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ELEMENTARY TOPOLOGY I

ALEX GONZALEZ

1. Introduction 32. Metric spaces 42.1. Continuous functions 82.2. Limits 102.3. Open subsets and closed subsets 122.4. Continuity, convergence, and open/closed subsets 153. Topological spaces 183.1. Interior, closure and boundary 203.2. Continuous functions 233.3. Basis of a topology 243.4. The subspace topology 263.5. The product topology 273.6. The quotient topology 313.7. Other constructions 334. Connectedness 354.1. Path-connected spaces 404.2. Connected components 415. Compactness 435.1. Compact subspaces of Rn 465.2. Nets and Tychonoff’s Theorem 486. Countability and separation axioms 526.1. The countability axioms 526.2. The separation axioms 547. Urysohn metrization theorem 618. Topological manifolds 658.1. Compact manifolds 66REFERENCES 67

Contents

1

2 ALEX GONZALEZ

ELEMENTARY TOPOLOGY I 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 absolute value, whichis essentially a rule to measure the distance between any two numbers in R. It wouldseem that continuity relies on the existence of a rule to measure distances, or moreprecisely, a metric.

4 ALEX GONZALEZ

2. Metric spaces

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).

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.

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 functions 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|

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 here? 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)

6 ALEX GONZALEZ

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.

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 it provides easy counterexamples to many statements.

Example 2.10. Consider the set of integer numbers Z, and let p be a prime number.The p-adic valuation of n ∈ Z is the maximum non-negative integer νp(n) such thatpνp(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.

8 ALEX GONZALEZ

2.1. Continuous functions. We can introduce now continuity of functions betweenmetric spaces.

Definition 2.11. Let (X, dX) and (Y, dY ) be metric spaces, and let f : X → Y be aset map. The map f is continuous at the point a ∈ X (with respect to the metricsdX , 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.

The map f is a homeomorphism if the inverse map f−1 exists (as a map of setsf−1 : Y → X) and both f and f−1 are continuous. Similarly, two metric spaces(X, dX) and (Y, dY ) are homeomorphic if there exists a homeomorphism f : 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.

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 2.12. 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 2.13. The discrete metric also illustrates the following (not intuitive) situa-tion: a bĳective map f : X → Y of sets which is continuous with respect to a choiceof metrics dX and dY may not have a continuous inverse! Indeed, the identity map(R, ddis)→ (R, dEuc) is continuous, but the inverse is not.


Example 2.14. 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 solution, 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 2.15. 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 2.16. Let d be the p-adic metric on the set of integers Z, and fix someα ∈ 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).

10 ALEX GONZALEZ

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 2.17. 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.

2.2. Limits. Convergence of sequences is another feature that depends on the exis-tence of a metric.

Definition 2.18. Let (X, d) be a metric space and let xnn∈N be a sequence ofelements of X. Then,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.


Theorem 2.19. 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 2.20. 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 2.21. 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.

12 ALEX GONZALEZ

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.

2.3. Open subsets and closed subsets. The idea of open/closed subset again reliesin the existence of a metric.

Definition 2.22. 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 2.23. Let dEuc and ddis be, respectively, the Euclidean and the discretemetrics on R2, and let a ∈ R2 be any point. Then, U = a ⊆ R2 is open with respectto ddis but not open with respect to dEuc.

Example 2.24. 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.

Example 2.25. In every metric space, a singleton is always a closed subset. Indeed,let (X, d) be a metric space, and let x ∈ X be a point. We have to check thatxc ⊆ X is an open subset. The complement of x ⊆ X can be described as thesubset U = y ∈ X | d(x, y) > 0. Fix some y ∈ U , and let ε = d(x,y)

2> 0. Then,

x /∈ By,ε, since d(x, y) > ε by assumption, and thus By,ε ⊆ U .

Lemma 2.26. 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 hints which is the right choice 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. Notice

that z 6= y by hypothesis. 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

The following is the most important and basic result about open subsets in a metricspace that we need in this course.

Theorem 2.27. Let then (X, d) be a metric space, and let O be the collection of all opensubsets of X with respect to d. Then,

14 ALEX GONZALEZ

(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,εi⊆ 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.

There is a similar result for closed subsets.

Proposition 2.28. 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.


2.4. Continuity, convergence, and open/closed subsets. The concepts that wehave introduced in the previous sections are not independent from each other. Inthis section we will study some of their interactions.

Theorem 2.29. 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 each 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.

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.

Theorem 2.30. 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).

16 ALEX GONZALEZ

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(U) = 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 2.31. 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 2.30. 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 isan 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 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).

The above result implies a crucial consequence: it is not the metric(s) what matters,but the collection(s) of open/closed subsets!

18 ALEX GONZALEZ

3. Topological spaces

We finished the previous chapter realizing that, when we are talking about continuityof functions, metrics are not so relevant as open subsets. We thus shift the focus tocollections of open subsets instead of metrics.

Definition 3.1. 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. The elements of T arecalled open subsets, while V ⊆ X is a closed subset if V c ∈ T.

Example 3.2. 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.

