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METRIC AND TOPOLOGICAL SPACES

ALEX GONZALEZ

1. Introduction 32. Metric spaces: basic definitions 52.1. Normed real vector spaces 92.2. Product spaces 103. Open subsets 123.1. Equivalent metrics 133.2. Properties of open subsets and a bit of set theory 163.3. Convergence of sequences in metric spaces 234. Continuous functions between metric spaces 264.1. Homeomorphisms of metric spaces and open maps 325. Interlude 356. Connected spaces 386.1. Path-connected spaces 426.2. Connected components 447. Compact spaces 457.1. Continuity improved: uniform continuity 538. Complete spaces 548.1. Properties of complete spaces 588.2. The completion of a metric space 619. Interlude II 6610. Topological spaces 6810.1. Set theory revisited 7011. Dealing with topological spaces 7211.1. From metric spaces to topological spaces 7511.2. A very basic metric-topological dictionary 7812. What topological spaces can do that metric spaces cannot 8212.1. One-point compactification of topological spaces 8212.2. Quotient topological spaces 85REFERENCES 89

Contents

1

2 ALEX GONZALEZ

METRIC AND TOPOLOGICAL SPACES 3

1. Introduction

When we consider properties of a “reasonable” function, probably the first thing thatcomes to mind is that it exhibits continuity: the behavior of the function at a certainpoint is similar to the behavior of the function in a small neighborhood of the point.What’s more, the composition of two continuous functions is also continuous.

Usually, when we think of a continuous functions, the first examples that come tomind are maps f : R→ R:

• the identity function, f(x) = x for all x ∈ R;

• a constant function f(x) = k;

• polynomial functions, for instance f(x) = xn, for some n ∈ N;

• the exponential function g(x) = ex;

• trigonometric functions, for instance h(x) = cos(x).

The set of real numbers R is a natural choice of domain to begin to study moregeneral properties of continuous functions. After all, we are familiar with many ofthe properties of the real line, and it is (relatively) easy to draw the graphs of functionsR→ R.

So, let’s review the definition of continuity for a function f : R→ R: the function f iscontinuous at the point a ∈ R if

limx→a

f(x) = f(a)

(in particular we are assuming that this limit exists!). Another way of putting theabove definition is the following: for each ε > 0 there exists some δ > 0 such that

|f(x)− f(a)| < ε for all x ∈ R such that |x− a| < δ.

The function f is then said to be continuous on all R if it is continuous for all a ∈ R.So, basically, the definition of continuity depends only on the notion of the distancebetween two points, codified in the expressions |x− a|, |f(x)− f(a)|.

Now consider functions f : Rn → Rm. In this case, as you know, there is also a well-defined notion of continuity, which again depends only on the metrics that we havepreviously defined in Rn and Rm. It is not a big step to say that the same applies forfunctions C→ C.

As it turns out, all these situations, and many more, fit together under the nameof metric spaces: sets (like R, N, Rn, etc) on which we can measure the distancebetween two points. The first goal of this course is then to define metric spaces andcontinuous functions between metric spaces.

4 ALEX GONZALEZ

A note of waning! The same set can be given different ways of measuring distances.Strange as it may seem, the set R2 (the plane) is one of these sets. We will seedifferent metrics for R2 pretty soon.

You may have noticed that metrics also have something to do with the notion ofconvergence of sequences. In fact, whenever a space has a metric, we can talk aboutsequences and investigate their convergence. Let’s see an interesting example of this:fix an interval [a, b], and consider the setX of all differentiable functions f : [a, b]→ R.We can define a metric (whatever this means) on X as follows: given f, g ∈ X,

d(f, g) = maxx∈[a,b]

|f(x)− g(x)|.

With this metric on the function space (i.e. notion of distance between functions), wecan define Taylor approximation in terms of converging sequences of functions on X.

Returning to continuous functions on one variable, there are a few more importantproperties of them which only depend on the distance between points. For example.

• Intermediate Value Theorem (a.k.a. Bolzano’s Theorem): if f : [a, b] → R iscontinuous and y ∈ R is a number between f(a) and f(b), then there existsc ∈ [a, b] such that f(c) = y.

• If f : [a, b]→ R is continuous, then there exists someK ∈ R such that |f(x)| ≤K for all x ∈ [a, b].

In the above statements, the domain of f is the interval [a, b], which is closed andbounded as a subset of R, and this is all we need to prove this results. As youprobably know already, the above results have their own versions for continuousfunctions f : Rn → Rm, since there is well stablished notions of closed and boundedsubsets in Rn, for all n.

The second main goal of this course is to generalize the notions of closed andbounded, so that we can generalize these results for continuous maps between metricspaces.

And what about topology? What is topology anyway? It turns out that we donâĂŹteven need a metric to talk about continuity. In order to consider continuous functionsbetween any spaces, all we must do is make clear from the beginning what the openand closed subsets of the spaces are. Topology gives us the rules for doing this. Thisis an abstract, powerful generalization of metric spaces, so be careful!


2. Metric spaces: basic definitions

Let X be a set. Roughly speaking, a metric on the set X is just a rule to measure thedistance between any two elements of X.

Definition 2.1. A metric on the set X is a function d : X ×X → [0,∞) such that thefollowing conditions are satisfied for all x, y, z ∈ X:

(M1) Positive property: d(x, y) = 0 if and only if x = y;

(M2) Symmetry property: d(x, y) = d(y, x); and

(M3) Triangle inequality: d(x, y) ≤ d(x, z) + d(z, y).

Let’s see some examples of metric spaces.

Example 2.2. The setX = R with d(x, y) = |x−y|, the absolute value of the differencex− y, for each x, y ∈ R. Properties (M1) and (M2) are obvious, and

d(x, y) = |x− y| = |(x− z) + (z − y)| ≤ |x− z|+ |z − y| = d(x, z) + d(z, y).

Let’s see now some examples of metrics on the set X = R2.

Example 2.3. The set X = R2 with

d((x1, x2), (y1, y2)

)=√

(x1 − y1)2 + (x2 − y2)2

for each (x1, x2), (y1, y2) ∈ R2. Again, properties (M1) and (M2) are easy to check.Property (M3), the triangle inequality, gets its name from this example. Write x =(x1, x2),y = (y1, y2) and z = (z1, z2). Then,

x

z

yd(x, z)

d(x, y)

d(z, y)

Can you think of other examples of metrics on R2? If we compare the metrics inExamples 2.2 and 2.3, we can see that the metric on R2 is some sort of “extension”of the metric on R1 to a higher dimension. But we can try other ways of extending it.

Example 2.4. The function d, d′ : R2 × R2 → [0,∞) defined by

d((x1, x2), (y1, y2)

)= |x1 − y1|+ |x2 − y2|

d′((x1, x2), (y1, y2)

)= max|x1 − y1|, |x2 − y2|

6 ALEX GONZALEZ

are also metrics on R2 (the details will be checked in Examples 2.5 and 2.8 below).

We have just see three different metrics on R2. But why stopping at R2? We canextend these metrics to Rn for all n ≥ 1.

Example 2.5. Let X = Rn and let d : Rn × Rn → [0,∞) be defined by

d((x1, x2, . . . , xn), (y1, y2, . . . , yn)

)=

n∑i=1

|xi − yi|.

Then, d is a metric on Rn (sometimes known by the name of Manhattan metric).Let’s check the details: conditions (M1) and (M2) follow easily. As for the triangleinequality, we have

d((x1, . . . , xn),(y1, . . . , yn)

)=

n∑i=1

|xi − yi| =

=n∑i=1

|(xi − zi) + (zi − yi)| ≤

≤n∑i=1

|xi − zi|+ |zi − yi| =

= d((x1, . . . , xn), (z1, . . . , zn)

)+ d((z1, . . . , zn), (y1, . . . , yn)

).

Example 2.6. The Euclidean metric on Rn is defined by the formula

d((x1, x2, . . . , xn), (y1, y2, . . . , yn)

)=

√√√√ n∑i=1

(xi − yi)2,

for each (x1, x2, . . . , xn), (y1, y2, . . . , yn) ∈ Rn. As usual (M1) and (M2) are easy tocheck, but (M3) is not trivial at all.

Indeed, let x = (x1, x2, . . . , xn),y = (y1, y2, . . . , yn) and z = (z1, z2, . . . , zn) ∈ Rn: Letalso ri = xi − zi and si = zi − yi, for i = 1, . . . , n. We have to prove that

d(x,y) =

√√√√ n∑i=1

(ri + si)2 ≤

√√√√ n∑i=1

r2i +

√√√√ n∑i=1

s2i = d(x, z) + d(z,y) (1)

Notice that both sides of the inequality are positive. Thus, by squaring the above, itis equivalent to prove that

n∑i=1

(ri + si)2 ≤

n∑i=1

r2i +

n∑i=1

s2i + 2

√√√√ n∑i=1

r2i

√√√√ n∑i=1

s2i . (2)


The left part of the inequality expands ton∑i=1

(ri + si)2 =

n∑i=1

r2i +

n∑i=1

s2i + 2

n∑i=1

risi.

Replacing this in (2) and simplifying, we deduce that (1) holds if and only if( n∑i=1

risi

)2

≤( n∑

i=1

r2i

)( n∑i=1

s2i

).

The last inequality is the Cauchy-Schwartz inequality, which we prove below asLemma 2.7.

Lemma 2.7 (Cauchy-Schwartz inequality). For each (a1, . . . , an), (b1, . . . , bn) ∈ Rn,( n∑i=1

aibi

)2

≤( n∑

i=1

a2i

)( n∑i=1

b2i

).

Proof. Consider the expression∑n

i=1

∑nj=1(aibj − ajbi)2. By expanding the brackets,

n∑i=1

n∑j=1

(aibj−ajbi)2 =( n∑i=1

a2i

)( n∑j=1

b2j

)+( n∑j=1

a2j

)( n∑i=1

b2i

)−2( n∑i=1

aibi

)( n∑j=1

ajbj

).

Now, collect the terms (and reindex the sums) to get

1

2

n∑i=1

n∑j=1

(aibj − ajbi)2 =( n∑i=1

a2i

)( n∑i=1

b2i

)−( n∑i=1

aibi

)2

.

Since the left part of the equality is positive, this proves the statement.

Example 2.8. Finally, the box metric on Rn is defined by

d((x1, . . . , xn), (y1, . . . , yn)

)= max|xi − yi|

∣∣ i = 1, . . . , n

Properties (M1) and (M2) are easily seen to hold. Let’s check property (M3). Let x =(x1, . . . , xn),y = (y1, . . . , yn) and z = (z1, . . . , zn) ∈ Rn. Then, for each i = 1, . . . , n,

|xi − yi| = |(xi − zi) + (zi − yi)| ≤ |xi − zi|+ |zi − yi| ≤ d(x, z) + d(z,y).

Thus, the triangle inequality holds. As an exercise, consider the set R2 with thismetric. Fix then a point a ∈ R2 and draw the set

Da = x ∈ R2∣∣ d(a,x) ≤ 1.

All these examples should serve as a warning of how flexible the notion of metric is:we should not be surprised to see that statements that go against all intuition are,in fact, true. The following is a good examples of this: it is a process to define ametric on every set. This procedure is not rather descriptive, but it is nonethelessimportant.

8 ALEX GONZALEZ

Example 2.9. Let X be any set. The discrete metric on X is defined by

d(x,y) =

1, if x 6= y0, if x = y

Properties (M1) and (M2) are obvious, so let’s check property (M3). Let x,y, z ∈ X.If x = y, then d(x,y) = 0, and there is nothing to check. Suppose then that x 6= y.Then, either z = x 6= y, or z = y 6= x, or z 6= x,y. Regardless the situation, we havethen

1 = d(x,y) and 1 ≤ d(x, z) + d(z,y) ≤ 2,

and the triangle inequality holds.

One could say that this metric is rather bad: if we think of a metric as a rule tomeasure the distance between points x,y in the set X, then in this case all pointsare far (if x 6= y) or really close (if x = y), but we cannot distinguish anything beyondthis. However, this metric is not useless: it becomes a nice tool to check if geometricintuition really works and provides counterexamples.

Exercise 2.10. Let X be a set and let d : X×X → [0,∞) be a function satisfying thefollowing conditions

(M1) d(x,y) = 0 if and only if x = y; and

(M3’) d(x,y) ≤ d(x, z) + d(y, z) for all x,y, z ∈ X.

Show that d is a metric on X.

Example 2.11. Consider the set of integer numbers Z, and let p be a prime number.The p-valuation of n ∈ Z is the maximum non-negative integer νp(n) such that pνp(n)

divides n.

We can now define the p-adic metric on Z by

d(a, b) =

0, if a = b;

1pνp(a−b)

, otherwise.

Property (M1) holds by definition of d, and property (M2) holds by definition of thep-adic valuation: νp(a − b) = νp(b − a). Let’s prove the triangle inequality. Let thena, b, c ∈ Z. We can assume that these are all different numbers. In this case, property(M3) looks like

1

pνp(a−b) ≤1

pνp(a−c) +1

pνp(c−b) ,

and this is a consequence of the following property of the p-adic valuation:

• νp(a− b) ≥ minνp(a− c), νp(c− b).

Let’s prove the above property. Let then k = νp(a− b), m = νp(a− c), n = νp(c− b)and l = minνp(a − c), νp(c − b). By definition of the p-adic valuation, there existx, y, z ∈ Z such that p does not divide any of them, and such that

pkx = a− b = a− c+ c− b = pmy + pnz = pl(pm−ly + pn−lz),


and now it is clear that k ≥ l.

Exercise 2.12. Consider Z with the 2-adic metric. For which elements n ∈ Z thedistance between n and 1 is 1

2?

Example 2.13. Let (X, d) be a metric space and let H ⊆ X be a non-empty subset.We can then restrict the metric d to the subset H: dH = d|H×H : H ×H → [0,∞). Inother words,

dH(x, y) = d(x, y) for all x, y ∈ H.It is clear that dH is a metric on H. We call dH the induced metric on H, and thismakes (H, dH) into a metric subspace of (X, d).

For example, for each n ≥ 1, Rn ∼= Rn × 1 ⊆ Rn+1. This way we can see theEuclidean metric on Rn as the induced metric from the Euclidean metric on Rn+1.

2.1. Normed real vector spaces. An special and important example of metric spacescomes from real vector spaces with a norm. We have already seen an example of this:R with the absolute value, but this example is way more general than that, and itdeserves its own section.

Definition 2.14. Let V be a real vector space. A norm on V is a function ρ : V → Rsatisfying the following conditions for any u,v ∈ V and any a ∈ R:

(1) ρ(u) = 0 if and only if u = 0;

(2) ρ(au) = |a|ρ(u); and

(3) (triangle inequality) ρ(u + v) ≤ ρ(u) + ρ(v).

Combining the three conditions on ρ, we deduce ρ(u) ≥ 0 for all u ∈ R:

0 = ρ(u− u) ≤ ρ(u) + ρ(−u) = ρ(u) + ρ(u) = 2ρ(u).

Given a normed vector space (V, ρ), we can now define the norm metric:

V × V d // [0,∞)

(u,v) // ρ(u− v).

(3)

Proposition 2.15. The norm metric is indeed a metric on the set V .

Proof. We have to check that the function d in (3) satisfies conditions (M1), (M2) and(M3) in Definition 2.1.

(M1) d(u,v) = 0 if and only if u = v. Since d(u,v) = ρ(u− v), this property holdsby property (i) of norms.

(M2) d(u,v) = d(v,u). This follows by property (ii) of norms:

d(u,v) = ρ(u− v) = ρ((−1)(v − u)) = | − 1|ρ(v − u) = d(v,u).

10 ALEX GONZALEZ

(M3) d(u,v) ≤ d(u,w)+d(w,v). This follows from the triangle inequality for norms:

d(u,v) = ρ(u−v) = ρ((u−w) + (w−v)) ≤ ρ(u−w) +ρ(w−v) = d(u,w) +d(w,v).

Let’s see an application of this.

Example 2.16. The set of complex numbers C is a real vector space of dimension 2.The modulus function, defined by

ρ(a+ bi) =√a2 + b2

induces a metric on C. Obviously, this metric is the same as the Euclidean metricon R2 ∼= C. In general, Rn with the norm

ρ(x1, . . . , xn) =

√√√√ n∑i=1

x2i

induces the Euclidean metric on Rn.

Example 2.17. Let m ≥ 2 be a fixed, natural number, and consider Rn as a realvector space. Define then a map ρ : Rn → R by

ρ(x1, . . . , xn) = m

√√√√ n∑i=1

|xi|m.

Then ρ is a norm on Rn, and the induced norm metric d is the lm-metric:

d(x,y) = m

√√√√ n∑i=1

|xi − yi|m.

Do not get the wrong impression, not all metrics on a real vector space V come froma norm.

Exercise 2.18. Let V be a real vector space of dimension at least 1. Prove that theredoes not exist any norm on V inducing the discrete metric.

2.2. Product spaces. Consider now the following situation: (X, dX) and (Y, dY ) aremetric spaces, and X × Y is the direct product set. It is natural now to ask if themetrics of X and Y can be put together to form a metric on X×Y , but the simplicityof this question hides a rather important problem: there is more than one way ofcombining dX and dY to form a metric for X × Y , and none of these constructions is“more natural” than the others!

Define then maps d1, d2, d∞ : (X × Y )× (X × Y )→ [0,∞) by the following formulas

d1

((x1, y1), (x2, y2)

)= dX(x1, x2) + dY (y1, y2) (4)


d2

((x1, y1), (x2, y2)

)=√dX(x1, x2)2 + dY (y1, y2)2 (5)

d∞((x1, y1), (x2, y2)

)= maxdX(x1, x2), dY (y1, y2) (6)

Proposition 2.19. Each of the maps d1, d2, d∞ is a metric on X × Y .

The metric d2 defined above is usually known as the Euclidean product metric, whiled∞ above is called the box product metric.

Proof. Properties (M1) and (M2) are easy to check in all three cases. Let’s check thetriangle inequality for d2, since the proof is similar for d1 and d∞. In this case, d2 isthe composition

(X × Y )× (X × Y )∼=−→ (X ×X)× (Y × Y )

dX×dY−→

−→ [0,∞)× [0,∞)incl−→ R× R ρEuc−→ [0,∞)

where ρEuc : R × R → R is the Euclidean norm on the real vector space R × R. Itfollows thend2

((x1, y1), (x2, y2)

)= ρEuc

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

)≤

≤ ρEuc

(dX(x1, x3) + dX(x3, x2), dY (y1, y3) + dY (y3, y2)

)≤

≤ ρEuc

(dX(x1, x3), dY (y1, y3)

)+ ρEuc

(dX(x3, x2), dY (y3, y2)

)=

= d2

((x1, y1), (x3, y3)

)+ d2

((x3, y3), (x2, y2)

)where the first inequality follows from the triangle inequality property of dX and dY ,and the second inequality follows from the triangle inequality for norms.

Corollary 2.20. For every (x1, y1), (x2, y2) ∈ X × Y ,

d∞((x1, y1), (x2, y2)

)≤ d2

((x1, y1), (x2, y2)

)≤ d1

((x1, y1), (x2, y2)

).

To understand why this poses a deep problem, consider the following situation. LetX = R = Y . In this case, the absolute value metric, the Euclidean metric, and thebox metric are all the same. So, which metric should we choose for R2 = X × Y ? Itis impossible to answer this question in this strict terms. As we will see in the nextchapter, the answer is that the above three metrics, (4), (5) and (6), are equivalentunder certain relation.

Remark 2.21. The same constructions and results apply to a finite collection ofmetric spaces (Xj, dj) with the obvious modifications.

12 ALEX GONZALEZ

3. Open subsets

In the previous chapter we left an important problem unsolved: given metric spaces(X, dX) and (Y, dY ), is there a right choice of a metric for X × Y ? The answer tothis question lies in the notion of open subsets of metric spaces: two metrics areequivalent if they define the same open subsets.

We will start by defining open subsets (with respect to a given metric). With this wecan then define what the right metric on a product means. All this work to decidewhich metric we give toX×Y will reveal a crucial fact: it is not so important the metricas it is the collection of open subsets that it defines! (in later chapters we will see howtrue this statement is). And so we will spend the second part of this chapter studyingopen subsets: we need to understand them before we can go further. In the last partof this chapter we will apply our recently acquired skills to study convergence ofsequences in metric spaces.

Let’s then define what open subsets of a metric space are. Consider first R2 withthe Euclidean metric. What do you consider an open subset in R2 with this metric?Intuitively, an open subset is a subset in which “one can get as close to the boundaryas possible, but never reach it”. This is a fairly illustrative point of view, but it is notstraightforward to formalize without getting tangled with technicalities. For example,how do we define the boundary of a subset?

Let’s analyze this situation from a different angle. The most characteristic exampleof open subset in R2 (with Euclidean metric) is the so-called (open) ball of radius raround a point a ∈ R2:

Ba,r = x ∈ R2∣∣ d(a,x) < ra

r

There is then a rather easy way to formalize what an open subset of R2 just in termsof balls of different radiuses. Indeed, this idea of “getting closer and closer to the


boundary without reaching it” can be stated as follows. A subset U ⊆ R2 is openwith respect to the Euclidean metric if, for each y ∈ U , there is some ε ∈ (0,∞) suchthat the open ball By,ε is contained in U (do you see any similarity with the notionof limit of a sequence?). It is clear now how to extend this definition to any metricspace.

Definition 3.1. Let (X, d) be a metric space. For each element a ∈ X and eachε ∈ (0,∞) ⊆ R, the open ball with center a and radius ε is the subset

Ba,ε = x ∈ X∣∣ d(a, x) < ε ⊆ X.

A subset U ⊆ X is open with respect to the metric d if for each y ∈ U there is someε(y) ∈ (0,∞) such that

By,ε(y) ⊆ U.

A subset U ⊆ X is closed with respect to the metric d if the set X \ U is open withrespect to the metric d.

The notation ε(y) above is there to stress the idea that the radius of the ball By,ε

depends on the element y. Note also that we have to specify with respect to whichmetric the subset U is open in X. The following example illustrates the reason.

Example 3.2. Let dEuc and ddis be, respectively, the Euclidean and the discrete met-rics on R2, and let a ∈ R2 be any point. Then, U = a ⊆ R2 is open with respect toddis but not open with respect to dEuc.

In the first exercise list we have already seen some examples of balls of radius 1 inR2 with respect to different metrics, so we should be aware by now that we cannottrust our senses on this.

3.1. Equivalent metrics. The definition of open subset provides a comparison cri-teria for different metrics over the same set.

