Topology

PART A TOPOLOGY COURSE: HT 2013

Contents

1. Metric spaces 21.1. What is Topology about? 21.2. Definitions and examples 31.3. Balls, open, closed and bounded sets 41.4. Maps between metric spaces 71.5. Convergent sequences 101.6. Complete metric spaces 121.7. The space of bounded continuous functions 141.8. Various strong versions of continuity 152. Topological spaces 172.1. Definitions and examples 172.2. Closure of a set 202.3. Interior of a set 232.4. Boundary of a set 242.5. Separation axioms 242.6. Subspace of a topological space 252.7. Basis for a topology 272.8. Product of topological spaces 293. Connected spaces 323.1. Definition and properties of connected spaces 323.2. Path-connected spaces 344. Compact spaces 374.1. Definition and properties of compact sets 384.2. Compact spaces and continuous maps 414.3. Sequential compactness 435. Quotient spaces, quotient topology 465.1. Definition and examples 465.2. Separation axioms 485.3. Quotient spaces and continuous maps 50

Recommended books in Topology:

(1) J. Dugundji, Topology ;(2) J.R. Munkres, Topology, A First Course;(3) W. Sutherland, Introduction to Metric and Topological Spaces (particularly recommended);(4) O. Viro, O. Ivanov, V. Kharlamov, N. Netsvetaev, Elementary Topology,

http://www.math.uu.se/∼oleg/educ-texts.html

For further reading see the list at the end of the synopsis on the Maths Institute web pages.

These notes are adapted from a set of lecture notes written by Cornelia Drutu and Wilson Suther-land. I wish to thank them both for having allowed me to use them.

2 PART A TOPOLOGY COURSE: HT 2013

1. Metric spaces

1.1. What is Topology about? You have already seen and used some topological notions:

(1) In Analysis, several concepts like

• convergent sequences;

• bounded sequences;

• continuous functions;

need a way of defining nearby points as well as a way of defining points not too far apart.

This can be done via a distance (or metric).

In order to differentiate, one needs to consider functions defined on open sets.

(2) In Linear Algebra, you have seen the notion of an inner product space. In such a space one candefine

• the length of a vector (hence of distance between two points);

• the angle between vectors: in particular one may speak of orthogonality, and use trigonometry.

In what follows we weaken that structure, and we only require the notion of the distance betweentwo points. In particular we give up the notion of angle.

This notion of distance is encapuslated in the following definition.

PART A TOPOLOGY COURSE: HT 2013 3

1.2. Definitions and examples.

Definition 1.1. A metric space is a non-empty set X endowed with a function d : X ×X → R withthe following properties:

(M1) d(x, y) > 0 for all x, y ∈ X ; d(x, y) = 0 if and only if x = y;

(M2) (Symmetry) for all x, y ∈ X, d(y, x) = d(x, y);

(M3) (Triangle inequality) for all x, y, z ∈ X, d(x, z) 6 d(x, y) + d(y, z).

The function d is called a metric or distance.

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y z

x

d(x, z)

A first example of a metric space is provided by the following familiar structure.

Definition 1.2. A normed vector space (NVS) is a vector space V over R or C together with afunction V → R called a norm and written v 7→ ||v|| such that

(N1) for all v ∈ V, ||v|| > 0; ||v|| = 0 if and only if v = 0;

(N2) for all v ∈ V and any scalar α, ||αv|| = |α| ||v||;

(N3) (subadditivity) for all u, v ∈ V, ||u+ v|| 6 ||u||+ ||v||.

Proposition 1.3. A norm || || defines a metric on V , d(x, y) = ||x− y|| , for every x, y ∈ V .

Proof. Property (M1) follows from (N1). Property (M2) follows from (N2) applied to α = −1 and tov = x− y .

Property (M3) follows from (N3) applied to u = x− y and v = y − z . �

Remarks 1.4. (1) A vector subspace U of a normed vector space (V, ‖ ‖) is naturally endowedwith a norm: the restriction of ‖ ‖ to U.

(2) A Cartesian product V ×W of two normed vector spaces (V, ‖ ‖V ) and (W, ‖ ‖W ) can be

endowed with several norms: ‖ ‖V + ‖ ‖W , max (‖ ‖V , ‖ ‖W ) and√‖ ‖2V + ‖ ‖2W .

Example 1.5. On Rn several norms can be defined. For x = (x1, x2, . . . , xn), the l2 , l1 and l∞norms are

||x||2 =

√√√√ n∑i=1

x2i , ||x||1 =

n∑i=1

|xi| , ||x||∞ = max1≤i≤n

|xi| .

See Sheet 1, Exercise 5.


Remark 1.6. The norm || ||2 is defined by the standard inner product (or dot product) on Rn , that

is ||x||2 =√〈x,x〉 , where

〈x,y〉 = x1y1 + x2y2 + · · ·+ xnyn .

The space Rn endowed with the above inner product is sometimes called Euclidean space, therefore|| ||2 is called the Euclidean norm and the metric it defines the Euclidean metric on Rn .

The other two norms || ||1 and || ||∞ are not defined by any inner product on Rn .

A more complicated example is the following.

Example 1.7. (Function spaces) Let S be any non-empty set and let X be the set B(S, R) of allbounded functions from S to R . Then X is a real vector space, where for f, g ∈ X and α ∈ R wedefine f + g and αf by:

(f + g)(s) = f(s) + g(s), (αf)(s) = αf(s) .

Moreover ||f ||∞ = sups∈S |f(s)| is a norm on B(S, R).

Proof. Property (N1) is immediate.

For any scalar α and function f ∈ X , ||αf ||∞ = sups∈S |αf(s)| = |α| sups∈S |f(s)| = |α|||f ||∞ .

For any two functions f, g ∈ X and any s ∈ S , |f(s) + g(s)| ≤ |f(s)| + |g(s)| ≤ ||f ||∞ + ||g||∞ .Whence ||f + g||∞ ≤ ||f ||∞ + ||g||∞ . �

There are many metric spaces with a metric that is not defined by a norm, as in the followingexample. This is also a rather counter-intuitive example: a metric space in which all distinct pointsare at the same distance.

Example 1.8 (Discrete metric spaces). The discrete metric on any non-empty set X is defined by

d(x, y) =

{0 if x = y,1 if x 6= y.

This space (X, d) is called a discrete metric space.

Proof. Properties (M1) and (M2) are clearly satisfied.

In (M3) it is enough to study the case when x 6= z . Then either x 6= y or z 6= y . Thus d(x, z) = 1and d(x, y) + d(y, z) ≥ 1. �

Examples 1.9. (1) A non-empty subset A of a metric space (X,d) is naturally endowed with ametric, written dA : the restriction of d to A×A.

(2) A Cartesian product X1 ×X2 of two metric spaces can be endowed with a metric (see Sheet1, Exercise 4).

1.3. Balls, open, closed and bounded sets.

Definition 1.10. Let (X, d) be a metric space, x0 a point in X and r > 0.

The open ball in X of radius r centred at x0 is the set B(x0, r) = {x ∈ X : d(x, x0) < r}.The closed ball in X of radius r centred at x0 is the set B′(x0, r) = {x ∈ X : d(x, x0) ≤ r}.The sphere in X of radius r centred at x0 is the set S(x0, r) = {x ∈ X : d(x, x0) = r}.If more than one metric is defined on X then we write Bd(x0, r), B

′d(x0, r), Sd(x0, r).

Examples 1.11. (a) X = R2, d = d2 : Bd2(x0, r) is the open disc of radius r centred at x0 .

(b) X = R2, d = d1 : Bd1(x0, r) is the inside of the square centred on x0 with diagonals of

length 2r parallel to the axes, as in the diagram below.


(c) X = R2, d the discrete metric: Bd(x0, r) =

{{x0} if r 6 1R2 if r > 1.

(d) X the set B([0, 1], R) of all bounded real-valued functions on [0, 1], d = d∞ the sup metric:for f0 ∈ X and r > 0, Bd∞(f0, r) is the set of all f ∈ X whose graphs lie inside a ribbon ofvertical width 2r centred on the graph of f0.

@@@

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(r, 0)

Definition 1.12. A subset A of a metric space X is bounded if A ⊆ B(x0, r) for some x0 ∈ X andr > 0.

Definition 1.13. Let (X, d) be a metric space and U ⊆ X. We say U is open in X if for everyx ∈ U , there exists εx > 0 such that B(x, εx) ⊆ U.

Examples 1.14. (a) X = R2, d = d2 : open sets are as in the MT Analysis course.

(b) X = R2, d = d1 or d∞ : open sets exactly as in (a) above (see Sheet 1, Exercise 6).

(c) X endowed with the discrete metric: any subset U ⊆ X is open.

(d) Any open ball B(x0, r) in a metric space (X, d) is open in X .

Let us explain why (d) is true: take an arbitrary point x ∈ B(x0, r) and εx = r − d(x, x0). Thenfor every y ∈ B(x, εx), d(x0, y) ≤ d(x0, x) + d(x, y) < r . Hence, B(x, εx) ⊆ B(x0, r).

Proposition 1.15. A subset in a metric space is open if and only if it is a union of open balls.

Proof. Consider a union of open balls A =⋃i∈I B(xi, ri). For every x ∈ A there exists j ∈ I such

that x ∈ B(xj , rj). According to the argument justifying Example (d) above, for εx = rj − d(x, xj)the ball B(x, εx) is contained in B(xj , rj), hence in A .

Conversely, let U be an open subset in a metric space (X,d). For every x ∈ U , there exists εx > 0such that B(x, εx) ⊆ U.

Then⋃x∈U B(x, εx) ⊆ U . Since x ∈ B(x, εx), it follows that U ⊆

⋃x∈U B(x, εx). Whence the

equality. �

Corollary 1.16. A subset in (R, | |) is open if and only if it is a union of open intervals.

Proposition 1.17. Suppose that X is a metric space. Then

(a) X and ∅ are open in X ;

(b) if U and V are open in X , then so is U ∩ V ;

(c) if Ui is open in X for all i ∈ I , then so is⋃i∈I

Ui.

Remark 1.18. Note that in Proposition 1.17, (c), I might be infinite, even uncountable.


Proof. (b) Consider an arbitrary point x ∈ U ∩ V .

As U is open there exists αx > 0 such that B(x, αx) ⊆ U . Likewise, as x ∈ V and V is openthere exists βx such that B(x, βx) ⊆ V .

Then for δx = min(αx , βx), B(x, δx) ⊆ B(x, αx) ⊆ U and B(x, δx) ⊆ B(x, βx) ⊆ V . ThereforeB(x, δx) ⊆ U ∩ V .

(c) For every x ∈⋃i∈I Ui there exists k ∈ I such that x ∈ Uk . Since Uk is open there exists

εx > 0 such that B(x, εx) ⊆ Uk ⊆⋃i∈I Ui . �

Remark 1.19. Property (b) in Proposition 1.17 is not true for infinite intersections:

⋂n∈N,n≥1

(− 1

n,

1

n

)= {0} .

Definition 1.20. A subset V of a metric space X is closed in X if X \ V is open in X .

Examples 1.21. (a) In Euclidean spaces, the closed sets are exactly as you saw last term.

(b) In a discrete metric space X , any subset of X is closed in X (since any subset is open in X ).

(c) Any closed ball B′d(x0, r) in a metric space (X, d) is closed in X .

Remark 1.22. (1) In a metric space there may exist subsets that are neither open norclosed. For instance intervals [a, b) in (R, | |).

(2) On the other hand in a discrete metric space X all subsets are both open and closed.

Proposition 1.23. Suppose that X is a metric space. Then

(a) X and ∅ are closed in X ;

(b) if A and B are closed in X , then so is A ∪B ;

(c) if Ai is closed in X for all i ∈ I , then so is⋂i∈I

Ai.

Proof. This follows from the fact that A is closed if and only if X \A is open, from Proposition 1.17,and from the De Morgan laws (see the Appendix). �

Remark 1.24. Property (b) in Proposition 1.23 is not true for infinite unions:⋃n∈N,n≥1

[1

n, 1− 1

n

]= (0, 1) .

Proposition 1.25. Let (X,d) be a metric space and let A ⊆ X be a non-empty subset endowed withthe restricted metric dA .

Then a subset Y ⊆ A is open (respectively, closed) in (A,dA) if and only if Y = A ∩ U , where Uis open (respectively, closed) in X .


Proof. We begin by noting that if a ∈ A and r > 0 then BdA(a, r) = B(a, r) ∩A .Assume that Y ⊆ A is open in (A,dA). Then for every y ∈ Y there exists εy > 0 such that

BdA(y, εy) ⊆ Y . As in the proof of Proposition 1.15 it follows that Y =⋃y∈Y BdA(y, εy).

Take

U =⋃y∈Y

B(y, εy) .

Then U ∩A =⋃y∈Y [B(y, εy) ∩A] = Y .

Conversely, let U be an open set in X and Y = U ∩A . For every y ∈ Y ⊆ U there exists εy > 0so that B(y, εy) ⊆ U . Then B(y, εy) ∩A = BdA(y, εy) ⊆ U ∩A = Y .

The statement on closed sets follows immediately from the statement on open sets, and from thefact that A \ (A ∩ U) = A ∩ (X \ U). �

1.4. Maps between metric spaces.

Definition 1.26. Let (X,dX) and (Y,dY ) be metric spaces. A function f : X → Y is an isometricembedding if, for every x1, x2 ∈ X ,

dY (f(x1), f(x2)) = dX(x1, x2).

If, in addition, f is surjective, f is said to be an isometry.

Example 1.27. The function i : (R,d) → (R2,d2), x 7→ (x, 0) is an isometric embedding, but not anisometry.

Remarks 1.28. (1) An isometric embedding is necessarily injective.

(2) If f is an isometry, then f−1 is defined and is also an isometry.

(3) The composition of two isometric embeddings is an isometric embedding. The composition oftwo isometries is an isometry.

For most purposes, isometric embeddings and isometries are too rigid. So, we come to a keydefinition.

Definition 1.29. Let (X, dX) and (Y, dY ) be metric spaces and let f : X → Y be a function.

(a) We say that f is continuous at x0 ∈ X if given any ε > 0, there exists δ > 0 such thatwhenever dX(x, x0) < δ

dY (f(x), f(x0)) < ε ;

equivalently, such that f(BdX (x0, δ)) ⊆ BdY (f(x0), ε).

(b) We say that f is continuous if f is continuous at every x0 ∈ X.

Examples 1.30. (1) The projection map p : (R2,d2) → (R,d), (x1, x2) 7→ x1 is continuous. Forin the above definition, we may set δ to be ε .

(2) If (Y,dY ) is any metric space and (X,ddisc) is a discrete metric space, then any functionf : (X,ddisc)→ (Y,dY ) is continuous.

To see this, x0 be any point of X , and let ε > 0. Setting δ = 1/2 gives B(x0, δ) = {x0} .So, f(Bddisc

(x0, δ)) ⊆ BdY (f(x0), ε), as required.

Now we come to another key definition.

Definition 1.31. Let (X,dX) and (Y, dY ) be metric spaces. A function f : X → Y is a homeomor-phism if it is a bijection, and both f and f−1 are continuous. We then say that (X,dX) and (Y, dY )are homeomorphic.

Examples 1.32. (1) (R,d) and ((0, 1),d) are homeomorphic. (See Sheet 1, Exercise 3.)


(2) The surface of a cylinder (without the ends)

X = {(x, y, z) ∈ R3 | x2 + y2 = 1 , 1 < z < 2}and an annulus lying in the xy -plane, without the boundary,

Y = {(x, y, 0) | 1 < x2 + y2 < 4}are homeomorphic.

Indeed, the maps

f : X → Y , f(x, y, z) = (zx, zy, 0)

and

f−1 : Y → X , f−1(x, y, 0) =

(x√

x2 + y2,

y√x2 + y2

,√x2 + y2

)are both continuous.

We leave it as an exercise to check that f(X) ⊆ Y , f−1(Y ) ⊆ X and that f ◦ f−1 =idY , f

−1 ◦ f = idX .

(3) Let S1 be the unit circle in R2 . Give it the metric that it inherits as a subset of R2 . Let Qbe the hollow square with side length 2 centred at the origin. Also give it the metric that itinherits as a subset of R2 . Define

r : Q → S1

(x, y) 7→

(x√

x2 + y2,

x√x2 + y2

)which is radial projection onto S1 . In Sheet 1, Exercise 4, you prove that r is a homeomor-phism.

Proposition 1.33. If A is a non-empty subset of the metric space (X,d) then the inclusion mapi : A→ X is continuous, when A is endowed with the restricted metric dA .

Proof. Let x0 ∈ A and let ε > 0. Then, set δ = ε and note that i(BdA(x0, δ)) ⊆ Bd(x0, ε). �

Remarks 1.34. (1) Every isometric embedding is continuous.

(2) Hence, every isometry is a homeomorphism.

(3) Not every homeomorphism is an isometry. Example: id : (Rn, ‖ ‖2) → (Rn, ‖ ‖1) (or id :(Rn, ‖ ‖2)→ (Rn, ‖ ‖∞)).


It is not true that if f : X → Y is a continuous bijection between metric spaces, then f−1 isnecessarily also continuous.