Example 3.3. Let (X, d) be a metric space, and let T be the collection of open subsetsof X with respect to the metric d. Then, T is a topology on X by Theorem 2.27, and(X,T) is a topological space. In other words, every metric on X defines a topologyon X.

Notice that different metrics on the same set X could give rise to the same topology.For example, let X = Rn. The Manhattan metric (Example 2.5), the Euclidean metric(Example 2.6) and the box metric (Example 2.8) all define the same topology on X.Another example of this situation is the following: all metrics on a given finite setdefine the same topology, the discrete metric.

Example 3.4. 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.


a b c

Notice the following: there are no open subsets B,C ∈ T such that b ∈ B, c /∈ Band c ∈ C, b /∈ C. In other words, the elements b and c are “at distance zero” inthis topology (of course the last statement does not make any sense). How manytopologies can you define on the set X = a, b?

Example 3.5. 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 3.6. This example is usually called the line with two origins. LetX = 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 \ 0) ∪ z∣∣ V ∈ T1 is such that 0 ∈ V .

20 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, thenA ∩B 6= ∅. In other words, it is impossible to separate 0 from z!

Example 3.7. This example is called the finite complement topology. Let X be a set,and let T be the collection of subsets U ⊆ X for which U c = X \ U is either empty,finite, or is all of X. Clearly Xc = ∅ and ∅c = X, and this takes care of condition (T1).Let Uαα∈Γ be a family of elements of T (we may assume that Uα 6= ∅, X without lossof generality). Then,

X \( ⋃α∈Γ

Uα)

=⋂α∈Γ

(X \ Uα)

is a finite set, since X \ Uα is finite for all α, and condition (T2) follows. Conditions(T3) follows similarly.

3.1. Interior, closure and boundary. This section reviews some basic concepts re-lated to open/closed subgroups.

Definition 3.8. Let (X,T) be a topological space, and let A ⊆ X.

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

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

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

Lemma 3.9. 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:

A =⋃

U∈T, U⊆A

U ;

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

A =⋂

V c∈T, A⊆V

V.

Proof. We prove part (1) and leave part (2) as an exercise. Let W = ∪U∈T, U⊆AU . SinceT is a topology, W ∈ T and W ⊆ A. This means that W ⊆ A. On the other hand,A ∈ T and A ⊆ A. By definition, A ⊆ W .

Lemma 3.10. Let (X,T) be a topological space. Then the following holds.

(1) A subset A ⊆ X is open if and only if A = A.

(2) For every A ⊆ X, if B = A then B = B.

(3) If A ⊆ B ⊆ A then A = B.


(4) Let Uαα∈Γ be a collection (possibly infinite) of subsets of X, and let U =⋃α∈Γ Uα and V =

⋂α∈Γ Uα; then⋃

α∈Γ

(Uα) ⊆ U V ⊆⋂α∈Γ

(Uα).

(5) Let U1, . . . , Un be a finite collection of subsets ofX and let V =⋂ni=1 Ui. Then,

V =n⋂i=1

(Ui).

Proof. Properties (1) and (2) are obvious by definition. Property (3) is proved asfollows. If A ⊆ B, then by definition A ⊆ B ⊆ B ⊆ A. Also, since B is an opensubset of A, it follows that B ⊆ A. The first part of property (4) is easily checked,while the second part follows by Lemma 3.9 (1), since V ⊆ Uα for each α. Finally,property (5) follows from property (4), since the intersection I = ∩ni=1(Ui)

is an opensubset of ∩ni=1Ui, and hence I ⊆ V .

A similar list of properties exists for the closure of a subset. We state it below withoutproof (the proof is similar to the result above, so it is left as an exercise).

Lemma 3.11. Let (X,T) be a topological space. Then the following holds.

(1) A subset A ⊆ X is closed if and only if A = A.

(2) For every A ⊆ X, if B = A then B = B.

(3) If A ⊆ B ⊆ A then A = B;

(4) Let Uαα∈Γ be a collection (possibly infinite) of subsets of X, and let U =⋃α∈Γ Uα and V =

⋂α∈Γ Uα; then⋃α∈Γ

Uα ⊆ U V ⊆⋂α∈Γ

Uα.

(5) Let U1, . . . , Un be a finite collection of subsets ofX and let V =⋃ni=1 Ui. Then,

V =n⋃i=1

Ui.

Lemma 3.12. Let (X,T) be a topological space, and let A ⊆ X be a subset. Then,

∂(A) = A \ A def= x ∈ A

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

Proof. Let us prove first that ∂(A) ⊆ A \ A. Fix x ∈ ∂(A), and note that x ∈ A since∂(A) = A ∩X \ A, so we have to show that x /∈ A.

22 ALEX GONZALEZ

Suppose for the sake of contradiction that x ∈ A. By definition, X \ A is the smallestclosed subset of X that contains X \ A. Since A is open, its complement X \ A isclosed, and X \ A ⊆ X \ A. In particular, we have

X \ A ⊆ X \ A.

However, if x ∈ A then x /∈ X \ A, and thus x /∈ X \ A, which is a contradiction.