Definition 3.3. Let X be a set, and let d, d′ be two metrics on X. We say that d andd′ are topologically equivalent, d ∼top d

′, if, for every subset U ⊆ X,

U is open in (X, d)⇐⇒ U is open in (X, d′).

In other words, two metrics are topologically equivalent if they define the same col-lections of open subsets on X. For example, the Euclidean metric and the discretemetric on R2 are not equivalent (see Example 3.2).

Lemma 3.4. The relation ∼top on the collection of all metrics of the set X is an equiv-alence relation.

Proof. It is obvious.

14 ALEX GONZALEZ

Let’s go back to the question of defining a metric on X × Y , given metric spaces(X, dX) and (Y, dY ). Let then d1, d2 and d∞ be the metrics on X × Y defined in (4),(5) and (6). We will now prove that these three metrics are topologically equivalent.First we need a little result.

Lemma 3.5. Let d1, d2 and d∞ be the following metrics on R2:

• d1

((x1, y1), (x2, y2)

)= |x1 − x2|+ |y1 − y2|;

• d2

((x1, y1), (x2, y2)

)=√

(x1 − x2)2 + (y1 − y2)2;

• d∞((x1, y1), (x2, y2)

)= max|x1 − x2|, |y1 − y2|.

Then, for each (x1, y1), (x2, y2) ∈ R2,

1

2d1

((x1, y1), (x2, y2)

)≤ 1√

2d2

((x1, y1), (x2, y2)

)≤ d∞

((x1, y1), (x2, y2)

).

Proof. By definition of the metric d2,

d2

((x1, y1), (x2, y2)

)=√

(x1 − x2)2 + (y1 − y2)2 ≤

≤√

max|x1 − x2|2, |y1 − y2|2+ max|x1 − x2|2, |y1 − y2|2 =

=√

2 max|x1 − x2|2, |y1 − y2|2 =

=√

2 d∞((x1, y1), (x2, y2)

).

That proves the inequality between d2 and d∞. Let’s now compare the metrics d1

and d2. Since everything is positive, proving the inequality between d1 and d2 in thestatement is equivalent to proving the resulting inequality after squaring both sides.In other words, we have to prove

1

4

(|x1 − x2|+ |y1 − y2|

)2 ≤ 1

2

((x1 − x2)2 + (y1 − y2)2

).

Let a = |x1 − x2| and b = |y1 − y2|. Then, the above inequality is the same as

(a+ b)2 ≤ 2(a2 + b2),

which is equivalent to a2 + b2 ≤ a2 + b2 + (a− b)2, and this last inequality is obviouslytrue. Therefore, the inequality between d1 and d2 in the statement is true.

Theorem 3.6. Let (X, dX) and (Y, dY ) be metric spaces, and let d1, d2 and d∞ be themetrics on X × Y defined in (4), (5) and (6). Then, d1 ∼top d2 ∼top d∞.

Proof. For each (x1, y1), (x2, y2) ∈ X × Y , let u =(dX(x1, x2), dY (y1, y2)

)∈ R2. Then,

we haved1

((x1, y1), (x2, y2)

)= dR

2

1 (u, 0)

d2

((x1, y1), (x2, y2)

)= dR

2

2 (u, 0)

d∞((x1, y1), (x2, y2)

)= dR

2

∞ (u, 0)


and combining Corollary 2.20 and Lemma 3.5 we have then1

2d1((x1, y1), (x2, y2)

)≤ 1√

2d2((x1, y1),(x2, y2)

)≤ d∞((x1, y1), (x2, y2)

)≤

≤ d2((x1, y1), (x2, y2))≤ d1((x1, y1), (x2, y2)

).

Let now Bd1a,ε, Bd2

a,ε and Bd∞a,ε be the open balls of radius ε around a ∈ X × Y with

respect to the metrics d1, d2 and d∞ respectively. The above inequalities imply that

Bd1a,ε ⊆ Bd2

a,ε ⊆ Bd∞a,ε ⊆ Bd2

a,√

2ε⊆ Bd1

a,2ε,

and the theorem follows from this.

Thus, even if d1, d2 and d∞ are different metrics on X × Y , they are all equivalent.Thus it seems that the “right” metric for X ×Y is any metric in the equivalence classof d1.

Exercise 3.7. Lemma 3.5 is stated only for R2 for simplicity, but there is a generalstatement for Rn. Namely, let d1, d2 and d∞ be the corresponding metrics on Rn

(see Example 2.17 if you are not sure about how they are defined). Then, for eachx,y ∈ Rn,

1

nd1(x,y) ≤ 1√

nd2(x,y) ≤ d∞(x,y).

Can you prove these inequalities? They play the same role as Lemma 3.5 whenproving Theorem 3.6 for a product of more than two metric spaces.

To finish the discussion about giving a metric to X × Y , we have to formalize whatwe understand by the “right metric”. Let’s start by defining an order relation in thecollection of metrics on a given set.

Definition 3.8. Let X be a set, and let d, d′ be metrics on X. We say that d is aweaker metric than d′ (and d′ is an stronger metric than d) if every open subset of Xwith respect to d is also open with respect to d′.

Theorem 3.9. Let (X, dX) and (Y, dY ) be metric spaces, and let d be a metric onX×Ysatisfying the following two conditions:

(1) for each w ∈ Y , the map iw : X → X × Y defined by iw(x) = (x,w) satisfies

d(iw(x), iw(x′)

)= dX(x, x′) for all x, x′ ∈ X

(2) for each z ∈ X, the map jz : Y → X × Y defined by jz(y) = (z, y) satisfies

d(jz(y), jz(y

′))

= dY (y, y′) for all y, y′ ∈ Y.

Then the metric d is weaker than the metric d1 (see (4)) defined by

d1

((x1, y1), (x2, y2)

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

16 ALEX GONZALEZ

Roughly speaking, conditions (1) and (2) above mean that the metrics induced byd on X and Y are dX and dY respectively. The theorem is then telling us whatthe expression “the right metric” means: given metrics dX and dY for X and Yrespectively, the metric d1 is the strongest metric on X × Y inducing dX and dY on Xand Y respectively.

Proof. Fix some u = (a, b) ∈ X × Y . Then, for every (x, y) ∈ X × Y we can form thetriangle below. The triangle inequality implies then the following.

u (x, b)

(x, y)d(u, (x, y)

)≤ d((a, b), (x, b)

)+ d((x, b), (x, y)

)=

= d(ib(a), ib(x)

)+ d(jx(b), jx(y)

)=

= dX(a, x) + dY (b, y) = d1

(u, (x, y)

)Thus, for all u ∈ X × Y and for all ε > 0, Bd1

u,ε ⊆ Bdu,ε, and d1 is stronger than d.

3.2. Properties of open subsets and a bit of set theory. The discussion aboutgiving a metric to a product of sets has pointed out the importance of open subsetswith respect to a given metric. In fact, open subsets are even more important thanwe yet know, and thus it is the purpose of this section to learn some basic propertiesand practice with open subsets.

A little hint for this section: a little drawing always help to illustrate and understanda statement about sets. Although many of the examples that we have seen alreadycontradict our intuition, a drawing can help see how to approach a possible proof.

Let’s start by describing the most important and basic properties of the collection ofopen subsets of a metric space.

Theorem 3.10. Let then (X, d) be a metric space, and let O be the collection of all opensubsets of X with respect to d. Then,

(1) the total subset, X, and the empty set, ∅, are elements of O;

(2) if Uαα∈Γ ⊆ O is any family (possibly infinite) of open subsets, then⋃α∈Γ

Uα ∈ O;

(3) if U1, . . . , Un ⊆ O is any finite family of open subsets, then

n⋂i=1

Ui ∈ O.


Proof. Property (1) is easy: since X contains all possible open balls, it is open. Also,the empty set ∅ contains no points at all, and thus it also satisfies the condition tobe an open subset.

Let now U = Uαα∈Γ ⊆ O be a family (possibly infinite) of open subsets of X, andlet V = ∪ΓUα ⊆ X. For each u ∈ V , there is at least one Uα ∈ U which contains u.Furthermore, since Uα is open, there exists some ε > 0 such that

Bdu,ε ⊆ Uα ⊆ V.

Hence V is open with respect to d.

Finally, let U = U1, . . . , Un ⊆ O be a finite family of open subsets. Let also V =∩ni=1Ui, and let u ∈ V . Since u ∈ Ui for all i, there exist ε1, . . . , εn > 0 such that

Bdu,di⊆ Ui

for all i. Let then ε = minε1, . . . , εn. Then, Bdu,ε ⊆ Bd

u,εifor all i, and thus

Bdu,ε ⊆ V

and V is open with respect to d.

Example 3.11. The following is an easy example of how property (iii) fails when weconsider the intersection of an infinite family of open subsets. Let X = R with theEuclidean metric, d(x, y) = |x− y|, and let Ui = (−1

i, 1i) for all i ≥ 1. Then,

∞⋂i=1

Ui = 0.

The properties describe in Theorem 3.10 can also be stated in terms of closed subsets.

Proposition 3.12. Let (X, d) be a metric space. Then, the following holds:

(1) X and ∅ are closed subsets;

(2) if U1, . . . , Un ⊆ O is any finite family of closed subsets, then the union⋃ni=1 Ui

is also a closed subset;

(3) if Uαα∈Γ ⊆ O is any family (possibly infinite) of closed subsets, then theintersection

⋂α∈Γ Uα is also a closed subset.

Example 3.13. Let X be a set with the discrete metric. Then, for each x ∈ X theone-element subset x ⊆ X is both open and closed. To check that it is open justnotice that x = Bx,1 (open ball of radius 1). To check that it is closed, notice thatthe complement

X \ x =⋃

y∈X\x

y =⋃

y∈X\x

By,1

is a (possibly infinite) union of open subsets, and hence is open.

18 ALEX GONZALEZ

Example 3.14. Consider R2 with the Euclidean metric. Then, for each u ∈ R2, thesubset u ⊆ R2 is closed but not open. Indeed, it is clear that u is not open. Tocheck that it is closed, consider the complement U = R2 \ u. We can describe U as

U = v ∈ R2∣∣ d(u,v) > 0.

Fix some v ∈ U and set ε = d(u,v)2

> 0. Then, for each w ∈ Bv,ε we have

d(u,w) ≥ |d(u,v)− d(v,w)| = d(u,v)− d(v,w) >d(u,v)

2> 0,

and thus w ∈ U (in the above list of (in)equalities we are using some result from thefirst exercise list).

Exercise 3.15. Let (X, d) be a metric space and let x ∈ X be any element. Provethat the subset x ⊆ X is closed with respect to d (in other words, regardless of themetric, a point is always a closed subset).

We have not given any particular name to the collection of closed subsets of a metricspace: since the property of being closed is defined in terms of the property of beingopen, it is enough to understand the collection of open subsets O. However, thisreasoning works also in the opposite direction: if we understand the collection ofclosed subsets then we understand the collection of open subsets, and in someparticular cases it will be easier to work with closed subsets.

Let (X, d) be a metric space. Which subsets of X do you know to be either openand/or closed? From Theorem 3.10 and Proposition 3.12 we know that

• both X and ∅ are open AND closed subsets.

Also, from Exercise 3.15 we know that, for each x ∈ X,

• x is a closed subset and X \ x is an open subset.

And what about open balls? after all, why calling them “open balls” if they turn outnot to be open? that would make no sense! Well, indeed open balls are open subsets,but this requires a little proof.

Lemma 3.16. Let (X, d) be a metric space. Then for each x ∈ X and for each ε > 0,

(1) Bx,ε = y ∈ X∣∣ d(x, y) < ε is an open subset of X with respect to d; and

(2) Dx,ε = y ∈ X∣∣ d(x, y) ≤ ε is a closed subset of X with respect to d.

Proof. To prove part (1), we have to show that, for each y ∈ Bx,ε there exists someδ > 0 such that By,δ ⊆ Bx,ε. The following sketch already hints which is the rightchoice of δ.


x

ε

y

δ d(x, y)

Let then δ = ε− d(x, y) > 0. Then, for eachz ∈ By,δ the triangle inequality implies

d(x, z) ≤ d(x, y) + d(y, z) <

< d(x, y) + δ =

= d(x, y) + ε− d(x, y) = ε

and hence z ∈ Bx,ε. This proves thatthe subset Bx,ε is open.

To prove (2), notice that X \Dx,ε = y ∈ X∣∣ d(x, y) > ε, and we have to prove that

this subset is open. Let then y ∈ X \Dx,ε, as in the following picture.

xε

y

ε

δ

d(x, y)

Let δ = d(x, y)− ε > 0, and suppose that forsome z ∈ By,δ, we have d(x, z) ≤ ε. Then

ε ≥ d(x, z) ≥ |d(x, y)− d(y, z)| =

= |ε+ δ − d(y, z)| > ε+ δ − δ = ε,

where the second inequality (starting from theleft) follows from Exercise 1 in List 1. But thisis impossible, and thus for each z ∈ By,δ wehave d(x, z) > ε. Thus Dx,ε is closed

A metric space (X, d) may contain many subsets which are neither open nor closed,as we have seen already in the second exercise list. What should we do with suchsubsets? We can always relate them to open and closed subsets using the followingdefinitions.

20 ALEX GONZALEZ

Definition 3.17. Let (X, d) be a metric space, and let A ⊆ X be a subset.

• The interior of A, Int(A), is the greatest open subset of X contained in A.

• The closure of A, Cl(A), is the smallest closed subset of X which contains A.

• The boundary of A, ∂A, is the intersection Cl(A) ∩ Cl(X \ A).

Thus, by definition, Int(A) ⊆ A ⊆ Cl(A) and ∂(A) = ∂(X \ A).

The following is an alternative characterization of the interior and the closure of asubset.

Lemma 3.18. Let (X, d) be a metric space, and let A ⊆ X. Then

(1) the interior of A is the union of all open subsets of X contained in A:

Int(A) =⋃

U open, U⊆AU ;

(2) the closure of A is the intersection of all closed subsets of X which contain A:

Cl(A) =⋂

V closed, A⊆V V.

Proof. We do the proof for (1) and leave the proof of (2) as an exercise. Let firstx ∈ Int(A). Since Int(A) is by definition the greatest open subset contained in A,there exists some ε > 0 such that Bx,ε ⊆ Int(A) ⊆ A. Thus,

Bx,ε ⊆⋃

U open, U⊆AU.

Conversely, let y be an element in the union of all open subsets contained in A. Inparticular, there exists such an open subset U containing y, and by definition ofInt(A) we have y ∈ U ⊆ Int(A).

The following is an example of the importance of the discrete metric.

Example 3.19. Intuition from the Euclidean metric on the plane R2 tells us that theclosure of an open ball of radius ε > 0 around an element x ∈ R2 should be thesubset of elements whose distance to x is smaller or equal than ε:

Cl(Bx,ε) = y ∈ R2∣∣ d(x,y) ≤ ε.

This is false in general, and the discrete metric is the perfect counterexample. Indeed,Let X be a set with the discrete metric, and let x ∈ X. Recall from Example 3.13that x = Bx,1, the open ball of radius 1. Furthermore, x is both open and closed.Thus,

x = Bx,1 = Cl(Bx,1) $ Dx,1 = X.


Exercise 3.20. Let V be a normed real vector space, with norm ρ, and let d be theinduced metric. Prove that, in this case for any v ∈ V and any ε > 0, the closure ofthe open ball of radius ε around v is the closed disk around v of radius ε,

Cl(Bv,ε) = Dv,ε.

Let’s see some properties of interior, closure and boundary of subsets in a metricspace.

Lemma 3.21. Let (X, d) be a metric space. Then the following holds:

(1) a subset A ⊆ X is open if and only if Int(A) = A;

(2) for every A ⊆ X, Int(Int(A)

)= Int(A);

(3) if Int(A) ⊆ B ⊆ A then Int(A) = Int(B);

(4) if Uαα∈Γ is a collection (possibly infinite) of subsets of X, then⋃α∈Γ

Int(Uα) ⊆ Int

( ⋃α∈Γ

Uα

)Int

( ⋂α∈Γ

Uα

)⊆⋂α∈Γ

Int(Uα);

(5) if U1, . . . , Un is a finite collection of subsets of X, then

Int

( n⋂i=1

Ui

)=

n⋂i=1

Int(Ui);

Proof. Properties (1) and (2) are obvious by definition. Property (3) is proved asfollows. If Int(A) ⊆ B, then by definition Int(A) ⊆ Int(B) ⊆ B ⊆ A. Also, sinceInt(B) is an open subset of A, it follows that Int(B) ⊆ Int(A). Property (4) followsfrom the fact that, for each γ ∈ Γ, we have ∩ΓUα ⊆ Uγ ⊆ ∪ΓUα. Finally, let’s proveproperty (5). The intersection I = ∩ni=1Int(Ui) is an open subset of ∩ni=1Ui, and henceI ⊆ Int(∩ni=1Ui). This combined with property (4) finishes the proof.

Example 3.22. This is an example about how the inclusion ∪ΓInt(Uα) ⊆ Int(∪Γ Uα

)may fail to be an equality. Consider then R with the metric d(x, y) = |x− y|, and letU1 = [0, 1] and U2 = [1, 2]. Then,

U = U1 ∪ U2 = [0, 2] and Int(U) = (0, 2),

while Int(U1) = (0, 1) and Int(U2) = (1, 2). Notice that 1 ∈ (0, 2)

1 /∈ (0, 1) ∪ (1, 2) $ (0, 2).

Note that, with the discrete metric, the inclusions in Lemma 3.21 (4) become equali-ties. Can you prove it?

A similar list of properties exists for the closure of a subset of a metric space. Westate it below without proof (the proof is similar to the result above, so it is left as anexercise).

22 ALEX GONZALEZ

Lemma 3.23. Let (X, d) be a metric space. Then the following holds:

(1) a subset A ⊆ X is closed if and only if A = Cl(A);

(2) for every A ⊆ X, Cl(Cl(A)

)= Cl(A);

(3) if A ⊆ B ⊆ Cl(A) then Cl(A) = Cl(B);


Cl(Uα) ⊆ Cl

( ⋃α∈Γ

Uα

)Cl

( ⋂α∈Γ

Uα

)⊆⋂α∈Γ

Cl(Uα);


Cl

( n⋃i=1

Ui

)=

n⋃i=1

Cl(Ui);

Finally, let’s see some interesting properties of the interaction of interior, closure andboundary.

Lemma 3.24. Let (X, d) be a metric space. Then, for any subsets A,B ⊆ X,

(1) the closure of the complement of A is the complement of the interior of A:

Cl(Ac) = Cl(X \ A) = X \ Int(A) =(Int(A)

)c;

(2) the interior of the complement of B is the complement of the closure of B:

Int(Bc) = Int(X \B) = X \ Cl(B) =(Cl(B)

)c;

(3) ∂(Cl(A)

)⊆ Cl(A).

(4) ∂(Int(B)

)∩ Int(B) = ∅; and

Proof. To prove (1), notice that Int(A) = A \ ∂(A) = A ∩(∂(A)

)c. Thus,(Int(A)

)c= X \ Int(A) = X \

(A ∩

(∂(A)

)c)= (X \ A) ∪

(X \

(∂(A)

)c)=

= Ac ∪((∂(A)

)c)c= Ac ∪ ∂(A) = Ac ∪ ∂(Ac) = Cl(Ac)

To prove (2) just replace A by Bc everywhere above. Property (3) follows by

∂(Cl(A)

)= Cl

(Cl(A)

)∩ Cl

(X \ Cl(A)

)= Cl(A) ∩ Cl

(X \ Cl(A)

)⊆ Cl(A).

To prove (4), notice that Y ∩ Y c = ∅ for any Y ⊆ X. Thus, we just need to check that∂(Int(B)

)⊆(Int(B)

)c. Indeed,

∂(Int(B)

)= ∂

(Int(B)c

)= ∂

(Cl(Bc)

)⊆ Cl(Bc) =

(Int(B)

)c,

where (from left to right) the first equality holds by definition, the second equalityholds by (2), the inequality holds by (3) and the last equality holds by (1).


Example 3.25. Consider the plane R2 with the Euclidean metric. For any pointu ∈ R2, we have

Int(u) = ∅ and Cl(u) = u.Thus, ∂(u) = u ∩ (∅)c = u ∩ R2 = u.

Still in the plane R2 with the Euclidean metric, we have

Int(Bu,ε) = Bu,ε and Cl(Bu,ε) = Du,ε,

and the boundary is then the circle of radius ε and center u:

∂(Bu,ε) = Du,ε ∩ (Bu,ε)c = v ∈ R2

∣∣ d(u,v) = ε.

If we consider now R2 with the discrete metric, then for each u ∈ R2,

Int(u) = u and Cl(u) = uand thus ∂(u) = u ∩ (u)c = ∅.

Exercise 3.26. Let (X, d) be a metric space. Prove that, for any subset A ⊆ X,

∂(A) = Cl(A) \ Int(A)def= x ∈ Cl(A)

∣∣ x /∈ Int(A).Deduce that, if A is an open and closed subset of X, then ∂(A) = ∅.

3.3. Convergence of sequences in metric spaces. The idea behind the notion oflimit of a sequence is that of approximating an object which is not completely knownor too complicated by simpler objects. For instance, the value of the numbers e andπ can be approximated by

e = limn→∞

(1 +

1

n

)nπ = 4 ·

∞∑n=1

(−1)n+1

2n− 1

(notice that the sum of an series can be interpreted as a limit).

Limits of sequences of real numbers are just a particular case of a more generalsituation. For instance, recall from the introduction (Chapter 1) the example aboutTaylor approximation. In this section we will generalize the notion of sequence andthe convergence of its limit to all metric spaces.

Definition 3.27. Let (X, d) be a metric space and let xnn∈N be a sequence ofelements of X. Then, the the sequence xnn∈N converges to the element x ∈ X if, forevery ε > 0 there exists some N ∈ N such that

d(x, xn) < ε for all n ≥ N.

When this is the case, the element x is called the limit of the sequence xnn∈N, andwe write x = limn→∞ xn.

Remark 3.28. The definition of convergence of a sequence in a metric space can berephrased in several equivalent ways, such as

24 ALEX GONZALEZ

(1) The sequence xnn∈N converges to v if the sequence of numbers d(v, xn)n∈Nconverges to zero.

(2) The sequence xnn∈N converges to v if for every ε > 0 the open ball Bv,ε

contains all but a finite number of terms of the sequence.

v

Bv,ε

x1x2

x3

x4

Theorem 3.29. Let (X, d) be a metric space. If the sequence xnn∈N in X has a limit,then it is unique.