Examples 1.35. (1) Let f be the identity map (R,ddisc) → (R,d) where ddisc is the discretemetric. By Example 1.30 (2), f is continuous. But f−1 is not continuous at any x ∈ R . To seethis, set ε = 1/2. Then, there is no δ > 0 such that f−1(Bd(x, δ)) ⊆ Bddisc(f−1(x), ε) = {x} .

(2) Consider f : [0, 1) → S1 , t 7→ (cos(2πt), sin(2πt)). It is not hard to prove that f is acontinuous bijection. But f−1 is not continuous at (1, 0).

Continuity of maps can also be defined using either open sets or closed sets.

Recall that if f : X → Y is a function, then for any subset A of Y , its inverse image (or pre-image)is

f−1(A) = {x ∈ X : f(x) ∈ A}.So, f−1(A) is defined even when f is not invertible.

Proposition 1.36. Let X, Y be metric spaces and let f : X → Y be a map. Then f is continuousif and only if the inverse image of every open set is open (i.e. whenever U is open in Y , theinverse image f−1(U) is open in X ).

Proof. Assume that f is continuous in the sense of Definition 1.29. Let U be an open set in Y . Weprove that f−1(U) is open in X . Let x0 be an arbitrary point in f−1(U). Then f(x0) ∈ U , andsince U is open there exists ε > 0 such that B(f(x0, ε)) ⊆ U .

The map f is continuous at x0 therefore for ε > 0 given above there exists δ > 0 such thatf(B(x0, δ)) ⊆ B(f(x0, ε)) ⊆ U . This implies that B(x0, δ) ⊆ f−1(U).

Conversely, assume that f−1(U) is open in X whenever U is open in Y . Let us prove that f iscontinuous in the sense of Definition 1.29.

Let x0 be an arbitrary point in X , and let ε > 0 be an arbitrary positive number. The setU = B(f(x0), ε) is open, therefore by hypothesis f−1(U) is an open set, moreover containing x0 . Itfollows that there exists δ > 0 such that B(x0, δ) ⊆ f−1(U). This implies that f (B(x0, δ)) ⊆ U =B(f(x0), ε).

We have thus proved that f is continuous at x0 , and x0 was arbitrary. �

Proposition 1.37. Let X, Y be metric spaces and let f : X → Y be a map.Then f is continuous if and only if the inverse image of every closed set is closed (i.e.

whenever V is closed in Y the inverse image f−1(V ) is closed in X ).

Proof. This follows from Proposition 1.36 and the formula for complements of inverse images:

f−1(Y \ V ) = X \ f−1(V )

�

Above, we saw that closed and open sets can used to verify continuity. Below, we show that somesets can be shown to be open or closed by using continuous maps.

Proposition 1.38 (defining open/closed sets via continuous functions). Let X be a metric space andlet f : X → R be a continuous function. Then for every λ ∈ R

(1) the sets {x ∈ X | f(x) > λ} and {x ∈ X | f(x) < λ} are open;

(2) the sets {x ∈ X | f(x) ≥ λ} , {x ∈ X | f(x) ≤ λ} and {x ∈ X | f(x) = λ} are closed.


Proof. (1) follows from Proposition 1.36 applied to U = (λ,+∞), and to U = (−∞, λ), open sets inR .

(2) follows from Proposition 1.37 applied to F = [λ,+∞), to F = (−∞, λ] , and to F = {λ} ,closed sets in R .

�

Examples 1.39. (1) The set {x ∈ R : ex > cosx} is open in R with the Euclidean metric.

(2) The planar set {(x, y) ∈ R2 : x3 ≤ y3 + y} is closed in R2 with the Euclidean metric (or themetric d1 or the metric d∞ ).

We investigate the behaviour of continuity with respect to composition of maps.

Proposition 1.40. If f : X → Y and g : Y → Z are continuous maps, where X,Y, Z are metricspaces, then so is g ◦ f .

Proof. Let U be an open set in Z .

Since g is continuous, g−1(U) is an open subset of Y .

Since f is continuous, f−1(g−1(U)

)is an open subset of X .

But the latter is the same as (g ◦ f)−1(U). �

1.5. Convergent sequences. In the setting of metric spaces, convergent sequences can be used todefine most of the important topological notions (closed sets, continuous functions etc).

Definition 1.41. A sequence (xn) in a metric space (X, d) converges to a point a ∈ X if for everyε > 0, there exists N ∈ N such that whenever n > N

d(xn, a) < ε :

equivalently, whenever n > N , xn ∈ B(a, ε).The sequence (xn) is called a convergent sequence; the element a is called the limit of the sequence

(xn).

Remark 1.42. In the above one can replace the inequality d(xn, a) < ε by the inequality d(xn, a) ≤ ε .

Indeed, consider the properties:

(strict) for every ε > 0, there exists N ∈ N such that whenever n > N

d(xn, a) < ε .

(non-strict) for every ε > 0, there exists N ∈ N such that whenever n > N

d(xn, a) ≤ ε .

We prove that these properties are equivalent.

Clearly (strict) implies (non-strict). Conversely, assume that (non-strict) is satisfied, andlet us prove (strict). Consider an arbitrary ε > 0. Property (non-strict) applied to ε

2 impliesthat there exists N such that for every n ≥ N , d(xn, a) ≤ ε

2 < ε .

Proposition 1.43. The limit of a convergent sequence in a metric space is unique.


Proof. Indeed, assume that (xn) converges to a , and that (xn) converges to b . Take ε > 0 arbitrary.There exists N1 such that for every n ≥ N1 , d(xn, a) < ε , and N2 such that for every n ≥ N2 ,d(xn, b) < ε . Then for any n ≥ max(N1, N2),

d(a, b) ≤ d(xn, a) + d(xn, b) < 2ε .

We have thus found that for every ε > 0, d(a, b) < 2ε . It follows that d(a, b) = 0, whencea = b . �

Examples 1.44. (a) X = R with the Euclidean metric: convergence is the usual convergence of(xn) to a .

(b) X with a discrete metric: a sequence (xn) is convergent if and only if there exists N suchthat xn = xN for every n ≥ N .

(c) X = B(A, R), where A ⊆ R , endowed with the metric

d∞(f, g) = ‖f − g‖∞ = supa∈A|f(a)− g(a)| .

The sequence (fn) of functions converges to f in (X, d∞) if and only if (fn) converges tof uniformly on A .

We can also rephrase convergence in terms of open sets.

Proposition 1.45. Let (xn) be a sequence of points in a metric space (X,d) . Then the followingare equivalent

(1) (xn) converges to a point a ∈ X ;(2) for every open set U containing a , there exists a natural number N such that

n ≥ N ⇒ xn ∈ U.

Proof. (1) ⇒ (2): Suppose (xn) converges to a , and that U is an open set containing a . Then thereis an ε > 0 such that B(a, ε) ⊆ U . By definition of convergence, there is an N ∈ N such that

n ≥ N ⇒ xn ∈ B(a, ε) ⊆ U,as required.

(2) ⇒ (1): For any ε > 0, B(a, ε) is open. So, assuming (2), there is an N ∈ N such that

n ≥ N ⇒ xn ∈ B(a, ε),

as required. �

One can use convergent sequences to check that a function between metric spaces is continuous.

Proposition 1.46. A map f : X → Y between two metric spaces is continuous if and only if forevery x ∈ X and every sequence (xn) in X converging to x , the sequence (f(xn)) converges to f(x) .

Proof. Assume that f is continuous. Let (xn) be a sequence that converges to x ∈ X . We want toprove that (f(xn)) converges to f(x). More precisely, we want to show that for every ε > 0 thereexists N such that f(xn) ∈ B(f(x), ε) whenever n ≥ N .

The map f is continuous at x , therefore there exists δ > 0 such that f(B(x, δ)) ⊆ B(f(x), ε).As (xn) converges to x , there exists N such that for any n ≥ N , xn ∈ B(x, δ), therefore f(xn) ∈f(B(x, δ)) ⊆ B(f(x), ε).

Conversely, assume that for every x ∈ X and every sequence (xn) in X converging to x , thesequence (f(xn)) converges to f(x). We want to prove that f is continuous.

Assume that f is not continuous at some point x . Then there exists ε > 0 such that for every δ > 0the set f(B(x, δ)) has some point outside B(f(x), ε). In particular for δ = 1

n there exists xn ∈ B(x, 1n )

such that d(f(xn), f(x)) ≥ ε . The sequence (xn) then converges to x while d(f(xn), f(x)) ≥ ε .


On the other hand, since (xn) converges to x by hypothesis f(xn) converges to f(x). Thiscontradicts the inequality d(f(xn), f(x)) ≥ ε .

We may then conclude that f is continuous. �

The notion of convergent sequence allows one to define closed subsets in metric spaces withouthaving to refer to their complementary subsets.

Proposition 1.47. A subset F in a metric space (X,d) is closed if and only if for every convergentsequence contained in F its limit is also in F .

Proof. Assume that F is closed and let (xn) be a convergent sequence in F with limit a .

If a ∈ X \ F , as X \ F is open, there exists ε > 0 such that B(a, ε) ⊆ X \ F .

Since (xn) converges to a it follows that there exists N such that for every n ≥ N , xn ∈ B(a, ε) ⊆X \ F . This contradicts the fact that the entire sequence (xn) is contained in F .

Therefore a cannot be in X \ F , it must be in F .

Conversely, assume that F has the property that for every convergent sequence contained in F itslimit is also in F . We want to prove that X \ F is open.

Assume, on the contrary, that X \ F is not open. Then there exists a ∈ X \ F such that no ballcentred in a is contained in X \ F . Thus, for every ε > 0, B(a, ε) ∩ F 6= ∅ .

In particular for every integer n ≥ 1 we can pick a point xn in B(a, 1

n

)∩ F . It is easy to check

that the sequence (xn) converges to a . Since (xn) is contained in F , by hypothesis a should belikewise in F . This contradicts the fact that a ∈ X \ F .

We conclude that X \ F must be open. �

1.6. Complete metric spaces. Complete metric spaces are spaces in which:

• convergent sequences can be defined without a priori knowing their limit (using the Cauchyproperty).

• the Contraction Map Theorem is true. This theorem is an important tool in Analysis, inparticular in solving various types of equations.

We notice a property satisfied by convergent sequences.

Definition 1.48. A sequence (xn) in a metric space (X, d) is Cauchy if for every ε > 0 there existsN ∈ N such that whenever m > n > N

d(xm, xn) < ε .

Remark 1.49. Inequality d(xm, xn) < ε may be replaced by d(xm, xn) ≤ ε without changing theproperty (see Remark 1.42).

Proposition 1.50. Any convergent sequence is Cauchy.

Proof. Let (xn) be a sequence in a metric space (X,d) that converges to a point a ∈ X . Let ε bean arbitrary positive number. There exists N such that for every n ≥ N , d(xn, a) < ε

2 . It followsthat for every m ≥ n ≥ N ,

d(xn, xm) ≤ d(xn, a) + d(xm, a) <ε

2+ε

2= ε .

�

Definition 1.51. A metric space X is complete if every Cauchy sequence in X converges (to a pointin X ).


Examples 1.52. (a) R is complete with the usual metric; so is Z , with the same metric;

(b) R \ {0} with the restricted Euclidean metric is not complete:(1n

)is a Cauchy sequence in

R \ {0} , but it is not convergent in R \ {0} .

Proposition 1.53. If (X,dX), (Y,dY ) are two isometric metric spaces and X is complete then Yis complete as well.

Proof. Let f : Y → X be an isometry. If (yn) is a Cauchy sequence then so is (f(yn)). SinceX is complete (f(yn)) converges to some point x ∈ X . Since f−1 is continuous, yn converges tof−1(x). �

Remarks 1.54. In Proposition 1.53 the hypothesis cannot be weakened to “there exists a homeomor-phism f : Y → X .” See for instance the function tan :

(−π2 ,

π2

)→ R .

Proposition 1.55. A subset A of a complete metric space is complete (when endowed with therestriction of the metric) if and only if A is closed.

Proof. Let (X,d) be a complete metric space and let A be a subset in X with the restricted metricdA .

Assume that (A,dA) is complete. We shall prove that A is closed using Proposition 1.47.

Let (xn) be a sequence in A convergent to a point x ∈ X . Then (xn) is Cauchy in (X,d), henceit is Cauchy in (A,dA). As (A,dA) is complete, it follows that (xn) converges to some a ∈ A .

The uniqueness of the limit of a convergent sequence (Proposition 1.43) implies that x = a ∈ A .

Conversely, assume that A is closed, and let us prove that (A,dA) is complete.

Let (xn) be a Cauchy sequence in (A,dA). Then (xn) is a Cauchy sequence in (X,d), which is acomplete metric space, therefore (xn) converges to a point x ∈ X .

The fact that A is closed and Proposition 1.47 imply that x ∈ A . Thus, (xn) converges in (A,dA)to a point x ∈ A . �

Examples 1.56. (a) the sets Q and R\Q with the restricted Euclidean metric are not complete;

(b) the subspace (0, 1) of R with the usual metric is not complete.

Proposition 1.57. A Cartesian product X ×Y of two complete metric spaces is complete (whateverthe product metric chosen on X × Y ).

Proof. Let (X,dX) and (Y,dY ) be two metric spaces. On X × Y the following three metrics can bedefined (see Exercise 4, Sheet 1):

d2((x1, y1), (x2, y2)) =√

dX(x1, x2)2 + dY (y1, y2)2 ,

d1((x1, y1), (x2, y2)) = dX(x1, x2) + dY (y1, y2) ,

d∞((x1, y1), (x2, y2)) = max (dX(x1, x2) , dY (y1, y2)) .

It is easy to prove that

(1) d∞ ≤ d2 ≤ d1 ≤ 2d∞ .

It suffices to show that X × Y is complete when endowed with one of the three metrics, for theother two metrics completeness will be deduced without difficulty via the sequence of inequalities (1).

If (xn, yn) is a Cauchy sequence in (X × Y, d1), then (xn) is a Cauchy sequence in (X,dX) and(yn) is a Cauchy sequence in (Y,dY ). Since X and Y are both complete, (xn) converges to a point


x ∈ X and (yn) converges to a point y ∈ Y . It follows that for every ε > 0 there exists N1 such thatdX(xn, x) ≤ ε

2 whenever n ≥ N1 , and there exists N2 such that dY (yn, y) ≤ ε2 whenever n ≥ N2 .

Then for every n ≥ max (N1, N2),

d1((xn, yn), (x, y)) = dX(xn, x) + dY (yn, y) < ε .

�

Theorem 1.58. For any non-empty set S , the function space (B(S, R), d∞) of bounded real-valuedfunctions on S with the sup metric is complete.

Remark 1.59. If S is finite, say S = {1, 2, .., n} , the map f 7→ (f(1), ..., f(n)) is a bijection fromB(S, R) to Rn , and via this map (B(S, R), d∞) is identified with (Rn, d∞). The latter metric spaceis known to be complete. Therefore what remains to be proved is the case when S is infinite. Thiscan be seen as an infinite dimensional version of the statement “(Rn, d∞) is complete.”

Proof. Let (fn) be a Cauchy sequence in (B(S, R), d∞). Then for an arbitrary ε > 0 there exists Nsuch that for any m ≥ n ≥ N ,

(2) ‖fn − fm‖∞ = sups∈S|fn(s)− fm(s)| < ε .

For an arbitrary s ∈ S , |fn(s) − fm(s)| ≤ ‖fn − fm‖∞ . This and the above imply that for anarbitrary s ∈ S , (fn(s)) is a Cauchy sequence in (R, | |). As (R, | |) is a complete metric space, itfollows that (fn(s)) converges to a real number `s .

We define the function f : S → R , f(s) = `s .

Consider an arbitrary ε > 0 and N such that whenever m ≥ n ≥ N , (2) holds.

Fix ε > 0 and n ≥ N . We have that for any m ≥ n , |fn(s)− fm(s)| < ε for every s ∈ S .

In the inequality corresponding to each s ∈ S we let m→∞ , and obtain that |fn(s)− f(s)| ≤ ε .

Then for each s ∈ S , |f(s)| ≤ |fn(s) − f(s)| + |fn(s)| ≤ ε + ‖fn‖∞ . This implies that f is abounded function.

The fact that |fn(s)− f(s)| ≤ ε for every s ∈ S implies that ‖fn − f‖∞ ≤ ε .We have thus obtained that for any ε > 0 there exists N such that for every n ≥ N , ‖fn−f‖∞ ≤ ε .This implies that (fn) converges to f in (B(S, R), d∞). �

1.7. The space of bounded continuous functions. Continuous functions form a very interestingmetric space.

Let (X, d) be a metric space. Consider the set Cb(X, R) of bounded continuous real-valued func-tions on X . It is a subset of B(X, R), the space of bounded real-valued functions on X . Recall thatthe latter is endowed with the sup-norm, inducing the sup-metric

d∞(f, g) = supx∈X|f(x)− g(x)|.

Theorem 1.60. The subspace Cb(X, R) is a closed subset in B(X, R) .

Proof. We prove that Cb(X, R) is closed in B(X, R) by proving that whenever it contains a convergentsequence it contains its limit as well.

Let (fn) be a sequence in Cb(X, R) convergent to f in B(X, R). Then for every ε > 0 there existsN such that for every n ≥ N ,

‖fn − f‖∞ = supx∈X|fn(x)− f(x)| < ε .