Let us prove now inclusion A\A ⊆ ∂(A). Fix x ∈ A\A, and notice that it is enoughto show that x ∈ X \ A.

Again, suppose that x /∈ X \ A. It follows that x ∈ X \ X \ A, and this is an opensubset of X. Furthermore, recall that A is the greatest open subset of X containedin A, and thus since X \ A ⊆ X \ A, it follows that

X \X \ A ⊆ A,

and x ∈ A, which is again a contradiction.

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

Lemma 3.13. 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:

Ac = X \ A = X \ (A) =(A)c

;

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

(Bc) = (X \B) = X \B =(B)c

;

(3) ∂(A)⊆ A; and

(4) ∂(B)∩B = ∅.

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

)c. Thus,(A)c

= X \ (A) = X \(A ∩

(∂(A)

)c)= (X \ A) ∪

(X \

(∂(A)

)c)=

= Ac ∪((∂(A)

)c)c= Ac ∪ ∂(A) = Ac ∪ ∂(Ac) = Ac

To prove (2) just replace A by Bc everywhere above. To check property (3), let B = A.Then

∂(B)

= B ∩X \B = B ∩X \B ⊆ B.

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

∂(B)

= ∂((B)c

)= ∂

(Bc)⊆ Bc = (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).

3.2. Continuous functions. Continuity of functions is also defined in terms of opensubsets.

Definition 3.14. 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 .

The map f : X → Y is a homeomorphism if f is bĳective (as a map between sets) andboth f and f−1 are continuous. The map f : X → Y is an imbedding if the restrictionto f : X → f(X) is a homeomorphism (in particular, f is injective as a map of sets).

Proposition 3.15. 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).

Exercise 3.16. Let X = 1, 2, 3, 4 and let Y = 1, 2, . . . , 8. Consider also thecollections

TX = ∅, 1, 4, 1, 3, 2, 4, 1, 2, 4, 1, 3, 4, XTY = ∅, 1, 1, 2, 1, 2, 3, . . . , X

Finally, let f : X → Y be the map defined by

f(n) =

2n, if n is evenn+ 1, if n is odd

Prove that TX and TY are topologies for X and Y respectively, and show that f is acontinuous function.

Theorem 3.17. Let (X,TX) and (Y,TY ) be topological spaces, and let f : X → Y bea map. Then the following are equivalent.

(1) f is a continuous map.

(2) For every subset A ⊆ X, we have f(A) ⊆ f(A).

(3) For every closed subset V ⊆ Y , the set f−1(V ) is closed in X.

(4) For each x ∈ X and each V ∈ TY containing f(x), there is some U ∈ TXcontaining x and such that f(U) ⊆ V .

Proof. We prove that (1) implies (2), (2) implies (3), and (3) implies (1). Afterwards weshow that (4) is equivalent to (1).

Assume first that (1) holds, and let A ⊆ X be a subset. For each x ∈ A we have tocheck that f(x) ∈ f(A). Fix such x and let V ∈ TY be any open subset containingf(x). Since f is continuous, the subset U = f−1(V ) is an open subset that contains

24 ALEX GONZALEZ

the element x. Note that U∩A 6= ∅, hence there exists y ∈ A∩U , and f(y) ∈ V ∩f(A).Since every open subset containing f(x) intersects f(A) nontrivially, this means thatf(A) ⊆ f(A).

Assume now that (2) holds. Let B ⊆ Y be a closed subset, and let A = f−1(B). Wehave to show that A = A. We have f(A) ⊆ B, and thus if x ∈ A then

f(x) ∈ f(A) ⊆ f(A) ⊆ B = B.

This implies that x ∈ A = f−1(B), and thus A = A.

Suppose next that (3) holds, and let V ⊆ Y be an open subset. Then W = V c = Y \Vis closed, and

f−1(W ) = f−1(Y \ V ) = X \ f−1(V )

By hypothesis f−1(W ) is closed, and thus f−1(V ) is open.

Next we prove that (1) implies (4). Le x ∈ X and let V ∈ TY containing f(x). Then theset U = f−1(V ) is an open subset containing x. Conversely, assume that (4) holds.Let V ∈ TY and let x ∈ f−1(V ). Then f(x) ∈ V , and by hypothesis there is someUx ∈ TX containing x and such that f(Ux) ⊆ V . Thus Ux ⊆ f−1(V ). It follows thatf−1(V ) =

⋃x∈f−1(V ) Ux, which is an element of TX .

3.3. Basis of a topology. Very often we will not be able to specify all the open subsetsof a given topology, but rather we will specify a subcollection that generates the wholetopology.

Definition 3.18. 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.

Example 3.19. 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 \Q

Both 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 3.20. 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.

(1) For each U ∈ T and each x ∈ U there exists some V ∈ B such that x ∈ V ⊆ U .

(2) Let V1, V2 ∈ B and let x ∈ V1 ∩ V2. Then there exists V ∈ B such that x ∈ V ⊆V1 ∩ V2.


Proof. Suppose first that B is a basis for T. In order to check condition (1), 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 (1), there exists V ∈ B such that x ∈ V ⊆ V1∩V2,and property (B2) holds.

