necessary condition ofdisconnectedness the S0-condition.

Theorem 4.5. The continuous image of a connected space is connected, i.e., con-tinuous surjection preserves the connectedness. As a consequence, the connectednessis a topological property, and the number of components is a topological invariant.

Proof. Let f : X → Y be a continuous surjection. If Y is disconnected, by Theo-rem 4.4, ∃ a continuous surjection g : Y → S0. Since the composition gf : X → S0

is a continuous surjection, by Theorem 4.4, X is disconnected, a contradiction.Since any homeomorphism is a continuous surjection, the connectedness is preservedby homeomorphisms and thus a topological property.

Question 4.6. Let Ω1,Ω2 be topologies on a set X s.t. Ω1 ⊆ Ω2.

1) Does the connectedness of (X,Ω1) imply the connectedness of (X,Ω2)?

Answer. No. For instance, take Ω1 to be indiscrete and take Ω2 to be discrete.

2) Does the connectedness of (X,Ω2) imply the connectedness of (X,Ω1)?

Answer. Yes. Assume that (X,Ω1) is disconnected. Then (X,Ω1) has anonempty proper clopen subset, say, A. Since Ω1 ⊆ Ω2, the set A is clopenin Ω2. Hence (X,Ω2) is disconnected.

When A ⊆ M ⊆ X, it is safe to say that “A is connected” withoutspecifying the ambient topology (ΩM or ΩX) from which ΩA is induced. Thisis because the connectedness of a space relies only on the topology of that space,while the topology of A induced from ΩM coincides with that from ΩX ; seeTheorem 2.11.

56 D.G.L. WANG

Theorem 4.7. Clopen sets “separate” connected sets. In other words, if A isclopen in X, and if C is connected in X, then either C ⊆ A or C ⊆ X \ A.

Proof. If not, then C ∩ A is a nonempty clopen subset of C, contradicting theconnectedness of C.

Theorem 4.8. Let Γ be a family of connected sets. If a connected set C meetsevery member in Γ, then ∪A∈ΓA ∪ C is connected.

Proof. Let X = ∪A∈ΓA ∪ C. Suppose that X = U t V where U and V are clopen.By Theorem 4.7, the connected set C is contained in either U or V . W.l.o.g. wecan suppose that C ⊆ U . For any A ∈ Γ, since A ∩ U 6= ∅, we infer that A ⊆ U byTheorem 4.7. Hence

V = V ∩X =⋃A∈Γ

(V ∩ A) = ∅.

This proves the connectedness of X.

An alternative proof of Theorem 4.8. If X is disconnected, by Theorem 4.4, ∃ acontinuous surjection f : X → S0. Let A ∈ Γ. Let xA ∈ C ∩ A. W.l.o.g., we cansuppose that f(xA) = 1. By Theorem 4.5, we infer that

f(x) = 1 ∀ x ∈ C ∪ A.For any other A′ ∈ Γ, let x′ ∈ C ∩ A′. Since x′ ∈ C, we have f(x′) = 0. ByTheorem 4.5, we find

f(x) = 1 ∀x ∈ A′.This proves that f(x) = 1 ∀ x ∈ X, contradicting that f is a surjection.

Corollary 4.9. Let Akk∈Z be a family of connected sets s.t. Ak meets Ak+1, ∀ k.Then the union ∪k∈ZAk is connected.

Proof. By Theorem 4.8, we can show by induction that the set Bn = ∪nk=−nAk isconnected, ∀ n ∈ N. Using Theorem 4.8 again, we obtain the connectedness of theunion ∪k∈ZAk = ∪n∈NBn.

Theorem 4.10. Any point in a topological space is contained in a unique component.As a consequence, two connected components either are disjoint or coincide. Inparticular, any connected space has only one component.

Proof. The union of connected components which contains p is the unique compo-nent. It is connected by Theorem 4.8.

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Theorem 4.11. A subset of R is connected ⇐⇒ it is an interval.

Proof. Let A ⊆ R. If A is not an interval, then ∃ x, z ∈ A and y ∈ Ac s.t. x < y < z.Then Eq. (4.1) holds true, and the set A is disconnected. Below we present twoproofs for the connectedness of any interval.

By the bisection method. Assume that an interval J is disconnected. Then ∃nonempty closed sets U and V s.t. J = U t V . W.l.o.g., we can suppose that ∃a0 ∈ U , b0 ∈ V s.t. a0 < b0. For n ∈ N, we write cn = (an + bn)/2 and define

(an+1, bn+1) =

(an, cn), if cn ∈ V ;

(cn, bn), if cn ∈ U.

We obtain a sequence an in U and a sequence bn in V . It is clear that an increasesand bn decreases, and that sup an = inf bn. Since both U and V are closed, thislimit lies in U ∩ V , a contradiction.

By the S0-condition. Assume that A is a disconnected interval. By Theorem 4.4,∃ a continuous surjection f : A→ S0. Then

f−1(±1) = A ∩⊔J∈Γ±

J,

where each Γ± is a nonempty collection of disjoint open intervals, and

J− ∩ J+ = ∅, ∀ J± ∈ Γ±.

It follows that

A = f−1(−1) ∪ f−1(1) = A ∩⊔

J∈ Γ−∪Γ+

J.

Since the collection Γ− ∪ Γ+ contains at least two disjoint open intervals, the set Ais not an interval, a contradiction.

Corollary 4.12. Let X be a connected space. Let f : X → R be a continuousfunction. Then the image f(X) is an interval of R.

Proof. By Theorems 4.5 and 4.11.

Corollary 4.13. Continuous functions map intervals to intervals.

Corollary 4.14 (Intermediate Value Theorem). Any continuous function

f : [a, b]→ Rtakes every value between f(a) and f(b).

58 D.G.L. WANG

Definition 4.15. A set A ⊆ Rn is convex: ∀ a, b ∈ A, we have

[a, b] = x ∈ Rn : x = (1− t)a+ tb for some t ∈ I ⊆ A.

Theorem 4.16. Any convex set in Rn is connected. In particular, the sets Rn, Bn

and Dn are connected.

Proof. Let A be a convex set in Rn with x0 ∈ A. Then the interval [a, x0] isconnected for any a ∈ A. Then the union A = ∪a∈A [a, x0] is connected byTheorem 4.8. In particular, the real line is convex and hence connected.

Question 4.17. Find all connected subspaces A, with |A| ≥ 2,

1) of the arrow;

Answer. Any subset. Assume that A ⊆ [ 0,∞) is disconnected. By Theorem 4.4,∃ a continuous surjection f : A→ S0. Note that any open set in A consists ofthe elements of A that are larger than some number M , and that any closedset in A consists of the elements of A that are smaller than or equal to somenumber m. Therefore, the clopen set f−1(1) in A has to be either the empty setor the whole set A, contradicting that f is surjective.

2) of the cofinite space;

Answer. Any infinite subset. On one hand, any finite set A with |A| ≥ 2 is theunion of two nonempty finite subsets. Since a subset is finite ⇐⇒ it is closed.By Theorem 4.4, we infer that A is disconnected. On the other hand, assumeto the contrary that an infinite set A is disconnected. By Theorem 4.4, ∃ acontinuous surjection f : A→ S0. Then each of the preimages f−1(±1) is closedin A. Hence A = f−1(−1) ∪ f−1(1) is finite, a contradiction.

Theorem 4.18. Every open set of R has a countable number of components.

Proof. Every open set of R is a union of disjoint open intervals, each of whichcontains a representative rational point.

Homework 4.2. Let A0 ⊇ A1 ⊇ · · · be an infinite decreasing sequence of closedconnected sets in the plane R2. Is ∩n≥0An connected?

Theorem 4.19. Suppose that both A ∪B and A ∩B are connected. Then both ofA and B are connected if either both A and B are open or both A and B are closed.

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Proof. Consider the ambient space C = A ∪B. From premise, C is connected. Bythe symmetry position of A and B, it suffices to show that A is connected. Assumeto the contrary that A is disconnected. Then

A = U t V,where U and V are nonempty clopen sets in A. Since A ∩B is a connected subsetof A, we infer that either A ∩ B ⊆ U or A ∩ B ⊆ V . W.l.o.g., we suppose thatA∩B ⊆ U . Since A∪B is connected, in order to obtain a contradiction, it sufficesto show that V is clopen.

In fact, if both A and B are open (resp., closed), then both U and V are open(resp., closed) by Theorem 2.10. On the other hand, the complement V c = U ∪Bis the union of open (resp., closed) sets and also open (resp., closed).

Remark 18. The condition of being open or closed is necessary. This can be seenby the example A = Q and B = R \Q.

Theorem 4.20. If A is connected and everywhere dense in X, then X is connected.

Proof. Let X = U t V where U and V are open. Then

A = A ∩X = A ∩ (U t V ) = (A ∩ U) t (A ∩ V ).

It follows from the connectedness of A that either U ∩ A = ∅ or V ∩ A = ∅. ByTheorem 2.25, any everywhere dense set meets every nonempty open set. ThereforeU = ∅ or V = ∅. This proves the connectedness of X.

An alternative proof of Theorem 4.20. Suppose not. By Theorem 4.4, ∃ a continu-ous surjection f : X → S0. By Theorem 4.5, we can suppose w.l.o.g. that f(a) = 1∀ a ∈ A. By Theorem 2.25, the open set f−1(−1) meets A, a contradiction.

Corollary 4.21. If A is connected and A ⊆ B ⊆ A, then B and A are connected.

Proof. Since A is everywhere dense in A, we infer that A is connected. Since theset A is everywhere dense in B. Thus B is connected by Theorem 4.20.

Corollary 4.22. Any component of a topological space is closed.

Proof. Let C be a component. By Corollary 4.21, the closure C is connected. ByTheorem 4.10, C = C is closed.

Example 4.23. Sn is connected.

60 D.G.L. WANG

Proof. By Homework 3.3 and Theorem 4.16, we have Sn \ 0 ∼= Rn is connected.By Corollary 4.21, Sn is connected.

Question 4.24. Are the following subsets of R2 connected?

(1) The set Q2 = (x, y) ∈ R2 : x, y ∈ Q.