We want to prove that f is continuous at an arbitrary point x0 ∈ X .


Let ε > 0. There exists N such that

‖fN − f‖∞ <ε

3.

By hypothesis fN is continuous at x0 , therefore there exists δ > 0 such that d(x, x0) < δ implies|fN (x)− fN (x0)| < ε

3 .Assume that d(x, x0) < δ . Then

|f(x)−f(x0)| ≤ |f(x)−fN (x)|+ |fN (x)−fN (x0)|+ |fN (x0)−f(x0)| < 2‖fN−f‖∞+ε

3<

2ε

3+ε

3= ε .

�

Corollary 1.61. The space (Cb(X, R), d∞) with the sup metric is complete.

Proof. This follows from:

• the fact that (B(X, R),d∞) is complete (Theorem 1.58);

• the fact that a closed subset in a complete metric space is complete (Proposition 1.55);

• Theorem 1.60.

�

Corollary 1.62. The space C[a, b] of continuous real-valued functions on [a, b] (where a < b arereal numbers) with the sup metric is complete.

Proof. We only need to recall that any continuous real-valued function on [a, b] is bounded, and applyCorollary 1.61. �

1.8. Various strong versions of continuity. We now define some strong versions of continuity.

Definition 1.63. Let (X, dX), (Y, dY ) be metric spaces.

A map f : X → Y is uniformly continuous if for every ε > 0 there exists δ > 0 such thatdX(x, x′) < δ ⇒ dY (f(x), f(x′)) < ε .

Remark 1.64. Uniform continuity is stronger than continuity.

Indeed ‘f continuous’ means that for every x0 ∈ X and for every ε > 0 there exists δx0> 0

(depending on x0 ) such that dX(x0, y) < δx0⇒ dY (f(x0), f(y)) < ε .

Uniform continuity requires the existence of a uniform δ , the same for all points x0 .

A particular case of a uniformly continuous map is the following.

Definition 1.65. Let (X, dX), (Y, dY ) be metric spaces. A map f : X → Y is K –Lipschitz (orsimply Lipschitz ), for some constant K ∈ (0,∞), if

dY (f(x), f(x′)) ≤ KdX(x, x′) , ∀x, x′ ∈ X .

Proposition 1.66. Any K –Lipschitz map is uniformly continuous.

Proof. For every ε > 0 one can take δ = εK . �

Examples 1.67. (a) Let f : I → R be a differentiable function, where I is an open interval in R ,such that f ′ is bounded, that is |f ′(x)| ≤M for every x ∈ I . Then f is M –Lipschitz.

Indeed, according to the Mean Value theorem, for every x < y in I , f(x) − f(y) =f ′(c)(x− y), where c ∈ (x, y).

(b) The function x 7→ xα with α ∈ [0, 1] is Lipschitz on [1,+∞).


(c) The function x 7→ 2x is Lipschitz on any interval [a, b] but not on R.

Indeed if the map were K –Lipschitz then for any x ∈ R , 2x+1 − 2x = 2x would have tobe at most K , which is impossible.

(d) Given a metric space (X,d) and a point a ∈ X , the function f : X → R , f(x) = d(x, a) is1–Lipschitz. This is an immediate consequence of the Triangle Inequality.

Example (d) and Proposition 1.38 imply that open balls and the complements of closed balls areopen sets. Indeed B(a, r) = f−1(−1, r) and X \B′(a, r) = f−1(r,+∞).

Remark 1.68. Any isometric embedding f : X → Y between metric spaces is 1-Lipschitz and henceuniformly continuous.

Definition 1.69. A map f : X → X of a metric space (X, d) to itself is called a K –contraction (orsimply a contraction) if it is K –Lipschitz for some constant K < 1.

Example 1.70. The cosine function on [0, 1].

According to the Mean Value Theorem, for every x, y ∈ [0, 1],

| cosx− cos y| ≤ sin 1 |x− y| .

Theorem 1.71 (Contraction Map Theorem). Suppose that X is a complete metric space andthat f : X → X is a K –contraction, for some constant K < 1 . Then

(a) f has a unique fixed point, i.e. there exists a unique p ∈ X such that f(p) = p ;

(b) for any x0 ∈ X , the sequence (xn) defined inductively by xn = f(xn−1) for all n > 1converges to p ;

(c) (error estimate) d(xn, p) 6Kn

1−Kd(x0, x1) for any n > 1 .

Remarks 1.72. (1) If X is not complete then Theorem 1.71 is not true. For instance thefunction f : (0, 1]→ (0, 1] , f(x) = x

2 is a 12 –contraction but it has no fixed point in (0, 1].

(2) The hypothesis ‘f is a contraction’ cannot be weakened to ‘f satisfies the inequal-ity:

(3) d(f(x1), f(x2)) < d(x1, x2) , for all x1 6= x2 in X .’

For instance the function f : R→ R , f(x) =√x2 + 1 satisfies (3), due to the Mean Value

Theorem, and R is complete, but f has no fixed point.

Remark 1.73. Theorem 1.71 is useful when trying to solve an equation x = f(x). When f is acontraction:

• Theorem 1.71, (a), implies that the equation admits a unique solution;

• Theorem 1.71, (b) and (c) give a way of approximating the solution of the equation x = f(x),and of estimating the error in the approximation.

See Exercise 9 on Problem Sheet 1, where an integral equation is solved using the contractionmapping theorem.


Proof of Theorem 1.71.Step 1: Existence of a fixed point. Let x0 be an arbitrary point in X , and define the sequence

(xn) inductively by xn+1 = f(xn).

We prove by induction that d(xn+1, xn) ≤ Knd(x1, x0). For n = 0 it is an equality.

Assume that the inequality is satisfied for n . Then

d(xn+2, xn+1) = d(f(xn+1), f(xn)) ≤ Kd(xn+1, xn) ≤ Kn+1d(x1, x0) .

For m ≥ n ≥ N we have that

d(xn, xm) ≤ d(xn, xn+1) + d(xn+1, xn+2) + · · ·+ d(xm−1, xm) ≤ (Kn +Kn+1 + · · ·+Km−1)d(x1, x0) .

The last term is equal to Kn 1−Km−n

1−K ≤ Kn

1−K ≤KN

1−K .

For any ε > 0 there exists N large enough so that KN

1−K < ε . It follows that (xn) is a Cauchy

sequence. Since X is a complete metric space, (xn) converges to some point p ∈ X .

The function f is continuous. In the equality f(xn) = xn+1 by letting n → ∞ we obtain thatf(p) = p .

Step 2: Uniqueness of the fixed point. Let q be another point such that f(q) = q . Then d(p, q) =d(f(p), f(q)) ≤ Kd(p, q). If p 6= q this yields a contradiction. It follows that p = q .

This in particular implies that any sequence constructed as in Step 1 converges to the same pointp , the unique fixed point of f .

Moreover, in Step 1 it was proved that for m ≥ n , d(xn, xm) ≤ Kn

1−K . If in the last inequality we

fix n and let m→∞ , we obtain d(xn, p) ≤ Kn

1−K . �

2. Topological spaces

2.1. Definitions and examples.

Definition 2.1. A topological space (X, T ) consists of a non-empty set X together with a family Tof subsets of X satisfying:

(T1) X, ∅ ∈ T ;

(T2) U, V ∈ T ⇒ U ∩ V ∈ T ;

(T3) Ui ∈ T for all i ∈ I ⇒⋃i∈I

Ui ∈ T .

The family T is called a topology for X . The sets in T are called the open sets of X . When T isunderstood we talk about the topological space X .

There are several good reasons to generalise the setting from metric spaces to topological spaces:

• it is clear that two isometric metric spaces are identical from the metrical point of view; anatural question to ask is what are the common properties of two homeomorphic metric spaces(like the cylinder and the annulus in (2) of Examples 1.32) ?

• some types of convergence cannot be defined in the setting of metric spaces, but can be definedin the setting of topological spaces; this is the case for point-wise convergence of a sequenceof functions;

• some constructions, like quotient spaces (defined in Section 5), may be endowed much moreeasily with a topology than with a metric;


• an important part of the theory of convergence and continuity, as well as other importantnotions, may be recovered using the three axioms (T1), (T2), (T3) only. This setting maysimplify some arguments and emphasize what is really essential to them.

Examples 2.2. (1) Any metric space (X, d) gives rise to a topological space (X, Td), where Tdconsists of those subsets of X which are open in X with respect to the metric d.

(2) (Discrete spaces) Let X be any non-empty set. The discrete topology on X is the set of allsubsets of X .

(3) (Indiscrete spaces) Let X be any non-empty set. The indiscrete topology on X is the familyof subsets {X, ∅} .

(4) Let X be any non-empty set. The co-finite topology on X consists of the empty set togetherwith every subset U of X such that X \ U is finite.

Definition 2.3. (1) We call a topological space (X, T ) metrizable if it arises as in Example 2.2,(1), from (at least one) metric space (X, d) i.e. there is at least one metric d on X such thatT = Td.

(2) Two metrics on a set are topologically equivalent if they give rise to the same topology.

Example 2.4. The metrics d1, d2, d∞ on Rn are all topologically equivalent - see Sheet 1, Ex. 6, (3).

We shall call the topology defined by the above three metrics the standard (or canonical) topologyon Rn .

Definition 2.5. Given two topologies T1, T2 on the same set, we say T1 is coarser than T2 if T1 ⊆ T2.

Remark 2.6. For any space (X, T ), the indiscrete topology on X is coarser than T which in turn iscoarser than the discrete topology on X .

Definition 2.7. Let (X, T ) be a topological space.

A subset V of X is closed in X if X \ V is open in X (i.e. X \ V ∈ T ).

Examples 2.8. (a) You have to be careful which space you’re saying a set is closed in.

For example in the space [0, 1) with the usual topology coming from the Euclidean metric,[1/2, 1) is closed.

(b) In a discrete space, all subsets are closed since their complements are open.

(c) In the co-finite topology on a set X , a subset is closed if and only if it is finite or all of X .

Proposition 2.9. Let X be a topological space. Then

(C1) X, ∅ are closed in X ;

(C2) if V1, V2 are closed in X then V1 ∪ V2 is closed in X ;

(C3) if Vi is closed in X for all i ∈ I then⋂i∈I

Vi is closed in X .

Proof. Properties (C1),(C2), (C3) follow from (T1), (T2), (T3) and from the De Morgan laws (seethe Appendix). �

Definition 2.10. A sequence (xn) in a topological space X converges to a point x ∈ X if given anyopen set U containing x there exists an integer N such that xn ∈ U for all n > N.


Examples 2.11. (a) In a metric space this is equivalent to the metric definition of convergence byProposition 1.45.

(b) In a (non-empty) indiscrete topological space X any sequence converges to any point x ∈ X .

(c) In an infinite space X with the co-finite topology any sequence (xn) of pairwise distinctelements (i.e. such that xn 6= xm when n 6= m) converges to any point x ∈ X .

Remark 2.12. Note that the uniqueness of the limit for a convergent sequence is not granted in atopological space.

In particular (b) and (c) show that both the indiscrete topology and the co-finite topology on aninfinite set are not metrizable, as in a metric space any convergent sequence has only one limit.

In order to have the uniqueness of the limit for a convergent sequence, one has to add an extraaxiom (Hausdorff). We shall discuss this axiom later on.

Proposition 2.13. If a subset F in a topological space X is closed then for any convergent sequencecontained in F any limit of it is also in F .

Proof. Let (xn) be a convergent sequence, contained in F , and let x ∈ X be a limit of (xn). Ifx ∈ X \ F then, as X \ F is open, it follows that there exists N such that xn ∈ X \ F for everyn ≥ N . This contradicts the hypothesis that (xn) is contained in F . �

Remark 2.14. Unlike in the case of metric spaces, the converse of the statement in Proposition 2.13might not be true.

Loose Remark. We shall see many implications of the following form:

topological property ⇒ property in terms of sequences.

In metric spaces the converse implication may sometime be true (i.e. the topological property maybe characterized in terms of sequences).

In the general topological setting the converse implication is never true.

Definition 2.15. Suppose that (X, TX) and (Y, TY ) are topological spaces and that f : X → Y isa map. We say that f is continuous if U ∈ TY ⇒ f−1(U) ∈ TX i.e. ‘inverse images of open sets areopen’.

If there are other topologies around we may say ‘f is continuous with respect to TX , TY ’ or ‘f is(TX , TY )-continuous’.

Proposition 2.16. A map f : X → Y of topological spaces is continuous if and only if f−1(V ) isclosed in X whenever V is closed in Y .

Proof. This follows from the definition of a continuous map and from the formula

f−1(Y \ V ) = X \ f−1(V ) .

�

Proposition 2.17. If f : X → Y and g : Y → Z are continuous maps, X,Y, Z topological spaces,then so is g ◦ f .

The proof is identical to the one of Proposition 1.40.


Definition 2.18. A homeomorphism between topological spaces X and Y is a bijection f : X → Ysuch that f and f−1 are continuous. The spaces X,Y are said to be homeomorphic.

Proposition 2.19. If a map f : X → Y is a continuous map between two topological spaces then forany sequence (xn) in X converging to a point x , (f(xn)) converges to f(x) .

Proof. Let U be an open set containing f(x). Then f−1(U) is an open set (f is continuous) con-taining x .

Since (xn) converges to x there exists N such that for n ≥ N , xn ∈ f−1(U). It follows thatf(xn) ∈ U . �

Remark 2.20. The converse of Proposition 2.19 might not be true.

The following simple result is often a really useful way of showing that a set is open.

Lemma 2.21. Let U be a subset of a topological space. Then the following are equivalent:

(1) U is open;(2) for every x in U , there is an open set Ux containing x such that that Ux ⊆ U .

Proof. (1) ⇒ (2): Suppose that U is open. For each x ∈ U , set Ux = U .(2) ⇒ (1): Suppose that for each x ∈ U , there is an open set Ux as in (2). Then clearly⋃x∈U Ux = U . So, U is a union of open sets, and hence is open. �

2.2. Closure of a set. Property (C3) of closed sets, that is:

‘if Vi is closed in X for all i ∈ I then⋂i∈I

Vi is closed in X .’

suggests the idea of approximating from above an arbitrary set by a closed set.

Definition 2.22. Consider a topological space X and a subset A ⊆ X .

The closure of A in X is the set

A =⋂

F closed,A⊆F

F .

A point in A is sometimes called an adherent point of A .

Proposition 2.23. The closure A is the smallest (with respect to inclusion) closed subsetof X containing A , that is:

(a) A is closed in X and it contains A ;

(b) if A ⊆ V where V is closed in X then A ⊆ V .

Proof. (a) is immediate from the definition of the closure and property (C3) of closed sets.

A closed subset V as in (b) appears in the intersection defining A , hence A ⊆ V . �

Proposition 2.24. Let A, B be subsets of a topological space X . Then

(a) A ⊆ B implies that A ⊆ B ;

(b) A is closed in X if and only if A = A ;

(c) A = A .


Proof. (a) We have A ⊆ B ⊆ B , we apply Proposition 2.23, (b).

(b) A ⊆ A . If A is closed then Proposition 2.23, (b), with V = A implies A ⊆ A , whence equality.

If A = A then A is closed.

(c) We apply (b) to A . �

The following is a way of characterising points in A that is often useful.

Proposition 2.25.

A = {x ∈ X : for every open U ⊆ X with x ∈ U , U ∩A 6= ∅}.

Proof. Consider A1 = {x ∈ X : for every open U ⊆ X with x ∈ U , U ∩A 6= ∅}.We first prove that A ⊆ A1 using Proposition 2.23, (b). Clearly A1 contains A . It remains to

prove that A1 is closed. We prove that X \A1 = B1 is open, by using Lemma 2.21. Note that

B1 = {x ∈ X : there exists an open set Ux such that x ∈ Ux ⊆ X \A} .

For every element y in some Ux one can take Uy = Ux and have y ∈ Uy ⊆ X \A . It follows thateach Ux is entirely contained in B1 . So, by Lemma 2.21, B1 is open.

We have thus proved that A1 is closed, which implies that A ⊆ A1 .

Assume that there exists x ∈ A1 \ A . Since x 6∈ A it follows that there exists a closed set Fcontaining A such that x 6∈ F . Then x ∈ X \ F , and X \ F is an open set which does not intersectA . This contradicts the fact that x ∈ A1 . �

Examples 2.26. (1) The closure of (a, b) in (R, | |) is [a, b] .

(2) The closure of Q in (R, | |) is R . Same for R \Q .

(3) The closure of any set A in a space X endowed with the co-finite topology is either X if Ais infinite, or A if A is finite.

Definition 2.27. A subset A in X is called dense if A = X .

We shall see in what follows that the closure of a set A contains two types of points:

• points in A that are isolated;

• points in A or outside A but near which points in A accumulate (i.e. accumulationpoints).

Definition 2.28. A point x ∈ X such that for any open U ⊆ X with x ∈ U , (U \ {x}) ∩ A 6= ∅ iscalled an accumulation point (or limit point) of A .

The set of accumulation points of A is sometimes denoted by A′ .

Note that A = A ∪ A′ . The points in A \ A′ are points x ∈ A such that for some open set Ucontaining x , U ∩A = {x} . Such points are called isolated points of A .