Suppose now that B satisfies properties (1) and (2) 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 (1), and we can write

U =⋃x∈U

Vx.

Hence B is a basis for T.

We can use Lemma 3.20 to formalize this idea of a topology generated by a basis.

Definition 3.21. Let X be a set. A basis for a topology on X is any collection B ofsubsets of X satisfying the following properties.

(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.

Given a basis B for a topology on X, 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 3.22. 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) X ∈ TB since every element in X is contained in some element of B. Theempty set is in TB since it is an element of B.

(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.

26 ALEX GONZALEZ

(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 3.20.

3.4. The subspace topology. Let (X,T) be a topological space, and let A ⊆ X be asubset. We can then define

TA = U ∩ A∣∣ U ∈ T.

Lemma 3.23. The collection TA is a topology on A.

This is known as the subspace topology for obvious reasons.

Proof. (T1) ∅ = ∅ ∩ A, A = X ∩ A ∈ TA.

(T2) Let Uγγ∈Γ ⊆ TA. By definition of TA, for each γ ∈ Γ there exists some Wγ ∈ T

such that Uγ = A∩Wγ. Since T is a topology, we have W =⋃γ∈ΓWγ ∈ 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.

Lemma 3.24. Let (X,T) be a topological space and let A ⊆ X, with the subspacetopology. Then, V ⊆ A is open if and only if there is some U ∈ T such that V = U ∩A.

Proof. The proof is immediate.

Lemma 3.25. Let (X,T) be a topological space, let A ⊆ X be a subspace. Then, theinclusion map incl : A→ X is continuous.

Proof. Let U ∈ T. We have to check that f−1(U) is an open subset of A. By definitionof the inclusion map, we have f−1(U) = U ∩ A, which is indeed open.

The subspace topology can be characterized by a universal property.


Theorem 3.26. Let (X,T) be a topological space, let A ⊆ X be a subset, and letincl : A→ X be the inclusion map. The subspace topology TA on A is the only topologysatisfying the following property.

(S) A map f : (Z,TZ) → (A,TA) is continuous if and only if the composition mapincl f : (Z,TZ)→ (X,T) is continuous.

Proof. Suppose first that f is a continuous map. The inclusion map incl : A → Xis continuous by Lemma 3.25, and it follows that incl f is a continuous map byProposition 3.15. Conversely, suppose that incl f is continuous. We have to checkthat f is continuous: for each V ∈ TA, f−1(V ) ∈ TZ . Let U ∈ TX be such thatV = U ∩ A. Then,

(incl f)−1(U) = f−1(U ∩ A) = f−1(V )

is open, and f is continuous.

Theorem 3.27 (Pasting Lemma). Let (X,TX) and (Y,TY ) be topological spaces, andsuppose that there exist open (closed) subsets A,B ⊆ X such that X = A

⋃B. Con-

sider A and B as topological subspaces of X. Then, the following holds.

(1) If h : X → Y is a continuous function, then f = h|A and g = h|B are continuousfunctions.

(2) If f : A→ Y and g : B → Y are continuous functions such that f(x) = g(x) forall x ∈ A ∩ B, then there exists a (unique) continuous function h : X → Y suchthat f = h|A and g = h|B.

Proof. Part (1) is an immediate consequence of Proposition 3.15 and Lemma 3.25:the map f (respectively g) is the composition of the inclusion map incl : A → X(respectively incl : B → X) and h.

To prove (2), let V ∈ TY . Then,

h−1(V ) = f−1(V ) ∪ g−1(V ).

Since f and g are continuous maps, f−1(V ) ∈ TA and g−1 ∈ TB. Since A and B areopen subsets of X, it follows that f−1(V ) and g−1(V ) are also open subsets of X, andthus h−1(V ) ∈ TX .

3.5. The product topology. Let now (X,TX) and (Y,TY ) be topological spaces, letX × Y be the product set, and consider the collection of subsets

B = U × V∣∣ U ∈ TX and V ∈ TY .

Lemma 3.28. The collection B is a basis for a topology on X × Y .

Proof. We have to check the conditions in Definition 3.21. Condition (B1) states thatfor each (x, y) ∈ X × Y there must exist some W ∈ B such that (x, y) ∈ W . LetU ∈ TX and V ∈ TY be such that x ∈ U and y ∈ V . Then W = U × V ∈ B

28 ALEX GONZALEZ

satisfies condition (B1). Condition (B2) states that, given U1 × V1, U2 × V2 ∈ B and(x, y) ∈ (U1 × V1) ∩ (U2 × V2), there exists some U × V ∈ B such that

(x, y) ∈ U × V ⊆ (U1 × V1) ∩ (U2 × V2).

Choose U = U1 ∩ U2 ∈ TX , V = V1 ∩ V2 ∈ TY . Then U × V clearly satisfies thecondition.

Let TX×Y be the topology generated by the above basis BX×Y , and let πX : X×Y → Xand πY : X × Y → Y be the obvious projection functions. The product topology isdetermined by the following universal property.

Theorem 3.29. 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,TZ)→ (X,TX) and πY f : (Z,TZ)→ (Y,TY ) are continuous.