Proof. Suppose the sequence xnn∈N converges to the points x, y ∈ X, and supposex 6= y. This means that, for each ε > 0 there exist Nx, Ny ∈ N such that

d(x, xn) < ε, for all n ≥ Nx and d(y, xn) < ε, for all n ≥ Ny.

In particular, let ε = d(x,y)2

> 0, and let N = maxNx, Ny, where Nx, Ny are asabove (with respect to this particular choice of ε). Then, for each n ≥ N , we haved(x, xn), d(y, xn) < ε. On the other hand, by the triangle inequality,

d(x, y) ≤ d(x, xn) + d(xn, y) < 2 · ε = 2 · d(x, y)

2= d(x, y),

which is impossible. Thus, x = y.

Example 3.30. Let X be a set with the discrete metric. Then, a sequence xnn∈N isconvergent if and only if there exists some N ∈ N such that xn = xN for all n ≥ N .Studying the convergence of sequences in a set with the discrete metric is not reallyinteresting.

Example 3.31. Let X be the set of all continuous functions f : [0, 1] → R, andconsider the following two metrics on X:

dint(f, g) =

∫ 1

0

|f(x)− g(x)|dx d∞(f, g) = max|f(x)− g(x)|∣∣ x ∈ [0, 1].

Now fix some α > 0 and let fnn∈N be the sequence where fn is the function withgraph

(0, 0) (1, 0)

(12, 0)

(0, α)

(12− 1

n, 0) (1

2+ 1

n, 0)


Then, the sequence fnn∈N converges to the function g = 0 (constant on [0, 1] withimage 0) with respect to the metric dint. Indeed, for each n

dint(0, fn) =

∫ 1

0

|0− fn(x)|dx =

∫ 1

0

fn(x) dx =1

nand lim

n→∞

1

n= 0.

However, the same sequence fn does not converge to the constant function g = 0with respect to the metric d∞, since for any n

d∞(0, fn) = max|0− fn(x)|∣∣ x ∈ [0, 1] = α > 0.

In fact, this sequence does not converge to any continuous function with respect tothe metric d∞, but we are not ready yet to prove this.

Sequences in a metric space provide a nice criterion to detect closed subsets.

Theorem 3.32. Let (X, d) be a metric space and let V ⊆ X be a subset. Then thefollowing statements are equivalent.

(1) V is a closed subset of X.

(2) the limit of every convergent sequence xnn∈N ⊆ V is an element of V .

Proof. Suppose first that (2) is true. We prove by contradiction that V c is open.Assume then that V c is not open, and let u ∈ V c be an element for which Bu,ε 6⊆ V c,for any ε > 0. This means, in particular, that for any n ∈ N the ball Bu, 1

ncontains

(at least) an element xn of V . We can form then the sequence xnn∈N ⊆ V , but thissequence clearly converges to the element u /∈ V , hence the contradiction.

Suppose now that (1) is true. We have to show that every convergent sequence ofelements of V has its limit in V . Let then xnn∈N ⊆ V be a convergent sequence,and suppose v = lim xn ∈ V c. Since V c is open by hypothesis, there exists someε > 0 such that Bv,ε ⊆ V c. On the other hand, since v = limxn, there exists someε′ < ε and some N ∈ N such that d(v, xn) < ε′ for all n ≥ N , and this implies that

xn ∈ Bv,ε′ ⊆ Bv,ε ⊆ V c,

which contradicts the assumption that xn ∈ V for all n.

We already know examples of different metrics on a set X which define the samecollection of open subsets of X. Thus we should not be surprised by the followingresult.

Corollary 3.33. Let X be a set and let d, d′ be two topologically equivalent metrics onX. Then, a sequence xnn∈N converges to xwith respect to d if and only if it convergesto x with respect to d′.

26 ALEX GONZALEZ

4. Continuous functions between metric spaces

We are finally ready to define and study continuous functions. Essentially, a functionbetween sets f : X → Y is just a relationship that we establish between the sets Xand Y , usually in order to express events happening in Y in terms of the elementsin X.

Sometimes we happen to haver a deeper understanding of the sets X and Y , whichtranslates into a richer, common structure, and then we want the function f to takeinto account such structure. For instance, X and Y could be groups, or rings, oralgebras over a given ring, and then we may want f to be a group homomorphism,or a ring homomorphism, or an algebra homomorphism.

And the same idea applies if we are dealing with metric spaces. That is, if (X, dX)and (Y, dY ) are metric spaces, we do not want to consider any map between the setsX and Y , but maps that take into consideration the corresponding metrics. So, whatabout considering functions f : X → Y satisfying dY (f(x), f(x′)) = dX(x, x′) for anyx, x′ ∈ X?

Definition 4.1. Let (X, dX) and (Y, dY ) be metric spaces. A map f : X → Y suchthat

dY (f(x), f(x′)) = dX(x, x′)

for all x, x′ ∈ X is called an isometry between X and Y .

We have encountered isometries already in these notes: the maps iw : X → X × Yand jz : Y → X × Y in Theorem 3.9 are isometries (check it!).

Example 4.2. Consider the sets R2 and C, both with the Euclidean metric. Then,the map f : C→ R2 defined by f(a+ bi) = (a, b) is an isometry.

It is clear that the condition on a map to be an isometry is rather restrictive. Forinstance, the map f : R→ R defined by f(x) = x2 is not an isometry:

d(3, 4) =√

9 + 16 = 5 but d(f(3), f(4)) = d(9, 16) =√

81 + 256 =√

337.

We want to consider more maps between metric spaces than just isometries, and thekey lies in the definition of continuity for functions f : R → R. Before stating theproper, rigorous definition of continuity of such a function, think of the idea behindcontinuity:

f is continuous at a ∈ R if f(x) is “close” to f(a) whenever x ∈ R is “close” to a.

This is what we want for functions between metric spaces to take into consideration.

The formalization of such idea is easy once we formalize continuity for functionsf : R → R: recall that such a function is continuous at a ∈ R if, for all eps > 0 thereexists some δ > 0 such that

if |a− x| < δ then |f(a)− f(x)| < ε.


Well, it is obvious now how to extend this definition to a map between metric spaces.

Definition 4.3. Let (X, dX) and (Y, dY ) be metric spaces, and let f : X → Y be a setmap. The map f is continuous at the point a ∈ X (with respect to the metrics dX , dY )if for each ε > 0 there exists some δ > 0 such that

if dX(a, x) < δ then dY (f(a), f(x)) < ε.

If f is continuous at every element a ∈ X, then we say that f is continuous.

Notice that the above definition depends on the given metrics dX and dY . In orderto specify that the map f is continuous with respect to the metrics dX and dY on Xand Y respectively, we may also write

f : (X, dX) −→ (Y, dY ).

This is actually rather important since the map f : X → Y can be continuous witha choice of metrics and discontinuous with another choice of metrics. Let’s see thiswith an example.

Example 4.4. Let X = R = Y , and let f : R→ R be the map

f(x) =

0, if x ∈ Q1, if x /∈ Q

Let then dEuc and ddis be the Euclidean and the discrete metrics on R. We have then

• f : (X, dEuc)→ (Y, dEuc) is not continuous;

• f : (X, dEuc)→ (Y, ddis) is not continuous;

• f : (X, ddis)→ (Y, dEuc) is continuous; and

• f : (X, ddis)→ (Y, ddis) is continuous.

Of course, this is an extreme example, but it perfectly describes the importance ofkeeping in mind which are the ambient metrics. As an exercise, fill in the details toshow each of the points above.

Remark 4.5. The discrete metric also illustrates the following (not intuitive) situation:a bĳective map f : X → Y of sets which is continuous with respect to a choice ofmetrics dX and dY may not have a continuous inverse! Indeed, the identity map(R, ddis)→ (R, dEuc) is continuous, but the inverse is not.

Let’s see some more examples of continuous functions.

Exercise 4.6. Let f : (X, dX)→ (Y, dY ) be an isometry. Prove that

(1) f is injective (as a map of sets); and

(2) f is a continuous function (with respect to dX and dY ).

28 ALEX GONZALEZ

Example 4.7. The map µ : R2 → R defined by µ(a, b) = a · b (multiplication of realnumbers) is a continuous function with respect to Euclidean metrics on R2 and R,which we will denote here by d2 and d1 respectively to avoid confusion.

Fix then some (a, b) ∈ R2. To prove that µ is continuous at (a, b) we have to checkthat for each ε > 0 there exists some δ > 0 such that

if d2((a, b), (u, v)) < δ then d1(µ(a, b), µ(u, v)) = d1(ab, uv) < ε.

Let’s ignore ε for a second, and suppose that

d2((a, b), (u, v)) =√

(a− u)2 + (b− v)2 < δ

for some δ > 0. Then, we have |a− u|, |b− v| < δ, and

d1(ab, uv) = |ab− uv| = |b(a− u) + a(b− v) + (a− u)(v − b)| ≤≤ |b| · |a− u|+ |a| · |b− v|+ |a− u| · |b− v| < δ2 + δ(|a|+ |b|).

Let’s go back now to the question of the continuity of µ at (a, b) ∈ R2. Let then ε > 0,and consider the quadratic equation on δ:

δ2 + (|a|+ |b|)δ − ε = 0.

This equation has always a positive solutions, namely δ =−(|a|+|b|)+

√(|a|+|b|)2+4ε

2, and

the above discussion implies that, for this particular choice of δ, if d2((a, b), (u, v)) < δthen d1(ab, uv) < ε as required.

Exercise 4.8. Let (X, d) be a metric space and let z ∈ X. We can then define a map

Xδz // [0,∞)

x // δz(x) = d(z, x)

Prove that the map δz is continuous with respect to the metric d on X and theEuclidean metric on [0,∞) ⊆ R.

Example 4.9. Let d be the p-adic metric on the set of integers Z, and fix some α ∈ Z.Then, the maps

(Z, d)f // (Z, d) (Z, d)

g // (Z, d)

k // α · k k

// k + α

are continuous.

Let’s prove first that f is continuous. For every a, b ∈ Z, we have

d(f(a), f(b)) = d(α · a, α · b) =1

pνp(α·a−α·b) =1

pνp(α(a−b)) =

=1

pνp(α) · pνp(a−b) =1

pνp(α)· d(a, b).


Since 1pνp(α)

≤ 1, it follows that, for each a ∈ Z and each ε > 0,

if d(a, b) < ε then d(f(a), f(b)) ≤ d(a, b) < ε

and hence f is a continuous map. In particular notice that if p does not divide αthen f is an isometry.

Let’s check now that the map g is continuous. In this case, for each a, b ∈ Z we have

d(f(a), f(b)) = d(a+ α, b+ α) =1

pνp(a+b−(b+α))=

1

pνp(a−b) = d(a, b),

so in fact this map is always an isometry (in particular, it is continuous).

Once we start working with functions, it is natural to consider compositions of func-tions.

Proposition 4.10. Let f : (X, dX) → (Y, dY ) and g : (Y, dY ) → (Z, dZ) be continuousmaps of metric spaces. Then, the composition map, (g f) : (X, dX) → (Z, dZ), is acontinuous map of metric spaces.

Proof. Set for simplicity h = (g f) : X → Z, and fix an element a ∈ X. We haveto check that, for each ε > 0 there is some δ > 0 such that, if dX(a, x) < δ, thendZ(h(a), h(x)) < ε.

Let b = f(a) ∈ Y . Since g is continuous, there exists some γ > 0 such that, ifdY (b, y) < γ, then

dZ(g(b), g(y)) = dZ(h(a), g(y)) < ε.

Similarly, since f is continuous, there exists some δ > 0 such that, if dX(a, x) < δ,then

dY (f(a), f(x)) = dY (b, f(x)) < γ.

Combining the above, we get the following:

dX(a, x) < δ =⇒ dY (f(a), f(x)) = dY (b, f(x)) < γ =⇒=⇒ dZ(g(b), g(f(x))) = dZ(h(a), h(x)) < ε

which proves that the composition is continuous at a ∈ X.

Let’s analyze the definition of continuity. Let then f : (X, dX)→ (Y, dY ) be a contin-uous function, and let a ∈ X. The map f is continuous at a if, for each ε > 0 thereis some δ > 0 such that

if dX(a, x) < δ then dY (f(a), f(x)) < ε.

The left part above says that x ∈ Ba,δ, while the right part says that f(x) ∈ Bf(a),ε.

This suggests some relation between continuity and open subsets, which we formu-late properly as the following result. You can take it as a set of equivalent definitionsof the notion of continuity.

30 ALEX GONZALEZ

Theorem 4.11. Let (X, dX) and (Y, dY ) be metric spaces, and let f : X → Y be a map.Then, the following statements are equivalent:

(1) f is a continuous function;

(2) for each open subset U ⊆ Y , f−1(U) is an open subset of X;

(3) for each closed subset V ⊆ Y , f−1(V ) is a closed subset of X;

(4) for any sequence xnn∈N in X, if xn converges to z ∈ X, then the sequencef(xn)n∈N in Y converges to f(z).

A note of warning: given a map of sets f : X → Y , the notation f−1(U) stands simplyfor the set of pre-images in X of the elements of U ,

f−1 = x ∈ X∣∣ f(x) ∈ U.

We are not assuming the existence of any inverse function! For instance considerthe function f : R→ R defined by f(x) = x2. Clearly, this function does not have aninverse, but the pre-image of the subset 1 ∈ R makes perfect sense:

f−1(1) = x ∈ R∣∣ f(x) = 1 = −1, 1.

Remark 4.12. With this set of equivalent definitions of continuity, we do not need todefine continuity at a point, we can define continuous functions directly: a functionbetween metric spaces f : X → Y is continuous if and only if for each open subsetU ⊆ Y the pre-image f−1(U) is an open subset of X.

Be careful with the following: condition (2) above concerns only the pre-image ofopen subsets of Y , but it says nothing about the image of open subsets of X. Thefollowing is, in general, false:

if U ⊆ X is an open subset, then f(U) ⊆ Y is an open subset.

Proof of Theorem 4.11. Since this proof is rather long, let’s split it in smaller parts,each proving an equivalence of statements.

• Statements (1) and (2) are equivalent.

First we prove that (1) implies (2). Suppose then that f is a continuous map, andlet U ⊆ Y . Let also a ∈ f−1(U), and let u = f(a). Since U is open, there existssome ε(u) > 0 such that Bu,ε(u) ⊆ U . Also, since f is continuous, there exists someδ(a) > 0 such that

if dX(a, x) < δ(a) then dY (f(a), f(x)) = dY (u, f(x)) < ε(u),

which implies that f(Ba,δ(a)) ⊆ U , and hence Ba,δ(a) ⊆ f−1(U). This proves thatf−1(U) is open with respect to dX .

Let’s see now that (2) implies (1). Suppose then that f−1(U) is an open subset of Xfor each open subset U ⊆ Y , and fix a ∈ X. For each ε > 0, the open ball Bf(a),ε is


an open subset of Y , and thus f−1(Bf(a),ε) is an open subset of X. Furthermore,

a ∈ f−1(Bf(a),ε),

and this implies that there exists some δ > 0 such that

Ba,δ ⊆ f−1(Bf(a),ε).

The above inclusion is equivalent to say that f is continuous at a ∈ X.


Let V ⊆ Y be a closed subset, and let U = V c, which is an open subset by hypothesis.By (2), f−1(U) is an open subset of X. Furthermore, we have

f−1(U) = f−1(V c) = f−1(Y \ V ) = X \ f−1(V ),

and thus f−1(V ) is a closed subset, and this proves that (2) implies (3). Replacing“open” by “closed” proves the converse.


First we prove that (1) implies (4). Let then xnn∈N be a sequence in X whichconverges to z ∈ X. Since f is continuous (at f(z)), for each ε > 0 there exists someδ > 0 such that

if dX(z, x) < δ then dY (f(z), f(x)) < ε.

Also, since the sequence is convergent, we know that there exists some N ∈ N suchthat dX(z, xn) < δ for all n ≥ N . Thus, for all n ≥ N , we have

dY (f(z), f(xn)) < ε,

which implies (4).

Finally, we have to check that (4) implies (1). Suppose otherwise that (1) is not trueon a fixed element a ∈ X: there exists some ε > 0 such that for each δ > 0 there issome x ∈ X such that

dX(a, x) < δ but dY (f(a), f(x)) ≥ ε.

By fixing such ε, we can then define the sets

An = x ∈ X∣∣ dX(a, x) <

1

nand dY (f(a), f(x)) ≥ ε,

which is non-empty for all n ∈ N. Next choose for each n an element xn ∈ An: thesequence xnn∈N clearly converges to a ∈ X, while the sequence f(xn)n∈N doesnot converge to f(a), contradicting (4).

Exercise 4.13. At the beginning of this chapter we saw several examples of continu-ous functions. Which of them are easier to check with these alternative definitions?

Exercise 4.14. Let f : X → Y be a map. Prove that f is continuous with respect tothe metrics dX and dY (on X and Y respectively) if and only if, for each y ∈ Y andeach ε > 0,

f−1(BdYy,ε) is an open subset of X.

32 ALEX GONZALEZ

In other words, it is enough to check condition (2) in Theorem 4.11 for all the openballs in Y .

A first consequence of Theorem 4.11 is that topologically equivalent metrics definethe same continuous maps. The following result state this idea rigorously.

Corollary 4.15. Let f : X → Y be a map. Let also dX , d′X be topologically equivalentmetrics on X and dY , d′Y be topologically equivalent metrics on Y . Then, the followingstatements are equivalent:

(1) the map f is continuous with respect to the metrics dX and dY ; and

(2) the map f is continuous with respect to the metrics d′X and d′Y .

Conversely, suppose that, for every map f : X → Y , f is continuous with respect tothe metrics dX and dY if and only if f is continuous with respect to the metrics d′X andd′Y . Then, the following holds:

(1) the metrics dX and d′X are topologically equivalent; and

(2) the metrics dY and d′Y are topologically equivalent.

Proof. Since topologically equivalent metrics define the same collections of open sub-sets, the result follows immediately by using (2) in Theorem 4.11.

Exercise 4.16. Prove the above result using only the epsilon-delta definition of con-tinuity. Do not expect the proof to be shorter than the above!

4.1. Homeomorphisms of metric spaces and open maps. We have see in Remark4.5 how a bĳective map of sets, which is in addition a continuous map betweenmetric spaces, does not have a continuous inverse. This section compiles a fewuseful notions related to the aforementioned remark, as well as some interestingexamples or exercises.

Definition 4.17. Let f : (X, dX)→ (Y, dY ) be a map of sets. Then, f is a homeomor-phism if the inverse map f−1 exists and both f and f−1 are continuous. Similarly, twometric spaces (X, dX) and (Y, dY ) are homeomorphic if there exists a homeomorphismf : X → Y .

Broadly speaking, the notion of homeomorphism is to metric spaces as the notionof isomorphism is to groups or other algebraic structures. Notice that the conditionof f−1 existing is already rather restrictive: you know of many continuous mapsf : R→ R which do not have an inverse function.

Example 4.18. Recall from Remark 4.12 that a continuous function f : X → Ydoes not necessarily map open subsets of X to open subsets of Y . As usual, a


counterexample to this statement arises from the discrete metric. Recall from Remark4.5 that the map

Id : (R, ddis) −→ (R, dEuc)

is continuous. Furthermore, we know that for any x ∈ R, the singleton x ⊆ Ris an open subset with respect to the discrete metric. On the other hand we haveId(x) = x ⊆ R, which is not an open subset with respect to the Euclidean metric.

Definition 4.19. A map f : (X, dX)→ (Y, dY ) is open (respectively, closed) if f(U) isopen (respectively, closed) for each open (respectively, closed) subset U ⊆ X.

Notice that, in the definition above, we do not require f to be continuous! For examplethe map Id : (R, dEuc)→ (R, ddis) is open but not continuous.

Example 4.20. The following example shows that open maps need not be closed,and vice-versa. For instance, let X be a finite set with the discrete metric. We canthen list the elements of X, namely X = x1, . . . , xn, and then define a map f fromX with the discrete metric to R with the Euclidean metric by

f(xk) = k for k = 1, . . . , n.

SinceX has the discrete metric it is immediate that this map is continuous (althoughsuch property is independent from f being open, closed, both or none).

Let’s check that this map is closed. First of all, every subset xk ⊆ X is mappedto k ⊆ R, which is also a closed subset (see Exercise 3.15). Since X is finite, anyclosed subset V is a finite union of singletons, and thus so is f(V ). Hence, f(V )is closed in R with the discrete metric. However, this map is clearly not open (seeExample 4.18 above).

Let now X = [0, 1], with the metric induced from R, and let f : X → X be defined asfollows

f(x) =

2x if 0 ≤ x ≤ 1

2x if 1

2< x ≤ 1

This map is not continuous at x = 12, but this is not relevant for the example. What

matters in this example is that it is not closed, since f([

12, 1])

=(

12, 1], but it is open.

Indeed, let x ∈ X and ε > 0. Then,

• if x ∈ [0, 12) and 1

2/∈ Bx,ε, then f

(Bx,ε

)= B2x,2ε;

• if x ∈ [0, 12) and 1

2∈ Bx,ε, then set x1 = 2x+2ε+1

4and δ1 = 2x+2ε−1

4, and

f(Bx,ε

)=(2(x− ε), 1

]⋃Bx1,δ1 ;

• if x = 12, then f

(B 1

2,ε

)=(

12, 1]⋃ (

12, 1

2+ ε);

• if x ∈ (12, 1] and 1

2∈ Bx,ε, then f

(Bx,ε

)= (1− x+ ε, 1]

⋃(1

2, x+ ε); and

• if x ∈ (12, 1] and 1

2/∈ Bx,ε, then f

(Bx,ε

)= Bx,ε.

34 ALEX GONZALEZ

Exercise 4.21. Let (X, dX) and (Y, dY ) be metric spaces, and consider X × Y withthe product metric. Then, the projection maps πX : X ×Y → X and πY : X ×Y → Yare continuous and open maps. Can you give an example of such a projection mapwhich is not closed?

Theorem 4.22. Let f : (X, dX)→ (Y, dY ) be a map. Then, the following are equivalent:

(1) f is a homeomorphism;

(2) f is bĳective, continuous and open;

(3) f is bĳective, continuous and closed;

In particular, every homeomorphism is an open and closed map.

Proof. Suppose first that f is a homeomorphism. Then f is continuous and bĳective,and we have to check that it is open (and closed). Since f is a homeomorphism,we know that the map g = f−1 is continuous. Then, for each open (closed) subsetU ⊆ X, we have f(U) = g−1(U), which is open (closed) since g is a continuous map.Thus, f is open (closed), and (1) implies (2) and (3).