Examples 2.29. In the metric space (R, | |)

(1) if A = (a, b) then A′ = A = [a, b] ;

(2) if A = Q or R \Q then A′ = A = R ;

(3) if A = Z then A = Z while A′ = ∅ .

(4) if A = (0, 1) ∪ {9, 10} then A = [0, 1] ∪ {9, 10} , A′ = [0, 1] and 9, 10 are isolated points.


Proposition 2.30. Let (X,d) be a metric space and let A ⊆ X .

(1) The closure A is the set of limits of convergent sequences (an) in A .

(2) The set of accumulation points A′ is the set of limits of convergent sequences (an) of pairwisedistinct elements in A (i.e. such that an 6= am if n 6= m).

Proof. (1) Let A` be the set of all limits of convergent sequences (an) in A .

Since A is closed and it contains A , it contains all the limits of convergent sequences (an) in A ,due to Proposition 2.13. Thus A` ⊆ A .

On the other hand, let x be a point in A . According to Proposition 2.25, for every integer n ≥ 1,the open ball B

(x, 1

n

)intersects A in a point an . The sequence (an) converges to x . Hence x ∈ A` .

We have thus proved that A ⊆ A` .

(2) Let A′` be the set of all limits of convergent sequences (an) in A such that an 6= am whenevern 6= m .

Let x be a point in A′` , the limit of a sequence (an). At most one element in the sequence can beequal to x , thus without loss of generality we may assume that an 6= x for all n .

For every open set U containing x there exists N such that an ∈ U for all n ≥ N . Moreoveran ∈ U \ {x} . In particular (U \ {x}) ∩A 6= ∅ . Thus x ∈ A′ .

We conclude that A′` ⊆ A′ .

Conversely let x be a point in A′ . We construct inductively a sequence (an) in A such thatd(x, an+1) < d(x, an) and 0 < d(x, an) < 1

n .

Since x ∈ A′ , B(x, 1) \ {x} intersects A in a point a1 .Assume that we found the elements in the sequence a1, ..., an satisfying the required hypotheses.

Let rn+1 = min(

1n+1 , d(x, an)

).

Since x ∈ A′ , B(x, rn+1) \ {x} intersects A in a point an+1 .

The sequence (an) thus constructed converges to x , and for n < m , d(x, am) < d(x, an), whenceam 6= an . Therefore x ∈ A′` .

We have proved that A′ ⊆ A′` , which together with the converse inclusion implies that A′ = A′` . �

Proposition 2.31. A map f : X → Y of topological spaces is continuous if and only if f(A) ⊆ f(A)for every A ⊆ X .

Proof. Assume that f is continuous. Since f(A) is closed, it follows that f−1(f(A)

)is closed. The

latter subset also contains A , therefore it contains A . We have thus proved that A ⊆ f−1(f(A)

),

which implies that f(A) ⊆ f(A).

Conversely, assume that f(A) ⊆ f(A) for every A ⊆ X .

Let F be a closed subset in Y and let A = f−1(F ). Then f(A) ⊆ F , whence f(A) ⊆ f(A) ⊆ F .It follows that A ⊆ f−1(F ) = A , therefore A = A . This implies that A is closed. �


2.3. Interior of a set. Property (T3) of open sets, that is:

‘if Ui are open for all i ∈ I then⋃i∈I Ui is open’

suggests the idea of approximating from below an arbitrary set by an open set.

Definition 2.32. Let X be a topological space X and A a subset in X . The interior of A in X isthe set

A =⋃

Uopen,U⊆A

U .

Note that A may be empty even if A is not. For instance consider the subset A = [0, 1]× {0} inR2 . Clearly A contains no open ball B((x, y), ε), therefore A contains no non-empty open subset.

Proposition 2.33. The interior A is the largest (with respect to inclusion) open subset ofX contained in A , that is:

(a) A is open and contained in A ;

(b) every open subset of X contained in A lies in A .

Both (a) and (b) follow easily from the definition of the interior.

Proposition 2.34. Let A, B be subsets of a topological space X . Then

(a) A ⊆ B implies that A ⊆ B ;

(b) A is open in X if and only if A = A ;

(c)˚A = A ;

Proof. (a) A ⊆ A ⊆ B , and A is open. Therefore A ⊆ B .

(b) If A is open, since A ⊆ A it follows that A ⊆ A , whence equality. The converse implication isimmediate.

(c) follows from (b) applied to A .�

Examples 2.35. • The interior of [a, b] (or (a, b]) in (R, | |) is (a, b);

Indeed (a, b) is open and contained in [a, b] , therefore in its interior.

If a was in the interior then there would exist ε > 0 such that (a − ε, a + ε) is in theinterior, therefore in [a, b] . This is impossible, hence a is not in the interior. Likewise for b .It follows that the interior is in (a, b), therefore it is equal to it.

• The interiors of Z , Q , R \Q in (R, | |) are ∅ .

None of the sets above contains an open interval, therefore none contains a non-empty opensubset.

• Let X be a space endowed with the co-finite topology, and let A ⊆ X . The interior A iseither equal to A if X \A is finite (i.e. A is open), or it is empty.

Indeed if A would be non-empty, then X \ A would be finite.

As A is contained in A , X \A is contained in X \ A , therefore it is also finite.

Next we note that the operation of taking the interior and of taking the closure are in some sensecomplementary to each other.


Proposition 2.36. Let A be a subset in a topological space X .

The closure of the complement of A is the complement of the interior, i.e.

X \A = X \ A .

Proof. Denote X \A by B .

As A is open, and A ⊆ A , its complement X\A is closed, and it contains B . Therefore B ⊆ X\A .

On the other hand B is closed and it contains B = X \ A . It follows that X \ B is open and

contained in A . Therefore X \B ⊆ A . This is equivalent to X \ A ⊆ B .

From the two opposite inclusions above we deduce that B = X \ A .�

If in Proposition 2.36 we denote X \A by B we obtain:

Proposition 2.37. Let B be a subset in a topological space X .

The interior of the complement of B is the complement of the closure, i.e.

˚X \B = X \B .

Note that the notation˚

X \B simply means the interior of X \B .

2.4. Boundary of a set.

Definition 2.38. The set ∂A = A \ A is called the boundary of A .

Proposition 2.39. The boundary of A equals the intersection A∩X \A , and it equals the boundaryof X \A .

Proof. Indeed ∂A = A \ A = A ∩ (X \ A) = A ∩ (X \A). The last inequality is due to Proposition2.36.

Then by the above ∂(X \A) = (X \A) ∩A .

We conclude that ∂A = ∂(X \A). �

Example 2.40. Let A = [a, b), where a < b . Then A = [a, b] .

On the other hand R \A = (−∞, a) ∪ [b,+∞), and its closure is R \A = (−∞, a] ∪ [b,+∞).

It follows that ∂A = {a, b} .

2.5. Separation axioms.

Definition 2.41. A topological space satisfies the first separation axiom if for any two distinct pointsa 6= b there exists an open set U containing a and not b .

Proposition 2.42. A topological space satisfies the first separation axiom if and only if singletonsare closed subsets in it.

Proof. If every singleton {x} is closed in X then for any two distinct points a 6= b consider U =X \ {b} .

Conversely, assume that X satisfies the first separation axiom. We prove that V = X \ {x} isopen, for any x ∈ X .

Let y ∈ V arbitrary. Then y 6= x , hence by the first separation axiom there exists Uy open,containing y , not containing x , hence contained in V .

So, by Lemma 2.21, V is open. �


Definition 2.43. A topological space X is said to be Hausdorff (or to satisfy the second separationaxiom) if given any two distinct points x, y in X , there exist disjoint open sets U, V with x ∈ U, y ∈V.

Examples 2.44. (a) Any metric space is Hausdorff.

(b) An indiscrete space X with more than one point does not satisfy the first separation axiom.

(c) An infinite set with the co-finite topology is not Hausdorff (consequently it is not a metrizabletopological space), but it satisfies the first separation axiom because singletons are closed init.

Proposition 2.45. If f : X → Y is an injective continuous map of topological spaces X, Y and Ysatisfies the first (respectively, second) separation axiom then so does X .

Proof. Assume that Y satisfies the first separation axiom. Let a, b be two distinct points in X .Since f is injective, f(a) 6= f(b). By the first separation axiom in Y there exists U open such thatf(a) ∈ U and f(b) 6∈ U . This implies that a ∈ f−1(U), and b 6∈ f−1(U). As f is continuous, f−1(U)is open. We have thus proved that X satisfies the first separation axiom.

A similar argument shows that if Y is Hausdorff then X is Hausdorff. Note that we must use thefact that U ∩ V = ∅ ⇒ f−1(U) ∩ f−1(V ) = ∅ . See the Appendix. �

Corollary 2.46. If spaces X, Y are homeomorphic then X satisfies the first (second) separationaxiom if and only if Y is satisfies the first (respectively the second) separation axiom.

Proposition 2.47. In a Hausdorff space, a sequence converges to at most one point.

Proof. Indeed assume that a convergent sequence (xn) has two distinct limits a 6= b . Then by theHausdorff property there exist U, V open and disjoint (i.e. U∩V = ∅) such that a ∈ U and b ∈ U . As(xn) converges to a there exists N such that xn ∈ U for every n ≥ N . Similarly, (xn) convergent tob implies that there exists N ′ such that xn ∈ V for every n ≥ N ′ . Then for every n ≥ max(N,N ′),xn ∈ U ∩ V , contradicting U ∩ V = ∅ .

�

Remark 2.48. Note that the first separation axiom does not suffice to ensure the uniqueness of thelimit of a convergent sequence. Indeed, an infinite space with the co-finite topology satisfies the firstseparation axiom (see above), but contains sequences with several limits (see Example 2.11, (c)).

2.6. Subspace of a topological space. We saw that a subspace of a metric space is naturallyendowed with a metric structure.

Now we shall see that the same thing happens for topological structures.

Definition 2.49. Let (X, T ) be a topological space and let A be a non-empty subset of X .

The subspace or induced topology on A is TA = {A ∩ U : U ∈ T } .

The three properties (T1), (T2), (T3) that a topology must satisfy are easily checked:

(T1) ∅ = ∅ ∩A ∈ TA ; A = X ∩A ∈ TA .

(T2) Given V1, V2 in TA , V1 = A ∩ U1 and V2 = A ∩ U2 , where U1 and U2 are both in T .

Then V1 ∩ V2 = U1 ∩ U2 ∩A ∈ TA .


(T3) Let (Vi)i∈I be such that Vi ∈ TA for every i ∈ I . Then for every i ∈ I there exists Ui opensubset in X such that Vi = A ∩ Ui .

It follows that⋃i∈I Vi = A ∩

⋃i∈I Ui which is in TA because

⋃i∈I Ui is in T .

The following is easy but useful.

Lemma 2.50. Let (X, T ) be a topological space, and let A be a non-empty subset of X . Give Athe subspace topology. Then the inclusion map i : A→ X is continuous.

Proof. Let U be an open subset of X . Then i−1(X) = A ∩ X , which is open in A . So, i iscontinuous. �

Proposition 2.51. A subset W in A is closed in TA if and only if there exists F closed in (X, T )such that W = A ∩ F .

Proof. Assume that W ⊆ A is closed in TA . Then A\W is open, i.e. there exists U open in X suchthat A \W = A ∩ U .

The last equality may be re-written as W = A \A∩U = A \U = A∩ (X \U). The set F = X \Uis closed in X .

Conversely, let F be closed in (X, T ), and let W = A ∩ F .

Then A \W = A \ F = A ∩ (X \ F ). Since F is closed in X , X \ F is open in X , hence A \Wis open in A . It follows that W is closed in A . �

Proposition 2.52. Let (A, TA) be a subspace of (X, T ) , and let B ⊆ A .

(a) The closure of B in (A, TA) , BA

, coincides with the intersection between A and the closure

of B in X , BX

.

(b) The interior of B in (A, TA) , BA , contains the interior of B in X , BX .

Proof. (a) The set BX

is closed in X , therefore BX∩A is closed in A , and it contains B . Therefore

BA ⊆ BX ∩A .

On the other hand BA

is a closed set in A .

It follows by Proposition 2.51 that BA

= F ∩ A , where F is closed in X . Since B ⊆ BA ⊆ F it

follows that BX ⊆ F , hence B

X ∩A ⊆ F ∩A = BA

.

We conclude that BX ∩A = B

A.

(b) Since BX is open in X and contained in B ⊆ A , it is an open set in A . Therefore BX ⊆BA . �

Remarks 2.53. (1) The inclusion BX ⊆ BA may be strict.

Consider for instance

B = (0, 1)× {0} ⊂ A = R× {0} ⊂ X = R2.

As B contains no open ball in R2 , BX is empty.

But B is an open set in A , therefore BA = B .


(2) If A is open then BX = BA for every B ⊆ A .

Indeed BA is open in A , hence BA = U ∩A for some U open in X .

Since A is also open, it follows that BA is open in X . It is also contained in B , thereforeBA ⊆ BX .

This and Proposition 2.52, (b), imply equality.

We finish our discussion of subspaces of topological spaces with some easy to see but importantremarks.

Remarks 2.54. (1) If A ⊆ B ⊆ X and (X, T ) is a topological space then the topology induced onA by T coincides with the topology induced on A by TB .

(2) If A ⊆ X and (X,d) is a metric space then the topology on A induced by Td coincides withthe topology on A induced by the restricted metric dA .

This follows immediately from the fact that for any a ∈ A , the ball centred in a and ofradius r > 0 in (A,dA) coincides with B(a, r)∩A , where B(a, r) is the ball centred in a andof radius r > 0 in (X,d).

(3) If A is open in X then TA is contained in T .

(4) If A is closed in X then any set closed in A is closed in X .

2.7. Basis for a topology. As in the case of vector spaces, we can consider a minimal amount ofdata (a basis) via which we may recover the whole space endowed with the considered structure.

This notion is particularly useful in two settings:

• that of topologies induced by metrics;

• that of products of topological spaces (to be studied further on).

Definition 2.55. Given a topological space (X, T ), a subfamily B of T is called a basis for T ifevery set in T can be expressed as a union of sets in B .

Example 2.56. In a metric space (X,d) the family of open balls

B = {B(x, r) : x ∈ X, r > 0}is a basis for Td .

In particular in R the family of open intervals B = {(a, b) | a, b ∈ R , a < b} is a basis for thestandard topology.

Moreover, the family of open intervals with rational endpoints BQ = {(a, b) | a, b ∈ Q , a < b} is abasis for the standard topology in R (Ex. 9, (2), Sheet 2).

Criterion 2.57. Let (X, T ) be a topological space, and B a basis for T .

A subset U is open in (X, T ) if and only if for every x ∈ U there exists B ∈ B such thatx ∈ B ⊆ U .

Proof. Indeed, if U is open then there exists (Bi)i∈I collection of subsets in the basis B such thatU =

⋃i∈I Bi . This accounts for the direct implication.

Conversely assume that for every x ∈ U there exists Bx ∈ B such that x ∈ Bx ⊆ U . ThenU ⊆

⋃x∈U Bx ⊆ U , whence U =

⋃x∈U Bx . �


Proposition 2.58. Let X,Y be topological spaces, and let B be a basis for the topology on Y .

A map f : X → Y is continuous if and only if for every B in B its inverse image f−1(B) is openin X .

Proof. The necessary part follows from the fact that B ⊆ T , i.e. a basis is composed of open sets.

The sufficient part follows from:

• the fact that every open set in Y can be written as⋃i∈I Bi , with Bi ∈ B for every i ;

• the fact that f−1(⋃

i∈I Bi)

=⋃i∈I f

−1 (Bi).

�

Proposition 2.59. Let (X, T ) be a topological space, B a basis for T and A ⊆ X . Then

(4) A = {x ∈ X : for every B ∈ B with x ∈ B , B ∩A 6= ∅} ,

(5) A = {x ∈ X : there exists B ∈ B such that x ∈ B ⊆ A}.

Proof. (4) Denote by AB the set

{x ∈ X : B ∩A 6= ∅ for any B ∈ B with x ∈ B} .Recall that according to Proposition 2.25,

A = {x ∈ X : U ∩A 6= ∅ for any U ∈ T with x ∈ U} .This and the fact that B ⊆ T imply that A ⊆ AB .

Let x be an arbitrary point in AB . Let U be an open set containing x . According to the Criterion2.57, there exists B ∈ B such that x ∈ B ⊆ U .

Since x ∈ AB , it follows that B ∩A 6= ∅ , whence U ∩A 6= ∅ .

We conclude that x ∈ A .

(5) We denote by IB the set

{x ∈ X : there exists B ∈ B such that x ∈ B ⊆ A}.Let x be an arbitrary point in IB . Then there exists B ∈ B such that x ∈ B ⊆ A . Note that the

whole of B is contained in IB . Thus x ∈ B ⊆ IB . This, according to the Criterion 2.57, implies thatIB is open.

Also IB is a subset of A by definition, therefore IB ⊆ A .

Let x be an arbitrary point in A , which is an open set. The Criterion 2.57 implies that there existsB ∈ B such that x ∈ B ⊆ A ⊆ A . It follows that x ∈ IB .

We may therefore conclude that A ⊆ IB , whence the two sets are equal. �

Remark 2.60. In a metric space, (5) can be made even more precise:

(6) A = {x ∈ X : there exists ε > 0 such that B(x, ε) ⊆ A}.