Answer. No. Any projection is continuous; the continuous image of any con-nected set is connected; the projection of Q2 onto one coordinate is Q, which istotally disconnected.

(2) The set (x, y) ∈ R2 : x, y ∩Q 6= ∅.

Answer. Yes! Any two points are joined by a polyline with at most 3 segments.

(3) The set Q2 ∪ (Qc)2, where (Qc)2 = (x, y) ∈ R2 : x, y 6∈ Q.

Answer. Yes! Let lk = (x, kx) : x ∈ R and X = ∪k∈Q\0lk. Then each line lkis connected. It is clear that X ⊂ Q2 ∪ (Qc)2 ⊂ X = R2. By Corollary 4.21, theset Q2 ∪ (Qc)2 is connected.

Simple constructions assigning homeomorphic spaces to homeomorphic ones, e.g.,deleting one or several points.

Example 4.25. R1 6∼= Rn for n > 1.

Example 4.26. The following spaces are pairwise non-homeomorphic:

I, [ 0, 1), R, S1.

Proof. The number of distinct points removing each of which yields a connectedspace for I, [0, 1), R and S1 are pairwise distinct, i.e., 2, 1, 0 and ∞ resp.

Homework 4.3. Let S be the set of points on the spiral

r = exp

(1

1 + φ2

), φ ≥ 0,

in the polar representation; see Fig. 4.1. We know that S is connected. Let T ⊆ S1.Prove or disprove that S ∪ T is connected.

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Figure 4.1. The spiral (r, φ).

Puzzle 4.27. Give a topological classification of the letters of the alphabet:

A,B,C,D,E, F,G,H, I, J,K, L,M,N,O, P,Q,R, S, T, U, V,W,X, Y, Z,

regarded as subsets of R2, where the arcs comprising the letters are assumed tohave zero thickness.

Hint. The answer depends on th graphics of the letters.

Theorem 4.28 (Extendability of connected sets). We have the following.

1) If A and B are connected and A ∩B 6= ∅, then A ∪B is connected.

Proof. See Homework 4.4.

2) Let A be a connected subset of a connected space X. Then A ∪B is connectedif B is a clopen subset of Ac.

Proof. Let B be a clopen subset of Ac and let C = A ∪ B. Assume to thecontrary that C is disconnected. Then ∃ open sets U and V s.t.

C ⊆ U ∪ V, U ∩ C 6= ∅, V ∩ C 6= ∅, and U ∩ V ∩ C = ∅.Let W = U ∪ V . By Theorem 2.10, both the sets U and V are clopen in W .Since A is connected, A ⊆ W , we infer from Theorem 4.7 that either A ⊆ U orA ⊆ V . W.l.o.g., let us suppose that A ⊆ U . Let

D = V ∩ C.Then D 6= ∅. We shall show that D is a clopen subset of X, which yields animmediate contradiction to the premise that X is connected.

62 D.G.L. WANG

In fact, consider the topological space Ac with a subset V . Since B is openin Ac from premise, the intersection D = V ∩B is an open subset of V . Since Vis open in X, we infer by Theorem 2.10 that D is open in X. On the other hand,consider the topological space Ac with a subset U c. Since B is closed in Ac frompremise, the intersection D = U c ∩ B is closed in U c. Since U c is closed in X,we infer by Theorem 2.10 that D is closed in X. This completes the proof.

Homework 4.4. If A and B are connected and A∩B 6= ∅, then A∪B is connected.

4.2. Path-connectedness.

Definition 4.29. Given (X,Ω).

• A path in X: a continuous map I → X, denoted γ.

The beginning or initial point of γ: the point γ(0).

The end or final point of γ: the point γ(1).

In this case, we say that γ connects γ(0) with γ(1).

A path joining x ∈ X and y ∈ X: a path γ s.t. γ(0) = x and γ(1) = y.

• The inverse path of γ: the path

γ−1 : I → X

t 7→ γ(1− t).

• A stationary path: a constant map I → X.

• X is path-connected: any two of its points can be connected by a path.

• A path-connected component or arcwise connected component of X: amaximal path-connected subset of X.

Remark 19. The terminology “inverse path” with its notation γ−1 are abused sincethey were used to denote the inverse map of γ. Fortunately, inverse maps rarelyappear in contexts involving paths, when we need to explain.

Remark 20. The elements of the fundamental group are constructed using paths.

Theorem 4.30. Any path-connected space is connected.

Proof. Let X be path-connected. Assume to the contrary that X is disconnected.By Theorem 4.4, ∃ a continuous surjection f : X → S0. Let x ∈ f−1(−1) andy ∈ f−1(1). Since X is path-connected, ∃ a continuous g : I → X connectingx and y. Then the composition h = f g : I → S0 is a continuous surjection,contradicting Theorem 4.5.

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Theorem 4.31. A continuous image of a path-connected space is path-connected.As a consequence, path-connectedness is a topological property, and the number ofpath-connected components is a topological invariant.

Proof. Let X be path-connected. Let f : X → Y be a continuous map. Letyi = f(xi) for i = 1, 2. Let u : I → X be a path connecting x1 and x2. Then thecomposition h = f u : I → Y is a continuous map connecting y1 and y2.

Theorem 4.32. If each point of a connected space X has a path-connected neigh-bourhood, then X is path-connected.

Proof. Let p ∈ X with a path-connected neighbourhood N . Let C be the path-connected component of X containing p. Then N ⊆ C. It follows that p is aninterior point of C, and C is open. If X has two path-connected components C1

and C2, then it is the union of at least two nonempty open sets, contradicting itsconnectedness.

Corollary 4.33. For open sets of Euclidean spaces, path-connectedness is equivalentto connectedness.

Example 4.34. Here are some examples for path-connected spaces.

(1) The image of a path, e.g., I.

Proof. Let f : I → X be a path. Let x, y ∈ I. Define g : I → f(I) by

g(t) = (1− t)f(x) + tf(y).

Then g is continuous. It follows that f(I) is path-connected.

(2) Rn.

Proof. Let x, y ∈ Rn. Define f : I → Rn by f(t) = (1− t)x+ ty.

(3) Sn.

Proof. Let x, y ∈ Sn. Let p ∈ Sn s.t. p 6= x and p 6= y. By Homework 3.3, ∃a homeomorphism h : Sn \ p → Rn. Since Rn is path-connected, ∃ a pathγ : I → Rn connecting h(x) and h(y). Then the map f = h−1 γ : I → Sn \ pis continuous s.t. f(0) = x and f(1) = y.

Example 4.35. S0 is not path-connected.

64 D.G.L. WANG

Proof. If not, then ∃ continuous map f : I → S0. By Corollary 4.13, the imagef(I) is an interval, a contradiction!

Example 4.36. Let X be a topological space with A ⊆ X. Let ι : A→ X be theinclusion. Then

a map u is a path in A ⇐⇒ the map ι u is a path in X.

As a consequence, A is path-connected

⇐⇒ any two points in A are connected by a path s : I → X with s(I) ⊆ A.

Theorem 4.37. Let x, y, z ∈ X. Let α be a path joining x to y, and β a pathjoining y to z. Then the map

γ : I → X

t 7→

α(2t), if 0 ≤ t ≤ 1/2

β(2t− 1), if 1/2 ≤ t ≤ 1

is a path. In other words, the map γ is continuous.

Proof. The cover [ 0, 1/2 ], [ 1/2, 1 ] of I is finite and closed, thus it is fundamentalby Theorem 3.18. Since α and β are continuous, by Theorem 3.16, the map γ iscontinuous.

Definition 4.38. Let x, y, z ∈ X. Let α be a path joining x to y, and β a pathjoining y to z. The path γ defined by in Theorem 4.37 is called the product of αand β, denoted αβ.

Corollary 4.39. The union of a family of pairwise intersecting path-connectedsets is path-connected.

Proof. Let x ∈ A, y ∈ B, and z ∈ A ∩ B. Then ∃ a path u connecting x and z,and a path v connecting z and y. By Theorem 4.37, the product uv is a pathconnecting x and y.

Corollary 4.40. Every point belongs to a path-connected component. Two path-connected components either coincide or are disjoint. Two points belong to thesame path-connected component ⇐⇒ they are connected by a path.

Some properties of connectedness do not carry over to the case of path-connectedness.

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Homework 4.5. Find an example of a path-connected set whose closure is notpath-connected, and an example of a path-connected component that is not closed.

For the properties that do carry over, proofs are usually easier in the case ofpath-connectedness.

Homework 4.6. Suppose that both A ∪B and A ∩B are path-connected. Thenboth of A and B are path-connected if either both A and B are open or both Aand B are closed.

66 D.G.L. WANG

5. Separation Axioms

We consider natural restrictions on the topological structure making the structurecloser to being metrizable. They are called “Separation Axioms”. As will be seen,metric spaces satisfy all of the separation axioms; but in fact, studyingspaces that satisfy only some axioms helps build up to the notion of full metrisability.The separation axioms that were first studied together in this way were theaxioms for accessible spaces, Hausdorff spaces, regular spaces, and normal spaces.Topologists assigned these classes of spaces the names T1, T2, T3 and T4, respectively,where the letter T originates from the German word Trennungsaxiom,which means separation axiom. Later this system of numbering was extendedto include T0, Te, Tπ, T5, and T6. The idea was supposed to be that every Ti spaceis a special kind of Tj space if i > j. But this is not necessarily true, as definitionsvary. For example, a T3 space may do not satisfy T2.

We will restrict ourselves to the axioms T0, T1, T2, T3 and T4.

5.1. Axioms T0, T1, T2, T3 and T4.

Definition 5.1. Given a topological space X.

• The Kolmogorov axiom or the zeroth separation axiom, or T0: at leastone of any two distinct points of this space has a neighbourhood that does notcontain the other point.

X is a Kolmogorov space or T0 space: X satisfies the Kolmogorov axiom.

Figure 5.1. Andrey Nikolaevich Kolmogorov (1903–1987) was aSoviet mathematician who made significant contributions to themathematics of probability theory, topology, intuitionistic logic, tur-bulence, classical mechanics, algorithmic information theory andcomputational complexity. Stolen from Wiki.