Suppose now that f is bĳective, continuous and open (closed). In particular, sincef is bĳective, we only have to check that the inverse map g = f−1 is continuous,i.e., that g−1(U) is open (closed) for every open (closed) subset U ⊆ X. But this isimmediate: g−1(U) = f(U) is open (closed) since f is an open (closed) map. Thus,(2) implies (1) (respectively (3) implies (1)).

Exercise 4.23. Consider R2 with the Euclidean metric, and let

S1 = (x, y)∣∣ x2 + y2 = 1 ⊆ R2

be the circle of radius 1 around the origin. Considering then S1 and D(0,0),1 (the diskof radius 1 around the point (0, 0)) as metric subspaces of R2, prove the followingstatement, known as Alexander’s trick: every homeomorphism f : S1 → S1 can beextended to a homeomorphism of the disk of radius 1, D0,1. Can you do the same inhigher dimensions?

Exercise 4.24. Consider X = (0, 1) and Y = (0,∞) as metric subspaces of R, andprove that X and Y are homeomorphic but not isometric.

Exercise 4.25. Let f : (X, dX)→ (Y, dY ) be a continuous map, and let

Gf = (x, f(x))∣∣ x ∈ X ⊆ X × Y

be the graph of the function. Prove that, as a metric subspace of X × Y (with theproduct metric), Gf is homeomorphic to X.

Now that we know about continuous functions, we want to study how different prop-erties of metric spaces behave with respect to continuous maps.


5. Interlude

During the previous chapters several notions have been introduced which were prob-ably new to you, and even some of them blatantly defied our previous intuition. Alot of work needed to be done in order to give the theory of metric spaces consistency(recall the question of defining a suitable metric for a product of metric spaces).

So, where is all this leading too? Why going through all the trouble? This interludeintends to partially answer this, with an absolute lack of rigor. The only purpose ofthis chapter is to motivate all the work we have done until this point and the workwe will do in the next chapters. Do not take anything in this chapter as rigorous!

Actually, there are many reasons to study metric and topological spaces. For in-stance, we have already seen in the tutorials that all circles in the Euclidean planeare homeomorphic to the circle of radius 1 and center (0, 0), and a similar statementis true for all disks in the Euclidean plane, and like this for many other geometricobjects in R2 (and not only in R2 but in Rn for any n ≥ 1). This is simply a matter ofform or shape, but not of size.

Let’s formalize this. Consider Rn with the Euclidean metric, and call a subset U ⊆ Rn

bounded if there exist x ∈ Rn and ε > 0 such that U ⊆ Bx,ε. We can define anequivalence relation among all bounded subsets of Rn: U is equivalent to V is theyare homeomorphic (check that this is an equivalence relation). And now we can saythat a shape is a class under this equivalence relation.

For instance you may have heard of the following example already: a mug and adoughnut have the same shape. Making the homeomorphism explicit can be a bitcomplicated, but here is an outline of how to go from the doughnut to the mug.

So we need to find ways of telling one shape from another: properties of metric spaceswhich do not change under homeomorphism, and which we can use to distinguishdifferent shaper. For instance, a cylinder and a Moebius band are the same shape?Let’s see. We can represent each of them as follows

Cylinder

a a

Moebius band

bb

where the arrows on each side of the bands determine how to glue one side with theother. What happens then with the boundary of the cylinder? It is formed by two

36 ALEX GONZALEZ

disjoint circles. And the boundary of the Moebius band? It is formed by a singlecircle (well, or something that resembles a circle), so here is something: cylinder andMoebius band cannot be the same shape if the boundaries are so different.

Related to the above discussion is another problem which boils down to metric (andtopological) spaces. Finding solutions to polynomial functions can get pretty difficultif the polynomial has more than one variable and the degree is high. For instance,what about solving the equation

5x2y3 − xy4 + y5 − x5 = 0?

Instead, let’s give this problem a different approach. Let P : Rn → R be a polynomialfunction (in one or more variables). The set

P−1(0) = (x1, . . . , xn) ∈ Rn∣∣ P (x1, . . . , xn) = 0 ⊆ Rn

determines a metric subspace of Rn. For instance, for the polynomial functionP (x, y) = x2 + y2 − 1 we get

P−1(0) = S1,

or, for something a bit fancier, consider P (x, y, z) = x2 + y2 + z2 − 1 and Q(x, y, z) =

(2−√x2 + y2)2 + z2 − 1. Then, P−1(0) is a sphere, while Q−1(0) is a torus:

Understanding the shape of P−1(0) helps finding solutions or discarding the exis-tence of them. Actually, believe or not, the study of such subsets leads to the p-adicmetric (but this is rather complicated to fit in this course, a shame).

The above example connects with the following situation, in the sense that againpolynomial functions are involved. Consider the continuous function f : R→ R givenby f(x) = x2+1. We know that, as such, there is no solution to the equation f(x) = 0.However, we can extend this function to the continuous function F : C → C, whereF (z) = z2 +1, and now the equation F (z) = 0 has solutions (notice that by extendingthe map we change the nature of the map dramatically: the map F is surjectivewheres f was not).

So, what happens in general? Can we extend any continuous function f : R→ R to acontinuous function F : C→ C? Here is a variation of this question. Let A ⊆ R be asubset, and let f : A→ R be a continuous function. Can we extend f to a continuousfunction F : Cl(A) → R? There are many more variations of this question, and the


idea behind all of them is the following: if the domain and codomain of the functionare missing some nice property, can we extend the function to bigger domain and orcodomain, and still keep the continuity property? Well, sometimes we can, sometimesit is not possible.

Also related to continuous functions is the following. Let A ⊆ R be a subspace, andlet C(A) be the set of continuous functions f : A → R. We can give this set thesupremum metric:

d(f, g) = sup|f(x)− g(x)|∣∣ x ∈ A.

Suppose we have a sequence of functions fn ⊆ C(A) satisfying that they get closedto each other when n grows larger and larger (formally we would say that it is aCauchy sequence, but let’s not worry about technicalities right now). What propertiesshould A satisfy so that such a sequence converge in C(A)? Maybe the way thisquestion was posed is a bit confusing, but you already know of an example of suchproblem: Taylor approximation of functions.

These and many more questions relate to the theory of metric and topological spaces.We have only uncovered the tip of the iceberg!

38 ALEX GONZALEZ

6. Connected spaces

Now that we have defined continuous maps between metric spaces, we can start com-paring different metric spaces, trying to find properties that will distinguish one spacefrom another. This chapter deals with the first of such properties: connectedness.

What defines connectivity for a metric space? This question turns out to be a bittricky, since intuition from the most obvious examples would suggest something inthe lines of “a metric space (X, dX) is connected if every two elements of X can belinked through a path in X” (think for instance of R2 with the Euclidean metric). Butwe know better than trusting intuition by now.

If we give it a bit more thinking, a connected metric space should be one that cannotbe formed out of smaller, disjoint pieces (compare (−4, 3) ⊆ R2 with (−4, 2)∪ (2, 3) ⊆R2, always with the Euclidean metric).

Let then (X, dX) be a metric space. Intuitively, the idea that any two points of Xcan be linked by a path in X implies this other idea of X not being made of smaller,disjoint pieces. However, is the converse true? we will see later in this chapter thatthe answer is negative, and so we are presented now with two ideas about how todefine connectivity for metric spaces, one of them seemingly stronger than the other.

Let’s start by formalizing what seems to be the weaker idea above. It actually turnsout to be the right idea to define what a connected metric space is. The other ideawill be discussed at the end of this chapter.

Definition 6.1. A metric space (X, dX) is connected if it satisfies the following.

(C) If a subset A ⊆ X is both open and closed then either A = X or A = ∅.

Let’s see some examples of connected and non-connected spaces.

Example 6.2. Let X be a set with the discrete metric. Then, X is connected if andonly if X is the empty set or a singleton (if X contains more than one element, theneach sub-singleton of X is open and closed).

Example 6.3. Consider R with the Euclidean metric, and let Q be the metric sub-space of rational numbers. Then, Q is not connected. Indeed, the subset

(−∞, π) ∩Q

is open and is also the complement of an open subset, hence closed.

Example 6.4. Consider the set Z with the p-adic metric. This space is not connected,since the subset A = n · p

∣∣ n ∈ Z ⊆ Z of integers divisible by p is both open andclosed. Indeed (see the exercise list 2), we have

A = B0,1 A = Z \( p−1⋃n=1

Bn,1

).


The equality on the left means that A is open, while the equality on the right saysthat A is the complement of an open subset, hence closed.

As we have seen already several times in this course, it is always useful to havealternative statements for the same definition, so let’s see some alternative ways ofdefining connectivity.

Theorem 6.5. Let (X, dX) be a metric space. Then, the following are equivalent.

(1) (X, dX) is connected.

(2) There does not exist open subsetsU, V ⊆ X such thatU∪V = X andU∩V = ∅.

(3) There does not exist any continuous map f : X → 0∪1 which is surjective.

Proof. Let’s check first that (1) is equivalent to (3). Suppose first that there is a sur-jective, continuous map f : X → 0 ∪ 1, and let A = f−1(0) and B = f−1(1).Since f is surjective, both A and B are non-empty subsets of X. Furthermore, since0 (respectively 1) is and open an closed subset of 0 ∪ 1 and f is continuous,it follows that A (respectively B) is an open and closed subset of X. Finally, sinceboth A and B are non-empty, it follows that none of them is either the empty set orX, and thus X is not connected.

Suppose now that X is not connected, and let U $ X be a non-empty, open andclosed subset. Let also V = U c, and notice that it is again a non-empty, open andclosed subset ofX. We can then define a map f : X → 0∪1 by setting f(U) = 0and f(V ) = 1, and it is clear that f is surjective and continuous.

Finally let’s prove that (2) is equivalent to (3). If there exists a map f as in (3), thenU = f−1(0) and V = f−1(1) are open subsets of X which clearly satisfy theconditions of (2). Conversely, if U, V ⊆ X satisfy the conditions of (2) then we justhave to define f by f(x) = 0 if x ∈ U and f(x) = 1 if x ∈ V .

Theorem 6.6. Let f : (X, dX) → (Y, dY ) be a continuous map of metric spaces, andsuppose X is connected. Then, f(X) ⊆ Y is a connected subspace of Y .

Proof. This is immediate using statement (2) in Theorem 6.5.

Corollary 6.7. Let h : (X, dX)→ (Y, dY ) be a homeomorphism of metric spaces. Then,X is connected if and only if Y is connected.

So, which spaces do we know to be connected so far? Looking at the examplesabove in this chapter, the only connected spaces mentioned are the empty set anda singleton, not too much to work with. In fact, most of the examples were of non-connected spaces! The first non-trivial examples of connected spaces that we canproduce need a bit of hard work, but after that everything becomes easier.

40 ALEX GONZALEZ

Theorem 6.8. The set R with the Euclidean metric is connected.

Proof. Suppose R = A∪B, whereA,B are non-empty open subsets such thatA∩B =∅. Then, A and B are both closed (they are the complement of each other), and thus

∂(A) = Cl(A) ∩ Cl(R \ A) = Cl(A) ∩ Cl(B) = A ∩B = ∅,and similarly for B.

On the other hand, we will now show that if U $ R is non-empty, then ∂(U) 6= ∅,hence contradicting the above. Let then a ∈ R \ U . If

U ∩ (a,∞) = ∅ = U ∩ (−∞, a),

then U = a, and hence U = ∂(U) 6= ∅. Suppose then that U ∩ (a,∞) 6= ∅ (the otherpossible case is similar), and let b = infx ∈ U ∩ (a,∞). Then we have

infd(b, u)∣∣ u ∈ U = 0 infd(b, v)

∣∣ v ∈ R \ U = 0,

and this implies that b ∈ ∂(U).

Now it is easy to deduce the following.

Example 6.9. For all a < b ∈ R, the open interval (a, b) ⊆ R is connected. Indeed,(a, b) is clearly homeomorphic to (−π

2, π

2) (give the explicit homeomorphism!), and the

map

tan:

(−π2,π

2

)−→ R

is a homeomorphism.

So open intervals in R are connected. What about closed intervals? Their connectivityis a consequence of the following result.

Proposition 6.10. Let (X, dX) be a metric space and let U,K ⊆ X be subsets. Sup-pose in addition that U is connected and U ⊆ K ⊆ Cl(U). Then K is connected.

Proof. Let f : K → 0∪ 1 be a continuous function. Since U is connected and therestriction of f to U is a continuous function, then either f(U) = 0 or f(U) = 1.Suppose for simplicity that f(U) = 0.

Now, for each k ∈ K, we have k ∈ Cl(U). Let then xn ⊆ U be a sequence convergingto k. Since f is continuous, it follows then that the sequence f(xn) converges tof(k). On the other hand, f(xn) = 0 for all n, and thus f(k) = 0. This prove thatf(K) = 0, and hence K is connected by Theorem 6.5.

Theorem 6.11. Consider R with the Euclidean metric, and let a < b ∈ R. Then,[a, b] ⊆ R is connected.

Proof. This follows immediately from Proposition 6.10 since [a, b] is the closure of(a, b), and the last is connected.


Once we know that R is connected, the next step should be to check that Rn isconnected for all n ≥ 2, but moving from dimension 1 to dimension higher or equalthan 2 is a tricky step.

Lemma 6.12. Let (X, dX) be a metric space, and let Aγγ∈Γ be a collection of con-nected subsets of X such that

Aγ ∩ Aω 6= ∅for all γ, ω ∈ Γ. Then, A = ∪ΓAγ is a connected subset of X.

Proof. Let f : A → 0 ∪ 1 be a continuous function. Since Aγ is connected for allγ ∈ Γ, we have

f(Aγ) = εγ,

where εγ is either 0 or 1. However, for any two γ, ω ∈ Γ, we have Aγ ∩ Aω 6= ∅, andthis implies that εγ = εω. Hence, the map f has to be constant on A: it cannot besurjective, and thus A is connected.

Theorem 6.13. Let (X, dX) and (Y, dY ) be metric spaces, and let (X × Y, d) be theproduct metric space. Then, X × Y is connected if and only if X and Y are connected.

Proof. If X × Y is connected, then both X and Y are connected, since the projectionfunctions πX : X × Y → X and πY : X × Y → Y are continuous by Exercise 4.21.

Suppose now that X and Y are connected. Fix then some element a ∈ X and letΓ = Y (as sets). Consider also the sets

• B = a × Y ⊆ X × Y ; and

• Cy = X × y ⊆ X × Y , where y ∈ Y .

Clearly, all the above subsets are connected subspaces of X × Y .

For each y ∈ Y , we have Cy ∩B = (a, y) 6= ∅, and hence the subset Uy = B ∪Cy ⊆X × Y is connected (for all y ∈ Y ) by Lemma 6.12. Now, consider the collectionUyy∈Y . For any y, y′ ∈ Y , we have Uy ∩ Uy′ 6= ∅, and hence⋃

y∈Y

Uy = X × Y

is connected by Lemma 6.12.

Corollary 6.14. For all n ≥ 1, Rn with the Euclidean metric is connected.

Let’s see now a less trivial example of connected space.

Lemma 6.15. Let f : (X, dX)→ (Y, dY ) be a continuous function, and suppose that Xis connected. Then, the graph of f , Gf = (x, y) ∈ X × Y

∣∣ y = f(x), is a connectedsubspace of X × Y .

Proof. By Exercise 4.25, Gf is homeomorphic to X, and thus is connected.

42 ALEX GONZALEZ

Example 6.16. This example is known as the topologist’s sine curve. Let

Ω = (x, sin(1/x)) ∈ R2∣∣ x > 0 ∪ (0, y) ∈ R2

∣∣ y ∈ [−1, 1] ⊆ R2.

The following is a sketch of Ω.

To show that this space is connected, let

Γ = (x, sin(1/x)) ∈ R2∣∣ x > 0.

First note that the subset Γ is connected because it is the graph of the continuousfunction f : (0,∞)→ R defined by f(x) = sin(1/x), and (0,∞) is connected. Finally,we have already seen in the tutorials that Ω = Cl(Γ), and hence by Proposition 6.10Ω is connected. We will come back to this example in the next section.

6.1. Path-connected spaces. Which other spaces do you know to be connected?Maybe a circle or a disk in R2 (always with the Euclidean metric)? Regular polygons?Spheres? But how to prove that these spaces are connected? By applying Definition6.1 or the alternative statements in Theorem 6.5, proving that any of these spaces isconnected can become a titanic task!

Let’s try something different then. At the beginning of this chapter it was suggestedto define connectivity of metric spaces in terms of linking any two points by a path.As mentioned, this is not the right way to define connectivity, but it is really helpful.

Definition 6.17. Let (X, dX) be a metric space, and let x, y ∈ X. A path in X fromx to y is a continuous map ω : [0, 1]→ X with ω(0) = x and ω(1) = y.

A metric space (X, dX) is path-connected if, for any two elements x, y ∈ X there is apath from x to y.

Theorem 6.18. Let (X, dX) be a path-connected space. Then (X, dX) is connected.

Proof. Suppose that X is not connected, and let f : X → 0 ∪ 1 be a continuous,surjective map. Let also x ∈ f−1(0) and y ∈ f−1(1). Since X is path-connected,


there is some path ω : [0, 1] → X with ω(0) = x and ω(1) = y. However, the compo-sition f ω : [0, 1]→ 0 ∪ 1 is both continuous (it is a composition of continuousfunctions) and is surjective, and this is impossible since [0, 1] is connected by Theo-rem 6.11.

Example 6.19. For all n ≥ 1, Rn is path-connected.

Example 6.20. A circle in R2 is the image of a path ω : [0, 1]→ R which satisfies theextra condition ω(0) = ω(1). Thus, a circle is connected.

Exercise 6.21. Let (X, dX) be a path-connected space, and let x, y, z ∈ X. Let alsoω, γ : [0, 1]→ X be paths from x to y and from y to z respectively. How can you definea path from x to z using ω and γ?

Corollary 6.22. A cylinder is not homeomorphic to a Moebius band.

Proof. Let M be a Moebius band and C be a cylinder, both subsets of R3. Let also∂(M) and ∂(C) be the boundaries of M and C are subsets of R3. Clearly, ∂(C) isnot connected, while ∂(M) is path-connected (hence connected). Hence, no maph : C →M can be a homeomorphism.

It turns out that not all connected spaces are path-connected, and we have alreadymet the first example of this behavior recently.

Example 6.23. Let Ω ⊆ R2 be the subset in Example 6.16, the topologist’s sinecurve, and recall that Ω is connected. Let’s prove now that Ω is not path-connected.Let then

A = (0, y) ∈ R2∣∣ y ∈ [−1, 1] Γ = (x, sin(1/x)) ∈ R2

∣∣ x > 0.If we choose elements a ∈ A and b ∈ Γ, then there is no path in Ω linking a to b! Toprove this, suppose otherwise and let ω : [0, 1] → Ω be a path with ω(0) = (0, 0) andω(1) = ( 1

π, 0). Let also F : [0, 1]→ R be the continuous map

F : [0, 1]ω−→ Ω

π1−→ R,where π1(x, y) = x. Since F (0) = 0 and F (1) = 1

π, we can use the Intermediate Value

Theorem to find some t1 ∈ (0, 1) such that F (t1) = 23π

. Using again the IntermediateValue Theorem we can then find some t2 ∈ (0, t1) such that F (t2) = 2

5π. Iterating the

process we can find a sequence tnn≥1, such that

• tn+1 ∈ (0, tn); and

• F (tn) = 2(2n+1)π

.

Since the sequence tn ⊆ [0, 1] is monotonic decreasing and bounded below, it isconvergent: there is some t ∈ [0, 1] such that lim tn = t. Finally, since ω is continuousthe sequence ω(tn) converges to ω(t). On the other hand, ω(tn) = ( 2

(2n+1)π, (−1)n),

and this sequence is not convergent, hence a contradiction.

44 ALEX GONZALEZ

6.2. Connected components. Connectivity constitutes then the first of a series ofproperties that we can use to characterize and distinguish metric spaces. Let’selaborate this idea a little bit more: a connected space is one which is not made ofsmaller, disjoint pieces, but what happens in general? That is, given a metric space(X, dX), how many pieces are needed to “construct” X?

Definition 6.24. Let (X, dX) be a metric space. A connected component of X is aconnected subset U ⊆ X such that

• if U $ K ⊆ X, then K is not connected.

The collection of connected components of X is denoted by π0(X).

Alternatively, the set π0(X) can be constructed as follows. Define a relation on theset X: x ∼ y if there is some connected subset U ⊆ X such that x, y ∈ U . This isclearly an equivalence relation, and then we can define π0(X) as the set of equivalenceclasses.

Proposition 6.25. Every continuous map f : (X, dX)→ (Y, dY ) induces a map of sets

f∗ : π0(X) −→ π0(Y ).

Proof. Let U ⊆ X be a connected component of X, and let x ∈ U . Let also V ⊆ Ybe a connected component which contains f(x) (each y ∈ Y has to be contained insuch a subset). We can define then f∗(U) = V .

To check that this map is well-defined, let z ∈ U be any other element, and letw = f(z) ∈ Y . We have to prove that f(z) ∈ V . Suppose otherwise: f(z) /∈ V . ThenV $ W = V ∪ f(z), and thus W is not connected. But this is impossible, sincef(U) has to be connected.

Corollary 6.26. Let f : (X, dX)→ (Y, dY ) be a homeomorphism of metric spaces. Thenf defines a bĳection of sets f∗ : π0(X)→ π0(Y ).

In other words, if X and Y are homeomorphic, they have to be made out of the samenumber of pieces.


7. Compact spaces

We will now study one of the most important notions of this course, and (not surpris-ingly) one of the most difficult notions too: compactness of metric spaces. The idea isthe following. If X is a finite set (with the discrete metric) then studying (continuous)maps out of X is reasonably easy, we just have to check what happens with thefunction on every element of X. But we want to consider other maps, not only mapsout of finite sets!

We can think of compactness as some sort of generalization of finiteness, in the sensethat maps out of a compact space behave nicely, as maps out of a finite set (this is,of course, rather rough, a proper justification will come soon in this chapter).

Before we define compactness for metric spaces, let’s review a few facts about metricspaces, subspaces and open subsets which will need in this chapter. Let then (X, dX)be a metric space: the metric dX is a function

dX : X ×X −→ [0,∞)

satisfying a certain list of conditions. Thus, given a subset A ⊆ X, we can automat-ically consider A as a metric subspace of X, just by restricting the above function toA×A ⊆ X ×X: the resulting function dA : A×A→ [0,∞) is a metric on A (all thisshould be clear by now, but it is worth recalling).