Proof. Let Ad be the set

{x ∈ X : there exists ε > 0 such that B(x, ε) ⊆ A}.

According to (5), Ad is contained in A .

The converse inclusion follows from the fact that if a point x is contained in some ball B(y, r) thenB(x, ε) ⊂ B(y, r) for ε = r − d(x, y).

�


Now we shall see:

• how to recognise when a family of subsets is a basis of some topology ;

• how to re-construct the topology when a basis is given.

Proposition 2.61. Let X be a set and B a family of subsets of X such that

(B1) X is a union of sets in B ;

(B2) the intersection B ∩B′ of any B,B′ ∈ B can be expressed as a union of sets in B .

Then the family TB of all unions of sets in B is a topology for X . Note that B is a basis for TB .

Proof. (T1) The set X is contained in TB due to (B1).

The empty set ∅ is contained in TB as the union of no sets from B .

(T2) Let U, V be two sets in TB . Then U =⋃i∈I Bi and V =

⋃j∈J B

′j , where Bj , B

′j ∈ B .

The intersection U ∩ V is equal to⋃i∈I, j∈J Bi ∩ B′j . According to (B2), Bi ∩ B′j =

⋃k∈Kij B

′′k ,

for some collection (B′′k )k∈Kij in B . It follows that

U ∩ V =⋃

i∈I, j∈J

⋃k∈Kij

B′′k ,

therefore U ∩ V is in TB .

Property (T3) is immediate from the definition of TB . �

2.8. Product of topological spaces. One of the main ways of constructing new topological spacesout of given ones is by taking Cartesian products. This is what we shall discuss now.

Proposition 2.62. Let (X, TX), (Y, TY ) be topological spaces. The family of subsets

BX×Y = {U × V : U ∈ TX , V ∈ TY } ,satisfies the conditions (B1) and (B2) in Proposition 2.61.

Proof. Indeed X × Y is itself in BX×Y . This accounts for (B1).

Also, consider U × V and U ′ × V ′ , with U,U ′ ∈ TX and V, V ′ ∈ TY . Then

(U × V ) ∩ (U ′ × V ′) = (U ∩ U ′)× (V ∩ V ′)is in BX×Y . This accounts for (B2). �

Definition 2.63. The family of all unions of sets from BX×Y is a topology TX×Y for X×Y (accordingto Proposition 2.61).

We call this topology the product topology.

The space (X × Y, TX×Y ) is called the topological product of (X, TX) and (Y, TY ).

NB One of the commonest errors about products is to assume that any open set in the product is a‘rectangular’ open set such as U × V .

Criterion 2.64. A subset W ⊆ X×Y is open in the product topology if and only if for any (x, y) ∈Wthere exist U ∈ TX , V ∈ TY such that (x, y) ∈ U × V ⊆W.

Proposition 2.65. Let (X × Y, TX×Y ) be the topological product of (X, TX) and (Y, TY ) .

If A is closed in X and B is closed in Y then A×B is closed in X × Y .


Proof. Since A is closed, its complement U = X \A is open.

Likewise B closed implies V = Y \B open.

The complement (X × Y ) \ (A×B) is not U × V , but

{(x, y) ∈ X × Y ; x 6∈ A or y 6∈ B} = (U × Y ) ∪ (X × V ) .

Therefore it is open. �

Remark 2.66. For finitely many topological spaces X1, ..., Xn one can define a topology on X1×· · ·×Xn either by considering a basis composed of products of open sets U1 × · · · × Un , or by inductionon n and by identifying X1 × · · · ×Xn with (X1 × · · · ×Xn−1)×Xn .

Proposition 2.67. Let (X,dX) and (Y, dY ) be two metric spaces.

The product topology of the two metric topologies TdX and TdY coincides with the topology inducedby any of the product metrics (recall that they are all topologically equivalent, by Ex. 4, Sheet 1).

Proof. Consider the product metric

d∞ ((x, y), (x′, y′)) = max [dX(x, x′) , dY (y, y′)] .

For every U ∈ TdX and V ∈ TdY , U × V is in Td∞ . It follows that BX×Y ⊆ Td∞ , whenceTX×Y ⊆ Td∞ , due to property (T3) of Td∞ .

Conversely, let W be a set in Td∞ . We prove that W is in TX×Y using the Criterion 2.64.

Let (x, y) ∈W . Then there exists ε such that Bd∞ ((x, y), ε) ⊆W . But

Bd∞ ((x, y), ε) = BdX (x, ε)×BdY (y, ε) ,

and the latter set is in BX×Y . �

Proposition 2.68. Let (X × Y, TX×Y ) be the topological product of (X, TX) and (Y, TY ) , and letA×B be a subset of X × Y . Then

(i) A×B = A×B ;

(ii)˚

A×B = A× B .

Proof. (i) According to Proposition 2.65, A×B is a closed set. Since it also contains A×B , we havethat A×B ⊆ A×B .

Conversely, let (x, y) be a point in A×B . Then for every U open in X containing x , U ∩A 6= ∅ .Likewise for every V open in Y containing y , V ∩B 6= ∅ . Then (U × V ) ∩ (A×B) 6= ∅ .

It follows that (x, y) satisfies the property described in Proposition 2.59, (4), for the set A×B withrespect to the basis BX×Y . Therefore (x, y) ∈ A×B . We have thus proved that A × B ⊆ A×B ,whence the equality.

(ii) The set A× B is open and contained in A×B , therefore it is contained in˚

A×B .

Conversely, let (x, y) be a point in˚

A×B . Proposition 2.59, (5), applied to the set A × B withrespect to the basis BX×Y implies that there exist U open in X and V open in Y such that

(x, y) ∈ U × V ⊆ A × B . This implies that x ∈ U ⊆ A and y ∈ V ⊆ B , whence x ∈ A and y ∈ B ,

that is (x, y) ∈ A× B .

We have proved that˚

A×B ⊆ A × B , which together with the opposite inclusion yields theequality. �


Proposition 2.69. (1) If X and Y satisfy the first separation axiom then so does their topolog-ical product X × Y .

(2) If X and Y are Hausdorff spaces, so is their topological product X × Y .

Proof. (1) Let (x, y) 6= (x′, y′). Then either x 6= x′ or y 6= y′ .

Assume that x 6= x′ . Then there exists U open in X containing x and not x′ . It follows thatU × Y contains (x, y) and not (x′, y′). The case y 6= y′ is similar.

The proof of (2) is done along the same lines, using the fact that

U ∩ V = ∅ ⇒ (U × Y ) ∩ (V × Y ) = ∅ .�

Proposition 2.70. (1) When X×Y is endowed with the product topology TX×Y , the projectionmaps pX : X ×Y → X , pX(x, y) = x, and pY : X ×Y → Y , pY (x, y) = y , are continuous.

(2) ( a second way of defining the product topology) The product topology TX×Y is the coarsesttopology making pX and pY continuous: any topology T on X × Y with respect to whichpX and pY are continuous contains TX×Y .

Proof. (1) Indeed for any open set U in X , p−1X (U) = U × Y ; likewise for any open set V in Y ,

p−1Y (V ) = X × V .

(2) According to the above any topology T with respect to which pX and pY are continuousmust contain all sets U × Y with U ∈ TX , and all sets X × V with V ∈ TY .

Therefore, by property (T2), T must contain all sets U×V = (U×Y )∩(X×V ). Thus BX×Y ⊆ Twhence, by property (T3) of T , TX×Y ⊆ T . �

Proposition 2.71. Let X,Y be two topological spaces.

(1) Let Z be a topological space, and let f : Z → X and g : Z → Y be two maps.

The map (f, g) : Z → X×Y is continuous, with X×Y endowed with the product topology,if and only if both f and g are continuous.

(2) ( a third way of defining the product topology) The product topology is the only topology forwhich (1) holds for all possible Z and f, g .

Remark 2.72. Proposition 2.71, (1) can be reformulated as follows:

A map F : Z → X×Y from a topological space Z into the topological product X×Y is continuousif and only if both pX ◦ F : Z → X and pY ◦ F : Z → Y are continuous.

Proof of Proposition 2.71. (1) Assume that the map (f, g) is continuous. Then f = pX ◦ (f, g)and g = pY ◦ (f, g) are also continuous, by Proposition 1.40.

Conversely, assume that both f and g are continuous.

For every U open in X and V open in Y ,

(f, g)−1(U × V ) = f−1(U) ∩ g−1(V )

is open.

This and Proposition 2.58 applied to the map (f, g) and the basis BX×Y allow us to conclude that(f, g) is continuous.

(2) Let T be another topology on X × Y for which (1) holds for all possible Z and f, g .


We take Z = X × Y with the product topology TX×Y and the maps f = pX and g = pYcontinuous. Then (pX , pY ) = idX×Y is a continuous map between (X ×Y , TX×Y ) and (X ×Y , T ).It follows that for every open set U ∈ T , its inverse image by idX×Y , which is U , is contained inTX×Y .

We have thus proved that T ⊆ TX×Y .

Clearly idX×Y = (pX , pY ) is a continuous map between (X × Y , T ) and (X × Y , T ). This andthe property (1) of T implies that pX and pY are continuous also when X × Y is endowed with thetopology T .

Proposition 2.70, (2), implies that TX×Y ⊆ T , whence equality. �

3. Connected spaces

3.1. Definition and properties of connected spaces. We now define the important notion ofconnected space.

• This notion allows one to define intervals in R in terms of the topology on R (instead of theorder on R). This was seen last term and shall be recalled here.

• In various problems in Analysis and Geometry one needs to work in a set A such that onecan move continuously inside A from an arbitrary point of A to another (i.e. a set A that ispath-connected).

• In real life too such questions naturally occur. If one can only travel by car for instance, oneis bound to consider the topological space X composed of all the points above the sea leveland make one’s travels plans within a path-connected component A of X .

• A weaker condition which is also much used in Analysis and Geometry is that A cannot bewritten as the disjoint union of two open non-empty sets (i.e. A is connected, as opposed todisconnected).

We shall see that the two conditions of connectedness and path-connectedness are not equivalentin general.

Definition 3.1. A topological space X is disconnected if there are disjoint open non-empty subsetsU and V such that U ∪ V = X .

If X is not disconnected, it is called connected.

Proposition 3.2. Let X be a topological space. The following properties are equivalent:

(1) the only subsets of X which are both open and closed are X and ∅ ;

(2) X is connected;

(3) any continuous map from X to {0, 1} (with the discrete topology) is constant.

Proof. (1)⇒ (2) If U, V are open sets in X such that U ∩ V = ∅ and U ∪ V = X , then U is bothopen and closed (as X \ U = V ). It follows by (1) that either U = ∅ or U = X , hence V = ∅ .

(2)⇒ (3) Let f : X → {0, 1} be a continuous map. Then U = f−1({0}) and V = f−1({1}) areopen, disjoint subsets of X , and U ∪ V = X . So, by definition of connectedness, one of U and V isempty, and therefore f is not surjective.

(3)⇒ (1) Let A be a subset of X both open and closed. Then the map f : X → {0, 1}such that f(x) = 1 if x ∈ A and f(x) = 0 if x 6∈ A is continuous: f−1({1}) = A which is open,f−1({0}) = X \A , which is open.


According to (3) either f is constant 1, which means that A = X , or f is constant 0, whichmeans that A = ∅ . �

Definition 3.3. A (non-empty) subset A of a topological space X is connected if A with the subspacetopology is connected.

Conventionally an empty set is assumed to be connected.

Remark 3.4. The connectedness of a subset A can be formulated in terms of the topology on X(rather than the subspace topology) as follows:

A is connected if and only if, whenever U and V are open subsets of X such that A ⊆ (U ∪ V )and U ∩ V ∩A = ∅ , then either U ∩A = ∅ or V ∩A = ∅ .

The following was proved in the Michaelmas term Analysis course.

Proposition 3.5. Let R be endowed with the standard topology and let A be a subset in R . Thefollowing statements are equivalent:

(1) A is a connected subset of R ;

(2) A is an interval, i.e. a set of the form

(−∞, b), (−∞, b], (a, b), [a, b), (a, b], [a, b], (a, ∞), [a, ∞), ∅, andR,

where b > a for intervals with endpoints a, b except for [a, b] , where b > a.

Proposition 3.6. If f : X → Y is a continuous map of topological spaces X, Y and A ⊂ X isconnected then so is f(A) . (“The continuous image of a connected set is connected.”)

Proof. Let U and V be open sets in Y such that f(A) ⊆ U ∪V and f(A)∩U ∩V = ∅ . Then f−1(U)and f−1(V ) are open in X . So, f−1(U) ∩ A and f−1(V ) ∩ A are disjoint and open in A . Since Ais connected, one of f−1(U)∩A and f−1(V )∩A is empty. Hence, one of U ∩ f(A) and V ∩ f(A) isempty. So, f(A) is connected. �

We saw that several topological properties (openness, closedness etc) are preserved under finiteintersections and unions. This is no longer true for connectedness: one can easily draw examplesdisproving the statement both for finite intersections and unions.

Still, under some extra conditions, connectedness can be granted for unions of connected sets.

Proposition 3.7. Suppose that {Ai : i ∈ I} is a family of connected subsets of a topological space Xwith Ai ∩Aj 6= ∅ for each pair i, j ∈ I .

Then⋃i∈I Ai is connected.

Proof. Let f :⋃i∈I Ai → {0, 1} be a continuous map. Its restriction to Ai , f |Ai , is also continuous.

Since Ai is connected it follows that f |Ai is constant equal to ci ∈ {0, 1} .

For each pair i, j ∈ I there exists a point x in Ai ∩ Aj , and f(x) = ci = cj . Therefore all theconstants ci are equal, and f is constant. �

Corollary 3.8. If {Ci : i ∈ I} and B are connected subsets of a topological space X such thatCi ∩B 6= ∅ for each i ∈ I , then B ∪

⋃i∈I Ci is connected.


Proof. Proposition 3.7 implies that Ai = Ci ∪B is connected for every i ∈ I .

Since Ai ∩ Aj contains B for every i, j , again by Proposition 3.7 it follows that⋃i∈I Ai =

B ∪⋃i∈I Ci is connected. �

Theorem 3.9. The topological product X×Y is connected if and only if both X and Y are connected.

Remark 3.10. An inductive argument allows one to extend Theorem 3.9 to any finite product oftopological spaces: a topological product X1 × · · · × Xn is connected if and only if X1, ..., Xn areconnected.

Proof of Theorem 3.9. If X × Y is connected then X = pX(X × Y ) and Y = pY (X × Y ) areconnected by Proposition 3.6.

Assume that X and Y are connected. Then X × {y} is connected for every y ∈ Y , and {x} × Yis connected for every x ∈ X .

Fix a point y0 ∈ Y . Corollary 3.8 applied to B = X × {y0} and to Cx = {x} × Y , connected andsuch that B ∩ Cx = {(x, y0)} implies that B ∪

⋃x∈X Cx = X × Y is connected. �

Theorem 3.11. Suppose that A is a connected subset of a topological space X and A ⊆ B ⊆ A .Then B is connected.

Proof. Let U and V be open subsets of X such that B ⊆ U ∪ V and B ∩ U ∩ V = ∅ . We will showthat one of U ∩B and V ∩B is empty.

Now, A ⊆ U ∪V and A∩U ∩V ⊆ B∩U ∩V = ∅ . So, by the connectedness of A , one of U ∩A andV ∩ A is empty. Say that U ∩ A is empty. Then, A lies in the closed set X\U . So, by Proposition2.23, A ⊆ X\U . So, U ∩A = ∅ . Since B ⊆ A , we deduce that U ∩B is empty, as required. �

3.2. Path-connected spaces.

Definition 3.12. A path connecting two points x, y in a topological space X is a continuous mapp : [0, 1]→ X with p(0) = x, p(1) = y.

Remark 3.13. If p : [0, 1]→ X is a path then p ([0, 1]) is a connected set.

Remark 3.14. If p : [0, 1]→ X is a path connecting x, y and q : [0, 1]→ X is a path connecting y, zthen we may construct a path connecting x, z as follows: r : [0, 1]→ X ,

r(t) =

p(2t) if t ∈[0, 12],

q(2t− 1) if t ∈[12 , 1].

The restriction of r to[0, 12]

is continuous because p is continuous and the map t 7→ 2t iscontinuous. We denote this restriction by r1 .

Likewise, the continuity of q and of the map t 7→ 2t− 1 implies that the restriction of r to[12 , 1]

is continuous. We denote the latter restriction by r2 .For any closed set F in X , r−1(F ) = r−11 (F ) ∪ r−12 (F ).We have that r−11 (F ) is closed in

[0, 12], which is closed in [0, 1], therefore r−11 (F ) is closed in

[0, 1]. A similar argument yields that r−12 (F ) is closed in [0, 1].We conclude that r−1(F ) is closed in [0, 1].


Definition 3.15. A topological space X is path-connected if any two points in X are connected bya path in X .

A subset A ⊆ X is path-connected if with the subspace topology it satisfies the previous condition.Equivalently, A is path-connected if any two points in A can be joined by a path p : [0, 1]→ X withimage in A . Conventionally the empty set is path-connected.

Examples 3.16. (1) Given a continuous function f : I → R , I ⊆ R an interval, the graphGf = {(x, f(x)) | x ∈ I} is path-connected.