Proof. The collection B is contained in TX×Y , and clearly satisfies the conditions (1)and (2) in Lemma 3.20, and thus TX×Y is the topology generated by B. We have tocheck then that it satisfies property (P), and that it is unique satisfying this condition.The rest 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 3.15: 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

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 3.30. 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.

The previous results can be stated actually for infinite cartesian products.

30 ALEX GONZALEZ

Definition 3.31. Let Xαα∈Γ be a collection of sets, and let X =⋃α∈ΓXα. The

cartesian product of Xαα∈Γ is∏α∈Γ

Xα = set functions ω : Γ→ X such that ω(α) ∈ Xα ∀α ∈ Γ.

The α-th coordinate of a function ω ∈∏

α∈Γ Xα is the element ω(α) ∈ Xα. The α-thprojection map is the function

πα :∏α∈Γ

Xα → Xα

defined by πα(ω) = ω(α).

Let (Xα,Tα)α∈Γ be a collection of topological spaces. For each α ∈ Γ, let

Sα = π−1α (U) | U ∈ Tα.

Let S =⋃α Sα, and let B be the collection of all finite intersections of elements of S.

Proposition 3.32. The collection B is a basis for a topology on∏

α∈ΓXα, called theproduct topology. With this topology, the map πα : (

∏α∈ΓXα,TB) → (Xα,Tα) is con-

tinuous for all α ∈ Γ. Moreover, the product topology is the only topology on∏

α∈Γ Xα

that satisfies the following condition.

(P) A map f : (Z,TZ)→ (∏

α∈ΓXα,TB) is continuous if and only if the compositionπα f : (Z,TZ)→ (Xα,Tα) is continuous for all α ∈ Γ.

Proof. That B is a basis for a topology is an easy exercise (compare with the proof ofLemma 3.28). Also, it is easy to check that the maps πα are continuous for all α ∈ Γ.Indeed, for a fixed α and some U ∈ Tα, we have π−1

α (Uα) ∈ B by definition.

To prove that property (P) holds for the product topology, suppose first that f : (Z,TZ)→∏α∈ΓXα,TB) is continuous. Then πα f is continuous for all α by part (i), since the

composition of continuous functions is always a continuous function. Conversely,suppose that πα f is continuous for all α ∈ Γ. Let U

∏α∈Γ Uα ∈ TB. If U ∈ B, then

U has the form ∏α∈Γ

Uα

where Uα = Xα for all but finitely many values of α, and

f−1(U) =⋂α∈Γ

(πα f)−1(Uα).

However, we know that Uα = Xα for all but finitely many indices γ ∈ Γ, namelyγ1, . . . , γn. Thus, it follows that

f−1(U) =n⋂i=1

(πγi f)−1(Uγi),


which is an open subset in Z since each πγi f is a continuous function and theintersection is over a finite set of indices. If U ∈ TB, then U is a union of elements ofB, and then f−1(U) is a union of elements of TZ by the above discussion. It followsthat f is continuous. Finally, to check that TB is the only topology satisfying (P), thesame arguments in the proof of 3.29 apply without modification.

Example 3.33. There are other, “natural” topologies that we can consider on theproduct

∏α∈ΓXα, but only one satisfies the universal property above. The box topol-

ogy on∏

α∈ΓXα is the topology generated by the basis

B = ∏α∈Γ

Uα | Uα ∈ Tα.

If the index set Γ is finite (i.e., if we are only considering a finite cartesian product)then the box product is the same as the product topology, but this is not the casewhen the index set Γ is infinite, in which case the box topology has more open subsetsthan the product topology. We will see more about the box topology in the homework.

3.6. The quotient topology. Both the subspace topology and the product topol-ogy are natural generalizations of constructions available for metric spaces. In thissection we study a new construction which is not available for metric spaces. Theidea behind the quotient topology is to create new spaces by “pasting” together otherspaces.

For example, we can construct a cylinder out of a stripe (a subset of R2), by gluingtogether 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), but we can use this example to analyzea more general construction.

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 (although

32 ALEX GONZALEZ

it 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 elementsthat 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 3.34. 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 thatthe 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 isan 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 3.35. 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 f : (X,T)→ (Y,TY ). If f is continuous, then clearly f π is continuous becauseit is a composition of continuous functions. Suppose that f π : (X,T) → (Y,TY )is continuous. We have to show that π−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 itfollows 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 = Idπ′ is continuous, it follows that Id′ is continuous, and hence T ⊆ T.

Example 3.36. As an example of the quotient topology, consider the Euclideanspaces (Rn, dEucl) for each n ≥ 1, and consider the following subsets

An = x ∈ Rn | dEucl(0, x) = 1 ≤ Dn = x ∈ Rn | dEucl(0, x) ≤ 1,with the subspace topologies. For each n ≥ 1, define an equivalence relation onDn by x ∼ y if x, y ∈ An, and no relation on x if x ∈ Dn \ An. The quotient spaceassociated to this relation is the space obtained by collapsing all the points of Antogether. As an exercise, you may try to show that Dn/ ∼ (with the quotient topology)is homeomorphic to An+1 (with the subspace topology).