Let then (X, dX) be a metric space and let (A, dA) be a metric subspace as above.Which is the relation (if any) between the set of open subsets of A and the set of opensubsets of X? It is as follows: for any open subset U of X, the subset V = U ∩ Ais an open subset of A (check it!), and conversely, any open subset V of A is of theform V = U ∩ A for some open subset U of X.

Exercise 7.1. Let (X, dX) be a metric space, and let A ⊆ X. Prove that x ∈ Cl(A) ifand only if infdX(x, a)

∣∣ a ∈ A = 0.

Definition 7.2. Let (X, dX) be a metric space and let A ⊆ X be any subset. An opencover of A in X is a collection Uγγ∈Γ of open subsets of X such that

A ⊆⋃γ∈Γ

Uγ.

A subcover of Uγγ∈Γ is a subcollection Vωω∈Ω ⊆ Uγγ∈Γ such that A ⊆ ∪ΩVω.

Thus, let (X, dX) be a metric space and let A ⊆ X. Then, it is exactly the same totalk about open covers of A in X, or open covers of A in (A, dA): if Uγγ∈Γ is anopen cover of A, then the collection

Wγ = Uγ ∩ Aγ∈Γ

is a collection of open subsets of (A, dA), and A = ∪ΓWγ.

46 ALEX GONZALEZ

Example 7.3. Consider the family of intervals Un = (n, n+2) ⊆ R (Euclidean metric),for n ∈ Z. The collection Unn∈Z is an open cover of R. Notice also that, if for somen ∈ Z we remove Un from the collection, then it stops being an open cover of R. Inother words, this open cover does not contain any proper subcover.

And we are now ready to define compactness for metric spaces. This definition israther complicated, so take your time to think about it.

Definition 7.4. Let (X, dX) be a metric space and let A ⊆ X. The metric subspace(A, dA) is

(1) bounded if there exist x ∈ X and ε > 0 such that A ⊆ Bx,ε;

(2) compact if every open cover of A in X contains a finite subcover.

Remark 7.5. Let (X, dX) be a metric space and A ⊆ X. Let also Uγγ∈Γ be an opencover of A in X. A finite subcover is then a finite subcollection Vωω∈Ω ⊆ Uγγ∈Γ

which is also an open cover of A. This is already a hint of the idea that compactnessgeneralizes finiteness of metric spaces.

WARNING!. As a note of caution, be extremely careful with the above definition.Specially with the bit every open cover... It is not enough to check that an open covercontains a finite subcover, we have to check ALL possible open covers! Indeed, noticethat, for any metric space (X, dX), the collection X is a finite open cover of X, butthis does not make X compact!

The best way to understand this notion is by looking at some examples. Let’s startwith example of spaces which are not compact.

Example 7.6. Consider R with the Euclidean metric. Then, the open cover in Exam-ple 7.3 does not contain any finite subcover, and hence R is not compact.

Example 7.7. Consider R with the Euclidean metric, and let (0, 1) ⊆ R be the openinterval. Then (0, 1) is not compact. Indeed, consider the open cover

Un =

(1

n, 1

)n∈N

It is easy to see that this open cover does not contain any finite subcover of (0, 1).This is actually not surprising since we already know that (0, 1) is homeomorphic toR.

And what about some example of compact space?

Example 7.8. Let X be a finite set with the discrete metric. Then it is clear that Xis compact: since X contains a finite number of elements, there is a finite number ofopen subsets of X, and hence any open cover of X is already finite.


Fine, but what about a nontrivial example of a compact space? Let’s start withsomething (relatively) easy: the compact subsets of R (with the Euclidean metric).

Theorem 7.9 (Heine-Borel Theorem for R). Consider R with the Euclidean metric, andlet A ⊆ R. Then, A is compact if and only it is closed and bounded.

Proof. Suppose first that A is compact. To see that it is bounded, consider thefollowing open cover of A:

Un = B0,n

∣∣ n ∈ N.Since it is an open cover of A and A is compact, there exists some finite subcover,namely Un1 , . . . , Unr. Let N = maxn1, . . . , nr. Clearly, Un1 , . . . , Unr ⊆ UN , andhence A ⊆ UN = B1,N , i.e. A is bounded.

Let’s see also that A is closed. Suppose otherwise, and let x ∈ Cl(A) \ A. For eachn ∈ N, let

Un =(Cl(Bx, 1

n))c

= R \ Cl(Bx, 1n).

By definition, each Un is the complement of a closed subset of R and thus is open.Furthermore, we have⋃

n∈N

Un =⋃n∈N

R \ Cl(Bx, 1n) = R \

⋂n∈N

Cl(Bx, 1n) = R \ x ⊇ A.

In other words, Unn∈N is an open cover of A in R, and since A is compact, itcontains a finite subcover, Un1 , . . . , Uns. Furthermore, by definition of the subsetsUn, we have Un1 , . . . , Uns ⊆ UN , where N = maxn1, . . . , ns, and also A ⊆ UN . Butthen we have

A ∩Bx, 1N

= ∅,which contradicts the hypothesis that x ∈ Cl(A). Hence A = Cl(A).

Let’s prove now the converse. Assume then that A is closed and bounded, and letUγγ∈Γ be an open cover of A. Since A is closed and bounded, there exist m andM ∈ A such that

m ≤ a ≤M ∀a ∈ A.

Now, for each x ∈ R, define the following sets:

• Wx = (−∞, x]; and

• Ax = A ∩Wx.

Define also B =x ∈ R

∣∣ Ax is covered by a finite subcover of Uγ

, and note thatB is non-empty, since m ∈ B. Indeed, Am = A ∩Wm = m by definition of m, andit is clear that Am is covered by a finite subcover of Uγ.

Notice also that, to finish the proof, it is enough to show that B is not bounded above:if this is the case, then M ≤ y for some y ∈ B, and then by definition of B we havethat A = AM = Ay is covered by a finite subcover of Uγ.

48 ALEX GONZALEZ

We will prove that B is not bounded above by contradiction. Suppose then thatu = supB exists: y ≤ u for all y ∈ B. If u /∈ A, then there exists some ε > 0 suchthat Bu,ε ∩ A = ∅ (this is because A is closed). This implies that

Au−ε = Au+ε.

But notice that u−ε < u, and hence u−ε ∈ B. The above implies then that u+ε ∈ B,and this contradicts the hypothesis that u = supB.

Thus we have to assume that u ∈ A. In this case, since Uγ is an open conver of A,there is some Uω ∈ Uγsuch that u ∈ Uω. Furthermore, since Uω is open, we canfind some δ > 0 such that

[u− δ, u+ δ] ⊆ Uω.

Now, since u − δ < u, the set Au−δ is covered by a finite subcover of Uγ, namelyUγ1 , . . . , Uγm, and then Uω, Uγ1 , . . . , Uγm is a finite subcover of Au+δ. Again, thisimplies that u + δ ∈ B, which contradicts the hypothesis that u = supB. Hence, Bis not bounded above, and A is compact.

We can use this theorem to deduce that some other subsets of R are compact. Thefollowing example is not intuitive at all.

Example 7.10. Let K ⊆ R be the Cantor set (see Exercise 6 in List 3): start fromI0 = [0, 1] ⊆ R, then form I1 = [0, 1

3] ∪ [2

3, 1] by removing the middle third (1

3, 2

3) of I0,

and keep iterating the process. We end up with a sequence Inn∈N of subsets of R,and

K =∞⋂n=0

In.

Since each In is closed and K is an (infinite) intersection of closed subsets, we knowthat K is closed. Also, it is clearly bounded, since K ⊆ I0, and hence by the Heine-Borel it is compact.

A version of the Heine-Borel Theorem is also true when we replace R by Rn, for anyn ≥ 1. However, the proof for Theorem 7.9 does not work in general (why?).

Thus, before attempting to generalize Theorem 7.9, let’s study the behavior of com-pact subsets under different situations: what about subsets of compact spaces? Andabout compact spaces and continuous maps? Finally, last but not least, what aboutthe product of two compact spaces? Each of these questions is answered below andwill be needed to prove the general version of the Heine-Borel Theorem.

If you check carefully the proof of the Heine-Borel Theorem for R, the proof that acompact subset of R is closed and bounded does not pose any complication. In fact,this turns out to be a general statement.

Proposition 7.11. Let (X, dX) be a metric space, and let A ⊆ X be a compact sub-space. Then, A is closed and bounded.


Proof. The same proof as in Theorem 7.9 (with minor modifications) works here, butlet’s reproduce it anyway. First let’s check that A is bounded. Fix some x ∈ X andlet Un = Bx,n for each n ∈ N. The collection Unn∈N is thus an open cover of A in X,and since A is compact there exists some finite subcover, Un1 , . . . , Unr. Let thenN = maxn1, . . . , nr: we have A ⊆ UN = Bx,N and hence A is bounded.

To see that A is also closed, suppose otherwise and let x ∈ Cl(A)\A. For each n ∈ Ndefine then

Un =(Cl(Bx, 1

n))c

= X \ Cl(Bx, 1n).

Then, Un is open for all n and we have⋃n∈N

Un =⋃n∈N

X \ Cl(Bx, 1n) = X \

⋂n∈N

Cl(Bx, 1n) = X \ x ⊆ A.

Thus, Unn∈N is an open cover of A, and since A is compact there is some finitesubcover, Un1 , . . . , Uns. Furthermore, by setting N = maxn1, . . . , ns, we haveA ⊆ UN , which implies

A ∩Bx, 1N

= ∅,contradicting the hypothesis that x ∈ Cl(A). Thus, A = Cl(A).

Lemma 7.12. Let (X, dX) be a metric space and letA ⊆ X be a compact subset. Then,each closed subset K ⊆ A is compact.

Proof. Let Uγγ∈Γ be an open cover of K in X. Since K is closed, its complementV = X \K is open, and the collection Uγγ∈Γ∪V is an open cover of A. Thus, sinceA is compact, there exists some finite subcover, W1, . . . ,Wn.

If W1, . . . ,Wn ⊆ Uγγ∈Γ we are finished, since this is already a finite subcover ofK. If the subset V is one of the elements of W1, . . . ,Wn, then by removing it fromthe collection we obtain a finite subcover of K.

Let’s see now how compact subsets behave when we apply to them continuous func-tions. The first property was to be expected: the image of a compact subset througha continuous function is again compact.

Proposition 7.13. Let f : (X, dX) → (Y, dY ) be a continuous map, and let A ⊆ X bea compact subset. Then, f(A) ⊆ Y is compact.

Proof. Let Uγγ∈Γ be an open cover of f(A) in Y . Since f is a continuous map, thecollection f−1(Uγ)γ∈Γ is then an open cover of A. Now, since A is compact we knowthat there is some finite subcover f−1(Uγ1), . . . , f

−1(Uγr), and thus Uγ1 , . . . , Uγris a finite subcover of f(A).

Let f : [0, 1]→ R be a continuous map. It is a well-known fact that the image of thismap is then bounded in R. But now that we know a bit about compact spaces, wecan analyze what is happening here: [0, 1] is a compact space, and thus the image

50 ALEX GONZALEZ

of f , Im(f), is a compact subset of R. Furthermore, since [0, 1] is connected, so isIm(f). Hence, Im(f) = [a, b] for some a < b ∈ R! It is clear now that Im(f) is boundedabove and below.

Corollary 7.14. Let f : (X, dX) → (Y, dY ) be a continuous map, and let A ⊆ X be acompact subset. Then, f(A) is bounded in Y .

Corollary 7.15. Let f : (X, dX) → (Y, dY ) be a homeomorphism. Then, X is compactif and only if Y is compact.

Example 7.16. Recall from the tutorials that the Cantor set, K, in Example 7.10, hasthe cardinality of the set R of real numbers. But now we know that K is compact,while R is not. Thus, these sets are not homeomorphic, despite having the samecardinality!

The last property of compact subsets that we need in order to prove the Heine-BorelTheorem for Rn turns out to be also the most complicated of all.

Theorem 7.17. Let (X, dX) and (Y, dY ) be metric spaces, and let A ⊆ X, B ⊆ Y becompact subsets. Then, A×B is a compact subset of X × Y .

Proof. Let Wγγ∈Γ be an open cover of A × B. In order to make this proof easier tofollow, let’s divide it in two main parts.

• We can assume that, for all γ ∈ Γ, Wγ = Uγ × Vγ, where Uγ ⊆ X and Vγ ⊆ Yare open subsets.

Recall that, with the product metric d on X × Y , we have

Bd(x,y),ε = BdX

x,ε ×BdYy,ε

for all (x, y) ∈ X × Y and all ε > 0.

Now, fix some γ ∈ Γ. Since Wγ ⊆ X × Y is open, for each (x, y) ∈ Wγ there existssome ε > 0 (which depends on (x, y)) such that Bd

(x,y),ε ⊆ Wγ, and

Wγ =⋃

(x,y)∈Wγ

Bd(x,y),ε.

We can then consider the following open cover of A×B

Bd(x,y),ε

∣∣ (x, y) ∈ Wγγ∈Γ

If this open cover contains a finite subcover then the original open cover will alsocontain a finite subcover (check it!).

• Let Wγγ∈Γ be an open cover of A × B, with Wγ = Uγ × Vγ for all γ. Then,this open cover contains a finite subcover.


For each a ∈ A, consider the subset a×B ⊆ A×B. This subset is homeomorphicto B, which is a compact subset of Y by hypothesis, and thus a×B is compact too,by Corollary 7.15. Since Wγ also covers a×B, there exists some finite subcover

Ωa = Wa,i1 , . . . ,Wa,ia.The following picture illustrates the situation.

A

B

Wa,1

Wa,2

Wa,3

Wa,4

Wa,5Wa,6

a

Recall that, for k = i1, . . . , ia, Wa,k = Ua,k × Va,k for some Ua,k and Va,k open in A andB respectively. Define then

Ua = Ua,i1 ∩ Ua,i2 ∩ . . . ∩ Ua,ia .This is a non-empty open subset of X, and thus the collection Uaa∈A is an opencover of A. Since A is compact, there exist a1, . . . , am ∈ A such that

Ua1 , . . . , Uam ⊆ Uaa∈Ais a finite subcover.

This finishes the proof, because the collection

Ωa1 ∪ . . . ∪ Ωam = Wa1,i1 , . . . ,Wa1,ia1 ∪ . . . ∪ Wam,i1 , . . . ,Wam,iam ⊆ Wγγ∈Γ

is a finite subcover of A×B.

Corollary 7.18. Let (X, dX) and (Y, dY ) be metric spaces, and let A ⊆ X, B ⊆ Y besubsets. Then, A and B are compact if and only if A×B is a compact subset of X×Y .

We can finally prove the Heine-Borel Theorem in its general version.

52 ALEX GONZALEZ

Theorem 7.19 (Heine-Borel Theorem for Rn). Consider Rn with the Euclidean metric,and let A ⊆ Rn be a subset. Then, A is compact if and only if it is closed and bounded.

Proof. By Proposition 7.11, if A is compact then it is closed and bounded. Supposethen that A is closed and bounded. In particular, there is some interval [m,M ] ⊆ Rsuch that

A ⊆ I = [m,M ]× . . .× [m,M ] ⊆ Rn.

Now, the subset I ⊆ Rn is compact by Theorem 7.17, since I is the direct productof compact subsets of R, and then by Lemma 7.12 A is compact since it is a closedsubset of I.

We can use now the Heine-Borel Theorem to prove that a certain subset of R iscompact.

Example 7.20. Consider R2 with the Euclidean metric. Then, every closed disk andevery circle is compact. More generally, in the Euclidean space Rn, the sphere Sn−1,

Sn−1 = (x1, . . . , xn) ∈ Rn∣∣ x2

1 + . . .+ x2n = 1

is a compact space.

Example 7.21. The cylinder and the Moebius band are both compact subsets ofR3. Thus compactness is not a good property to distinguish these two subsets.Fortunately connectivity has done the job for us already.

The Heine-Borel Theorem is really nice: it gives an easy-to-apply criterion to decidewhetherA ⊆ Rn is compact or not. It would be great if such a tool was available for allmetric spaces, wouldn’t it? However, things are not that nice, and a counterexamplearises immediately.

Example 7.22. Consider the set Z with the discrete metric. Then, Z is closed andbounded, but not compact. Indeed, Z is closed as a subset of itself, and it is boundedbecause Z is contained in the open ball of radius 2 around any element. However,every singleton n in Z is an open subset, and hence the collection

n

n∈Z is an

open cover which does not contain any finite subcover.

Compactness is a rather strong property, as we already know: think for instance ofthe difference between continuous functions from (0, 1) (which is not compact) to Rand continuous functions from [0, 1] (which is compact) to R. In general, it is betterto work with compact spaces, but as we have seen in this section it is difficult ingeneral to show that a space is compact. The goal is then to find some criterion forcompactness. This will be done in the next chapter.


7.1. Continuity improved: uniform continuity. In this short section we show howcompactness affects to continuity of function.

Definition 7.23. A map f : (X, dX) → (Y, dY ) between metric spaces is uniformlycontinuous if, for all ε > 0 there exists some δ > 0 such that

if dX(x, y) < δ then dY (f(x), f(y)) < ε.

Notice the difference with the definition of a continuous function (Definition 4.3):here, δ may depend on ε, but not on the elements x and y. Thus, this is a strongernotion than usual continuity.

Proposition 7.24. Let f : (X, dX)→ (Y, dY ) be a continuous function of metric spaces,and suppose in addition that X is compact. Then f is uniformly continuous.

Proof. Let ε > 0. Since f is a continuous map, we know that for each x ∈ X thereexists some δ(x) > 0 such that

if dX(x, y) < 2δ(x) then dY (f(x), f(y)) <ε

2.

Clearly, the collection Bx,δ(x)x∈X is an open cover of X, and since X is compactwe know that there exist x1, . . . , xm ∈ X such that Bx1,δ(x1), . . . , Bxm,δ(xm) is a finitesubcover of X.

Let then δ = minδ(x1), . . . , δ(xm), and let x, y ∈ X be such that dX(x, y) < δ. Sincethe above is a finite subcover of X, we have x ∈ Bxk,δ(xk) for some xk as above, andthen by the triangle inequality

dX(xi, y) ≤ dX(xi, x) + dX(x, y) < δ(xi) + δ ≤ 2δ(xi).

By the continuity of f at the element xi, we havedX(xi, x) < δ(xi) < 2δ(xi) =⇒ dY (f(xi), f(x)) < ε

2

dX(xi, y) < 2δ(xi) =⇒ dY (f(xi), f(y)) < ε2

and hence using the triangle inequality and the symmetry property of dYif dX(x, y) < δ then dY (f(x), f(y)) ≤ dY (f(xi), f(x)) + dY (f(xi), f(y)) < ε

and the proof is finished.

54 ALEX GONZALEZ

8. Complete spaces

The main purpose of this chapter is to give conditions for a space to be compact.Surprisingly enough, in this chapter sequences and their convergence become themain character. First of all, let’s recall the notion of Cauchy sequence. This notionhas already appeared briefly in some tutorials and in the continuous assessment.

Definition 8.1. Let (X, dX) be a metric space. A sequence xnn∈N in X is a Cauchysequence if, for all ε > 0, there exists some M ∈ N such that

dX(xn, xm) < ε for all n, m ≥M.

Recall from the tutorials that every convergent sequence is a Cauchy sequence, butthe converse is not necessarily true (the p-adic metric on Z provides examples ofthis).

Definition 8.2. A metric space (X, dX) is complete if every Cauchy sequence is con-vergent.

Thus, in a complete space a sequence is convergent if and only if it is Cauchy.

Example 8.3. Let X be a set with the discrete metric. Then X is a complete space.Indeed, let xnn∈N be a Cauchy sequence, and let ε = 1

2. Then, there exists some

N ∈ N such that, for all n,m ≥ N ,

d(xn, xm) <1

2,

which implies that d(xn, xm) = 0. In other words, the sequence is constant forall n ≥ N , and hence it is convergent. This is also an example of the fact thatcompleteness does not imply compactness!

Before studying other examples, notice that the definition of Cauchy sequence onlyinvolves the terms of the sequence, whereas the notion of convergent sequence in-volves also the limit of the sequence (which we may not know a priori).

Example 8.4. Consider the sequence xnn∈N, where xn = Σnk=1

1k2

. Let also n,m ∈ N,and suppose n ≥ m for simplicity. Then,

|xn − xm| = xn − xm = Σnk=m+1

1

k2≤

≤ Σnk=m+1

( 1

k − 1− 1

k

)=

1

m− 1

n≤ 1

m+

1

n

Now it is easy to check that xn is a Cauchy sequence. Actually, this sequenceis also convergent (we will show soon that R is complete). What is the limit of thissequence?

The next result is probably known to you already, but it is nevertheless worth re-viewing.


Theorem 8.5. The set R with the Euclidean metric is complete.

Proof. Let xnn∈N be a Cauchy sequence of real numbers. First, let’s check that thesequence must be bounded. By applying the Cauchy condition with ε = 1 it followsthat there is some N ∈ N such that, for all n,m ≥ N ,

|xn − xm| < 1.

In particular, the above inequality is true for m = N and all n ≥ N . We have

|xn| − |xN | ≤ |xn − xN | < 1,

and hence |xn| < 1 + |xN |. Thus, it is clear that the sequence is bounded, since forall n ∈ N we have

|xn| ≤ max|x1|, |x2|, . . . , |xN−1|, 1 + |xN |.

Next, out of xnn∈N we construct a new sequence, ynn∈N, which is convergent andbounds xn in a particular way. Let

An = xm∣∣ m ≥ n,

and note that A1 = xnn∈N, and An+1 ⊆ An for all n. Furthermore, since the set A1

is bounded, so is An for each n. We may consider then the supremum of An:

yn = sup(An).

The sequence ynn∈N is monotonic decreasing, since An+1 ⊆ An, and is boundedbelow because A1 is bounded below and yn ≥ xn for all n. Thus, this new sequenceis convergent. Let

L = limn∈N

yn.

The proof is finished if we show that xnn∈N also converges to L. Given ε > 0, thereexist

(1) N1 ∈ N such that, for all n,m ≥ N1, |xn − xm| < ε;

(2) N2 ∈ N such that, for all k ≥ N2, |L− yk| < ε.

Let N = maxN1, N2. Since yN = sup(AN), there must exist some M ≥ N such that

xM > yN − ε and xM ≤ yN .

Thus, for all n ≥ N , we have

|L− xn| = |L− yN + yN − xM + xM − xn| ≤ |L− yN |+ |yN − xM |+ |xM − xn| < 3ε

which proves that L is the limit of xnn∈N.

We can use this example to study other examples.