For instance consider the function f : R→ R ,

f(x) =

{x sin 1

x if x 6= 0 ,0 if x = 0 .

(2) Any normed vector space (V , ‖ ‖) is path-connected.

Indeed, consider u, v two vectors in V . Let p : [0, 1]→ V be defined by

p(t) = (1− t)u+ tv .

Clearly p(0) = u and p(1) = v . Moreover

‖p(s)− p(t)‖ = ‖(t− s)u− (t− s)v‖ = |t− s|‖u− v‖ .

It follows that p is K –Lipschitz, with K = ‖u− v‖ .

(3) Any ball in a normed vector space is path-connected.

According to Remark 3.14 it suffices to prove that any vector u in a ball B(a,R) isconnected by a path to a .

We consider the path p : [0, 1]→ V defined by p(t) = (1− t)u+ ta .

For every t ∈ [0, 1], ‖p(t) − a‖ = |1 − t| ‖u − a‖ ≤ ‖u − a‖ < R . Thus the image of thepath p is entirely contained in B(a,R).

Proposition 3.17. Any path-connected space is connected.

Proof. Let X be a path-connected space and let a be a fixed point in X . For every x ∈ X thereexists a path px : [0, 1]→ X such that px(0) = a and px(1) = x . By Remark 3.13, Px = px[0, 1] is aconnected set.

We apply Corollary 3.8 to the collection of connected sets {Px : x ∈ X} and to the connected set{a} , and conclude that

⋃x∈X Px ∪ {a} = X is connected. �

Remark 3.18. There exist connected spaces that are not path-connected.

Example 3.19. Let Gf be the graph of the function f : (0, ∞)→ R , f(x) = sin 1x , that is the set

Gf =

{(x, sin

1

x

)| x > 0

}.

Clearly Gf is path-connected, hence connected.

The closure of Gf in R2 with the standard topology, Gf , is equal to Gf ∪ ({0} × [−1, 1]) . It isconnected but not path-connected.

Remark 3.20. The example 3.19 also shows that ‘A path-connected’ does not in general imply ‘Apath-connected’.


In the above statement two things must be proved:

(i) that Gf = Gf ∪ ({0} × [−1, 1]);

(ii) that Gf = Gf ∪ ({0} × [−1, 1]) is not path-connected.

(i) First we prove that Gf ⊆ Gf ∪ ({0} × [−1, 1]).

Let (a, b) be an arbitrary point in Gf .

Proposition 2.30, (1), implies that (a, b) is the limit of a sequence(xn , sin 1

xn

)with xn > 0 (we

are in the metric space R2 ). This implies that (xn) converges to a , therefore a must be in [0,+∞).

If a ∈ (0,+∞) then by the continuity of the map x 7→ sin 1x we have that the limit b of sin 1

xn

must be sin 1a . Thus in this case (a, b) ∈ Gf .

Assume that a = 0. The sequence(

sin 1xn

)converges to b and it is contained in [−1, 1] closed in

R , therefore b ∈ [−1, 1].

We have thus found that Gf ⊆ Gf ∪ ({0} × [−1, 1]).

For the converse inclusion, we take an arbitrary point (0, b) ∈ {0} × [−1, 1].

The sequence(xn , sin 1

xn

)n∈N

in Gf with xn = 1arcsin b+2nπ converges to (0, b). Therefore (0, b) ∈

Gf , hence {0} × [−1, 1] ⊆ Gf .

(ii) Assume that Gf = Gf ∪ ({0} × [−1, 1]) is path-connected. In particular there is a path

p : [0, 1]→ G connecting (0, 0) and (1, sin 1).

Since p is continuous and F = {0} × [−1, 1] is closed, we have that p−1(F ) is closed in [0, 1]. LetT be the supremum of p−1(F ). This in particular implies that p(T, 1] ∩ F = ∅ .

As p−1(F ) is closed, its supremum T must be contained in it, therefore p(T ) = (0, b) for someb ∈ [−1, 1].

The path p is continuous at T hence for ε = 14 there exists δ > 0 such that

∀s ∈ [T, T + δ] , ‖p(s)− p(T )‖ < 1

4.

We may assume that δ is such that T + δ < 1 and we denote T + δ by S .

We have that p(S) ∈ Gf , thus p(S) =(x0 , sin 1

x0

)for some x0 > 0.

Let π : R2 → R be the projection π(x, y) = x . It is a continuous map, and so is p , therefore π ◦ pis continuous. The interval [T, S] is a connected set, therefore π ◦ p[T, S] is connected.

It follows that π ◦ p[T, S] is an interval containing 0 and x0 > 0, thus containing [0, x0] .

Then for every x ∈ (0, x0] there exists a point in p[T, S] contained in π−1(x).

On the other hand p[T, S] is contained in Gf and the latter intersects π−1(x) in a unique point,(x, sin 1

x

).

We then conclude that p[T, S] contains{(x, sin

1

x

): x ∈ (0, x0]

}.

According to the choice of S , p[T, S] should be contained in the ball B(p(T ) , 1

4

). In particular

any two points in p[T, S] should be at distance ≤ 12 , by the triangle inequality.

On the other hand, for n large enough xn = 1π2 +2nπ and yn = 1

3π2 +2nπ

are both in (0, x0] .


It follows that p[T, S] contains the points(xn, sin

1xn

)= (xn, 1) and

(yn, sin

1yn

)= (yn,−1),

which are at distance at least 2. This gives a contradiction.

Nevertheless the implication ‘connected ⇒ path-connected’ holds for a particular type of topologicalspace.

Proposition 3.21. A connected open subset U of a normed vector space is path-connected.

Remark 3.22. The hypothesis ‘U open’ in Proposition 3.21 cannot be removed, as shown by theExample 3.19 (U = Gf is a connected subset of the normed vector space (R2, ‖ ‖)).

Proof. Let a be a fixed point in U and let C be the set of elements x ∈ U connected to a by a pathin U .

The set C contains a , therefore it is non-empty. We prove that C is both open and closed in U .

First, we show that C is open. Indeed, consider a point x in C . Since x ∈ U which is open,there exists ε > 0 such that B(x, ε) ⊆ U . All points in B(x, ε) can be connected by a path to x , byExample 3.16, (3). This and Remark 3.14 imply that B(x, ε) ⊆ C . Hence, C is open.

Now we prove that C is closed in U by proving that U \ C is open.

Let x be a point in U \C . Since x is in U , there is some ε > 0 such that B(x, ε) ⊆ U . We claimthat B(x, ε) is disjoint from C . This will show that B(x, ε) ⊆ U \C , and hence that U \C is open.

Suppose that there were a point y in B(x, ε)∩C . Then, x can joined to y by a path in B(x, ε) ⊆ U ,using Example 3.16, (3). Also, y can be joined to a by a path in U , since y is in C . Hence, Remark3.14 implies that there is a path in U joining x to a , which contradicts the assumption that x is notin C .

Thus C is a non-empty, closed and open subset in U connected, therefore C = U . This andRemark 3.14 imply that U is path-connected. �

Proposition 3.23. If f : X → Y is a continuous map of topological spaces X, Y and A ⊂ X ispath-connected then so is f(A) . (“The continuous image of a path-connected set is path-connected.”)

Proof. For every two elements x, y ∈ f(A) let a, b be two elements in A such that f(a) = x andf(b) = y .

The set A is path-connected, therefore there exists a path p : [0, 1]→ A connecting a and b . Thenf ◦ p is a path in f(A) connecting x and y . �

4. Compact spaces

Compact sets are a particular type of bounded closed set, very much used in all branches ofmathematics. Here are two reasons why this notion is interesting:

• in Analysis it is important to know whether there exists a certain object minimizing a givenfunctional or not (for instance in calculus of variations problems);

• from the topological point of view, as it will become clear from the theory, compact sets are‘the next best thing, after singletons and finite sets’.


4.1. Definition and properties of compact sets.

Definition 4.1. A family {Ui : i ∈ I} of subsets of a space X is called a cover if X =⋃i∈I

Ui.

If each Ui is open in X then U is called an open cover for X .

Definition 4.2. A subcover of a cover {Ui : i ∈ I} for a space X is a subfamily {Uj : j ∈ J} forsome subset J ⊆ I such that {Uj : j ∈ J} is still a cover for X .

We call it a finite (or countable) subcover if J is finite (or countable).

Definition 4.3. A topological space X is compact if any open cover of X has a finite subcover.

Proposition 4.4. Let X be a topological space. The following are equivalent:

(1) X is compact;

(2) if {Vi : i ∈ I} is an indexed family of closed subsets of X such that⋂j∈J

Vj 6= ∅ for any finite

subset J ⊆ I then⋂i∈I

Vi 6= ∅.

Proof. (1) ⇒ (2) Assume that {Vi : i ∈ I} is a family of closed subsets of X such that⋂j∈J

Vj 6= ∅

for any finite subset J ⊆ I , while⋂i∈I

Vi = ∅. Then⋃i∈I(X \ Vi) = X .

According to (1) there exists J finite subset in I such that⋃j∈J(X \ Vj) = X . This is equivalent

to⋂j∈J Vj = ∅ , contradicting the hypothesis.

(2) ⇒ (1) Let {Ui : i ∈ I} be an open cover of X . Then X =⋃i∈I Ui , whence

⋂i∈I(X \Ui) = ∅ .

According to property (2) applied to the family of closed sets {X \ Ui ; i ∈ I} there exists somefinite subset J of I such that

⋂j∈J(X \ Uj) = ∅ . Equivalently

⋃j∈J Uj = X . �

Definition 4.5. A subset A of a topological space X is compact if it is compact when endowed withthe subspace topology.

Remark 4.6. In view of the way in which the subspace topology is defined, a subset A of a topologicalspace X is compact if and only if for every family {Ui : i ∈ I} of open sets in X such that A ⊆

⋃i∈I Ui

there exists a finite subset J ⊆ I such that A ⊆⋃j∈J Uj .

This can be rephrased in terms of open covers of A , as follows.

Definition 4.7. A family {Ui : i ∈ I} of subsets of a space X is called a cover for A if A ⊆⋃i∈I

Ui.

We also sometimes say that the sets Ui, i ∈ I , cover the set A .

Thus, a subset A of X is compact if and only if every family of open sets in X which form a coverfor A have a finite subcover.

Examples 4.8. (1) Any finite space X is compact.


(2) If (xn)n∈N is a sequence converging to x in a topological space then the set

L = {xn | n ∈ N} ∪ {x}is compact.

Indeed consider {Ui : i ∈ I} , an open cover of L . There exists k ∈ I such that x ∈ Uk .

The sequence (xn) converges to x therefore there exists N such that xn ∈ Uk for everyn ≥ N . For every n ∈ {1, ..., N} , xn is contained in some Uin , in ∈ I . Therefore L iscontained in Ui1 ∪ · · · ∪ UiN ∪ Uk .

(3) Any (non-empty) set with the cofinite topology is compact. Likewise for the indiscrete topology.Thus any set X can be endowed with a topology T such that (X, T ) is compact.

The statement for the indiscrete topology follows from the fact that there are only twoopen sets in that topology.

The statement for the co-finite topology is easier to check with Proposition 4.4, (2). Indeedconsider a family {Fi : i ∈ I} of closed sets in X such that any intersection of finitely manyof those sets is non-empty.

Pick k ∈ I and consider Fk = {x1, ..., xm} . Assume that for every xj there exists Fij notcontaining it. Then Fk ∩ Fi1 ∩ · · · ∩ Fim is empty. This contradicts the hypothesis.

Therefore some xj is contained in all Fi with i ∈ I . It follows that⋂i∈I Fi contains xj .

The following was proved in the Michaelmas term Analysis course.

Proposition 4.9. (the Heine-Borel Theorem) Any interval [a, b] in R is compact.

Remark 4.10. An open interval (a, b), where a < b , is not compact in R (with the Euclidean topology).Nor are the intervals [a,∞) or (−∞, b] .

Indeed (a, b) =⋃n∈N

(a+ 1

n , b−1n

). Any finite subfamily

(a+ 1

n1, b− 1

n1

), ......,

(a+ 1

nk, b− 1

nk

)(for which we may assume that n1 < ... < nk , otherwise we change the order) covers only

(a+ 1

nk, b− 1

nk

).

A similar argument can be made for [a,∞) =⋃n∈N[a, n) and for (−∞, b] =

⋃n∈N(−n, b] .

Proposition 4.11. Any closed subset A of a compact space X is compact.

Proof. Indeed let {Vi : i ∈ I} be an indexed family of closed subsets of A such that⋂j∈J

Vj 6= ∅ for

any finite subset J ⊆ I .

As A is closed in X , Vi are also closed in X , which is compact. Therefore⋂i∈I

Vi 6= ∅. �

Remark 4.12. The converse of Proposition 4.11 may not be true: a singleton in X is always compactbut may not be closed.

As the example suggest, a separation axiom must be added in order to have the converse ofProposition 4.11.

Proposition 4.13. Let X be a Hausdorff space.

If K is a compact subset of X and x ∈ X \K then there exists U, V disjoint open sets such thatK ⊆ U and x ∈ V .


Proof. For every y ∈ K , according to the Hausdorff property, there exists a pair of disjoint open setsUy and Vy , y ∈ Uy , x ∈ Vy .

The collection {Uy : y ∈ K} is an open cover for K . Therefore there exist y1, ..., ym such thatK ⊆ Uy1 ∪ · · · ∪ Uym . Let U = Uy1 ∪ · · · ∪ Uym .

The set V = Vy1 ∩ · · · ∩ Vym is open, it contains x , and V ∩ U = V ∩⋃mj=1 Uyj = ∅ . �

Corollary 4.14. Any compact subset K of a Hausdorff space X is closed in X .

Proof. According to Proposition 4.13, for every x ∈ X \K , there exists V open such that x ∈ V ⊆X \K . This implies that X \K is open, by Lemma 2.21. �

Proposition 4.15. Let X be a topological space.

(1) If K1, ...,Kn are compact subsets in X then K1 ∪ · · · ∪Kn is compact in X .

(2) If {Ki : i ∈ I} is a non-empty family of compact subsets and X is Hausdorff then⋂i∈I Ki

is compact.

Proof. (1) Let {Ui : i ∈ I} be an open cover of K1 ∪ · · · ∪Kn . In particular for every t ∈ {1, ..., n} ,Kt ⊆

⋃i∈I Ui . As Kt is compact, there exists a finite subset Jt of I such that Kt ⊆

⋃j∈Jt Uj .

It follows that K1 ∪ · · · ∪Kn ⊆⋃nt=1

⋃j∈Jt Uj .

(2) Since X is Hausdorff, every Ki is a closed set according to Corollary 4.14. ThereforeL =

⋂i∈I Ki is also closed. It is also contained in some (each) compact space Ki , hence by Proposition

4.11, L is compact. �

Remark 4.16. (1) Property (1) is no longer true for infinite unions.

Example:⋃n∈N[0, n] = [0,∞).

(2) Property (2) is no longer true if X is not Hausdorff (see Ex. 5, (2), Sheet 4).

Proposition 4.17. A compact subset A of a metric space X is bounded.

Proof. Let x be a point in X . We have that A ⊆ X =⋃n∈NB(x, n).

It follows that there exist n1 < ... < nk such that A ⊆ B(x, n1) ∪ · · · ∪B(x, nk) = B(x, nk). �

Corollary 4.18. A compact subset of a metric space Y is bounded and closed in Y .


4.2. Compact spaces and continuous maps.

Proposition 4.19. If f : X → Y is a continuous map of topological spaces and A ⊂ X is compactthen so is f(A) . (‘The continuous image of a compact set is compact.’)

Proof. Let {Ui : i ∈ I} be an open cover of f(A).

Then {f−1(Ui) : i ∈ I} is an open cover of A .

The compactness of A implies that there exists a finite subcover {f−1(Uj) : j ∈ J} , J ⊆ I .

It follows that {Uj : j ∈ J} is a finite subcover for f(A). �

Corollary 4.20. Suppose that X is a compact space, Y is a metric space and f : X → Y iscontinuous.

Then f(X) is bounded and closed in Y . (When f(X) is bounded we say ‘f is bounded’.)

Corollary 4.21 (the extreme value theorem). If f : X → R is a continuous real-valued function ona compact space X then f is bounded and attains its bounds, i.e. there exist m,M ∈ X such thatf(m) ≤ f(x) ≤ f(M) for every x ∈ X .

Proof. Since X is compact, f(X) is compact, in particular it is bounded and closed.

Let a,A be the highest lower bound, respectively the least upper bound of f(X).

Since f(X) is closed in R , it follows that both a and A are in f(X). �

Remark 4.22. Another consequence of Proposition 4.19 is the following: if A ⊂ X ⊂ Y , where Y is atopological space and X is endowed with the subspace topology, then A compact in X ⇒ A compactin Y . This follows by using Proposition 4.19 for the inclusion map i : X → Y .

The same implication does not hold when ‘compact’ is replaced by ‘closed’: (0, 1] is closed in (0, 2)but not in R .

The converse implication holds both for ‘closed’ and for ‘compact’, i.e. A compact (closed) inY ⇒ A compact (closed) in X .

Theorem 4.23. The product X × Y is compact if and only if both X and Y are compact.

Proof. If the product X × Y is compact then X = pX(X × Y ) and Y = pY (X × Y ) are compact,due to Proposition 4.19.