3.7. Other constructions. Combining the subspace topology, the product topologyand the quotient topology we can create many new topologies (or topological spaces).Here are some of the most relevant constructions in topology.

Example 3.37. The mapping cylinder. Let f : (X,TX)→ (Y,TY ) be a map of topolog-ical spaces. To construct the mapping cylinder of f , let

Mf =(([0, 1]×X)

∐Y)/ ∼,

where ∼ is the equivalence relation generated by (1, x) ∼ f(x) for all x ∈ X. Thetopology of Mf is a combination of the product topology and the quotient topology.

Example 3.38. The mapping cone. Let f : (X,TX)→ (Y,TY ) be a map of topologicalspaces. The mapping cone is a slight variation of the mapping cylinder:

Cf =(([0, 1]×X)

∐Y)/ ≡,

where ≡ is the equivalence relation generated by (1, x) ≡ f(x) and (0, x) ≡ (0, z)for all x, z ∈ X. Again, the topology of Cf is a combination of the product and thequotient topologies.

Example 3.39. The suspension of a topological space. The suspension of (X,T) is

SX = ([0, 1]×X)/ ∼,where ∼ is the equivalence relation generated by (0, x) ∼ (0, z) and (1, x) ∼ (1, z)for all x, z ∈ X. The topology on SX is again a combination of the product and thequotient topologies.

34 ALEX GONZALEZ

Example 3.40. The join of two spaces. The join of (X,TX) and (Y,TY ) is

X ∗ Y = ([0, 1]×X × Y )/ ≡,where ≡ is the equivalence relation generated by (0, x, y) ≡ (0, x, u) and (1, x, y) ≡(1, z, y) for all x, z ∈ X and all y, u ∈ Y .


4. Connectedness

Now that we understand the basics of topological spaces, we can start comparingdifferent topological spaces, trying to find properties that will distinguish one spacefrom another. This chapter deals with the first of such properties: connectedness.Essentially, a topological space is connected if it cannot be formed out of smaller,disjoint pieces.

Definition 4.1. 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 = ∅.

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

Example 4.2. Let X be a set with the discrete topology. Then, X is connected if andonly if X is the empty set or a singleton (if X contains more than one element, theneach singleton of X is open and closed).

Example 4.3. Every set with the indiscrete topology is connected.

Example 4.4. Consider R with the topology corresponding to the Euclidean metric,and let Q be the subspace of rational numbers. Then, Q is not connected. Indeed,the subset

(−∞, π) ∩Qis open and is also the complement of an open subset, hence closed.

Example 4.5. Consider the set Z with the topology associated to the p-adic metric.This space is not connected, since the subset A = n · p

∣∣ n ∈ Z ⊆ Z of integersdivisible by p is both open and closed. Indeed, 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 connectedness.

Theorem 4.6. Let (X,TX) be a topological space. Then, the following are equivalent.

(1) (X,TX) is connected.

(2) There do not exist open subsets U, V ⊆ X such that U ∪V = X and U ∩V = ∅.

(3) There does not exist any surjective continuous map f : (X,TX)→ (0, 1,Tdisc).

36 ALEX GONZALEZ

Proof. Let’s check first that (1) is equivalent to (3). Suppose first that there is asurjective, continuous map f : (X,TX) → (0, 1,Tdisc), and let A = f−1(0) andB = f−1(1). Since f is surjective, both A and B are non-empty subsets of X.Furthermore, since 0 (respectively 1) is and open an closed subset of 0, 1 andf is continuous, it follows that A (respectively B) is an open and closed subset of X.Finally, since both A and B are non-empty, it follows that none of them is either theempty set or X, 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 of X. We can then define a map f : (X,TX)→ (0, 1,Tdisc) by settingf(U) = 0 and 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 4.7. Let f : (X,TX) → (Y,TY ) be a continuous map of topological spaces,and suppose X is connected. Then, f(X) ⊆ Y is a connected subspace of Y .

Proof. Exercise.

Corollary 4.8. Let h : (X,TX) → (Y,TY ) be a homeomorphism of topological spaces.Then, X is connected if and only if Y is connected.

Theorem 4.9. The set R with the topology associated to the Euclidean metric is con-nected.

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) = A ∩ R \ A = A ∩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. This part of the proof uses the Euclidean metric onR. Let then a ∈ R \ U . If

U ∩ (a,∞) = ∅ = U ∩ (−∞, a),

then U = ∅, contrary to the assumption that U is non-empty. Suppose then thatU ∩ (a,∞) 6= ∅ (the other possible 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 4.10. 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.

Proposition 4.11. Let (X,TX) be a topological space and let U,K ⊆ X be subsets.Suppose in addition that U is connected and U ⊆ K ⊆ U . Then K is connected.

Proof. Suppose that K = A⋃B, where A,B are open, non-empty subsets of X, and

such that A ∩ B = ∅, and such that K ∩ A,K ∩ B 6= ∅. Since U is connected andU ⊆ K, it follows that either U ⊆ A or U ⊆ B, since otherwise we would contradictthe connectivity of U . Suppose for simplicity that U ⊆ A. Then we have

U ⊆ K ⊆ U ⊆ A.