56 ALEX GONZALEZ

Example 8.6. The set of rational numbers Q with the Euclidean metric is not com-plete. For instance, let

xn =b10n · πc

10n,

where byc denotes the greatest integer which is smaller or equal to y. In this case,xn keeps the first n decimal terms of π and ignores the rest. Clearly, xnn∈N is aCauchy sequence, but it is not convergent in Q.

Example 8.7. The open interval (0, 1) ⊆ R with the Euclidean metric is not complete.For instance the sequence xn = 1

2nn∈N is Cauchy but not convergent.

The above example is actually rather important: recall from Example 6.9 that (0, 1)is homeomorphic to R. On the other hand, R is complete while (0, 1) is not.

Corollary 8.8. Completeness is not preserved by homeomorphisms.

Following the usual program of previous chapters, we next study whether Rn is alsocomplete, for n ≥ 2.

Theorem 8.9. For all n ≥ 1, the Euclidean space Rn is complete.

Proof. By Theorem 8.5 we can assume that n ≥ 2. Let then xkk∈N be a Cauchysequence in Rn, where

xk = (x1,k, . . . , xn,k)

for all k ∈ N. Now, the sequence xi,kk∈N is a Cauchy sequence in R for eachi = 1, . . . , n (check the details), and hence it is convergent. For each i write

Li = limk∈N

xi,k.

To finish the proof we have to check that xkk∈N converges to the point L =(L1, . . . , Ln). Since each of the sequences xi,kk∈N is convergent, for each ε > 0there exist N1, . . . , Nn ∈ N such that

if k ≥ Ni then |Li − xi,k| < ε.

Let then N = maxN1, . . . , Nn. If k ≥ N , then

d(L,xk) =√

Σni=1(Li − xi,k)2 <

√n · ε.

Hence, the sequence xkk∈N is convergent, and Rn is complete.

The following, more general result will be proved in the tutorials (the proof above is agood guideline to prove the theorem below).

Theorem 8.10. Let (X1, d1), . . . (Xn, dn) be metric spaces. Then, X1 × . . . ×Xn (withthe product metric) is complete if and only if (Xi, di) is complete for all i = 1, . . . , n.


Do you know of other complete metric spaces? Well, in fact you do: rememberExercise 3 in the continuous assessment?

Example 8.11. Let C[0, 1] be the set of continuous functions from [0, 1] to R, withthe metric

d(f, g) = max|f(x)− g(x)|∣∣ x ∈ [0, 1].

Then, C[0, 1] is complete, as was proved in the continuous assessment.

However, the same set C[0, 1] with the following metric is not complete any more:

d′(f, g) =

∫ 1

0

|f(x)− g(x)|dx.

Indeed, consider the sequence of functions fnn∈N, where

fn(x) =

0, x ∈

[0, 1

2− 1

n

]n ·(x+ 1

n− 1

2

), x ∈

[12− 1

n, 1

2

]1, x ∈

[12, 1]

As an exercise, check that this sequence is indeed a Cauchy sequence with respectto the metric d′ above. We have to prove that this sequence does not converge (againwith respect to the metric d′).

Be careful with the following: it seems that if the sequence fnn∈N converges, itshould converge to the function

f(x) =

0, x ∈

[0, 1

2

)1, x ∈

[12, 1]

which is not continuous. But this is NOT a proof that the sequence fnn∈N is notconvergent with respect to the metric d′: it could still be the case that there is somecrazy (i.e, impossible to describe with a formula as above), continuous function towhich the sequence converges. We have to discard this possibility.

Suppose the sequence fnn∈N converges to a function g ∈ C[0, 1] with respect to themetric d′. Then the following holds:

(1) for all n ∈ N and all x ∈ [0, 1],

|f(x)− g(x)| = |f(x)− fn(x) + fn(x)− g(x)| ≤ |f(x)− fn(x)|+ |g(x)− fn(x)|;

(2) for all n ∈ N,∫ 1

0|f(x)− fn(x)|dx = 1

2n; and

(3) if αn =∫ 1

0|g(x) − fn(x)|dx for each n ∈ N, then the sequence αnn∈N is

convergent with limit 0.

Thus, the sequence βn = 12n

+ αnn∈N also converges to 0, and we have for all n

0 ≤∫ 1

0

|f(x)− g(x)|dx ≤∫ 1

0

|f(x)− fn(x)|dx+

∫ 1

0

|g(x)− fn(x)|dx = βn

58 ALEX GONZALEZ

This means that g(x) = f(x) for all x ∈ [0, 1] such that both f and g are continuousat x. In other words, g(x) = f(x) for all x ∈ [0, 1

2) ∪ (1

2, 1], and this is impossible

because g is continuous.

8.1. Properties of complete spaces. We have already seen some examples of com-plete and non complete metric space, and now it is time to study general propertiesof complete spaces and their subsets.

Lemma 8.12. Let (X, dX) be a metric space and let A ⊆ X be a complete metricsubspace. Then A is closed in X.

Proof. Let x ∈ Cl(A). Then, there exists a sequence xnn∈N ⊆ A converging to x.Since every convergent sequence is a Cauchy sequence and A is complete, it followsthat the sequence xnn∈N is convergent to an element of A, and hence x ∈ A.

Lemma 8.13. Let (X, dX) be a complete metric space, and let A ⊆ X be a closedsubset. Then A is a complete metric subspace.

Proof. Let xnn∈N be a Cauchy sequence in A. In particular, this is also a sequencein X, and since X is complete it follows that the sequence xn is convergent, withlimit x. Since A is closed, then x ∈ A.

A more important result is the following, relating the notions of compactness andcompleteness.

Proposition 8.14. Let (X, dX) be a compact metric space. Then (X, dX) is complete.

Proof. Let xnn∈N be a Cauchy sequence in A, and let S ⊆ A be the following subset

S = x1, x2, . . . , xn, . . . ⊆ A.

We may assume that S is infinite (if S is finite, then it follows immediately that thesequence xn is convergent). The proof is a bit long, so it is better to divide it inshorter steps.

(1) There exists an “accumulation point” a ∈ A: for each ε > 0,

Ba,ε ∩ S 6= ∅ or a.

Suppose otherwise that no such element a ∈ A exists. Then, for each x ∈ A thereexists some ε(x) > 0 such that

Bx,ε(x) ∩ S = ∅ or x.The collection of open balls Bx,ε(x)x∈A is an open cover of A, and it must contain afinite subcover since A is compact: there exist x1, . . . , xk such that

S ⊆ A ⊆ Bx1,ε(x1) ∪Bx2,ε(x2) ∪ . . . ∪Bxk,ε(xk).

But we are assuming that S is infinite, and also that the intersection Bx,ε(x) ∩ S iseither empty or x for all x ∈ A, and hence, the above is impossible!


(2) Let a ∈ A be an “accumulation point” as in (1). Then, xn contains a subse-quence xn(k)k∈N which converges to a.

Note that we can assume that a /∈ S (otherwise, we can remove all the terms xn inthe sequence xn satisfying xn = a). This part of the proof is done by induction, solet first n(1) ∈ N be such that

xn(1) ∈ Ba,1 ∩ S.Such an integer must exist by part (1). Suppose now that we have chosen integersn(1) < n(2) < . . . < n(k) such that

xn(i) ∈ Ba, 1i∩ S i = 1, . . . , k,

and let ε = min 1k+1

, dX(a, x1), dX(a, x2), . . . , dX(a, xn(k)). Since xn 6= a for all n, itfollows that ε > 0, and then by part (1) there exists some n(k + 1) ∈ N such that

xn(k+1) ∈ Ba,ε ∩ S ⊆ Ba, 1k+1∩ S.

Furthermore, the choice of ε implies that n(k + 1) > n(k), and this completes theinduction step.

Now, it is clear that the sequence xn(k)k∈N converges to a since, for all N ∈ N, ifn(k) ≥ N then dX(a, xn(k)) <

1k≤ 1

N.

(3) The sequence xnn∈N converges to a because it is a Cauchy sequence.

Let ε > 0. Then there exist

(1) N ∈ N such that, if n,m ≥ N , then dX(xn, xm) < ε;

(2) R ∈ N such that, if r ≥ R, then dX(a, xn(r)) < ε.

Choose then r ∈ N such that n(r) ≥ N,R. Then

if n ≥ N , then dX(a, xn) ≤ dX(a, xn(r)) + dX(xn(r), xn) < 2 · ε

and hence the sequence xnn∈N converges to a.

Recall from Proposition 7.11 that, if A ⊆ X is a compact subset of the metric space(X, dX), then A is also closed and bounded. Hence, we deduce the following.

Corollary 8.15. Let (X, dX) be a metric space and let A ⊆ X be a compact subset.Then A is closed, bounded and complete.

The converse of the above corollary is not true, and a counter-example has alreadyappeared in this chapter: the set N with the discrete metric is closed (every metricspace is closed), it is bounded (since N is contained in the open ball of center 1 andradius 2), and it is complete (because it has the discrete metric); on the other hand,we have already seen in the tutorials that N with the discrete metric is not compact.Nevertheless, this is almost what we are after.

60 ALEX GONZALEZ

Definition 8.16. Let (X, dX) be a metric space. A subset A ⊆ X is totally boundedif for each ε > 0 there exists a finite set Sε ⊆ A such that

A ⊆⋃x∈Sε

Bx,ε.

Exercise 8.17. Let (X, dX) be a metric space. Show that if A ⊆ X is totally boundedthen it is bounded. Give an example of a metric space which is bounded but nottotally bounded.

The following result is a refinement of something that we already knew: if a subsetis compact, then it is bounded.

Lemma 8.18. Let (X, dX) be a metric space, and let A ⊆ X be a compact subset.Then, A is totally bounded.

Proof. Let ε > 0, and consider the open cover Ba,εa∈A. Since A is compact, thisopen cover must contain a finite subcover. In other words, there exist a1, . . . , ak suchthat A ⊆ Ba1,ε ∪ . . . ∪Bak,ε.

Corollary 8.19. Let (X, dX) be a metric space, and let A ⊆ X be a compact subset.Then, A is complete and totally bounded.

Proof. A is complete (Proposition 8.14), and A is totally bounded (Lemma 8.18).

Theorem 8.20. Let (X, dX) be a metric space, and let A ⊆ X be a complete, totallybounded subset. Then A is compact.

Proof. This proof is also done by contradiction, so suppose that Uγγ∈Γ is an opencover of A which does not contain any finite subcover.

Since A is totally bounded, in particular there exists a finite set S1 ⊆ A such that

A ⊆⋃x∈S1

Bx,1.

Suppose that, for each x ∈ S1, the subset A∩Bx,1 could be covered by a finite numberof subsets Uγ in the open cover. Then, Uγ should contain a finite subcover, andwe are assuming this is not the case. Hence, there exists y0 ∈ S1 such that A∩By0,1

cannot be covered with a finite number of subsets Uγ.

Since A is totally bounded and A∩By0,1 ⊆ A, then A∩By0,1 is again totally bounded.In particular, there exists some finite subset S2 ⊆ A ∩By0,1 such that

A ∩By0,1 ⊆⋃x∈S2

Bx, 12.

Note also that dX(y0, x) ≤ 1 + 12

for all x ∈ S2 by the triangle inequality.

Again, suppose that, for each x ∈ S2, the subset A∩Bx, 12

could be covered by a finitenumber of subsets Uγ in the open cover. Then, A ∩ By0,1 would also be covered by


a finite number of elements Uγ, which contradicts our assumptions. Hence, thereexists some y1 ∈ S2 such that A ∩ By1,

12

cannot be covered by a finite number ofelements Uγ.

The assumption that Uγ does not contain any finite subcover implies that thisprocess goes on forever. In particular, at the n-th step we can choose the radius ofthe open balls to be 1

2n, and this for each n ≥ 0. This way, we construct a sequence

ynn≥0 such that

dX(xn, xn+1) ≤ 1

2n+

1

2n+1,

and in particular the sequence yn is Cauchy (as an exercise, check that this isindeed the case).

Now, since A is complete by hypothesis, the sequence yn must converge to anelement a ∈ A. Let then Uω ∈ UγΓ be such that a ∈ Uω.

Since Uω is an open subset, there exists some δ > 0 such that

Ba,δ ⊆ Uω.

Furthermore, there exists some n ∈ N such that

dX(yn, a) <δ

2

1

2n<δ

2.

Putting everything together, we have

A ∩Byn,12n⊆ Byn,

12n⊆ Ba,δ ⊆ Uω,

which is impossible since we chose each yn so that A ∩ Byn,12n

cannot be covered afinite number of elements Uγ.

8.2. The completion of a metric space. There are many situations where workingin a complete metric space makes life much nicer. Fortunately for us, given a metricspace (X, d) we can always produce a new metric space (X, d) which is complete andunique with respect to X in a very sensible way. This process is called completing themetric space (X, d). Let’s formalize this.

Definition 8.21. Let (X, d) be a metric space. A completion of (X, d) is a metricspace (X, d), together with a map ι : X → X, such that

(1) (X, d) is complete;

(2) ι is an isometry; and

(3) Cl(ι(X)) = X.

The task is double: we have to prove that such a completion always exists, and thatit is unique up to homeomorphism. But first, and example.

62 ALEX GONZALEZ

Example 8.22. Consider the set Q with the Euclidean metric. Then, R with Eu-clidean metric is a completion of Q: R is complete, the inclusion of Q in R is anisometry, and the closure of Q in R is the whole set of real numbers R.

Let now (X, d) be a metric space. We can then consider the set of all Cauchy se-quences in X, which we will call C(X) (so an element of C(X) is a Cauchy sequencexnn∈N in X), and we can define a relation in C(X):

xnn∈N ∼ ynn∈N ⇔ limn→∞

d(xn, yn) = 0.

Exercise 8.23. Check that this is an equivalence relation.

We can define the set X = C(X)/ ∼ of equivalence classes of Cauchy sequences inX. Also, define the function d : X × X → R by the formula

d(x,y) = limn→∞

d(xn, yn), (7)

where xnn∈N and ynn∈N are representatives of x and y in C(X).

Before we do anything else, there is something we have to check: it is not clear that dis a well-defined function. Thus, we have to make sure that the limit limn→∞ d(xn, yn)in the definition exists and that it does not depend on a particular choice of repre-sentatives.

Lemma 8.24. The function d in (7) is well-defined.

Proof. Let x,y ∈ X, and let xnn∈N, ynn∈N ∈ C(X) be some representatives. Thenwe can consider the sequence of real numbers d(xn, yn)n∈N: it is a Cauchy se-quence! Indeed, for any n,m ∈ N, we have

|d(xn, yn)− d(xm, ym)| = |d(xn, yn)− d(yn, xm) + d(yn, xm)− d(xm, ym)| ≤≤ |d(xn, yn)− d(yn, xm)|+ |d(yn, xm)− d(xm, ym)| ≤≤ d(xn, xm) + d(yn, ym),

and thus d(xn, yn)n∈N is a Cauchy sequence because both xn and yn areCauchy sequences. Since R is complete, it follows that d(xn, yn) is a convergentsequence and its limit exists.

Let’s check now that the definition of d does not depend on a particular choice ofrepresentatives for x and y. Let then x′n, y′n ∈ C(X) be different representatives.Then the triangle inequality implies that, for all n,

d(x′n, y′n) ≤ d(x′n, xn) + d(xn, yn) + d(yn, y

′n).

By hypothesis, xn ∼ x′n and yn ∼ y′n, and hence we have

limn→∞

d(x′n, y′n) = lim

n→∞d(x′n, xn) + lim

n→∞d(xn, yn) + lim

n→∞d(yn, y

′n) = lim

n→∞d(xn, yn).


Proposition 8.25. The function d defined above is a metric on X.

Proof. Clearly, d(x,y) ≥ 0 for all x,y ∈ X. As an exercise, check that d(x,y) = 0 ifand only if x = y. Symmetry and the triangle inequality follow just because d is ametric and already satisfies these properties.

Before we can prove that (X, d) is a completion of (X, d) we are still missing one pieceof data: a reasonable map ι : X → X! How can we map X into X? The simpler thebetter: by using the most basic of all Cauchy (and convergent) sequences: for eachx ∈ X, define

ι : X −→ X

by mapping each x ∈ X to the class of the constant sequence xn = xn∈N.

Theorem 8.26. Let (X, d) be a metric space. Then, (X, d), together with the map ιabove, is a completion of (X, d).

Proof. Let’s check the conditions as they appear in Definition 8.21.

(i) (X, d) is complete.

Let xnn∈N be a Cauchy sequence in X. We have to check that it is convergent. Foreach n ∈ N, fix xn,kk∈N ∈ C(X) be a representative of xn, and let yk = xk,kk∈N.The following picture may help you in following the proof.

k ∈ N

n ∈ N

x1,1

x1,2

x1,3

...

x1,k

x1

x2,1

x2,2

x2,3

...

x2,k

x2

x3,1

x3,2

x3,3

...

x3,k

x3

. . .

. . .

. . .

. ..

. . .

. . .

xn,1

xn,2

xn,3

...

xn,k

xn

x1,1

x2,2

x3,3

...

xk,k

x?

64 ALEX GONZALEZ

By the triangle inequality we have

d(yk,k, yr,r) ≤ d(xk,k, xk,r) + d(xk,r, xr,r),

and it follows that yk is a Cauchy sequence because xk,rr∈N is a Cauchy sequencein X and xnn∈N is a Cauchy sequence in X. Let then y ∈ X be the class of yk.

Again using the triangle inequality, we have

d(xk,k, xn,k) ≤ d(xkk, xn,n) + d(xn,n, xn,k),

and this implies that

d(y,xn) = limk→∞ d(xk,k, xn,k)n→∞ // 0

In other words, y is the limit of the sequence xn, and thus X is complete.

(ii) The map ι(x) is an isometry.

Recall that ι : X → X sends each x ∈ X to the class of the constant sequencexn = xn∈N. Let then x, y ∈ X. We have

d(ι(x), ι(y)) = limn→∞

d(xn, yn) = limn→∞

d(x, y) = d(x, y).

(iii) Cl(ι(X)) = X.

We have to check that, for each x ∈ X, there is a sequence in ι(X) that converges tox. Let ynn∈N ∈ C(X) be a sequence representing x, and let xn = ι(yn) for each n.The sequence xn is clearly a sequence in ι(X).

This way, we have

d(x,xn) = limk→∞ d(yk, yn)n→∞ // 0

since yn is a Cauchy sequence. Hence, the closure of ι(X) is X.

We have just proved that for each metric space there is always a completion, andnow we have to make sure that this completion is unique in a sensible way.

Theorem 8.27. Let (X, d) be a metric space. Then, the completion (X, d) is unique upto homeomorphism.

Proof. Let (X ′, d′), together with a map ι′ : X → X ′, be another completion of (X, d).We have to construct a homeomorphism between (X, d) and (X ′, d′).

For each x ∈ X let ι(xn)n∈N be a sequence in ι(X) which converges to x (sucha sequence exists because Cl(ι(X)) = X). We can consider then the sequenceι′(xn)n∈N in X ′.

Since ι′ is an isometry and the sequence xn is a Cauchy sequence, it follows thatι′(xn) is a Cauchy sequence in X ′. Furthermore, (X ′, d′) is complete by hypothesis,


and hence the sequence ι′(xn) is convergent, and we can define a map h : X → X ′

by h(x) = limn→∞ ι′(xn).

To check that this is well-defined, let ynn∈N is another sequence in X such thatι(yn)n∈N converges to x. Then, since ι is an isometry, we have

limn→∞


d(ι(xn), ι(yn)) = d( limn→∞

ι(xn), limn→∞

ι(yn)) = d(x,x) = 0,

and then, since ι′ is also an isometry, we have limn→∞ ι′(xn) = limn→∞ ι

′(yn).

In fact, the map h : X → X ′ is an isometry. Indeed, let x,y ∈ X, and let xn, ynbe sequences in X such that ι(xn), ι(yn) converge to x,y respectively. Then,

d′(h(x), h(y)) = d′( limn→∞

ι′(xn), limn→∞

ι′(yn)) = limn→∞

d′(ι′(xn), ι′(yn)) =

= limn→∞


d(ι(xn), ι(yn)) = d(x,y).

This proves that h is a continuous map. Notice that the definition of h only uses theproperties of (X, d) as a completion of (X, d), but not the explicit definition of (X, d).This means that we can define the inverse h−1 just as we defined h itself, and it willbe an isometry again. Thus, h is bĳective, and both h and h−1 are continuous: h isa homeomorphism.

66 ALEX GONZALEZ

9. Interlude II

One way of introducing what comes next , perhaps a bit too strong, is that we havebeen doing too much work in the previous chapters. Well of course this is not true, itwas absolutely necessary to understand how a metric determines certain properties,such as continuity, connectivity, etc.

However, maybe you have noticed this already, the metric has been playing more andmore a decorative role. If it did not cross your mind before, think of it now: conti-nuity has been defined in terms of open/closed subsets, connectivity is also definedin terms of open/closed subsets, compactness is defined in terms of open/closedsubsets...

Furthermore, since practically the beginning of the course we know lots of examplesof different metrics that define the same open/closed subsets. What is the point thenin fixing a metric and having to stick to it? Continuity of functions, convergence ofsequences, connectivity... all these properties do not depend on the metric but onthe equivalence class of a metric.

Thinking more geometrically, let me introduce the Klein bottle. This is a subset ofR4, defined as the solutions of the equation(x2 +y2 + z2 + 2y−1

)[(x2 +y2 + z2−2y−1

)2−8z2]

+ 16xz(x2 +y2 + z2−2y−1

)= 0.

This way, the Klein bottle becomes a metric space, with the Euclidean metric. Youmay have heard of this amazing surface, where inside cannot be told apart fromoutside (as opposed to an sphere, where it is pretty clear what inside and outsidemean), perhaps you have seen this picture before

However, from the point of view of metric spaces this is not very useful. For instance,how can we use the above description to define continuous maps to other metricspaces? Even checking that this space is connected is already a rather complicatedtask.

In addition, the above picture is really nice, but it is hard to manipulate. Let’s finda simpler way of representing the Klein bottle. You may remember that a cylinder


can be constructed from a stripe by glueing together two opposite sides following thesame direction, while a Moebius band can be constructed from the same stripe byglueing together two opposite sides following opposite directions.

Cylinder

a a

Moebius band

bb

We can represent the Klein bottle in a similar way, as represented in the followingpicture.

a

a

b b

This representation is easier to manipulate, but still it does not make it much easierto define a metric on the Klein bottle. The natural thing to do seems to be to inducea metric from the Euclidean metric on the square, but since we are glueing together(identifying) parts of the square that were unrelated, this leads to all sorts of trouble.