67

• The Tychonoff axiom, or first separation axiom, or T1: each one of anytwo distinct points has a neighbourhood that does not contain the other point.

X is an accessible space or T1 space: X satisfies the Tychonoff axiom.

Figure 5.2. Andrey Nikolayevich Tychonoff (also Tikhonov, Ty-chonov, Tihonov, and Tichonov) (1906–1993) was a Soviet and Rus-sian mathematician and geophysicist known for important contri-butions to topology, functional analysis, mathematical physics, andill-posed problems. Stolen from Wiki.

• The Hausdorff axiom, or T2: any two distinct points have disjoint neighbour-hoods.

X is a Hausdorff space: X satisfies the Hausdorff axiom.

Figure 5.3. Felix Hausdorff (1868–1942) was a German mathemati-cian who is considered to be one of the founders of modern topologyand who contributed significantly to set theory, descriptive set theory,measure theory, and functional analysis. Stolen from Wiki.

• The third separation axiom, or T3: every closed set and every point of itscomplement have disjoint neighbourhoods.

X is a regular space: X satisfies both T1 and T3.

68 D.G.L. WANG

• The fourth separation axiom, or T4: any two disjoint closed sets have disjointneighbourhoods.

X is a normal space: X satisfies both T1 and T4.

Remark 21. We do not use the terminology “a Tychonoff space” to denote a T1

space, since it stands for a T3 12

space in some textbooks, in which any closed set

and any point in its complement can be separated by a continuous function; seeTheorem 5.19 for an analogue.

Remark 22. In some textbooks a regular space is defined to be a topological spacesatisfying T3, and a normal space is defined to be a topological space satisfying T4.

Theorem 5.2. Let X be a topological space. Then X satisfies T0

⇐⇒ any two different points of X have different closures

⇐⇒ X contains no indiscrete subspaces consisting of two distinct points

⇐⇒ X contains no indiscrete subspaces consisting of more than one point.

Theorem 5.3. Let X be a topological space. Then X satisfies T1

⇐⇒ every singleton in X is closed

⇐⇒ every finite set in X is closed

⇐⇒ every point is the intersection of all its neighbourhoods.

The last condition in Theorem 5.3 can be compared with a characterisation ofHausdorff spaces; see Theorem 5.4.

Theorem 5.4. A space X is Hausdorff

⇐⇒ every point is the intersection of the closures of all its neighbourhoods.

Proof. Let y 6= x.

⇒. Since X is Hausdorff, ∃ Nx∩Ny = ∅, where Nw is a neighbourhood of w ∈ x, y.It follows that y 6∈ Nx. Hence y 6∈ ∩x∈NN .

⇐. Since y 6∈ ∩x∈NN , ∃N s.t. x ∈ N and y 6∈ N . Then the open set Nc

is aneighbourhood of y, without intersection with the neighbourhood N of x.

This completes the proof.

Theorem 5.5. Normal ⇒ Regular ⇒ T2 ⇒ T1 ⇒ T0.

Proof. By definition or Theorem 5.3.

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Theorem 5.6. Regular spaces and normal spaces can be characterized as follows.

• A topological space is regular ⇐⇒ it satisfies T2 and T3.

• A topological space is normal ⇐⇒ it satisfies T2 and T4.

Example 5.7. Here are some threshold examples subject to Theorem 5.5.

1) Non-T0: any indiscrete space with at least two points.

2) T0 but not T1: the Sierpinski space.

3) T1 but not T2: the cofinite space.

4) T2 but not regular: the space X = R with the coarsest topology containing thestandard topology on R and the closed set A = 1/n : n ∈ Z+.

Proof. The point 0 and the set A cannot be separated by neighbourhoods.

5) Regular but not normal: The Moore space; see Theorem 5.16.

6) Normal: any discrete space and any metric space.

Proof. Let A and B be disjoint closed sets in a metric space (X, d). Consider

U = x ∈ X : d(x,A) < d(x,B) and V = x ∈ X : d(x,A) > d(x,B).

It is clear that A ⊆ U , B ⊆ V and U ∩ V = ∅. By Theorem 3.13, both U and Vare open. Hence X is normal.

7) T2 but not T3: *******

8) T2 but not T4: *******

9) T3 but not T1: any indiscrete space. Another example is the real line withtwo zeros, denoted X = R ∪ 0′, with a base consisting of all open intervals(a, b) ⊆ R and modified intervals (a, b)′ = (a, 0) ∪ 0′ ∪ (0, b), where a < 0 < b.

10) T3 but not T2: see Theorem 5.6 and Item 9).

11) T3 but not T4: see Item 5).

12) T4 but not T1: an indiscrete space or the arrow.

13) T4 but not T2: see Theorem 5.6 and Item 12).

14) T4 but not T3: the Sierpinski space.

70 D.G.L. WANG

5.2. Hausdorff spaces.

Definition 5.8. Let A = ann be a sequence of points of a topological space X.A limit of A is a point each of whose neighbourhoods contains all members of Athat have sufficiently large indices, i.e., a point p ∈ X s.t.

∀ neighbourhood U of p, ∃ N > 0 s.t. an ∈ U , ∀n ≥ N.

In this case, we say that the sequence converges or tends to p as n tends to ∞.See Definition 2.12 for the definition of a limit point of X.

Theorem 5.9. Any sequence in a Hausdorff space has at most one limit.

Proof. Let A = ann be a sequence in a Hausdorff space X. Assume that A hasdistinct limits p and q. Since X is Hausdorff, ∃ a neighbourhood U of p and aneighbourhood V of q s.t.

(5.1) U ∩ V = ∅.Since p is a limit, ∃ N1 > 0 s.t. an ∈ U for n > N1. Since q is a limit, ∃ N2 > 0 s.t.an ∈ V for n > N2, contradicting Eq. (5.1).

Example 5.10. Each point in the cofinite space is a limit of the sequence ofnatural numbers.

Proof. Let x ∈ R with a neighbourhood U . Since the set U c is finite, the numberMx = maxy : y ∈ U c is well defined. Therefore, we have n ∈ U ∀n > Mx. Inother words, the point x is a limit of the sequence N.

Definition 5.11. Let f, g : X → Y be maps.

• The coincidence set of f and g: C(f, g) = x ∈ X : f(x) = g(x).• A fixed point of f : x ∈ X s.t. f(x) = x, if Y = X.

Theorem 5.12. Let f, g : X → Y be continuous maps, where Y is Hausdorff.Then the coincidence set of f and g is closed.

Proof. We shall show that the complement of the coincidence set is open. Let x ∈X \C(f, g), i.e., f(x) 6= g(x). Since Y is Hausdorff, ∃ an open neighbourhood U off(x) and an open neighbourhood V of g(x) s.t. Eq. (5.1). Let W = f−1(U)∩g−1(V ).Then x ∈ W ⊆ X. Since f and g are continuous, we infer that W is open. Itsuffices to show that W ⊆ X \ C(f, g). Let y ∈ W . Then f(y) ∈ U and g(y) ∈ V .By Eq. (5.1), we infer that f(y) 6= g(y). This completes the proof.

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Corollary 5.13. Let f, g : X → Y be continuous maps, where Y is Hausdorff.Suppose that X has an everywhere dense set A s.t. fA = gA. Then f = g.

Proof. By Theorem 5.12, the coincidence set C(f, g) is closed. Since fA = gA,we have A ⊆ C(f, g). By Theorem 2.19, we infer that X = A ⊆ C(f, g). HenceC(f, g) = X, i.e., f = g.

Corollary 5.14. The fixed-point set of a continuous map from a Hausdorff spaceto itself is closed.

Proof. Consider the identity map together, and use Theorem 5.12.

Remark 23. See Corollary 3.9.

Homework 5.1. Find an example showing that the Hausdorff condition in Corol-lary 5.14 is essential.

Answer. Let X be the arrow. The fixed-point set of the transformation on Xdefined by x 7→ x+ sinx is 2kπ : k ∈ N, which is not closed.

5.3. Regular spaces & normal spaces.

Theorem 5.15. Let X be a topological space. We have the following.

• X satisfies T3 ⇐⇒ every neighbourhood of every point p ∈ X contains theclosure of a neighbourhood of p.

• X satisfies T4 ⇐⇒ every neighbourhood of every closed set F ⊆ X containsthe closure of some neighbourhood of F .

Proof. First we prove the equivalence for T3. ⇒. Let W be a neighbourhood ofp. Then ∃ open set O s.t. p ∈ O ⊆ W . Applying T3 to the point p and the closedset Oc, we obtain disjoint open sets U and V s.t. p ∈ U and Oc ⊆ V . Since U ⊆ V c

and V c is closed, we infer by Theorem 2.16 that U ⊆ V c. In conclusion, we have

p ∈ U ⊆ U ⊆ V c ⊆ W.

⇐. Let F be a closed set and p ∈ F c. Then F c is a neighbourhood of p. Frompremise, ∃ an open set U s.t. p ∈ U ⊆ U ⊆ F c. Then the open set (U)c is aneighbourhood of F which is disjoint with the neighbourhood U of p.

Replacing p by a closed set gives a proof for the other equivalence.

72 D.G.L. WANG

Space 10. Let H be the upper half plane, and N = H ∪ R the closed upper halfplane. The Moore space or Nemytskii space is the space N with the topologygenerated by the base

β = ΩH ∪By(x, y) ∪ (x, 0) : x ∈ R, y > 0

,

where ΩH is the subspace topology of H induced from the standard topology of R2,and By(x, y) is the open ball centred at (x, y) with radius y.

Figure 5.4. Robert Lee Moore (1882–1974) was an American math-ematician known for his work in general topology and the Mooremethod of teaching university mathematics. The right part is an openneighbourhood of the Moore plane, tangent to the x-axis. Stolenfrom Wiki.

Theorem 5.16. The Moore space is regular but not normal.

Proof. By Theorem 5.3, the Moore space satisfies T1 since every point is theintersection of all its neighbourhoods. By Theorem 5.15, it satisfies T3 since

By/2(x, y/2) ∪ (x, 0) ⊂ Dy(x, y) = By(x, y) ∪ (x, 0),

where Dy(x, y) is the disk centred at (x, y) with radius y. Now, from definition,the Moore space is regular. Below we shall show that it is not normal.