Moreover, since A and B are disjoint open subsets, it follows that A ∩ B = ∅, andthus K ∩ B = ∅, which is a contradiction with the assumptions above. Thus K isconnected.

Theorem 4.12. Consider R with the topology associated to the Euclidean metric, andlet a < b ∈ R. Then, [a, b] ⊆ R is connected.

Proof. This follows immediately from Proposition 4.11 since [a, b] is the closure of(a, b), and the latter is connected.

Lemma 4.13. Let (X,TX) 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. 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 4.14. Let (X,TX) and (Y,TY ) be topological spaces, and let (X × Y,T) bethe product topological space. Then, X × Y is connected if and only if X and Y areconnected.

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 maps.

38 ALEX GONZALEZ

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 .

Note that B is homeomorphic to Y , while Cy is homeomorphic to X for all y ∈ Y .Thus, 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 4.13. 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 4.13.

Corollary 4.15. A finite cartesian product of connected topological spaces is connected.In particular, the space Rn (with the topology associated to the Euclidean metric) isconnected.

Theorem 4.16. Let (Xα,Tα)α∈Γ be a collection of topological spaces, and considerthe product space (

∏α∈ΓXα,Tprod) with the product topology (see 3.31 for details).

Then, (∏

α∈ΓXα,Tprod) is connected if and only if (Xα,Tα) is connected for all α ∈ Γ.

Proof. For each element b = (bα)α∈Γ ∈∏

α∈ΓXα and each γ ∈ Γ, let Yb,γ ⊆∏

α∈ΓXα

be the subset of elements of the for y = (yα)α∈Γ, with yα = bα for all α 6= γ. Then,Yb,γ, with the subspace topology, is homeomorphic to Xγ (check the details!). Inparticular, Yb,γ is connected since Xγ is connected.

Suppose for the sake of contradiction that∏

α∈ΓXα is not connected, and let U, V ∈Tprod be nonempty subsets such that∏

α∈Γ

Xα = U⋃

V and U⋂

V = ∅.

Fix some element b ∈∏

α∈ΓXα, and consider the subspaces Yb,αα∈Γ. For eachα ∈ Γ, we have Yb,α ⊆ U or Yb,α ⊆ V , because Yb,α is connected. However, sinceYb,α ∩ Yb,γ 6= ∅ for all α, γ ∈ Γ, it follows that either⋃

α∈Γ

Yb,α ⊆ U or⋃α∈Γ

Yb,α ⊆ V.

Suppose for simplicity that⋃α∈Γ Yb,α ⊆ U .

Let B be the usual basis for the product topology. Since V is open, there is someW ∈ B such thatW ⊆ V . By definition of B, we haveW =

∏α∈Γ Wα, whereWα = Xα

for all but finitely many indices. Let γ1, . . . , γn ∈ Γ be the indices such thatWγi 6= Xγi,and let b′ = (b′α)α∈Γ ∈ W , so that b′γi ∈ Wγi for each i = 1, . . . , n.


Fix γ ∈ Γ be such that γ 6= γ1, . . . , γn. Let b1 = (b1α)α∈Γ be the element with coordinates

b1γ1

= b′γ1 and b1α = bα otherwise. Then,

Yb1,γ ∩ Yb,γ1 6= ∅.

Since Yb1,γ ∼= Xγ is connected and Yb,γ ⊆ U , it follows that Yb1,γ ⊆ U . Moreover, wededuce now that Yb1,α ⊆ U for all α ∈ Γ, since Yb1,α ∩ Yb1,γ 6= ∅.

Next, let b2 = (b2α)α∈Γ be the element with coordinates b2

γ1= b′γ1, b

2γ2

= b′γ2 and b2α = bα

otherwise. Then,Yb2,γ ∩ Yb1,γ2 6= ∅.

Again, we see that Yb2,γ ⊆ U , since Yb2,γ is connected and Yb1,γ2 ⊆ U as shown above.As a consequence, Yb2,α ⊆ U for all α ∈ Γ. Inductively, we can see that Yb′,γ ⊆ U . Onthe other hand, we have Yb′,γ ∩ V 6= ∅ since b′ ∈ W ⊆ V . Thus, U ∩ V 6= ∅, which isa contradiction.

Example 4.17. Let X = RN be a countable cartesian product of copies of R. Let alsoT be the box topology (see Example 3.33). Then, (X,T) is not connected. To seethis, let

• A ⊆ X be the set of all bounded sequences of real numbers (i.e., if a ∈ A thenthere exists some N ∈ R such that |an| ≤ N for all n).

• B ⊆ X be the set of all unbounded sequences of numbers.

These two subsets are clearly disjoint, and X = A⋃B. We claim that they are both

open. Let a ∈ X be a point, and consider the open subset

U = (a1 − 1, a1 + 1)× (a2 − 1, a2 + 1)× . . .