There must be some easier way... or perhaps the problem is that we are trying sodesperately to keep a metric that we have not considered that maybe we do not needthe metric any more.

Before we start formalizing all this, let me give you another reason why we shoulddefinitely consider something more general than metric spaces. we have recently seenhow to construct a complete metric space out of any metric space, a process that wecalled completion. The idea for completion was to create the smallest possible metricspace which has the missing property (in this case, completeness) and contains theoriginal space.

Well, compactness is an important property, so why not doing the same for non-compact metric spaces? Simply because there is no reasonable way to do it in thelanguage of metric space (here, reasonable means that can be done without fillingpages and pages of calculations and proofs).

Time to leave the nest...

68 ALEX GONZALEZ

10. Topological spaces

Roughly speaking, a metric d on a set X is a rule to say whether two points are closeor far from each other, by means of an exact scalar. From this point of view, in atopological space we still want to tell whether two points are close or far from eachother, but in a vaguer way. Here is the proper definition.

Definition 10.1. Let X be a set. A topology on X is a collection T of subsets of Xsatisfying the following conditions

(T1) the total set, X, and the empty set, ∅, are elements of T;

(T2) if Uγγ∈Γ is a (possibly infinite) family of elements of T, then⋃γ∈Γ

Uγ ∈ T;

(T3) if U1, . . . , Un ⊆ T is a finite family of elements of T, thenn⋂i=1

Ui ∈ T.

A topological space is a pair (X,T) where T is a topology on X.

Does this definition remind you of anything?

Example 10.2. Let (X, d) be a metric space, and let T be the collection of opensubsets of X with respect to the metric d. Then, T is a topology on X by Theorem3.10, and (X,T) is a topological space. In other words, every metric on X defines atopology on X.

Furthermore, by definition, two metrics d and d′ on X are topologically equivalentif they define the same open subsets. Hence, topologically equivalent metrics on Xdefine exactly the same topology on X!

Definition 10.3. Let (X,T) be a topological space. Then, U ⊆ X is an open subsetif U ∈ T and is a closed subset if U c = X \ U ∈ T.

Perhaps the best way to see that topological spaces are a more general notion thanmetric spaces is the following example.

Example 10.4. Let X be any set, and let T1 be the collection of all the subsets of X,and let T2 = ∅, X. Then, both T1 and T2 are topologies on X. More precisely,

(1) T1 is called the discrete topology on X, while

(2) T2 is called the indiscrete topology on X.


The topology T1 is induced by the discrete metric, whereas the T2 is as far from thediscrete topology as it can be, hence the name, and it is not induced by a metric.

For instance, if X is a finite set, then we know that any metric on X is topologicallyequivalent to the discrete metric (Exercise 3 in List 3). Thus, the only topology on Xthat is induced by a metric is the discrete topology.

Example 10.5. Let X = a, b, c and let T = ∅, a, a, b, X. Then, T is a topologyon X which is not induced by any metric. Notice the following: there don’t exist opensubsets B,C ∈ T such that b ∈ B, c /∈ B and c ∈ C, b /∈ C. In other words, theelements b and c are “at distance zero” in this topology (of course the last statementdoes not make any sense).

Example 10.6. Let T be the collection of subsets of R which, in addition to ∅ and R,contains all the subsets of the form (x,∞), for x ∈ R. Clearly, this is a topology onR, with less open subsets than the Euclidean topology, since for example the subset(0, 1) is not open in the topology T.

This is another example of a topology which is not induced by a metric on R. Indeed,suppose that there exists a metric d on R which induces the above topology. Sinceopen balls are in particular open subsets, for each x ∈ R and each ε > 0, we have

Bx,ε = R or Bx,ε = (y,∞)

for some y ∈ R. Furthermore, if Bx,ε = R for all x ∈ R and all ε > 0, then d cannotinduce the topology T on R (why?).

Thus, there exists at least one x ∈ R and some ε > 0 such that

Bx,ε = (y,∞)

for some y ∈ R. This way, for each n ∈ N, we have

Bx, ε2n+1

= (yn+1,∞) ⊆ Bx, ε2n

= (yn,∞)

for some yn, yn+1 ∈ R. This means that, for any z > x and all n ∈ N,

d(x, z) <ε

2n,

which is impossible since z 6= x.

Let’s see another example of all the unexpected situations that can happen whendealing with topological spaces.

Example 10.7. This example is usually called the line with two origins. Let X =R ∪ z (here z denotes simply an extra element), and let

(1) T1 = U ⊆ R∣∣ U is open with the Euclidean metric;

(2) T2 = ∅ ∪ W = V ∪ z∣∣ V ∈ T1 is such that 0 /∈ V .

70 ALEX GONZALEZ

Finally, let T = T1 ∪ T2. In other words, an element of T is a union U ∪ W , withU ∈ T1 and W ∈ T2. This is a topology in X (check the details), and it has thefollowing property: if A,B ∈ T are such that

• 0 ∈ A, z /∈ A; and

• z ∈ B, 0 /∈ B,

then A ∩B 6= ∅. In other words, it is impossible to separate 0 from z!

It turns out that almost all the notions that we have defined for metric spaces can betranslated to topological spaces, with a major exception (can you guess which?).

10.1. Set theory revisited. Let’s start by reviewing the basic notions about set the-ory that we already know for metric spaces. There are no surprises in this section,everything behaves as it already does for metric spaces, so most of the proofs areomitted.

Definition 10.8. Let (X,T) be a topological space, and let A ⊆ X.

• The interior of A, Int(A), is the greatest open subset of X contained in A.

• The closure of A, Cl(A), is the smallest closed subset of X which contains A.

• The boundary of A, ∂(A), is the intersection Cl(A) ∩ Cl(X \ A).

Lemma 10.9. Let (X,T) be a topological space and let A ⊆ X. Then,

(1) the interior of A is the union of all the open subsets which are contained in A:

Int(A) =⋃

U∈T, U⊆A

U ;

(2) the closure of A is the intersection of all the closed subsets which contain A:

Cl(A) =⋂

V c∈T, A⊆V

V.

Proof. We prove part (i) and leave part (ii) as an exercise (you will need the equivalentof Proposition 3.12 for topological spaces). Let W = cupU∈T, U⊆AU . Since T is atopology, W ∈ T and W ⊆ A. This means that W ⊆ Int(A). On the other hand,Int(A) ∈ T and Int(A) ⊆ A. By definition, Int(A) ⊆ W .

The following three lemmas describe the basic properties of interior and closure ofsubsets, and related both notions. Their proofs are left as an exercise (again, youjust have to check the corresponding results for metric space).

Lemma 10.10. Let (X,T) be a topological space. Then the following holds:

(1) a subset A ⊆ X is open if and only if Int(A) = A;


(2) for every A ⊆ X, Int(Int(A)

)= Int(A);

(3) if Int(A) ⊆ B ⊆ A then Int(A) = Int(B);


Int(Uα) ⊆ Int

( ⋃α∈Γ

Uα

)Int

( ⋂α∈Γ

Uα

)⊆⋂α∈Γ

Int(Uα);


Int

( n⋂i=1

Ui

)=

n⋂i=1

Int(Ui);

Lemma 10.11. Let (X,T) be a topological space. Then the following holds:

(1) a subset A ⊆ X is closed if and only if A = Cl(A);

(2) for every A ⊆ X, Cl(Cl(A)

)= Cl(A);

(3) if A ⊆ B ⊆ Cl(A) then Cl(A) = Cl(B);


Cl(Uα) ⊆ Cl

( ⋃α∈Γ

Uα

)Cl

( ⋂α∈Γ

Uα

)⊆⋂α∈Γ

Cl(Uα);


Cl

( n⋃i=1

Ui

)=

n⋃i=1

Cl(Ui);

Lemma 10.12. Let (X,T) be a topological space. Then, for any subsets A,B ⊆ X,

(1) the closure of the complement of A is the complement of the interior of A:

Cl(Ac) = Cl(X \ A) = X \ Int(A) =(Int(A)

)c;

(2) the interior of the complement of B is the complement of the closure of B:

Int(Bc) = Int(X \B) = X \ Cl(B) =(Cl(B)

)c;

(3) ∂(Cl(A)

)⊆ Cl(A).

(4) ∂(Int(B)

)∩ Int(B) = ∅; and

72 ALEX GONZALEZ

11. Dealing with topological spaces

We want to study properties of topological spaces, the same way we wanted to under-stand how metric spaces behave. Many of the notions that we have studied for metricspaces make perfect sense when we consider topological spaces, and the only thingrequired is to translate properly the notion to the language of topological spaces. Theidea of this section is to see which properties generalize to topological spaces, andhow do they generalize.

This is, however, a difficult task without any further tools, so before we start analyzinggeneral properties of topological spaces, let us develop a little bit this notion. Keepin mind Example 10.2, since it will inspire most of the definitions and results, andwill give deeper meaning to anything said in this chapter.

First of all, if the objects to study are topological spaces, how do we compare twotopological spaces? When we want to compare two groups or two rings, we use thenotion of group or ring homomorphism. Thinking this way, we need to define what a“morphism” between topological spaces is.

Definition 11.1. Let (X,TX) and (Y,TY ) be topological spaces. A map f : X → Y isa continuous map if, for all U ∈ TY ,

f−1(U) ∈ TX .

Remark 11.2. As already happened in Theorem 4.11, the notation f−1(U) standsfor the following subset of X:

f−1(U) = x ∈ X∣∣ f(x) ∈ U.

No inverse function is assumed! Check Theorem 4.11 and the comments and remarksafter it, because they apply to this situation as well. In particular, it is important toemphasize the following: if f : (X,TX)→ (Y,TY ) is a continuous map, then

W ∈ Tx 6=⇒ f(W ) ∈ TY .

Proposition 11.3. Let f : (X,TX) → (Y,TY ) and g : (Y,TY ) → (Z,TZ) be continuousmaps. Then, the composition g f : (X,TX)→ (Z,TZ) is continuous.

Proof. For any W ∈ TZ , we have U = (g f)−1(W ) = f−1(g−1(W )) ∈ TX becauseV = g−1(W ) ∈ TY (g is continuous) and U = f−1(V ) ∈ TX (f is continuous).

If continuous maps are the equivalent of group homomorphisms for topologicalspaces, then the following correspond to isomorphisms between topological spaces.

Definition 11.4. Let (X,TX) and (Y,TY ) be topological spaces. A map f : X → Y isa homeomorphism if f is bĳective (as a map between sets) and both f and f−1 arecontinuous.


Remark 11.5. And what happens with isometries? Recall that an isometry is a mapf : (X, dX) → (Y, dY ) such that dY (f(x), f(y)) = dX(x, y) for all x, y ∈ X. Well, thisnotion simply does not make much sense when we deal with topological spaces.

The definition of topology is a natural generalization of the notion of metric, but it isnot easy to check conditions (T1) - (T2) - (T3) in all cases. Perhaps it is a good ideato carefully analyze Example 10.2.

Let (X, d) be a metric space, and let Od be the collection of open subsets of X withrespect to the metric d. Within Od there is a sub-collection of special importance:the open balls in X. It is true that the notion of open ball cannot be generalized totopological spaces, since the notion of radius means nothing now.

WARNING!. As a consequence of this, the notion of bounded subset has little mean-ing when working with topological spaces.

In any case it is worth analyzing the behavior of the collection of open balls as a subsetof Od. Recall the definition of open subset: U ∈ Od if, for each x ∈ U there existsε(x) > 0 such that U contains the open ball of radius ε(x) around x: Bx,ε(x) ⊆ U .Another way of putting this is

U =⋃x∈U

Bx,ε(x),

so it makes sense to say that the collection of all open balls in X generate all theopen subsets in Od.

Definition 11.6. Let (X,T) be a topological space. A basis for T is a collection B ⊆ T

such that each open set is a union of sets in B.

WARNING!. The notion of basis of a topology does NOT have the same kind of prop-erties as, for instance, the basis of a vector space. As we will see in the followingexample, the same topology may have many, different bases!

Example 11.7. Consider R with the topology induced by the Euclidean metric, andlet

B1 = (x0, x1) ⊆ R∣∣ x0, x1 ∈ Q B2(x0, x1) ⊆ R

∣∣ x0, x1 ∈ R \QBoth collections above are bases for the same topology, their intersection is empty,and while B1 has countably many elements, B2 does not!

An alternative characterization of the basis of a topology is as follows.

Lemma 11.8. Let (X,T) be a topological space, and let B be a subset of T. Then, Bis a basis for T if and only if it satisfies the following properties.

(B1) For each U ∈ T and each x ∈ U there exists some V ∈ B such that x ∈ V ⊆ U .

(B2) Let V1, V2 ∈ B and let x ∈ V1 ∩ V2. Then there exists V ∈ B such that x ∈ V ⊆V1 ∩ V2.

74 ALEX GONZALEZ

Proof. Suppose first that B is a basis for T. In order to check condition (B1), let U ∈ T

and x ∈ U . By definition of basis, there exists some collection Vγγ∈Γ ⊆ B such that

U =⋃γ∈Γ

Vγ.

In particular, there is some Vγ such that x ∈ Vγ ⊆ U .

Let now V1, V2 ∈ B and let x ∈ V1 ∩ V2. Since V1, V2 ∈ T and T is a topology, it followsthat V1∩V2 ∈ T. Thus, by property (B1), there exists V ∈ B such that x ∈ V ⊆ V1∩V2,and property (B2) holds.

Suppose now that B satisfies properties (B1) and (B2) above. We have to check thateach element U ∈ T is a union of elements of B. Fix an open subset U ∈ T. For eachx ∈ U there exists some Vx ∈ B satisfying condition (B1), and we can write

U =⋃x∈U

Vx.

Hence B is a basis for T.

We can use Lemma 11.8 to formalize this idea of a topology generated by a basis. LetX be a set and let B be a collection of subsets of X satisfying

(B1’) For each x ∈ X there exists some V ∈ B such that x ∈ V .

(B2) Let V1, V2 ∈ B and let x ∈ V1 ∩ V2. Then there exists V ∈ B such thatx ∈ V ⊆ V1 ∩ V2.

Now we can define the following collection of subsets:

TB = U ⊆ X∣∣ for each x ∈ U there exists V ∈ B such that x ∈ V ⊆ U.

Proposition 11.9. The collection of subsets TB is a topology on X with basis B.

We call TB the topology generated by B. Again, keep in mind that the same topologymay be generated by more than one basis.

Proof. First, let’s prove that TB is a topology on X.

(T1) Both ∅ and X satisfy the condition to be elements of TB.

(T2) Let Uγγ∈Γ ⊆ TB. We have to check that U = ∪ΓUγ ∈ TB. Each elementx ∈ U is contained in Uγ, for some γ ∈ Γ. Now Uγ ∈ TB, and this means thatthere exists some V ∈ B such that

x ∈ V ⊆ Uγ ⊆ U.

Thus, U ∈ TB.


(T3) Let U1, . . . , Un ⊆ TB. We have to show that U = ∩iUi ∈ TB. Let then x ∈ U .This means that x ∈ Ui for all i, and thus there exist V1, . . . , Vn ∈ B such that

x ∈ Vi ⊆ Ui for each i = 1, . . . , n.

Since B satisfies condition (B2), it follows that there exists some V ∈ B suchthat

x ∈ V ⊆ Ui for each i = 1, . . . , n.

Thus, x ∈ V ⊆ U and U ∈ TB.

The fact that B is a basis for TB is a consequence of Lemma 11.8.

The following exercises were already suggested for metric spaces.

Exercise 11.10. Let f : (X,TX) → (Y,TY ) be a map, and let B ⊆ TY be a basis.Show that f is continuous if and only if f−1(V ) ∈ TX for each V ∈ B.

Exercise 11.11. Let f : (X,TX) → (Y,TY ) be a continuous map, and let U ⊆ X.Show that if f is a homeomorphism, then the following holds:

(1) if U ∈ TX then f(U) ∈ TY ; and

(2) if U c ∈ TX then f(U)c ∈ TY .

11.1. From metric spaces to topological spaces. We have now the tools we need,and it is time to start translating from metric to topological spaces. As an exampleof what we are going to do in this section, consider the following situation. If (X, d)is a metric space and A ⊆ X is any subset, then the metric d restricts to a metric onA, and this makes (A, d|A) into a metric space. The metric d|A is usually called thesubspace metric.

Is it possible to mimic this process of restricting the metric d to a subset for topologicalspaces? In other words, does it make sense to restrict a topology to a subset? And ifso, how do we do it? Understanding the example of the metric space (X, d) and thesubspace (A, d|A) gives the right hints.

Indeed, let OX and OA be the collections of open subsets of X and A respectively,with respect to the metrics d and d|A respectively. Then, recall that each open subsetof A is the intersection of an open subset of X with A:

U ∈ OA if and only if ∃W ∈ OX such that U = W ∩ A.

Now, let (X,T) be a topological space, and let A ⊆ X be a subset. We can then define

TA = U ∩ A∣∣ U ∈ T,

and it is not difficult to check that this is a topology.

(T1) ∅ = ∅ ∩ A, A = X ∩ A ∈ TA.

76 ALEX GONZALEZ

(T2) Let Uγγ∈Γ ⊆ TA. By definition of TA, for each γ ∈ Γ there exists someWγ ∈ T

such that Uγ = A ∩Wγ. Since T is a topology, we have W =⋃γ∈Γ Wγ ∈ T,

and hence

U =⋃γ∈Γ

Uγ =⋃γ∈Γ

(Wγ ∩ A) =( ⋃γ∈Γ

Wγ

)∩ A = W ∩ A ∈ TA.

(T3) Let U1, . . . , Un ⊆ TA. Again, for each i there exists some Wi ∈ T such thatUi = Wi ∩ A. Since T is a topology, we have

U =n⋂i=1

Ui =n⋂i=1

(Wi ∩ A) =( n⋂i=1

Wi

)∩ A ∈ T.

This is known as the subspace topology for obvious reasons.

Now we can try something slightly more complicated. Let (X,TX) and (Y,TY ) betopological spaces, and consider the set X × Y . As happened with metric spaces, wewant to put a topology on this product, in a sensible way.

Now, if you check the corresponding situation for metric spaces (Theorem 3.9) youwill see that the “right” metric to put on a product was the strongest possible metricwhich induces the original metrics on each of the factors of the product. Somethingsimilar could be done for topological spaces, but we do not have the time.

Instead, let (X,TX) and (Y,TY ) be the topological spaces above, let X × Y be theproduct set, and let TX×Y be the collection of subsets of X × Y generated by

BX×Y = U × V∣∣ U ∈ TX and V ∈ TY .

Let also πX : X × Y → X and πY : X × Y → Y be the obvious projection functions.

Theorem 11.12. The collection TX×Y is the only topology on X × Y satisfying thefollowing condition.

(P) A map f : (Z,TZ)→ (X × Y,TX×Y is continuous if and only if the compositionsπX f : Z → X and πY f : Z → Y are continuous.

Proof. The collection BX×Y satisfies the conditions (B1) and (B2) in Lemma 11.8, andthus TX×Y is the topology generated by BX×Y (so it is a topology). We have to checkthen that it satisfies property (P), and that it is unique satisfying this condition. Therest of the proof is divided into several smaller parts.

1. The maps πX : (X×Y,TX×Y )→ (X,TX) and πY : (X×Y,TX×Y )→ (Y,TY ) arecontinuous.

For instance, let’s check this for πX : we have to show that π−1X (U) ∈ TX×Y for all

U ∈ TX , and this is immediate because π−1X (U) = U × Y .

2. If f : (Z,TZ) → (X × Y,TX×Y ) is continuous then so are the compositionsπX f and πY f .


This is an immediate consequence of Proposition 11.3: the composition of continuousmaps is continuous.

3. If the compositions πX f : Z → X and πY f : Z → Y are continuous thenso is the map f : Z → X × Y .

We have to check that, for each W ∈ TX×Y , f−1(W ) ∈ TZ . Let’s assume first thatW = U × V , for some U ∈ TX and V ∈ TY . Then we have

f−1(W ) = f−1(U × V ) = f−1((U × Y ) ∩ (X × V )) = f−1(U × Y ) ∩ f−1(X × V ) =

= f−1(π−1X (U)) ∩ f−1(π−1

Y (V )) = (πX f)−1(U) ∩ (πY f)−1(V ) ∈ TZ .

This means that the preimage of any element of the basis BX×Y is in TZ . Now, eachW ∈ TX×Y is a union of elements of BX×Y , and this shows that f is continuous.

4. The topology TX×Y is unique satisfying property (P).

Let T be another topology on X × Y satisfying property (P), and consider the identitymap Id : (X×Y,T)→ (X×Y,T). This is clearly continuous, and hence πX = πX Idand πY = πY Id are continuous too. This means that, for each U ∈ TX and V ∈ TY ,we have

π−1X (U) = U × Y ∈ T π−1

Y (V ) = X × V ∈ T

and hence U × V = (U × Y ) ∩ (X × V ) ∈ T. In other words, T contains the basisBX×Y , and thus TX×Y ⊆ T.

On the other hand, we know from part 1. that the maps

πX : (X × Y,TX×Y )→ (X,TX) πY : (X × Y,TX×Y )→ (Y,TY )

are continuous. Notice that the compositions

πX : (X × Y,TX×Y )Id // (X × Y,T)

πX // (X,TX)

πY : (X × Y,TX×Y )Id // (X × Y,T)

πY // (Y,TY )

are continuous maps, and thus by property (P) the map Id : (X × Y,TX×Y ) → (X ×Y,T) is continuous, and this means that T ⊆ TX×Y .

WARNING!. The definition of the product topology TX×Y on X×Y does NOT say thatall the open subsets in TX×Y are of the form U × V for U ∈ TX and V ∈ TY !

For instance, let TR be the Euclidean topology (induced by the Euclidean metric) onR, and let TR×R be the product topology on R × R. The subsets (0, 2), (1, 3) ⊆ R areopen in this topology, and we may consider the subset

78 ALEX GONZALEZ

W = ([0, 2]× [0, 2]) ∪ ([1, 3]× [1, 3]) ⊆ R× R

which is not of the form U × V for any U, V ∈ TR!

Exercise 11.13. Let f : (X,TX)→ (Y,TY ) be a continuous map, and let

Gf = (x, f(x)) ∈ X × Y∣∣ x ∈ X ⊆ X × Y

be the graph of the function, with the subspace topology induced from the producttopology on X × Y . Show that Gf is homeomorphic to X.

11.2. A very basic metric-topological dictionary. In the previous sections youhave got a taste of how metric spaces “evolve” into topological spaces. Let’s see nowhow the main properties of metric spaces can be adapted to topological spaces. Thefirst of these properties is connectivity. You can compare the following definition withDefinition 6.1, see how many differences you find.