From definition, the subspace topology of R induced from N is discrete. Itfollows that any subset of R is closed in R. Since R is a closed subset of N , weinfer by Theorem 2.10 that any subset of R is closed in N . In particular, the sets Qand R \Q are closed. In order to show that N is not normal, it suffices to showthat any neighbourhood of R \Q in N meets any neighbourhood of Q in N .

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Let U be an open neighbourhood of R\Q. For each x ∈ R\Q, fix a number r(x)s.t. Br(x)(x, r(x)) ⊆ U . Define

Zn = x ∈ R \Q : r(x) > 1/n.It is clear that ∪n∈Z+Zn = R \Q. Consider the standard topology on R. Note that

Q =⋃n∈Z+

k/n : k ∈ Z

is the union of a countable number of nowhere dense sets. By Theorem 2.29, ∃ m s.t.the set Zm is not nowhere dense in R. By Theorem 2.28, ∃ a set O, which is openin R, s.t. O′ ∩ Zm 6= ∅, ∀ open set O′ ⊆ O. By Theorem 2.25, ∃ an open interval Js.t. Zm is everywhere dense in J . Let q ∈ J ∩Q. For any open set V containing Q,∃ rq > 0 s.t. Brq(q, rq) ⊂ V . Since Zm is everywhere dense in J , ∃ x ∈ Zm ∩ J s.t.|x−q| < min(rq, 1/m). Then B1/m(x, 1/m) ⊆ U and B1/m(x, 1/m)∩Brq(q, rq) 6= ∅.It follows that U ∩ V 6= ∅. This proves that N is not normal.

Theorem 5.17. The Moore space can be embedded in a normal space s.t. thecomplement of the image is a singleton.

Proof. Let N∗ = N ∪ x∗. Define the topology Ω∗ on N∗ to be obtained from thetopology Ω on N by adding the sets of the form x∗ ∪ U , where U ∈ Ω containsall but a finite number of points in R. One may verify that (N∗, Ω∗) is a normalspace.

Urysohn characterized normal spaces as spaces in which any two disjoint closedsubsets can be separated by a continuous function; see Theorem 5.19.

Lemma 5.18. Let A and B be nonempty disjoint closed subsets of a normalspace X. Then ∃ a collection Orr∈Λ of open subsets of X, where

Λ = k/2n ∈ I : k, n ∈ Z,s.t. A ⊆ O0, B = Oc

1, and Op ⊆ Oq, ∀ p < q.

Proof. By Theorem 5.15, A has an open neighbourhood O0 s.t. O0 ⊆ O1. Nowlet O0 plays the role of A, the closed set O0 has an open neighbourhood O1/2

s.t. O1/2 ⊆ O1. Along the same line, the set O0 has a neighbourhood O1/4 s.t.

O1/4 ⊆ O1/2, and the set O1/2 has a neighbourhood O3/4 s.t. O3/4 ⊆ O1. Continuingin this way, one obtains a desired collection Orr∈Λ.

Theorem 5.19 (Urysohn’s Lemma). Let A and B be nonempty disjoint closedsubsets of a normal space X. Then ∃ a continuous function f : X → I s.t.

f(A) ⊆ 0 and f(B) ⊆ 1.

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Figure 5.5. Pavel Samuilovich Urysohn (1898–1924) was a Sovietmathematician of Jewish origin who is best known for his contribu-tions in dimension theory, and for developing Urysohn’s MetrizationTheorem and Urysohn’s Lemma, both of which are fundamentalresults in topology. Stolen from Wiki.

Proof. Keeping the notation in Lemma 5.18, we define

f(x) =

1, if x 6∈ Or, ∀ r,infλ ∈ Λ: x ∈ Oλ, otherwise.

Then f(A) ⊆ 0 and f(B) ⊆ 1. We shall show that f is continuous; seeHomework 5.2.

Remark 24. One clever part of the proof is the indexing by dyadic fractions foropen sets.

Homework 5.2. Show that the function f in the proof of Theorem 5.19 is contin-uous.

Hint. x : f(x) < s = ∪t<sOt and x : f(x) ≤ s = ∩t>sOt.

Theorem 5.20 (Tietze Extension Theorem). Let F be a closed subset of a normalspace X. Let f : F → [−1, 1] be a continuous function. Then ∃ a continuousfunction g : X → [−1, 1] s.t. g|F = f .

Proof. For convenience, we trisect the closed interval In = [−2n/3n, 2n/3n] to theunion (−Jn) ∪Kn ∪ Jn of three closed intervals of the equal length |In|/3, where

J−n =

[−2n

3n, − 2n

3n+1

], Kn =

[− 2n

3n+1,

2n

3n+1

], and J+

n =

[2n

3n+1,

2n

3n

].

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Note that I0 = I. Let f0 = f . By Theorem 3.5, the sets

A−0 = f−10 (J−0 ) and A+

0 = f−10 (J+

0 )

are disjoint and closed in F . By Theorem 2.10, they are closed in X. By Theo-rem 5.19, ∃ a continuous function g0 : X → K0 s.t.

g0(A−0 ) ⊆ −1/3 and g0(A+0 ) ⊆ 1/3.

Let f1 = f0 − g0 : F → R. By Theorem 3.11, the function f1 is continuous. It isclear that f1(F ) ⊆ I1.

Again, the sets A±1 = f−11 (J±1 ) are closed in X. By Theorem 5.19, ∃ a continuous

function g1 : X → K1 s.t. g1(A±1 ) ⊆ ±2/9. Then the function f2 = f1− g1 : F →R is continuous and f2(F ) ⊆ I2.

Continuing in this way, ∃ a continuous function gn : X → Kn s.t.

gn(A±n ) ⊆ ±2n/3n+1,where A±n = f−1

n (J±n ). We obtain a continuous function fn+1 = fn − gn : F → Rs.t. fn+1(F ) ⊆ In+1.

Now, the function fn = f−g1−g2−· · ·−gn−1 : F → In satisfies |fn(F )| ≤ 2n/3n.The limit limn→∞

∑ni=1 gi converges to a continuous function g : X → [−1, 1] and

g|F = f .

5.4. Countability axioms. In this subsection, we study topological propertiesthat are additionally imposed on a topological structure in order to make theabstract situation under consideration closer to special situations. The restrictionsrequire something to be countable.

Definition 5.21. Given (X,Ω) with p ∈ X.

• A (neighbourhood) base at p: a collection Σ of neighbourhoods of p s.t. eachneighbourhood of p contains a neighbourhood from Σ.

• X is first countable or satisfies the first axiom of countability: X has acountable neighbourhood base at each point.

• X is second-countable or satisfies the second axiom of countability: Ωhas a countable base.

• X is separable if X contains a countable dense subset.

Theorem 5.22. The second axiom of countability implies both the first one andthe separability.

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Proof. From definition, it is straightforward that the second countability impliesthe first countability. Let β = Bii∈N be a countable base of a space X. Letpi ∈ Bi. We claim that the set A = pii∈N is dense. In fact, any open set O isthe union of some members Bi. Suppose that Bj ⊆ O. Then pj ∈ O ∩ A. ByTheorem 2.25, the set A is dense.

Homework 5.3. Find a first countable space and a separable space which are notsecond countable.

Answer. An uncountable discrete space is first countable but not second countable.The cofinite space is separable but not second countable.

Example 5.23. The arrow is second countable.

Theorem 5.24. Any metric space is first countable.

Theorem 5.25. For metrizable spaces, separability ⇐⇒ the second axiom ofcountability.

Proof. By Theorem 5.22, it suffices to show that any separable metric space issecond countable. Let A be an everywhere dense set of a metric space (X, d). ByTheorem 2.32, the collection Br(x) : x ∈ A, r ∈ Q+ is a countable base of X.

Homework 5.4. Find a metric space which is not second countable.

Corollary 5.26. Rn is second-countable.

Proof. The set Qn is countable and everywhere dense in Rn.

Theorem 5.27. The continuous image of a separable space is separable.

Proof. Direct from Theorem 3.6.

Homework 5.5. Construct an example proving that a continuous image of asecond countable space may be not second countable.

Answer. Consider id : R→ RT1 .

Definition 5.28. A Lindelof space is a topological space in which every opencover has a countable subcovering.

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Figure 5.6. Ernst Leonard Lindelof (1870–1946) was a Finnishmathematician, who made contributions in real analysis, complexanalysis and topology. Stolen from Wiki.

Theorem 5.29 (Lindelof ’s lemma). Any second-countable space is Lindelof.

Proof. Let X be a second countable space, with a countable base B and an opencover Γ. Let Λ = B ∈ β : B ⊆ Vi for some Vi ∈ Γ. For each B ∈ Λ, take a setVi ∈ Γ s.t. Bi ⊆ Vi. Then Λ is a cover of X, and Vii∈N is a countable subcoveringof Γ.

Corollary 5.30. Each base of a second countable space contains a countablesubbase.

Proof. Let X be a second countable space with a countable base β0 and a base β.By Theorem 5.29, each set in β0 is the union of a countable number of sets in β.

78 D.G.L. WANG

6. Compactness

6.1. Compact spaces. Compactness is a sort of topological counterpart for theproperty of being finite in the context of set theory.

Definition 6.1. A topological space is compact if its every open cover has a finitesubcovering.

Remark 25. The word compactness was used for the weaker property that anycountable open cover contains a finite subcovering. The modern notion of com-pactness was introduced by Alexandrov and Urysohn in 1923; see Fig. 6.1. They

Figure 6.1. Pavel Sergeyevich Alexandrov (1896–1982) was a Sovietmathematician, sometimes romanised Paul Alexandroff or Aleksan-drov. He wrote about 300 papers, making important contributionsto set theory and topology. Stolen from Wiki.

suggested for it the term bicompactness. This notion turned out to be fortunate; ithas displaced the original one and even took its name, i.e., ”compactness”. Anotherdeviation from the terminology used here comes from Bourbaki: we do not includethe Hausdorff property in the definition of compactness, while Bourbaki does.

Theorem 6.2. A subset A ⊆ X is compact ⇐⇒ each cover of A with sets openin X contains a finite subcovering.