Conversely, assume that X and Y are compact.

Let {Wi : i ∈ I} be an open cover of X × Y .

Since the rectangular open sets U × V compose a basis for the product topology, we may write

X × Y =⋃i∈I

Wi =⋃j∈J

Uj × Vj ,

where each Uj × Vj is contained in some Wi .

Therefore it is enough if we prove that the open cover {Uj × Vj : j ∈ J} has a finite subcover.

For every y ∈ Y , the compact set X × {y} is covered by {Uj × Vj : j ∈ J} . Therefore there existsa finite subset Fy ⊆ J such that

X × {y} ⊆⋃j∈Fy

Uj × Vj .


The set Vy =⋂j∈Fy Vj is an open set containing y . The family {Vy : y ∈ Y } is an open cover of

Y , which is compact.

It follows that there exist y1, ..., ym such that Y = Vy1 ∪ · · · ∪ Vym .

We state that X × Y =⋃mk=1

⋃j∈Fyk

Uj × Vj . Indeed consider an arbitrary element (x, y) in the

product. Then there exists k ∈ {1, ...,m} such that y ∈ Vyk . In particular y ∈ Vj for all j ∈ Fyk .On the other hand X × {yk} ⊆

⋃j∈Fyk

Uj × Vj , whence there exists j0 ∈ Fyk such that x ∈ Uj0 . It

follows that (x, y) ∈ Uj0 × Vj0 . �

Theorem 4.24. (general Heine-Borel) Any closed bounded subset of (Rn , ‖ ‖i) , where i ∈{1, 2,∞} , is compact.

Proof. Let A be a closed bounded subset of Rn . Since A is bounded, there exists x = (x1, ..., xn)and r > 0 such that A ⊆ Bd∞(x, r).

The ball Bd∞(x, r) in (Rn , ‖ ‖∞) can be rewritten as

Bd∞(x, r) = (x1 − r, x1 + r)× · · · × (xn − r, xn + r) ⊆ [x1 − r, x1 + r]× · · · × [xn − r, xn + r] .

Each interval [xi−r, xi+r] is compact by Proposition 4.9, and their product is compact by Theorem4.23.

It follows that A is a closed subset contained in a compact space, therefore it is compact byProposition 4.11. �

N.B. Theorem 4.24 fails in general when (Rn , ‖ ‖i) is replaced by an arbitrary metricspace (X,d) :

• an easy example of this is (0, 1] - it’s bounded, and closed in (0, 2), but it’s not compact;

• another such example, in which the ambient space is a normed vector space, moreover complete,will be given in Exercise 2, Sheet 4.

The inverse of a continuous map is not necessarily continuous, as Examples 1.35 shows. Still, thestatement is true under some extra conditions.

Theorem 4.25. Suppose that X is a compact space, that Y is a Hausdorff space and that f : X → Yis a continuous bijection. Then f is a homeomorphism (i.e. f−1 is continuous too).

Proof. Let V be a closed subset of X . Then V is compact, by Proposition 4.11. It follows that f(V )is compact, by Proposition 4.19. Since Y is Hausdorff, this implies that f(V ) is closed, according toCorollary 4.14. �

Corollary 4.26. (a) Let X be a space, and let TH ⊆ TK be two topologies such that (X, TH) isHausdorff and (X, TK) compact. Then TH = TK .

(b) Let (X, T ) be a Hausdorff compact space. Then any strictly coarser topology T ′ ( T is nolonger Hausdorff, and any strictly finer topology T ′′ ) T is no longer compact.

Proof. (a) We apply Theorem 4.25 to the map id : (X, TK)→ (X, TH).

(b) is a reformulation of (a). �

Quirky application 4.27. No topology on [0, 1] which is strictly coarser than the Euclidean topologycan be Hausdorff. No topology which is strictly finer than the Euclidean topology can be compact.


Corollary 4.28. Suppose that X is a compact space, that Y is a Hausdorff space and that f : X → Yis injective and continuous.

Then f determines a homeomorphism of X onto f(X) .

Remark 4.29. This generalises the Mods result: if f : [0, 1] → R is continuous and monotonic thenits inverse function f−1 : f([0, 1])→ [0, 1] is continuous.

An important property of continuous maps defined on compact metric spaces is the following.

Theorem 4.30. If f : X → Y is a continuous map of metric spaces and X is compact then f isuniformly continuous.

Proof. Let ε be an arbitrary positive number (which we consider as fixed in the sequel).

Let x be an arbitrary point in X . The map f is continuous at x . Therefore, for the given ε > 0there exists δx such that

(7) d(y, x) < δx ⇒ d(f(y), f(x)) <ε

2.

We have that X =⋃x∈X B

(x, δx2

). The space X being compact, there exist x1, ..., xn such that

X =⋃ni=1B

(xi,

δxi2

).

Let δ = min{δx12 , ....,

δxn2

}. We shall prove that d(y, z) < δ ⇒ d(f(y), f(z)) < ε .

Indeed, let y, z be two points such that d(y, z) < δ . As y ∈⋃ni=1B

(xi,

δxi2

), there exists

k ∈ {1, ..., n} such that y ∈ B(xk,

δxk2

), that is d(y, xk) <

δxk2 .

Then d(z, xk) ≤ d(z, y) + d(y, xk) < δ +δxk2 < δxk .

It follows according to (7) that d(f(y), f(xk)) < ε2 and d(f(z), f(xk)) < ε

2 , whence

d(f(y), f(z)) ≤ d(f(y), f(xk)) + d(f(z), f(xk)) < ε .

�

4.3. Sequential compactness. Loose Remark. We already saw that in a metric space manytopological properties can be characterised in terms of sequences.

In what follows we show that one of these properties is compactness.

Definition 4.31. A topological space X is sequentially compact if every sequence in X has a subse-quence converging to a point in X .

A (non-empty) subset A of a topological space X is sequentially compact if, with the subspacetopology, A is sequentially compact. (Conventionally, the empty set is sequentially compact.)

Theorem 4.32. (1) (Bolzano-Weierstrass theorem) In a compact topological space X everyinfinite subset has accumulation points.

(2) Every compact metric space is sequentially compact.


Proof. (1) We prove that if, for a subset A of X , the set of accumulation points A′ is empty then Ais finite.

Recall that the closure A is equal to A ∪A′ . Therefore if A′ is empty then A = A = A \A′ .The set A\A′ is the set of isolated points of A , hence all points of A are isolated: for every a ∈ A

there exists Ua open set such that Ua ∩A = {a} .

We have that A = A ⊆⋃a∈A Ua . The set A is a closed set in a compact metric space, therefore

it is compact by Proposition 4.11. The Borel-Lebesgue property of compact sets implies that thereexist a1, ..., an in A such that A = A ⊆

⋃ni=1 Uai .

We may then write A =⋃ni=1 (Uai ∩A) =

⋃ni=1{ai} = {a1, ..., an} .

(2) Consider (xn) a sequence in a compact metric space X . Let A = {xn : n ∈ N} .

Assume that A is finite, A = {a1, ..., ak} . For every ai we consider the set Ni = {n ∈ N : xn = ai} .Since N = N1 t N2 t · · · t Nk , at least one of the sets Ni is infinite. It follows that (xn) has a sub-sequence which is constant, therefore convergent.

Assume that A is infinite. Then according to (1), A′ 6= ∅ . Let a be a point in A′ . For every openset U containing a , (U \ {a}) ∩A 6= ∅ .

In particular there exists

• a term xn1 of the sequence contained in [B(a, 1) \ {a}] ∩A ,

• a term xn2 in[B(a, 12

)\ {a, x1, ..., xn1}

]∩A , etc.

• a term xnk in[B(a, 1k

)\ {a, x1, ..., xnk−1

}]∩A .

In other words, we construct inductively a subsequence of (xn) converging to a . �

Theorem 4.33. Any compact metric space is complete.

Proof. Let (xn) be a Cauchy sequence in a compact metric space (X,d).

According to Theorem 4.32, (2), (xn) has a subsequence (xϕ(n)) convergent to a point a , whereϕ : N→ N is a one-to-one increasing map. We prove that (xn) converges to a .

Consider an arbitrary ε > 0. The sequence (xn) is Cauchy, therefore there exists N such that forevery n,m ≥ N , d(xn, xm) < ε

2 .

The subsequence (xϕ(n)) converges to a , therefore there exists M such that for every n ≥ M ,

d(xϕ(n) , a

)< ε

2 .

Let k ≥M large enough so that ϕ(k) ≥ N . Then for every n ≥ N ,

d(xn, a) ≤ d(xn , xϕ(k)

)+ d

(xϕ(k) , a

)<ε

2+ε

2= ε .

�

Theorem 4.34. Any sequentially compact metric space is compact.

N.B. This and Theorem 4.32, (2), imply that a metric space is compact if and only if it issequentially compact.

The proof is done in two steps, which we formulate as two separate Propositions.

Proposition 4.35. Let X be a sequentially compact metric space. For any open cover U of X thereexists ε > 0 such that for every x ∈ X, B(x, ε) is entirely contained in some single set from U .


Definition 4.36. The number ε in Proposition 4.35 is called a Lebesgue number for the cover U .

Proof. Assume that for every n ∈ N there exists a point xn ∈ X such that the ball B(xn ,

1n

)is

contained in no subset U ∈ U .

The sequence (xn) has a subsequence (xϕ(n)) converging to some point a . There exists a subsetU in the cover U containing a . As U is open, it also contains a ball B(a, 2ε), for ε > 0.

The subsequence (xϕ(n)) converges to a , hence there exists N such that if n ≥ N then xϕ(n) ∈B(a, ε). This implies that B

(xϕ(n), ε

)⊆ B(a, 2ε) ⊆ U .

For n large enough 1ϕ(n) < ε , whence B

(xϕ(n),

1ϕ(n)

)⊆ B

(xϕ(n), ε

)⊆ U .

The last inclusion contradicts the choice of the sequence (xn). �

Definition 4.37. Given a real number ε > 0, an ε -net for a metric space X is a subset N ⊆ Xsuch that {B(x, ε) : x ∈ N} is a cover of X .

Example 4.38. For every ε >√n2 , Zn is an ε -net for Rn with the Euclidean norm ‖ ‖2 .

Proposition 4.39. Let X be a sequentially compact metric space and let ε be an arbitrary positivenumber.

Then there exists a finite ε-net for X .

Remark 4.40. Thus compact metric spaces can be approximated by finite sets.

This illustrates once more our remark that from the topological point of view compact sets are ‘thenext best thing, after singletons and finite sets’.

Proof. Assume that there exists no finite ε -net for X .

Let x1 be a point in X . According to the assumption there exists x2 ∈ X \B(x1, ε).

We argue by induction and assume that there exist x1, ..., xn points in X such that d(xi, xj) ≥ εfor every i 6= j in {1, 2, ..., n} .

Since {x1, ..., xn} is finite, it cannot be a ε -net. Therefore there exists a point xn+1 in X \⋃ni=1B(xi, ε).

In this manner we construct a sequence (xn) such that

(8) d(xi, xj) ≥ ε ∀i 6= j, i, j ∈ N .

The space X is sequentially compact, therefore (xn) has a convergent subsequence (xϕ(n)), whereϕ : N→ N is a strictly increasing function.

In particular (xϕ(n)) is a Cauchy sequence. It follows that there exists N such that if m ≥ n ≥ N ,d(xϕ(n), xϕ(m)) < ε . This contradicts (8). �

We now finish the proof of the fact that a sequentially compact metric space is compact.

Indeed, let U be an open cover of X and let ε be the Lebesgue number for the cover U providedby Proposition 4.35.

Proposition 4.39 implies that there exists a finite ε -net, {x1, ..., xn} .

By the definition of the Lebesgue number, each ball B(xi, ε) is contained in some set Ui ∈ U .

Thus X =⋃ni=1B(xi, ε) ⊆

⋃ni=1 Ui . �


5. Quotient spaces, quotient topology

In many cases, a space X on which an equivalent relation R is defined is a topological space (e.g.the group R and the equivalence relation defined by a subgroup, say Z or Q).

Natural questions to ask are:

• whether the quotient space X/R is a topological space too;

• which of the topological properties of X are inherited by X/R .

We shall see that a quotient space always has a natural topology, but that this topology may nothave good properties - not even the basic separation properties, even though the ambient space X isa very good space, e.g. R .

5.1. Definition and examples. First we recall some basic notions on equivalence relations andpartitions.

Let X be a space. Recall that an equivalence relation R on X is a relation that is

• reflexive: xRx for every x ∈ X ;

• symmetric: xRy ⇒ yRx ;

• transitive: if xRy and yRz then xRz .

The graph of the equivalence relation R is the set

(9) GR = {(x, y) ∈ X ×X : xRy} .

The equivalence class of an element x ∈ X is the set

[x] = {y ∈ X : yRx} .

The set of equivalence classes is denoted by X/R , and it is called the quotient space of X withrespect to R .

Example 5.1. (from Mods)

(a) Let (G, ∗) be a group and H a subgroup of it.

The relation xRy ⇔ x−1 ∗ y ∈ H is an equivalence relation.

An equivalence class is a left coset gH , g ∈ G .

The quotient space is G/H .

(b) For instance we may consider the group (R,+) and the subgroup Z . Or the subgroup Q .

Definition 5.2. A cover {Ui : i ∈ I} of a space X is called a partition if

• Ui 6= ∅ for every i ∈ I ;

• Ui ∩ Uj = ∅ for every i 6= j , i, j ∈ I .

Remark 5.3. (from Mods) Defining an equivalence relation on a space X is equivalent to defininga partition of X .


Indeed, given an equivalence relation on a space X , the equivalence classes compose a partitionof X .

Conversely, given a partition {Ui : i ∈ I} of X , one can define an equivalence relation by

xRy ⇔ ∃ i ∈ I such that {x, y} ⊆ Ui .

Let p : X → X/R be the map that assigns to each point x in X the equivalence class [x] (the setin the partition containing x).

Proposition 5.4. Let (X, T ) be a topological space, and R an equivalence relation on X .

The family T of subsets U in X/R such that p−1(U) ∈ T is a topology for X/R , called thequotient topology.

Proof. (T1) p−1(∅) = ∅ and p−1 (X/R) = X are both in T .

(T2) and (T3) follow from the fact that

p−1(U ∩ V

)= p−1

(U)∩ p−1

(V)

and that

p−1

(⋃i∈I

Ui

)=⋃i∈I

p−1(Ui) .

�

Proposition 5.5. A subset V in X/R endowed with the quotient topology is closed if and only if

p−1(V ) is closed in X .

Proof. This follows from the fact that

p−1(

(X/R) \ V)

= X \ p−1(V).

�

Example 5.6. Consider the quotient space R/Q defined in Example 5.1, (b).

Let U be an open non-empty set in R/Q . Then p−1(U) is open and non-empty in R .

For every x ∈ R , the set −x+ p−1(U) is open and non-empty in R . Since Q is dense in R , thereexists q ∈ (−x+ p−1(U)) ∩Q .

So, x + q ∈ p−1(U). The set p−1(U) contains any equivalence class that it intersects, thereforex+ Q ⊆ p−1(U)⇔ p(x) ∈ U .

We have thus proved that for every x ∈ R , p(x) ∈ U . It follows that U = R/Q .

Thus the quotient topology on R/Q is the indiscrete topology.


5.2. Separation axioms.

Proposition 5.7. A quotient topological space satisfies the first separation axiom if and only if everyequivalence class is closed.

Proof. The first separation axiom is equivalent to the statement that every singleton {[x]} in X/Ris closed.

According to Proposition 5.5, the latter is equivalent to the fact that the inverse image of {[x]} byp , which is [x] seen as a subset of X , is closed. �

In order to discuss the Hausdorff property we need to define a particular type of sets.

Lemma 5.8. Let R be an equivalence relation on a space X , let p : X → X/R be the map x 7→ [x]and let A be a subset in X . The following are equivalent:

(1) if A intersects an equivalence class [x] (i.e. A ∩ [x] 6= ∅) then A contains [x] ;

(2) whenever x ∈ A and yRx the element y is also in A ;

(3) A = p−1(p(A)) .

The proof of the equivalence is easy and left as an exercise.

Definition 5.9. A subset A of X is saturated with respect to the equivalence relation R if it satisfiesone of the equivalent conditions in Lemma 5.8.

Remark 5.10. If A is a saturated subset of X with respect to R then its complement X \ A is alsosaturated with respect to R .

Proof. Let x ∈ X \A and let yRx . If y ∈ A then x ∈ A , contradiction.

Therefore y ∈ X \A , hence X \A is saturated. �

Example 5.11. For every subset B in X/R , p−1(B) is saturated.

Example 5.12. (a) A map f : X → Y defines the equivalence relation xRfy ⇔ f(x) = f(y).The equivalence classes are f−1(y) , y ∈ f(X).

For instance, given the map f : R→ C , f(t) = e2iπt = cos(2πt)+i sin(2πt) the equivalencerelation Rf is the same as that in Example 5.1, (b), defined by the subgroup Z of R .

(b) We may say that a set A is saturated with respect to f or f -saturated if it is saturated withrespect to Rf .

This is equivalent to the fact that A = f−1(f(A)).