Definition 11.14. A topological space (X,TX) is connected if it satisfies the following.

(C) If a subset A ⊆ X is both open and closed then either A = X or A = ∅.Example 11.15. Every set X with the indiscrete topology is connected.

Example 11.16. Let X = a, b, c and let T = ∅, a, a, b, X. Then (X,T) isconnected: let A $ X be a proper subset, A 6= ∅. Then either A has one or twoelements. Suppose in addition that A is open. Then, either A = a or A = a, b.In any case, A cannot be closed because

ac = b, c /∈ T a, bc = c /∈ T.

The following result is a generalization of Theorem 6.5. This is another example of aresult in which the metric played a rather decorative role, and the proof will be leftas a (really easy) exercise.

Theorem 11.17. Let (X,T) be a topological space. Then, the following are equivalent.

(1) (X,T) is connected.

(2) There does not exist subsets U, V ∈ T such that U ∪ V = X and U ∩ V = ∅.

(3) There does not exist any continuous function f : (X,T)→ (0, 1,Tdis) which issurjective.


Proof. As promised, the proof is left as an exercise.

Corollary 11.18. Let f : (X,TX)→ (Y,TY ) be a continuous map of topological spaces,and suppose (X,TX) is connected. Then, f(X) ⊆ Y is connected with the subspacetopology.

Corollary 11.19. Let f : (X,TX)→ (Y,TY ) be a homeomorphism of topological spaces.Then, (X,TX) is connected if and only if (Y,TY ) is connected.

The relation between connectivity and products of topological spaces behave exactlyas happened for metric spaces. Although the proofs for topological spaces requirelittle modification from the proof of Theorem 6.13, we will do them again to practiceproving statements on topological spaces.

Lemma 11.20. Let (X,T) be a topological space, and let Aγγ∈Γ be a collection ofconnected subsets of X such that

Aγ ∩ Aω 6= ∅for all γ, ω ∈ Γ. Then, A = ∪ΓAγ is a connected subset of X.

Proof. Let f : A → 0, 1 be a continuous function, where 0, 1 has the discretetopology. Since Aγ is connected for all γ ∈ Γ, we have

f(Aγ) = αγ ∈ 0, 1.Let now Aω be another set in the collection. Since Aγ ∩ Aω 6= ∅ it follows thatf(Aω) = f(Aγ). Hence, the map f cannot be surjective, since it has to be constant,and thus A is connected.

Theorem 11.21. Let (X,TX) and (Y,TY ) be topological spaces. Then, the product(X × Y,TX×Y ) is connected if and only if both (X,TX) and (Y,TY ) are connected.

Proof. If the product space is connected, then so are (X,TX) and (Y,TY ), since theprojections πX : X × Y → X and πY : X × Y → Y are both continuous maps.

Suppose otherwise that (X,TX) and (Y,TY ) are connected. Fix some element a ∈ Xand consider the sets

• B = a × Y ⊆ X × Y ; and

• Cy = X × y ⊆ X × Y , where y ∈ Y .

Then B is homeomorphic to Y , while Cy is homeomorphic to X for all y ∈ Y . Inparticular, each of the above subsets is connected.

For each y ∈ Y we have Cy ∩B 6= ∅, and hence the subset Uy = B ∪ Cy is connectedfor all Y , by the previous Lemma. Consider then the collection of subsets Uyy∈Y .For any y, y′ ∈ Y , the intersection Uy ∩ Uy′ is non-empty, and thus, again by theprevious Lemma, ∪y∈YUy = X × Y is a connected topological space.

80 ALEX GONZALEZ

Remark 11.22. We could talk about path-connected topological spaces, just as wedid for metric spaces (again, the definition for topological spaces requires barely anychange with respect to the definition for metric spaces). The results (and proofs) thatwe saw for path-connected metric spaces, in particular Theorem 6.18, generalize inthe most obvious way to topological spaces

Definition 11.23. Let (X,T) be a topological space. A connected component of X isa subset of X which is maximal with respect to the property of being connected. Theset of connected components of X is denoted by π0(X).

Let’s see now how compactness is defined for topological spaces. As we did for con-nectivity, the missing proofs are left as exercises (where probably the only difficultythat you will find is to locate in the previous chapters the corresponding result aboutmetric spaces).

Be careful now because some basic results about compact subsets of metric spaces donot generalize to topological space. For instance, recall that bounded is a meaninglessword in this context!

Definition 11.24. Let (X,T) be a topological space and let A ⊆ X be any subset.An open cover of A in X is a collection Uγγ∈Γ ⊆ T such that

A ⊆⋃γ∈Γ

Uγ.

A subcover of Uγγ∈Γ is a subcollection Vωω∈Ω ⊆ Uγγ∈Γ such that ⊆ ∪ΩUω.

The subset A is compact if every open cover of A in X contains a finite subcover.

WARNING!. As happened already for metric spaces, showing that a subset of a topo-logical space is compact means checking ALL possible open covers!

Example 11.25. Let X be any set. Then, X is compact with the indiscrete topology.

Example 11.26. Let X = a, b and T = ∅, a, X. Since X is finite, any of itssubsets is compact. In particular, a ⊆ X is a compact subset, but it is not closed,since b 6⊆ X.

Proposition 11.27. Let f : (X,TX) → (Y,TY ) be a continuous map of topologicalspaces, and let A ⊆ X be a compact subset of X. Then f(A) ⊆ Y is compact.

Corollary 11.28. Let f : (X,TX)→ (Y,TY ) be a homeomorphism of topological spaces.Then, X is compact if and only if Y is compact.

Theorem 11.29. Let (X,TX) and (Y,TY ) be topological spaces, and let A ⊆ X andB ⊆ Y . Then, A × B ⊆ X × Y is compact with respect to the product topology if andonly if both A and B are compact.

It is a good exercise to do it again in the context of topological spaces: the prooffollows the same lines as the proof of Theorem 7.17, but we have to replace the openballs by elements of the basis for the product topology.


Proof. If A × B is compact, then both A and B are compact since the projectionsπX : X × Y → X and πY : X × Y → Y are continuous.

Suppose now that A and B are compact as subsets of (X,TX) and (Y,TY ) respec-tively, and let Wγγ∈Γ ⊆ TX×Y be an open cover of A× B. Recall also that TX×Y isthe topology generated by the basis

BX×Y = U × V ⊆ X × Y∣∣ U ∈ TX and V ∈ TY .

• We can assume that, for all γ ∈ Γ, Wγ = Uγ × Vγ ∈ BX×Y .

Let Wγ be any term of the open cover. Since BX×Y is a basis of TX×Y , for each(x, y) ∈ Wγ there exists Kx ×Ky ∈ BX×Y such that

(x, y) ∈ Kx ×Ky ⊆ Wγ,

and we can write Wγ = ∪(x,y)Kx ×Ky.

Doing this for each γ ∈ Γ, we end up with the open cover

Kx ×Ky

∣∣ (x, y) ∈ Wγγ∈Γ.

If this new open cover contains a finite subcover then it follows that the original opencover Wγ contains a finite subcover.

• If Wγ = Uγ × Vγ ∈ BX×Y for all γ ∈ Γ, then the open cover Wγ contains afinite subcover.

For each a ∈ A, consider the set a × B ⊆ A× B. This subset is homeomorphic toB, which is a compact subset of Y by hypothesis, and thus a × B is compact too.Since Wγ is also an open cover of a × B, it follows that there exists some finitesubcover

Ωa = Wa,i1 , . . . ,Wa,ia,where Wa,ik = Ua,ik × Va,ik ∈ BX×Y .

Still for a ∈ A, define the subset Ua = Ua,i1 ∩ Ua,i2 ∩ . . . ∩ Ua,ia ⊆ A. This is aopen subset of X which contains the element a, and hence the collection Uaa∈Ais an open cover of A. Since A is compact there exist a1, . . . , am ∈ A such thatUa1 , . . . , Uam is a finite subcover.

To finish the proof, notice that collection Ωa1∪. . .∪Ωam is a finite subcover ofA×B.

There is another property of metric spaces that we have studied in detail: complete-ness. Is there any sensible way to define completeness for topological spaces? Theshort answer is no: completeness requires that sequences behave very sensibly, andwe have already seen in the tutorials that in general sequences in topological spacesbehave pretty badly. A longer answer to the question would be, in fact, really long,so let’s take it as simply “no”.

82 ALEX GONZALEZ

12. What topological spaces can do that metric spaces cannot

In the previous chapter we did not see too many examples, it was a matter of redefin-ing properties of metric spaces in the language of topological spaces, and there wasfew place for examples. The importance of topological spaces (and thus the interest-ing examples) lie in what we can do with them which we could not do with metricspaces.

In this chapter we will see two constructions which did not make much sense in thecontext of metric spaces, but which are natural in the context of topological spaces:quotient topological spaces and compactification of topological spaces. You can thinkof these constructions as geometric manipulations, and in this sense they are evenmore important since they are natural also as geometrical manipulations.

12.1. One-point compactification of topological spaces. The first construction tostudy has to do with compactness, and it has already appeared in the continuousassessment. Roughly speaking, the idea is the following: the property of compactnessis missing in the topological space (X,TX), so how can we modify (X,TX) to make itcompact? It turns out that it is enough to add an extra element to X!

Let’s formalize this. Let then (X,TX) be a topological space, and define X = X∪∞,where ∞ simply designates an extra element which does not belong to X. Now wehave to define a topology on X, and we want to do it so that X is compact and therestriction to X ⊆ X is the original topology TX :

TX = TX ∪ T′,

where T′ = ∅ ∪ W = V ∪ ∞ ⊆ X∣∣V ⊆ X and X \ V is compact and closed.

This means that each element U ∈ TX can be written as

U = Y ∪ (V ∪ ∞),

where Y ∈ TX and V ∪ ∞ ∈ T′. The first thing to check is that TX is indeed atopology on X.

Lemma 12.1. Let (Y,TY ) be a topological space, and let B ⊆ Y be a compact subset.Then, each closed subset K ⊆ B is compact.

Proof. The proof is identical to the proof of Lemma 7.12.

Lemma 12.2. The collection TX defined above is a topology on the set X.

Proof. It is clear that ∅ ∈ TX , and X ∈ TX since X = ∅∪(X∪∞) ∈ TX ∪T′ = TX .That proves property (T1).


Let now Uγγ∈Γ ⊆ TX . For each γ ∈ Γ, we can write Uγ = Yγ ∪Wγ, with Yγ ∈ TXand Wγ ∈ T′, and thus ⋃

γ∈Γ

Uγ =( ⋃γ∈Γ

Yγ)∪( ⋃γ∈Γ

Wγ

).

Since TX is a topology it follows that ∪γYγ ∈ TX , and we have to check that ∪γWγ ∈ T′.For simplicity, we can assume that Wγ 6= ∅ for all γ ∈ Γ. Thus, for each γ,

Wγ = Vγ ∪ ∞,where Vγ ⊆ X is such that X \ Vγ is compact and closed, and we have to check thatX \ (∪ΓVγ) is closed and compact.

Note that X \ Vγ is closed for each γ ∈ Γ, and hence Vγ ∈ TX . Thus, ∪ΓVγ ∈ TX , soX \ (∪ΓVγ) is closed. To show that X \ (∪ΓVγ) is compact, recall the equality

X \( ⋃γ∈Γ

Vγ)

=⋂γ∈Γ

(X \ Vγ).

In particular, X \ (∪ΓVγ) ⊆ X \Vγ for every γ ∈ Γ, and by Lemma 12.1 it follows thatX \ (∪ΓVγ) is compact. Thus, ∪γWγ ∈ T′, and⋃

γ∈Γ

Uγ ∈ TX .

Finally, let U1, . . . , Un ⊆ TX . We have to prove that the intersection ∩ni=1Ui ∈ TX .Again, write Ui = Yi ∪Wi, with Yi ∈ TX and Wi ∈ T′ for each i. We have

n⋂i=1

Ui =( n⋂i=1

Yi)∪( n⋂i=1

Wi

).

Since TX is a topology, we have ∩ni=1Yi ∈ TX , and we have to check that ∩ni=1Wi ∈ T′.If Wi = ∅ for any i, then this is obvious, so suppose that Wi 6= ∅ for all i. We have

n⋂i=1

Wi =n⋂i=1

(Vi ∪ ∞) =( n⋂i=1

Vi)∪ ∞,

where Vi ⊆ X is such that X \Vi is closed and compact, for each i. We have to checkthat X \ (∩ni=1Vi) is closed and compact.

Note that X \ Vi is closed, and thus Vi ∈ TX . Hence, ∩ni=1Vi ∈ TX , and X \ (∩ni=1Vi)is closed. To show that X \ (∩ni=1Vi) is compact, let Aαα∈Λ be an open cover ofX \ (∩ni=1Vi), and recall the equality

X \( n⋂i=1

Vi)

=n⋃i=1

(X \ Vi).

In particular, Aαα∈Λ is an open cover of X \ Vi, for each i. Since X \ Vi is compact,there must exist a finite subcover, namely Ωi. Putting all these finite subcovers

84 ALEX GONZALEZ

together we get Ω1 ∪ . . . ∪ Ωn ⊆ Aα, which is a finite subcover of X \ (∩ni=1Vi).Hence, ∩ni=1Wi ∈ T′, and thus

n⋂i=1

Ui ∈ TX .

Thus, TX is a topology on X.

Theorem 12.3. Let (X,TX) be a topological space, and let (X, TX) be as above. Then,the following holds.

(1) (X, TX) is a compact space.

(2) The restriction of TX to X is TX .

For obvious reasons, (X, TX) is called the one-point compactification of (X,TX). Itis sometimes called also Alexandroff’s compactification.

Proof. To prove (1), let Uγγ∈Γ ⊆ TX be an open cover of X, and write

Uγ = Yγ ∪Wγ

with Yγ ∈ TX and Wγ ∈ T′. For simplicity assume that Wγ 6= ∅ for all γ. Write thenWγ = Vγ ∪ ∞, where Vγ ⊆ X is such that X \ Vγ is closed and compact.

Now, since Uγ is an open cover of X, there must exist some ω ∈ Γ such that∞ ∈ Uω = Yω ∪ Vω ∪ ∞.

Notice that Yγ ∪ Vγγ∈Γ is now an open cover of X \ Vγ, and the latter is compact byhypothesis. Thus, there is a finite subcover, namely Yγ1 ∪ Vγ1 , . . . , Yγn ∪ Vγn.

It follows then that Uγ1 , . . . , Uγn , Uω is a finite subcover of X, and hence it is com-pact.

To prove (2), letT′X = U ∩X

∣∣ U ∈ TX,

which is the topology on X induced by TX . We have to show that TX = T′X . Clearly,TX ⊆ T′X , because given Y ∈ TX , we have Y ∪ ∅ ∈ TX .

Let then U ∈ TX . For simplicity we can assume that U = Y ∪(V ∪∞), with Y ∈ TXand V ∪ ∞ ∈ T′. Then,

X ∩ U = (X ∩ Y ) ∪ (X ∩ V ) ∈ TX ,

since Y, V,X ∈ TX . Thus T′X ⊆ TX .

Let’s see some examples of this construction.


Example 12.4. Let X = R, with the topology induced from the Euclidean metric.Then, the one-point compactification of X is the topological space in Exercise 3 ofthe second part of the continuous assessment. In particular, the one-point compact-ification of R is homeomorphic to a circle.

More generally, letX = Rn, with the topology induced by the Euclidean metric. Then,the one-point compactification of X is homeomorphic to the n-dimensional sphere

Sn = (x1, . . . , xn+1) ∈ Rn+1∣∣ x2

1 + . . .+ x2n+1 = 1.

In view of the exercise in the continuous assessment, can you construct a homeo-morphism from the n-dimensional sphere to the one-point compactification?

Example 12.5. The one-point compactification of the open interval (0, 1) is, again,homeomorphic to the circle S1. This is because (0, 1) is homeomorphic to R (hencetheir one-point compactifications are homeomorphic).

12.2. Quotient topological spaces. This is the most geometrical of the two con-structions, and it is actually rather important. Recall the idea of constructing acylinder out of a stripe (a subset of R2), by gluing together two non-adjacent sides,as described in the following sketch.

x

y

x

y

Stripe

(0, 0)

(0, 1) (2, 1)

(2, 0)

quotient

Cylinder

aa

(2, 0)

(2, 1)(0, 1)

(0, 0)

In this case, it is easy to provide the cylinder with a metric (just consider the cylinderas a subset of R3 with the Euclidean metric). However, the example of the cylinder ispart of a more general construction, so let’s see in more detail what is going on withthe cylinder.

As a set, the cylinder is the quotient set of the stripe by the following equivalencerelation:

(x, y) ∼ (x, y) if x 6= 0, 1 (0, y) ∼ (1, y).

The question is then which topology do we put on this quotient set. Notice that thequotient set of the stripe by this equivalence relation is not a subset of R3 (althoughit can be identified with a subset of R3). As happened with the product of topologicalspaces, not any topology is acceptable.

Let (X,T) be a topological space, and let ∼ be an equivalence relation on the set X.You may think of the elements of X in the same equivalence class as the elements

86 ALEX GONZALEZ

that we are going to glue together, just as we did with the stripe above. Let then Xbe the set of equivalence classes, and let

π : X −→ X

be the map that sends each x ∈ X to its equivalence class. Notice that the map π issurjective (as a map of sets). Define also

T = U ⊆ X∣∣ π−1(U) ∈ T.

Lemma 12.6. The collection T is a topology on X.

Proof. Since ∅ = π−1(∅) ∈ T and X = π−1(X), condition (T1) to be a topology isclearly seen to hold.

To check condition (T2), let Uγγ∈Γ ⊆ T. We have to check that the union ∪ΓUγ ∈ T.Notice that there is an equality

f−1( ⋃γ∈Γ

Uγ)

=⋃γ∈Γ

f−1(Uγ).

Since f−1(Uγ) ∈ T by hypothesis, and T is a topology, we have ∪Γf−1(Uγ) ∈ T, and

thus ∪ΓUγ ∈ T.

Finally, to check condition (T3), let U1, . . . , Un ⊆ T. Again, there is an equality

f−1( n⋂i=1

Ui)

=n⋂i=1

f−1(Ui),

and condition (T3) follows immediately.

For obvious reasons, the topological space (X,T) is called the quotient space of (X,T)by the equivalence relation ∼, and the map π : X → X is called the quotient map.

Theorem 12.7. Let (X,T) be a topological space, and let ∼ be an equivalence relationon X. Let also (X,T) be the quotient space by ∼. Then, T is the unique topology on Xsatisfying the following conditions.

(Q1) The map π : (X,T)→ (X,T) is continuous.

(Q2) A map f : (X,T)→ (Y,TY ) is continuous if and only if the composition f π iscontinuous.

Proof. First we have to check that the topology T satisfies conditions (Q1) and (Q2).The map π : X → X is continuous since, by definition, π−1(U) ∈ T for all U ∈ T.

Let then f : (X,T) → (Y,TY ). If f is continuous, then clearly f π is continuousbecause it is a composition of continuous functions.


Suppose then that f π : (X,T) → (Y,TY ) is continuous. We have to show thatπ−1(V ) ∈ T for all V ∈ TY . Since f π is continuous, we have (f π)−1(V ) =f−1(π−1(V )) ∈ T, and hence by definition of T it follows that π−1(V ) ∈ T.

Now, suppose that T′

is another topology on X which also satisfies properties (Q1)and (Q2) with respect to the map π′ : (X,T)→ (X,T). We have to prove that T

′= T.

Consider first the identity map Id : (X,T)→ (X,T′). Since T

′satisfies condition (Q1),

the map π′ = Id π is continuous and thus Id is continuous by property (Q2). Thisimplies that T

′ ⊆ T.

Similarly, we can consider the map Id′ : (X,T′)→ (X,T). Again, since π = Id π′ is

continuous, it follows that Id′ is continuous, and hence T ⊆ T.

Example 12.8. How many ways do you know of “constructing” a topological spacewhich is homeomorphic to the 2-dimensional sphere

S2 = (x, y, z) ∈ R3∣∣ x2 + y2 + z2 = 1?

One way is the one-point compactification of the Euclidean plane. This way we obtaina huge 2-dimensional sphere, but a sphere all the same.

Another way is by gluing together two discs through their boundary, as sketched inthe following picture.

a a

Actually, it is enough to use a single circle.

a

a

88 ALEX GONZALEZ

The above constructions, using circles, are examples of quotient spaces. In the firstcase, the set X is a disjoint union of two disc, and the equivalence relation is givenby identifying the boundary of one circle with the boundary of the other circle. Inthe second case, you can think of X as the complex disc of radius 1, and we identifyeach element of the boundary with its complex conjugate.

Yet another way of constructing a 2-dimensional sphere, again using quotient spaces,is to “crush to a point” each of the lids of a cylinder.

And what about the Platonic solids? They are also homeomorphic to a sphere. Infact, any convex polyhedron is homeomorphic to a 2-dimensional sphere, and notethat each polyhedron is in fact a quotient space, where X is the disjoint union of allthe faces, and we glue the faces together to form the polyhedron.


REFERENCES

The main published references for this course are the following.

1. M. O. Searcoid, Metric Spaces, Springer Undergraduate Mathematics Series,ISBN 978-1-84628-369-7.

2. W. A. Sutherland, Introduction to Metric and Topological Spaces, 2nd Edition,Oxford University Press, ISBN 978-0-19-956308-1.

Other general, electronic sources are the following.

1. http://www-history.mcs.st-and.ac.uk/~john/MT4522/index.html

2. http://en.wikibooks.org/wiki/Topology/Metric_Spaces

References about the p-adic metric.

1. http://scientopia.org/blogs/goodmath/2013/01/13/define-distance-differently-the-p-adic-norm/

2. http://en.wikipedia.org/wiki/P-adic_number

References about the Cantor set.

1. http://planetmath.org/cantorset

2. http://mathworld.wolfram.com/CantorSet.html

3. http://en.wikipedia.org/wiki/Cantor_set

References about the Heine-Borel Theorem.

1. http://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem

References about Hausdorff spaces (this has not been explicitly defined in the course,but the notion has appeared several times in examples).

1. http://en.wikipedia.org/wiki/Hausdorff_space#Examples_and_counterexamples

References about the classification of surfaces.

1. http://www.open.edu/openlearn/science-maths-technology/mathematics-and-statistics/mathematics/surfaces/content-section-0

2. http://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Huang.pdf

3. http://www.cis.upenn.edu/~jean/surfclass-n.pdf

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