Proof. ⇒: Let A be a compact subset of X. Let O be a cover of A with sets openin X. Then the collection O ∩ A : O ∈ O is an open cover of A, and has a finitecover, say, O ∩ A : O ∈ O′, where O′ is a finite subset of O. In other words, wehave A ⊆ ∪O∈O′(O ∩ A) = A ∩ ∪O∈O′O. It follows that A ⊆ ∪O∈O′O.

⇐: Let O be an open cover of A. By the subspace topology, each member in Ocan be written as O ∩ A, where O is an open set of X. Let O′ be the collection ofthese open sets O. From premise, ∃ a finite subset O′′ of O′ s.t. A ⊆ ∪O∈O′′O. Itfollows that A ⊆ A ∩ ∪O∈O′′O = ∪O∈O′′(O ∩ A).

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Example 6.3. Give (X,Ω).

(1) X is compact if Ω is finite. In particular, every subset of a finite or indiscretespace is compact.

(2) If X is discrete, then X is compact ⇐⇒ X is finite.

(3) Every subset of the cofinite space is compact.

(4) A subset A of the arrow is compact ⇐⇒ inf A ∈ A.

(5) R is not compact. The set [ 0, 1) is not compact in R.

(6) The set 0 ∪ 1/n : n ∈ Z+ is compact in R.

Proof. Any open set containing the singleton 0 does not contain only a finitenumber of elements in 1/n : n ∈ Z+.

(7) I.

Proof. Let O be an open cover of I. Assume that I is not compact. Then Ihas no finite subcovering of O. Let I1 be an interval of [0, 1/2] and [1/2, 1]without finite subcovering of O. Then we bisect I1 and let I2 be one of the twosubintervals which does not have a finite subcovering of O. Continuing in thisway, we obtain a sequence Inn∈N of nested closed intervals no of which havea finite subcovering of O, where I0 = I. It is clear that ∩n∈NIn is a singleton,say, s. Let O ∈ O which covers s. Then the set O covers all but a finite numberof the intervals In. In particular, O covers an interval In with sufficiently largesubscript n, a contradiction.

(8) Any cube Jn = (x1, . . . , xn) ∈ Rn : xi ∈ J, where J ∈ R is a closed interval.

Proof. Same to the previous. Subdivide the cube into 2n smaller ones at Step n.

(9) Suppose that X has topologies Ω1 and Ω2 s.t. Ω1 ⊆ Ω2. If (X,Ω2) is compact,then so is (X,Ω1). It is not true in general vice versa.

Theorem 6.4. If A and B are compact, then A ∪B is compact.

Homework 6.1. Find an example that both A and B are compact but A ∩B isnot compact.

Theorem 6.5. Any closed subset of a compact space is compact.

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Theorem 6.6. A continuous image of a compact space is compact. As a conse-quence, compactness is a topological property.

Theorem 6.7. A second countable space is compact ⇐⇒ any of its countableopen cover contains a finite subcovering.

Proof. Immediate from Theorem 5.29.

Definition 6.8. A collection of subsets of a set is said to have the finite inter-section property if each finite subcollection has a nonempty intersection.

Lemma 6.9. A collection Σ of subsets of a set X has the finite intersection property⇐⇒ @ finite Σ′ ⊆ Σ s.t. the complements of sets in Σ′ cover X.

Theorem 6.10. A space is compact ⇐⇒ every collection of its closed subsetswith the finite intersection property has a nonempty intersection.

Proof. Let X be a topological space.

⇒: Suppose that X is compact. Let F be a collection of closed subsets of X withthe finite intersection property. If ∩F∈FF = ∅, then the collection F c : F ∈ F isan open cover of X, and it has a finite subcovering, say, F c : F ∈ F ′, where F ′ isa finite subset of F . In other words, we have ∪F∈F ′F c = X. Applying de Morgan’slaw, we obtain ∩F∈F ′F = ∅, contradicting the finite intersection property.

⇐: Assume that X is not compact. Then ∃ an open cover O of X which doesnot contain a finite subcovering. By Lemma 6.9, the collection Oc : O ∈ O ofclosed sets has the finite intersection property. From premise, we have ∩O∈OOc 6= ∅.Applying de Morgan’s law, we obtain ∪O∈OO 6= X, contradicting the assumptionthat O is a cover of X.

Theorem 6.11. Any compact subset of a metric space is bounded.

Proof. Let p be a point of the subset. Then the open cover Bn(p)n∈Z+ has nofinite subcovering.

Homework 6.2. Find a closed and bounded, but noncompact set in a metricspace.

Answer. Let X = [0, 1) with the subspace topology induced from R. Then Xinduces the metric from the standard metric of R. Then the set X is closed andbounded. It is not compact; see Eg. 6.3.

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6.2. Interaction of compactness with other topological properties.

Lemma 6.12. Let A be a compact subset of a Hausdorff space X, and let b ∈ Ac.Then ∃ disjoint open sets U and V of X s.t. A ⊆ U and b ∈ V .

Proof. ∀ a ∈ A, since X is Hausdorff, ∃ an open neighbourhood Ua of a and anopen neighbourhood Va of b s.t. Ua ∩ Va = ∅. Since A is compact, the open cover∪a∈AUa of A contains a finite subcovering, say, Ua : a ∈ A′, where A′ is a finitesubset of A. Let

U = ∪a∈A′Ua and V = ∩a∈A′Va.Then the sets U and V and open and disjoint s.t. A ⊆ U and b ∈ V .

Theorem 6.13. Any compact subset of a Hausdorff space is closed.

Proof. Let A be a compact subset of a Hausdorff space X. By Lemma 6.12, everypoint b ∈ Ac is an interior point. Hence Ac is open and A is closed.

Remark 26. In Hausdorff spaces, the compactness is stronger than closedness.

Corollary 6.14. The intersection of any family of compact subsets of a Hausdorffspace is compact.

Proof. LetX be a Hausdorff space with a collection Kλ : λ ∈ Λ of compact subsets.By Theorem 6.13, every set Kλ is closed. Thus the intersection J = ∩λ∈ΛKλ isclosed. Let o ∈ Λ. Now J is a closed subset of the compact set Ko. By Theorem 6.5,J is compact.

Remark 27. Cf. Homework 6.1.

Theorem 6.15. Any compact Hausdorff space is normal.

Proof. Let X be a compact Hausdorff space with disjoint closed sets A and B. ByTheorem 6.5, both sets A and B are compact. Since A is compact, by Lemma 6.12,∀ b ∈ B, ∃ an open neighbourhood Ub of A and an open neighbourhood Vb of bs.t. Ub ∩ Vb = ∅. Since B is compact, the open cover ∪b∈BVb of B has a finitesubcovering, say, Vb : b ∈ B′ where B′ is a finite subset of B. Let

U = ∩b∈B′Ub and V = ∪b∈B′Vb.Then both sets U and V and open, A ⊆ U , B ⊆ V and U ∩ V = ∅. This provesthe normality of X.

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Theorem 6.16. Let X be a Hausdorff space, let Kλ : λ ∈ Λ be a family of itscompact subsets, and let U be an open set containing ∩λ∈ΛKλ. Then ∃ a finiteΛ′ ⊆ Λ s.t. ∩λ∈Λ′Kλ ⊆ U .

Proof. Let λ0 ∈ Λ. Let J = ∩λ∈ΛKλ. By Theorem 6.5, the closed set Kλ0 \ U iscompact. On the other hand, applying de Morgan’s law, we infer that

Kλ0 \ U ⊆ U c ⊆ J c = ∪λ∈ΛKcλ.

It follows that the open cover Kcλ : λ ∈ Λ of the set Kλ0\U has a finite subcovering,

say, Kcλ : λ ∈ Λ′, where Λ′ is a finite subset of Λ. Hence ∩λ∈Λ′Kλ ⊆ U .

Corollary 6.17. Let Knn∈N be a decreasing sequence of nonempty compactconnected sets in a Hausdorff space. Then the intersection ∩n∈NKn is nonemptyand connected.

Proof. By Theorem 6.5, every set Kn is a closed subset of the compact set K1.It is clear that the collection Kn satisfies the finite intersection property. ByTheorem 6.10, the intersection J = ∩n≥1Kn is nonempty. Assume to the contrarythat J is disconnected. Then J = F1 t F2 for some closed sets F1 and F2. ByTheorem 6.15, the set K1 is normal. Therefore, ∃ an open neighbourhood Ui of Fi(i = 1, 2) s.t. U1 ∩ U2 = ∅. Let U = U1 ∪ U2. Now we have J ⊆ U .

By Theorem 6.16, ∃ a finite subset A ⊂ N s.t. ∩n∈AKn ⊆ U , that is, Km ⊆ U ,where m = maxn∈A n. Therefore, the set Km is the disjoint union of open setsU1 ∩ Km and U2 ∩ Km of Km. If U1 ∩ Km = ∅, then F1 ⊆ J ⊆ Km ⊆ U2, acontradiction. Hence U1 ∩Km = ∅, and U2 ∩Km = ∅ for the same reason. Thiscontradicts the connectedness of Km and completes the proof.

Theorem 6.18. A subset of Rn is compact ⇐⇒ it is closed and bounded.

Proof. Let A ⊆ Rn.

⇒. Suppose that A is compact. Since Rn is a metric space, by Theorem 6.11, Ais bounded. Since Rn is Hausdorff, by Theorem 6.13, A is closed.

⇐. Let A be closed and bounded. Then ∃ a cube Jn containing A. From Eg. 6.3,we see that any cube is compact. By Theorem 6.5, A is compact.

Question 6.19. Which of the following sets are compact?

(1) [0, 1).

(2) rayR+ = x ∈ R : x ≥ 0.

83

(3) Sn.

(4) any ellipsoid.

(5) Q ∩ I.

Theorem 6.20. A continuous map from a compact space to a Hausdorff space isclosed.

Proof. Let F be a closed set in a compact space X. Let f : X → Y be a continuousmap, where Y is Hausdorff. By Theorem 6.5, the set F is compact. By Theorem 6.6,the image f(F ) is compact. By Theorem 6.13, the set f(F ) is closed.