Proposition 5.13. A quotient topological space is Hausdorff if and only if any two distinct equivalenceclasses are contained in two disjoint open saturated sets.

Proof. The proof is Exercise 4, Sheet 4. �

Examples 5.14. (1) Let S ={

1n : n ∈ N

}. Consider the partition of R composed of S and all

the singletons {x} with x ∈ R \ S , and the corresponding equivalence relation R .

The quotient space X/R does not satisfy the first separation axiom.


(2) Let X = [0, 1]× [0, 1], and let ℘ be the partition composed of

{(0, y), (1, 1− y)} with y ∈ [0, 1], {(x, y)} with x ∈ (0, 1) .

The quotient X/℘ is called the Mobius band; it is Hausdorff.

Indeed, consider two distinct equivalent classes {(0, y), (1, 1 − y)} and {(a, b)} with a ∈(0, 1).

For ε > 0 and δ > 0 small enough, the open saturated sets

[B((0, y), ε) ∪B((1, 1− y), ε)] ∩X and B((a, b), δ) ⊆ Xare disjoint.

Similar arguments work for any two distinct equivalent classes in X/℘ .

Proposition 5.15. (1) If X/R is Hausdorff then the graph of R is closed.

(2) If the graph of R is closed and for every U open in X , p(U) is open then X/R is Hausdorff.

Proof. (1) Let (x, y) ∈ X ×X be a point outside the graph GR . It follows that [x] 6= [y] , hence there

exist U , V open disjoint subsets of X/R such that [x] ∈ U and [y] ∈ V . Then p−1(U) and p−1(V )are open and contain x , respectively y .

Assume that p−1(U)× p−1(V ) intersects GR . Then there exist a ∈ p−1(U) and b ∈ p−1(V ) such

that aRb . It follows that [a] = [b] is in U ∩ V , contradiction.

We have thus found (x, y) ∈ p−1(U)× p−1(V ) ⊆ (X ×X) \ GR .

(2) Consider two distinct points [x] 6= [y] in X/R .

The point (x, y) is in (X ×X) \ GR which is open, therefore there exist U, V open in X such that

(10) (x, y) ∈ U × V ⊆ (X ×X) \ GR .

Then [x] ∈ p(U) and [y] ∈ p(V ), p(U), p(V ) open.

If p(U) ∩ p(V ) contains some element [z] then there exists u ∈ U and v ∈ V such that uRz andvRz . By transitivity uRv , therefore (u, v) ∈ (U × V ) ∩ GR . This contradicts the inclusion in (10).

Thus p(U) ∩ p(V ) = ∅ . We have proved that X/R is Hausdorff. �

Remark 5.16. The converse of Proposition 5.15, (2), is not true.

Indeed, consider the Mobius band X/℘ , a Hausdorff quotient.

The graph of the equivalence relation corresponding to the partition ℘ is closed, as it is composedof the diagonal

∆ = {(v,v) ∈ X ×X : v ∈ X}and of the two graphs

{((0, x), (1, 1− x)) : x ∈ [0, 1]} and {((1, x), (0, 1− x)) : x ∈ [0, 1]} .

But it is not true that for every U open in X , p(U) is open in X/℘ (equivalently, p−1(p(U)) isopen in X ).

Indeed for U = B((0, 1/2), 1/4) ∩X , p−1(p(U)) is

p−1(p(U)) = [B((0, 1/2), 1/4) ∩X] ∪ [{1} × (1/4, 3/4)] .

The set p−1(p(U)) is not open, because it does not contain an open ball centred in (1, 1/2).


Examples 5.17. R/Z is Hausdorff.

Indeed, the graph of the equivalence relation is the union of parallel lines⋃n∈Z{(x, y) : y = x+ n} ,

and it is closed.

For any open set U in R , p−1(p(U)) =⋃n∈Z(n+U) which is open as union of open sets. It follows

that p(U) is open in R/Z .

Proposition 5.15, (2), implies that R/Z is Hausdorff.

A similar argument shows that Rn/Zn is Hausdorff.

Definition 5.18. The real n-dimensional projective space PRn is the quotient space of Rn+1 \{0} with respect to the equivalence relation R where xRy if and only if there exists λ 6= 0 such thatx = λy .

In other words PRn is the set of lines in Rn+1 through 0.

Example 5.19. For any n ≥ 1, PRn is Hausdorff (Ex. 6, Sheet 4).

5.3. Quotient spaces and continuous maps.

Proposition 5.20 (a second way of defining the quotient topology). Let p : X → X/R be x 7→ [x] .Then quotient topology on X/R is the largest topology on X/R making p continuous.

Proof. Let T ′ be a topology on X/R such that p : X →(X/R , T ′

)is continuous. Then for every

U ∈ T ′ , p−1(U) is in T .

It follows that U ∈ T . We have thus proved that T ′ ⊆ T . �

Proposition 5.21. A map g from X/R to another space Z is continuous if and only if g ◦ p iscontinuous.

Proof. If g is continuous then g ◦ p is continuous.

Conversely, assume that g ◦ p is continuous. For every open subset V of Z , (g ◦ p)−1(V ) =p−1

(g−1 (V )

)is open in X .

According to the definition of the quotient topology, this implies that g−1 (V ) ∈ T .

We conclude that g is continuous. �

We now define the notion of a quotient map, which is useful when trying to prove that aquotient space is homeomorphic to another topological space.

Definition 5.22. A map p : X → Y between topological spaces is a quotient map if

(1) p is surjective, and(2) for every U ⊆ Y , U is open if and only if p−1(U) is open.

The following gives an alternative way of viewing quotient maps.

Proposition 5.23. Let p : X → Y be a map between topological spaces. The following conditions areequivalent:

(1) p is a quotient map;


(2) p is continuous and surjective, and the image under p of every p-saturated open set in X isopen in Y .

Proof. The implication (1) ⇒ (2) follows from the fact that a p-saturated set A satisfies p−1(p(A)) =A .

Indeed if A is an open saturated set then p−1(p(A)) is also an open set, therefore by (1), p(A) isan open set.

We prove the implication (2) ⇒ (1).Assume that p−1(U) is open. It is also p -saturated therefore according to (2) p

(p−1(U)

)= U is

open. In the last equality we used the fact that p is surjective. �

There is an important relationship between quotient maps and quotient spaces, as explained by thefollowing proposition.

Proposition 5.24. (1) Let R be an equivalence relation on a space X , endow X/R with thequotient topology, and let p : X → X/R be the map sending each point of X to its equivalenceclass. Then p is a quotient map.

(2) Suppose that p : X → Y is a quotient map and that R is the equivalence relation on Xcorresponding to the partition {p−1(y) : y ∈ Y } . Then X/R and Y are homeomorphic.

Proof. (1) The map p is a quotient map due to the way in which the quotient topology is defined.

(2) Consider the map p : X/R → Y , p(p−1(y)

)= y . It is surjective because p is surjective, and

it is injective by construction.

If π : X → X/R is the standard quotient map, then by the definition of p we have that p ◦ π = p .In particular, as p is continuous, p is continuous according to Proposition 5.21.

Let U be an open set in X/R . Then π−1(U) is an open p -saturated set in X . According toProposition 5.23, (2), p(π−1(U)) is open.

The relation p = p ◦ π implies that p(π−1(U)) = p(U), therefore p(U) is open.

We have thus proved that p is a continuous bijection, with continuous inverse, therefore a homeo-morphism. �

Examples 5.25. (1) R/Z is homeomorphic to the planar unit circle S1 .

Consider the map f : R→ S1 , f(t) = e2iπt = cos(2πt) + i sin(2πt) .

It defines the same equivalence relation as the subgroup Z .

The map f is surjective and continuous.

We now show that, for every open set U in R , f(U) is open in S1 . This will imply, inparticular, that f is a quotient map.

For every x ∈ U , there exists ε > 0 such that (x − ε, x + ε) ⊆ U . We may pick ε < 12 .

Hence the image f(x− ε, x+ ε) is the intersection of an open half-plane with the unit circleS1 and therefore f(x − ε, x + ε) is open. So, for every point in f(U), there is an open setcontaining that point, lying within f(U). Hence, f(U) is open.

Proposition 5.24, (2), implies that R/Z is homeomorphic to S1 .


(2) Rn/Zn is homeomorphic to S1 × · · · × S1︸︷︷︸n times

. This space is denoted by Tn and it is called the

n-dimensional torus. Note that it is a compact space.

To prove the homeomorphism it suffices to consider the quotient map

F (t1, .., tn) =(e2iπt1 , ..., e2iπtn

)and argue as in (1).

In many cases, there is an easier way to establish that a space is a quotient space, using the followingproposition.

Proposition 5.26. Let f : X → Y be a surjective continuous map between topological spaces. Let Rbe the equivalence relation on X defined by the partitition {f−1(y) : y ∈ Y } . If X is compact and Yis Hausdorff, then X/R and Y are homeomorphic.

Proof. Define f : X/R → Y by [x] 7→ f(x). This is well-defined by the definition of R . Now, f ◦p = fwhere p : X → X/R is x 7→ [x] . So, Proposition 5.21, f is continuous. It is injective by the definitionof R , and it is surjective because f is. Since X is compact, so is X/R , by Proposition 4.19. So,f : X/R → Y is a continuous bijection from a compact space to a Hausdorff space. By Theorem 4.25,f is a homeomorphism. �

Examples 5.27. (1) Let X = [0, 1]× [0, 1], and let ℘ be the partition composed of:

• {(x, 0), (x, 1)}, with x arbitrary number in the interval (0, 1),

• {(0, y), (1, y)}, with y arbitrary in the interval (0, 1),

• {(0, 0), (1, 0), (0, 1), (1, 1)}• and {(x, y)} with x, y two arbitrary numbers in the interval (0, 1).

The quotient space [0, 1]× [0, 1]/℘ is homeomorphic to the 2-dimensional torus.

Let f : [0, 1]× [0, 1]→ T2 be (t1, t2) 7→(e2iπt1 , e2iπt2

). The partition {f−1(y) : y ∈ T2} is

precisely ℘ . Hence, by Proposition 5.26, [0, 1]× [0, 1]/℘ is homeomorphic to T2 .

(2) Consider the closed planar unit disk

D2 = {(x, y) ∈ R2 : x2 + y2 ≤ 1} .

Let ℘ be the partition composed of:

• all the singletons in the open disk {(x, y)} with x2 + y2 < 1

• and the boundary circle S1 .

The quotient space D2/℘ is homeomorphic to the 2-dimensional sphere

S2 = {(x, y, z) : x2 + y2 + z2 = 1} .

This is proved by applying Proposition 5.26 to the map f : D2 → S2 , f(0, 0) = (0, 0, 1),and for (x, y) 6= (0, 0)

f(x, y) =

(x√

x2 + y2sin(π√x2 + y2

),

y√x2 + y2

sin(π√x2 + y2

), cos

(π√x2 + y2

)).

Proposition 5.28. There is a (topological) embedding of the 2-dimensional torus T2 in R3 , i.e. T2

is homeomorphic to a subset of R3 with the subspace topology.


Proof. Consider the map p : R2 → R3 ,

p(x, y) = ((2 + cos(2πy)) cos(2πx) , (2 + cos(2πy)) sin(2πx) , sin(2πy)) .

One can easily check that p(x, y) = p(x′, y′) if and only if (x, y)− (x′, y′) ∈ Z2 .

Indeed (2 + cosα) cosβ = (2 + cosα′) cosβ′ and (2 + cosα) sinβ = (2 + cosα′) sinβ′ imply thatthe sums of the squares are equal therefore (2 + cosα)2 = (2 + cosα′)2 .

The latter is equivalent to |2 + cosα| = |2 + cosα′| , but since 2 + cosα ≥ 0 for every α we obtainthat cosα = cosα′ .

Therefore p(x, y) = p(x′, y′) implies that cos(2πy) = cos(2πy′) and sin(2πy) = sin(2πy′), whencey − y′ ∈ Z . Also cos(2πx) = cos(2πx′) and sin(2πx) = sin(2πx′), whence x− x′ ∈ Z .

The map p therefore induces a bijective map p : R2/Z2 → p(R2) defined by p((x, y) + Z2

)=

p(x, y). If we denote by π : R2 → R2/Z2 the standard quotient map then p ◦ π = p . Since p iscontinuous it follows by Proposition 5.21 that p is continuous.

The quotient R2/Z2 is compact and p(R2) is a subspace of R3 , therefore Hausdorff. Theorem 4.25implies that p is a homeomorphism.

�

Definition 5.29. The Klein bottle K is the quotient space of the rectangle [0, 2π]× [0, π] by theequivalence relation which identifies the points (x, 0), (2π−x, π) for each x ∈ [0, 2π] and the points(0, y), (2π, y) for each y ∈ [0, π].

Note that, since [0, 2π]× [0, π] is compact, the Klein bottle is compact.

Proposition 5.30. There is an embedding of the Klein bottle K in R4 .

Proof. This embedding is defined by the map g : [0, 2π]× [0, π]→ R4 ,

g(x, y) = ((2 + cosx) cos 2y , (2 + cosx) sin 2y , sinx cos y , sinx sin y) .

Indeed g is continuous.

Assume that g(x, y) = g(x′, y′).

Then the same calculation as in the proof of Proposition 5.28 implies that cosx = cosx′ .

It follows that cos 2y = cos 2y′ and sin 2y = sin 2y′ whence y = y′ modulo π .

If y = y′ ∈ (0, π) then sinx = sinx′ hence either x = x′ or x = 0, x′ = 2π or x = 2π, x′ = 0.

If y = 0 and y′ = π then sinx cos y = sinx′ cos y′ implies that sinx = − sinx′ . Together withcosx = cosx′ this implies x′ = 2π − x .

Thus we proved that the equivalence relation on [0, 2π]× [0, π] defined by g coincides to the onewith quotient space the Klein bottle.

Consider the continuous bijection g : K → g(K) ⊆ R4 defined by g .

The Klein bottle K is compact and g(K) is a subspace of R4 , therefore it is Hausdorff. Theorem4.25 implies that g is a homeomorphism. �

For the next proposition we use the following notation:

(a) The unit sphere in R3 is S2 = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1} ;

(b) The upper hemisphere in S2 is D+ = {(x, y, z) ∈ S2 : z > 0} ;

(c) The closed unit disc in R2 is D2 = {(x, y) ∈ R2 : x2 + y2 6 1} .


Proposition 5.31. The following are all homeomorphic to the projective plane PR2 .

(a) The quotient space S2/R where R identifies each pair of antipodal points of S .

(b) The quotient space D+/R where R identifies each pair of antipodal points on the boundaryof D+ .

(c) The quotient space D2/R where R identifies each pair of antipodal points on the boundary ofD .

(d) The quotient space of the square [0, 1] × [0, 1] by the equivalence relation which identifies(s, 0), (1− s, 1) for each s ∈ [0, 1] and (0, t), (1, 1− t) for each t ∈ [0, 1].

Remarks 5.32. (1) Proposition 5.31, (a), implies that the projective plane is compact.

(2) A proposition similar to Proposition 5.31 can be formulated for every projective space PRn ,n ≥ 2.

Proof. To prove (a) and (b) it suffices to restrict the quotient map p : R3 → PR2 to S2 , respectivelyD+ , and apply Proposition 5.24, (2).

(c) The stereographic projection π : R2 → S2 ,

π(x) =

(2x

1 + x2 + y2,

2y

1 + x2 + y2,

1− x2 − y2

1 + x2 + y2

),

restricted to the unit disk D2 defines a homeomorphism between D2 and the upper hemisphere.We apply Proposition 5.24, (2) to p|D+ ◦ π|D2 .

(d) follows from the fact that the unit disk D2 and the unit square [0, 1]× [0, 1] are homeomorphic.�

Proposition 5.33. There is an embedding of the real projective plane PR2 in R4 , i.e. PR2 ishomeomorphic to a subset of R4 .

Proof. See Exercise 7, Sheet 4. �

APPENDIX A: USEFUL IDENTITIES

Let X and Y be sets, {Ui}i∈I a set of subsets of X and {Vj}j∈J a set of subsets of Y .

(1) De Morgan laws

X \⋂i∈I

Ui =⋃i∈I

(X \ Ui)

X \⋃i∈I

Ui =⋂i∈I

(X \ Ui)

(2) Distributivity of⋂

over⋃

A⋂(⋃

i∈IUi

)=⋃i∈I

(A⋂Ui

)


(3) Images and inverse imagesLet f : X → Y be a map. Recall that for any subset A in Y its inverse image or pre-image

isf−1(A) = {x ∈ X ; f(x) ∈ A} .

Then

f(U) ⊆ V if and only if U ⊆ f−1(V )

f−1(Y \ V ) = X \ f−1(V )

f

(⋃i∈I

Ui

)=⋃i∈I

f(Ui)

f

(⋂i∈I

Ui

)⊆⋂i∈I

f(Ui)

f−1

⋃j∈J

Vj

=⋃j∈J

f−1 (Vj)

f−1

⋂j∈J

Vj

=⋂j∈J

f−1 (Vj)

If A ∩B = ∅ , A,B subsets of Y , then f−1(A) ∩ f−1(B) = ∅ .Indeed assume that there exists x ∈ f−1(A) ∩ f−1(B). Then f(x) ∈ A and f(x) ∈ B ,

which contradicts A ∩B = ∅ .

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