Theorem 6.21. A continuous bijection of a compact space onto a Hausdorff spaceis a homeomorphism.

Proof. Direct from combining Theorems 3.23 and 6.20.

Theorem 6.22 is a corollary of Theorem 6.21. We write it as a theorem forits significance in identifying a quotient space with a space; see the proof ofTheorem 7.40.

Theorem 6.22. Any continuous injection from a compact space to a Hausdorffspace is an embedding.

Corollary 6.23. A continuous real-valued function on a compact space is boundedand attains its maximal and minimal values.

Corollary 6.24. Let A ⊆ Rn. Then A is compact ⇐⇒ iff each continuousreal-valued function on A is bounded.

Proof. ⇒. By Corollary 6.23.

⇐. By Theorem 6.18, it suffices to show that A is bounded and closed. In fact,by Theorem 3.13, the function d(0, x) is continuous. Since it is bounded frompremise, A is bounded. On the other hand, assume that A is not closed. Then∃ x0 ∈ A \ A. It follows that the continuous function 1/d(x0, x) is unbounded, acontradiction.

Theorem 6.25. In any metric space, a compact subset and a closed subset havezero distance ⇐⇒ they meet.

84 D.G.L. WANG

Proof. Let (X, d) be a metric space with a compact subset A and a closed subset Fs.t. A∩F = ∅. By Theorem 3.13 and Corollary 6.23, the continuous function dF (x)attains its minimum, say, m. In other words, ∃ x ∈ A s.t. dF (x) = m. Note thatm = d(A,F ) from definition of the distance between two subsets. It is clear thatm ≥ 0. If m = 0, by Theorem 2.46, we infer that x ∈ F , contradicting A ∩ F = ∅.Hence m > 0.

Corollary 6.26. Any open set containing a compact set A of a metric spacecontains a neighbourhood of A.

Theorem 6.27 will be applied several times in later sections.

Theorem 6.27 (Lebesgue’s number lemma). Let (X, d) be a compact metricspace with an open cover F . Then ∃ δ > 0 (called a Lebesgue number of F) s.t.any subset of X of diameter less than δ is contained in some member of F .

Figure 6.2. Henri Leon Lebesgue (1875–1941) was a French math-ematician most famous for his theory of integration, which was ageneralization of the 17th century concept of integration. Stolenfrom Wiki.

Proof. If X ∈ F , then any δ > 0 is qualified. Below we can suppose that X 6∈ F .Let A1, . . . , An be a finite subcovering of F . Then Aci 6= ∅ for each i. ByTheorems 3.11 and 3.13, the function

f : X → R

x 7→ 1

n

n∑i=1

d(x, Aci)

is continuous.

85

Since X is compact, by Corollary 6.23, f attains its minimum, say, m. Letδ = m/2. By Theorem 6.5, the set Aci is compact. By Theorem 6.25, we have

f(x) = 0 ⇐⇒ d(x, Aci) = 0, ∀ i ⇐⇒ x ∈ ∩ni=1Aci .

But A1, . . . , An is a cover of X, @x ∈ ∩ni=1Aci . In other words, δ > 0.

Let A ⊆ X s.t. diamA < δ. Take a ∈ A. Then A ⊆ Bδ(a). Since f(a) ≥ 2δ, ∃1 ≤ i ≤ n s.t. d(a,Aci) ≥ 2δ. It follows that A ⊆ Bδ(a) ⊆ Ai.

Puzzle 6.28. Is the closure of a compact set always compact?

Answer. No. For example, consider the topology O ∪ 0 : O ∈ ΩR ∪ ∅ of R,

where ΩR is the standard topology on R. It is clear that 0 = R.

86 D.G.L. WANG

7. Product Spaces & Quotient Spaces

7.1. Product spaces.

Definition 7.1. Given two sets X and Y . Let f : X → Y .

• The Cartesian product or direct product or product of X and Y :

X × Y = (x, y) : x ∈ X, y ∈ Y .One simply denotes a × Y by a× Y , and denotes X × b by X × b.• A fibre of X × Y : a subset of X × Y which has the form a× Y or X × b.• The diagonal map on X: the map

∆: X → X ×Xx 7→ (x, x).

The diagonal of X ×X: the set ∆(X) = (x, x) : x ∈ X.• The graph of f : the set Γf (X) = (x, f(x)) : x ∈ X ⊆ X × Y .

Remark 28. The terminology “graph” here is somehow different from one’s intuition.It requires that the set Γf ∩ (x× Y ) is a singleton ∀x ∈ X.

Question 7.2. Is it true that (A1 ×B1) ∩ (A2 ×B2) = (A1 ∩ A2)× (B1 ×B2)?

Answer. Yes.

Definition 7.3. Given (X,ΩX) and (Y,ΩY ). Let p = (x, y) ∈ X × Y .

• An elementary open set of X × Y : a set U × V where U ∈ ΩX and V ∈ ΩY .

An elementary neighbourhood of p: an elementary open set containing p.

• The product topology in X × Y : the topology generated by the set ofelementary open sets.

• The product space or product of X and Y : the set X × Y with the producttopology.

Remark 29. For any subspaces A and B of spaces X and Y , the product topologyon A×B coincides with the subspace topology induced from X × Y .

Remark 30. The space Y ×X is canonically homeomorphic to X × Y .

Theorem 7.4. If A ⊆ X and B ⊆ Y are closed, then so is A×B ⊆ X × Y .

Proof. (A×B)c = (Ac × Y ) ∪ (X ×Bc).

87

Theorem 7.5. The product of connected spaces is connected.

Proof. Let X, Y be connected spaces with pi = (xi, yi) ∈ X × Y for i = 1, 2. Byusing classical homeomorphisms, we infer that the fibres X × y2 and x1 × Y areconnected. By Theorem 4.8, the set W = (X×y2)∪ (x1×Y ) ⊆ X×Y is connected.Since pi ∈ W for i = 1, 2, the product X × Y is connected.

By a similar argument, we obtain Theorem 7.6.

Theorem 7.6. Let X, Y be connected spaces with proper subsets A and B resp.Then (A×B)c is connected.

Theorem 7.7. The product of Hausdorff spaces is Hausdorff.

Proof. Let X and Y be Hausdorff spaces with xi ∈ X and yi ∈ Y (i = 1, 2). Letpi = (xi, yi). Suppose that the points p1 and p2 are distinct points inX×Y . W.l.o.g.,we supposes that x1 6= x2. Since X is Hausdorff, ∃ an open neighbourhood U of x1

and an open neighbourhood V of x2 s.t. U ∩ V = ∅. Then the disjoint sets U × Yand V × Y are open neighbourhoods of p1 and p2 resp.

Lemma 7.8. ∀A ⊆ X and B ⊆ Y , we have

A×B = A×B and (A×B) = A ×B.

Remark 31. Recall from Remark 8 that the closure and interior resemble dual toeach other.

Theorem 7.9. The product of regular spaces is regular.

Proof. Let X and Y be regular spaces. Let p = (x, y) ∈ X × Y with an elementaryneighbourhood U × V .

Since X is regular, by Theorem 5.15, ∃ an open neighbourhood S of x s.t. S ⊆ U .For the same reason, ∃ an open neighbourhood T of y s.t. T ⊆ V . It follows thatthe set W = S × T is an elementary neighbourhood of p. By Lemma 7.8, we haveW = S × T ⊆ U × V . By Theorem 5.15, the space X × Y is regular.

Remark 32. Cf. Theorem 7.7. The product of normal spaces is not necessarilynormal.

Theorem 7.10. The product of separable spaces is separable.

88 D.G.L. WANG

Proof. Let A and B be countable dense subset of X and Y resp. Then A× B iscountable and dense in X × Y .

Theorem 7.11. The product of second countable spaces is second countable.

Proof. Let X, Y be spaces with bases U and V resp. It suffices to show that the set

U × V : U ∈ Σ, V ∈ Γis a base of X × Y . In fact, this is true if each elementary open set of X × Y is aunion of sets of such form. Indeed,⋃

α

Uα ×⋃β

Vβ =⋃α,β

(Uα × Vβ).

This completes the proof.

Theorem 7.12. The product of path-connected spaces is path-connected.

Proof. Let X and Y be path-connected spaces with (xi, yi) ∈ X × Y (i = 0, 1).Let u be a path joining x0 and x1. Let v be a path joining y0 and y1. Then themap

f : I → X × Yt 7→

(u(t), v(t)

)is a path joining the points (x0, y0) and (x1, y1).

Theorem 7.13. The product of compact spaces is compact.

Proof. Let X and Y be compact spaces. Let Uλ×Vλ : λ ∈ Γ be a cover of X×Yconsisting of elementary open sets. Since Y is compact, each fibre x×Y has a finitesubcovering Ux

γ×V xγ : γ ∈ Γx, where Γx is a finite subset of Γ. Put W x = ∩λ∈ΓxU

xλ .

Since X is compact, the cover W x : x ∈ X has a finite subcovering W x : x ∈ X ′,where X ′ is a finite subset of X. Then Ux

λ × V xλ : x ∈ X ′, λ ∈ Γx is the required

finite subcovering.

Theorem 7.14. The product space (R1)n coincides with the space Rn. The productspace I × I × · · · × I (n times) coincides with the subspace In of Rn.

Proof. A base of R1 × R1 is the collection of open rectangles, which is also a baseof R2. One may show by induction that (R1)n = Rn. As subspaces we haveI × I × · · · × I( n times) = In.

89

Theorem 7.15. A map f : X → Y is continuous ⇐⇒ the restriction

prX∣∣Γf (X)

: Γf (X)→ X

is a homeomorphism.

Proof. The restriction h = prX∣∣Γf (X)

is obviously a bijection. The inverse map

h−1 : X → Γf (X) is continuous ⇐⇒ so is the map

g : X → X × Yx 7→

(x, f(x)

),

which is true because g−1(U × V ) = U ∩ f−1(V ). Conversely, the map

f = prY (prX

∣∣Γf (X)

)−1

is the composition of continuous maps.

Remark 33. See Theorem 3.5 for six characterisations of continuous maps.

Theorem 7.16. The projection pr : X × Y → X is closed if Y is compact.

Proof. Let F be a closed set in X × Y . Let x ∈ pr(F )c. ∀ y ∈ Y , the point(x, y) ∈ F c has an open neighbourhood Uy × Vy. Since the fibre x× Y is compact,the open cover Uy × Vy : y ∈ Y has a finite subcovering Uyi × Vyi : i = 1, . . . , n.It is clear that the set U = ∩ni=1Uyi is an open neighbourhood of x, and that it isdisjoint with pr(F ). Hence pr(F )c is open and pr(F ) is closed.

Corollary 7.17. Let f : X → Y be a map, where Y is compact. If the graph Γf (X)is closed, then f is continuous.

Proof. By Theorem 7.16, the projection prX : X × Y → X is a closed map. Sincethe set Γf (X) is closed, the restriction prX

∣∣Γf (X)

is a closed map. By Theorem 3.23,

it is a homeomorphism. By Theorem 7.15, the map f is continuous.

Remark 34. The compactness condition in Corollary 7.17 is essential, which can beseen from the function

f : R→ R

x 7→

0, if x = 0;

1/x, otherwise.

90 D.G.L. WANG

Definition 7.18. Let gi : Xi → Yi (i = 1, 2) be maps. The Cartesian productof g1 and g2 is the map

g1 × g2 : X1 ×X2 → Y1 × Y2

(x1, x2) 7→(g1(x1), g2(x2)

).

Theorem 7.19. The Cartesian product of continuous maps is continuous.

Hint. (g1 × g2)−1(B1 ×B2) = g−11 (B1)× g−1

2 (B2).

Theorem 7.20. The Cartesian product of open maps is open.

Hint. (g1 × g2)(A1 × A2) = g1(A1)× g2(A2).

Theorem 7.21. A space X is Hausdorff

⇐⇒ the diagonal ∆(X) is closed in X ×X.

Proof. ⇒. Let (x, y) ∈ ∆(X)c. Then x 6= y. Since X is Hausdorff, ∃ an openneighbourhood U of x and an open neighbourhood V of y s.t. U ∩ V = ∅. Then(U × V ) ∩ ∆(X) = ∅. This proves that ∆(X)c is open. ⇐. Let x and y bedistinct points in X. Then (x, y) ∈ ∆(X)c. Since ∆(X) is closed, ∃ an elementaryneighbourhood U × V of (x, y) s.t. U × V ⊆ ∆(X)c, that is, U ∩ V = ∅. Thisproves that X is Hausdorff.

Remark 35. See Theorem 5.4 for a characterisation of Hausdorff spaces.

Corollary 7.22. Let f : X → Y be a continuous map where Y is a Hausdorffspace. Then the graph Γf (X) is closed in X × Y .

Proof. The map f × idY is a composition of continuous maps. By Theorem 7.21,the diagonal ∆(Y ) is closed, and so is the graph

Γf (X) = (f × idY )−1(∆(Y )

).

Remark 36. Cf. Theorem 5.12 and Corollary 7.17.

Theorem 7.23. Any metric d : X×X → R is continuous w.r.t. the metric topology.

91

Proof. By the subadditivity, we have

d(Bε/2(x), Bε/2(y)

)⊆(d(x, y)− ε, d(x, y) + ε

).

By Theorem 3.5, the metric d is continuous at every point (x, y), and hencecontinuous.

Definition 7.24. Let X, Y, Z be sets and let f : Z → X × Y . The compositionsprXf and prY f are called factors or components of f , where prX : X×Y → Xand prY : X × Y → Y are the projections.

Theorem 7.25. f : Z → X × Y is continuous ⇐⇒ so is each component of f .

Proof. ⇒. For W ∈ X, Y , the function prW f is a composition of continuousmaps. ⇐. Let A ∈ ΩX and B ∈ ΩY . Then

f−1(A×B) = (prX f)−1(A) ∩ (prY f)−1(B)

is open in Z.

Theorem 7.26. We have R2 \ 0 ∼= S1 × R.

Proof. Define a map

φ : S1 × R→ R2 \ 0((s, t), r

)7→ (ser, ter).

It is continuous with the continuous inverse

φ−1 : R2 \ 0 → S1 × R

(x, y) 7→(( x√

x2 + y2,

y√x2 + y2

), ln

√x2 + y2

).

Definition 7.27. Given a collection X = (Xλ,Ωλ) : λ ∈ Λ of topological spaces.

• The Cartesian product or direct product or product of spaces in X :

∏ =∏λ∈Λ

Xλ =

f : Λ→

⋃λ∈Λ

Xλ

∣∣∣∣ f(λ) ∈ Xλ, ∀λ ∈ Λ

.

• The λth projection: the function

prλ : ∏→ Xλ

f 7→ f(λ).

92 D.G.L. WANG

• The product topology (or Tychonoff topology) on ∏, denoted Ω∏:

(1) the initial topology w.r.t. the projections prλ; i.e.,

(2) Ω∏ = 〈 pr−1λ (Ωλ) | λ ∈ Λ 〉; or

(3) every member in Ω∏ is a union of sets of the form∏

λOλ where Oλ ∈ Ωλ andOλ 6= Xλ for only finitely many λ.

• The product space of X : (∏, Ω∏).

• The box topology on ∏: ΩB = 〈∏

λ Ωλ 〉.

Remark 37. The product topology makes the product space a categorical productof its factors, whereas the box topology is too fine; this is the sense in which theproduct topology is a “natural topology”.

Remark 38. The definitions of product of two sets in Defs. 7.1 and 7.27 arecompatible.

7.2. Quotient spaces.

Definition 7.28. Let X be a set.

• An equivalence relation on X: a binary relation ∼ on X which is reflexive,symmetric and transitive.

• A setoid: a set Y with an equivalence relation ∼ on it, denoted (Y,∼).

• An equivalence class of X w.r.t. ∼: [a] = x ∈ X : x ∼ a.• The quotient set or set of equivalence classes of X w.r.t. ∼:

X/∼ = [x] : x ∈ X.

Remark 39. The bracket symbol depends on the equivalence relation which isnot indicated from the symbol itself. We will make it clear when more than oneequivalence relations are considered at the same time on a set.

Recall that a partition of a set determines a unique equivalence relation ∼ onthat set and vice versa.

Definition 7.29. Let X be a set with a partition S. Let A ⊆ X.

• The quotient set X/S: the quotient set X/∼, where ∼ is the equivalencerelation determined by S uniquely.

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• The canonical projection or factorisation map on X w.r.t. S: the map

pr : X → X/S

x 7→ [x],

where [x] is an equivalence class of X w.r.t. S.

• A is saturated w.r.t. S: ∃B ⊆ X/S, s.t.

A =⋃

[x]∈B

pr−1([x]).

• The saturation of A w.r.t. S: the set pr−1(pr(A)

), i.e., the smallest saturated

set containing A w.r.t. S.

• S is open: the saturation of each open set is open, i.e., the canonical projectionX → X/S is a open.

• S is closed: the saturation of each closed set is closed, i.e., the canonicalprojection X → X/S is a closed.

Proposition 7.30. A set A is saturated ⇐⇒ A is equal to its saturation.

Definition 7.31. Given (X, Ω) with a partition S of X.

• The quotient space or identification space of X w.r.t. S: (X/S, ΩS), where

ΩX/S = U ⊆ X/S : pr−1(U) ∈ Ωis the quotient topology or identification topology of X/S.

Remark 40. From definition, the quotient topology is the final topology w.r.t. thecanonical projection pr : X → X/S. More explicitly, a set A in X/S is a collectionof equivalence classes [x]. It is open ⇐⇒ its “flattened set” is open in X.

Question 7.32. Give an explicit description of the quotient space of I by thepartition consisting of the intervals [0, 1/3], (1/3, 2/3] and (2/3, 1].

Answer. Denote X/S = a, b, c, where

pr−1(a) = [0, 1/3], pr−1(b) = (1/3, 2/3] and pr−1(c) = (2/3, 1].

Then ΩX/S = 0, c, b, c, a, b, c.

Proposition 7.33. A set in a quotient space X/S is closed ⇐⇒ it is the preimageof a closed set in X under the projection pr : X → X/S.

94 D.G.L. WANG

Theorem 7.34. Any quotient space of a connected space is connected. Any quotientspace of a path-connected space is path-connected. Any quotient space of a separablespace is separable. Any quotient space of a compact space is compact.

Proof. Any quotient space of one of the spaces under consideration is the continuousimage of a space of that kind. See Theorems 4.5, 4.31, 5.27 and 6.6.

Homework 7.1. Is it true that any quotient space of a Hausdorff space is Haus-dorff?

Answer. The quotient space of R by the partition R≥0, R<0 is not Hausdorff.

Definition 7.35. Let f : X → Y be a map, where X and Y are sets with partitionsS and T resp.

• If f is constant on each member of S, then ∃! a map

f/S : X/S → Y

[x] 7→ f(x),

called the quotient map, or identification map, or factor map of f w.r.t. S.

• If ∀ [x] ∈ S, ∀x′ ∈ [x], ∃! [y] ∈ T s.t. f(x′) ∈ [y], then ∃! a map

f/(S, T ) : X/S → Y/T

[x] 7→ [f(x)],

called the quotient map, or identification map, or factor map of f w.r.t.S and T . In other words, the map f/(S, T ) is the one makes the diagram

Xf−−−−−−−−→ Y

pr1

y ypr2

X/Sf/(S,T )−−−−−−−−→ Y/T

commutative.

Proposition 7.36. Keep the notation in Def. 7.35. If f is open, then so is thequotient map f/S.

Proof. For each open set U ⊆ X/S, the image f/S(U) = f(pr−1(U)

)is open as

the image of the open set pr−1(U) under the open map f .

95

Theorem 7.37. Keep the notation in Def. 7.35. If f : X → Y is continuous, thenf/(S, T ) : X/S → Y/T is continuous.

Proof. Using the commutative diagram, we see that an open set in Y/T correspondsto an open set in Y , whose preimage under f is open in X, and determines anopen set in X/S.

Corollary 7.38. Keep the notation in Def. 7.35. If f : X → Y is continuous, thenf/S : X/S → Y is continuous.

Definition 7.39. A map f : X → Y determines a partition of the set X intononempty preimages of the elements of Y . This partition is denoted by S(f). Themap

f/S(f) : X/S(f)→ Y

[x] 7→ f(x)

is injective, and called the injective quotient or injective factor of f .

Theorem 7.40 (Identifying a quotient space with a known space). If f : X → Yis a continuous map from a compact space X to a Hausdorff space Y , then theinjective factor f/S(f) is a topological embedding.

Proof. By Theorem 7.34, the quotient space X/S(f) is compact. By Corollary 7.38,the map f/S(f) is continuous. By Theorem 6.22, the factor f/S(f) is an embedding.

An accurate literal description of a partition can often be somewhat cumbersome,but usually it can be shortened and made more understandable by a more flexiblevocabulary. For instance, by phrasing as factorise, pass to a quotient, attach, gluetogether, identify, contract, paste, and so on.

Definition 7.41. Let A ⊆ X. Let S be the partition of X consisting the set Aand all singletons in Ac. Then the quotient set X/S is denoted simply as X/A,and called the contraction of A to a point.

Proposition 7.42. Keep the conditions in Def. 7.41. If A is closed, then S isclosed.

Proof. Let pr : X → X/S be the canonical projection. If A∩S = ∅, then pr(A) = Ais open in X/S. Otherwise A ∩ S 6= ∅, then pr(A)c = (A ∪ S)c is open, whichimplies that pr(A) is closed.

96 D.G.L. WANG

Definition 7.43. Let A,B ⊆ X s.t. A∩B = ∅ with a homeomorphism f : A→ B.Then passing to the quotient of X by the partition into singletons in (A ∪B)c

and two-element sets x, f(x) where x ∈ A, we glue or identify the sets A and Bvia the homeomorphism f . The quotient space can be denoted as

X/[a ∼ f(a) ∀ a ∈ A] or X/[a ∼ f(a)].

Some partitions are easily described by a picture, especially if the original spacecan be embedded in the plane. In such a case, we draw arrows on the segments tobe identified to show the directions to be identified.

Example 7.44. We have I/[0 ∼ 1] ∼= S1. In other words, the quotient space

∼=

Figure 7.1. Illustration for the homeomorphism I/[0 ∼ 1] ∼= S1.

of I by the partition consisting of 0, 1 and singletons for all other points in theinterval (0, 1) is homeomorphic to a circle; see Fig. 7.1.

Proof. By Theorem 7.40, the quotient map f/[0 ∼ 1] of the map

f : I → S1

t 7→ e2πti

is a homeomorphism.

Example 7.45. We have Dn/Sn−1 ∼= Sn. In other words, contracting the boundaryof an n-dimensional ball to a point, we obtain an n-dimensional sphere.

Hint. Consider the map

f : Dn/Sn−1 → Sn

[(x1, . . . , xn)] 7→(x1

rsin(rπ), . . . ,

xnr

sin(rπ), − cos(rπ)

),

with r =√∑n

i=1 x2i .

97

Figure 7.2. Illustration for the homeomorphism

I2/[(0, t) ∼ (1, t) ∀ t ∈ I] ∼= S1 × I.

Example 7.46. We have I2/[(0, t) ∼ (1, t) ∀ t ∈ I] ∼= S1 × I. In other words,gluing the side edges of a square by identifying points on the same hight producesa cylinder; see Fig. 7.2.

Example 7.47. We have (S1 × I)/[(z, 0) ∼ (z, 1) ∀ z ∈ S1] ∼= S1 × S1. In otherwords, gluing the base circles of a cylinder by identifying pairs of points on thesame element gives a torus.

Hint. Consider the product map f × idI where f is defined in the proof of Eg. 7.44.

Example 7.48. We have I2/[(0, t) ∼ (1, t), (t, 0) ∼ (t, 1)] ∼= S1 × S1. In otherwords, identifying the sides of a square according to the left part of Fig. 7.3 producesa torus.

Theorem 7.49 (Transitivity of factorisation). Let S be a partition of a space X,and let S ′ be a partition of the space X/S. Then the quotient space (X/S)/S ′ iscanonically homeomorphic to X/T , where T is the partition of X into preimagesof elements of S ′ under the projection X → X/S.

Proof. It follows from investigating the topologies of spaces in the commutativediagram

Xpr1−−−−−−→ X/S

pr2

y ypr3

X/Tq−−−−−−−→ (X/S)/S ′

where q is defined by an obvious bijection.

98 D.G.L. WANG

Figure 7.3. Illustration for the homeomorphism

I2/[(0, t) ∼ (1, t), (t, 0) ∼ (t, 1)] ∼= S1 × S1.

Definition 7.50. The Mobius strip or Mobius band: the quotient spaceI2/[(0, t) ∼ (1, 1− t)]. In other words, this is the quotient space of the square I2

by the partition into centrally symmetric pairs of points on the vertical edges of I2,and singletons that do not lie on the vertical edges. The Mobius strip is obtained,so to speak, by identifying the vertical sides of a square in such a way that thedirections shown on them by arrows are superimposed; see Fig. 7.4.

Figure 7.4. The square to be glued and a Mobius strip.

Definition 7.51. The Klein bottle: I2/[(t, 0) ∼ (t, 1), (0, t) ∼ (1, 1− t)].

Remark 41. The Klein bottle cannot be embedded in R3.

Definition 7.52. The projective plane: identifying each boundary point of thedisk D2 with its antipodal point.

Remark 42. The projective plane cannot be embedded in R3. Gluing togethera disk and a Mobius strip via a homeomorphism between their boundary circlesresults in a projective plane.

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Appendix A. Some elementary inequalities

Theorem A.1 (Jensen’s inequality). Let x1, . . . , xn ∈ R be in the domain ofa real function f . Let wi ≥ 0 s.t.

∑ni=1 wi = 1. Then we have

f

( n∑i=1

wixi

)≤

n∑i=1

wif(xi), if f is convex and

f

( n∑i=1

wixi

)≥

n∑i=1

wif(xi), if f is concave,

where each of the equalities hold if and only if either x1 = x2 = · · · = xn, or thefunction f is linear.

Theorem A.2 (Weighted power mean inequality). Let p > 0. Suppose thatwi ≥ 0 s.t.

∑ni=1wi = 1. Denote by

Mp =

( n∑i=1

wixpi

)1/p

the p-th weighted power mean of positive numbers xi > 0. Then the function Mp isincreasing in p, i.e., Mp ≤Mq whenever p < q, where the equality holds if and onlyif x1 = x2 = · · · = xn.

Proof. Consider the function f(x) = xq/p. Since

f ′′(x) =q

p·(q

p− 1

)· xq/p−2 > 0,

the function f(x) is convex. By Jensen’s inequality we find( n∑i=1

wixpi

)q/p= f

( n∑i=1

wixpi

)≤

n∑i=1

wif(xpi ) =n∑i=1

wixqi .

Raising both sides to the power of 1/q yields the desired inequality. The equalityholds if and only if all the numbers xi are equal because that is the equalitycondition for Jensen’s inequality.

Definition A.3. Let H = x ∈ R : x ≥ 1 ∪ +∞.

• Holder conjugate of p ∈ H: q ∈ H s.t. 1/p+ 1/q = 1, i.e., p+ q = pq.

100 D.G.L. WANG

Lemma A.4 (Young’s inequality for products). Let p, q > 1 be Holder con-jugates of each other. Suppose that a, b ≥ 0. Then

ab ≤ ap/p+ bq/q,

where the equality holds if and only if ap = bq.

Proof. Let t = 1/p. Then 1 − t = 1/q. By using the concavity of the logarithmfunction, we infer that

log(tap + (1− t)bq

)≥ t log

(ap)

+ (1− t) log(bq) = log(ab),

with the equality holds if and only if ap = bq. Young’s inequality follows byexponentiating.

Theorem A.5 (Holder’s inequality). Let p, q > 1 be Holder conjugates of eachother. For any x = (x1, . . . , xn) ∈ Cn and y = (y1, . . . , yn) ∈ Cn, we have

n∑i=1

|xiyi| ≤ ‖x‖p · ‖y‖q,

where the equality holds if and only if the vectors

xp = (xp1, . . . , xpn) and yq = (yq1, . . . , y

qn)

are linearly dependent.

Proof. It is w.l.o.g. to suppose that ‖x‖p = ‖y‖q = 1. Lemma A.4 gives

(A.1) |xiyi| ≤ |xi|p/p+ |yi|q/qfor any i, where the equality holds if and only if |xi|p = |yi|q. Adding Ineq. (A.1)for i = 1, . . . , n, we obtain

∑|xiyi| ≤ 1 as desired.

The p = 2 case of Holder inequality is the Cauchy-Schwarz inequality.

Theorem A.6 (Minkowski’s inequality). For x, y ∈ Rn and p ≥ 1, we have

‖x+ y‖p ≤ ‖x‖p + ‖y‖p.

Proof. If p = 1, then Minkowski’s inequality reduces to the subadditivity of theabsolute value function. If ‖x+ y‖p = 0, then Minkowski’s inequality holds for thenonnegativity of any norm. Below we can suppose that

p > 1 and ‖x+ y‖p > 0.

Let q be the Holder conjugate of p. Then (p− 1)q = p and 1/q = (p− 1)/p. Let

z = (|x1 + y1|p−1, . . . , |xn + yn|p−1).

101

Then we have

‖z‖q =(∑

|xi + yi|(p−1)q)1/q

=(∑

|xi + yi|p)(p−1)/p

= ‖x+ y‖p−1p .

By Theorem A.5, we can deduce that

‖x+ y‖pp =∑|xi + yi|p ≤

∑|xi| · |xi + yi|p−1 +

∑|yi| · |xi + yi|p−1

≤ ‖x‖p · ‖z‖q + ‖y‖p · ‖z‖q= (‖x‖p + ‖y‖p) · ‖x+ y‖p−1

p .

By canceling the positive factor ‖x+ y‖p−1p we obtain Minkowski’s inequality.

School of Mathematics and Statistics, Beijing Institute of Technology, 102488Beijing, P. R. China

Email address: [email protected]