If a ∈ A (if a is bounded), then all the points in U are bounded sequences, thusU ⊆ A. On the other hand, if a ∈ B (if a is unbounded), then every point of U is anunbounded sequence, so U ⊆ B. This way, we can describe A and B as unions ofopen subsets in the product topology, and thus they are both open.

Lemma 4.18. 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. Prove that the space Gf is homeomorphic to X (exercise). Then the statementfollows.

Example 4.19. 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.

40 ALEX GONZALEZ

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. Thespace Ω is the closure of Γ, i.e. Ω = Γ (prove it!), and by Proposition 4.11 Ω isconnected. We will come back to this example in the next section.

4.1. Path-connected spaces. Intuition may tell us that a space is connected if wecan “link together” any two elements by a path. Although this idea is not exactlyright, it is the basis of an important concept that we analyze in this section.

Definition 4.20. Let (X,TX) be a topological space, and let x, y ∈ X. A path in Xfrom x to y is a continuous map ω : [0, 1]→ X with ω(0) = x and ω(1) = y.

A topological space (X,TX) is path-connected if, for any two elements x, y ∈ X thereis a path from x to y.

Theorem 4.21. Let (X,TX) be a path-connected space. Then (X,TX) 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 com-position 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 4.12.

Example 4.22. For all n ≥ 1, Rn is path-connected.

Example 4.23. 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 4.24. Let (X,TX) 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 4.25. A cylinder is not homeomorphic to a Moebius band.

Proof. Let M be a Moebius band and C be a cylinder, both seen as subsets of R3. Letalso ∂(M) and ∂(C) be the boundaries of M and C are subsets of R3. Clearly, ∂(C)is not connected, while ∂(M) is path-connected (hence connected). Thus, no maph : M → C can be a homeomorphism (as an exercise, show why should h map ∂(M)to ∂(C)).

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

Example 4.26. Let Ω ⊆ R2 be the subset in Example 4.19, 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) = 1π, 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+1) =2

(2n+ 3)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.

4.2. Connected components. Connectivity constitutes then the first of a series ofproperties that we can use to characterize and distinguish topological 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 topologicalspace (X,TX), how many pieces are needed to “construct” X?

Definition 4.27. Let (X,TX) be a topological space. A connected component of X isa connected subset U ⊆ X such that

42 ALEX GONZALEZ

• 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 4.28. Every continuous map f : (X,TX)→ (Y,TY ) 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 w ∈ V . Suppose otherwise: w /∈ V . ThenV $ W = V ∪ w, and thus W is not connected, since V is a connected componentof Y . But this is impossible, since f(U) has to be connected.

Corollary 4.29. Let f : (X,TX) → (Y,TY ) be a homeomorphism. Then f defines abĳection 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.


5. Compactness

Compactness can be thought of as a generalization of finiteness. If X is a finite set(with the discrete topology) then studying (continuous) maps out of X is reasonablyeasy, we just have to check what happens with the function on every element of X.But we want to consider other maps, not only maps out of finite sets! The idea is thatmaps out of compact spaces behave nicely, as maps out of finite spaces (of coursethis is rather imprecise).

Definition 5.1. Let (X,TX) be a topological space and let A ⊆ X be any subset. Anopen cover 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.

Example 5.2. Consider the family of intervals Un = (n, n+ 2) ⊆ R (Euclidean topol-ogy), for n ∈ Z. The collection Unn∈Z is an open cover of R. This open cover doesnot contain any proper subcover: if we remove Un for some n ∈ Z then the union ofthe remaining subsets Um, m 6= n, does not cover R.

Definition 5.3. Let (X,TX) be a topological space. A subset A ⊆ X is compact ifevery open cover of A contains a finite subcover.

Remark 5.4. Let (X,TX) be a topological space and A ⊆ X. Let also Uγγ∈Γ be anopen cover of A in X. A finite subcover is 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.

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 topological space (X,TX), the collection X is a finite open cover of X,but this 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 5.5. Consider R with the Euclidean topology. Then, the open cover inExample 5.2 does not contain any finite subcover, and hence R is not compact.

Example 5.6. Consider R with the Euclidean topology, 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).

44 ALEX GONZALEZ

Example 5.7. Let X = 0 ∪ 1/n |n ∈ N, with the subspace topology of a subsetof R. Then X is a compact. Indeed, let Uγγ∈Γ be an open cover of X, and let α ∈ Γbe such that 0 ∈ Uα. Then Uα must contain all but finitely many elements of X. Foreach x ∈ X \ Uα, choose Ux ∈ Uγγ∈Γ containing x. This forms a finite subcover.

Theorem 5.8. Let (X,TX) be a compact space. If A ⊆ X is closed then A is compact.

Proof. Let A ⊆ X be a closed subset, and let Uγγ∈Γ be an open cover (recall that thesubsets Uγ are open in X). Then the collection Ac

⋃Uγγ∈Γ is an open cover of X,

and there exists some finite subcover, W1, . . . ,Wn, since X is compact. But clearlythis subcover also covers A, and upon removing Ac from W1, . . . ,Wn if necessarythis constitutes a finite subcover of the original cover Uγγ∈Γ. Thus A is compact.

Theorem 5.9. Let f : (X,TX)→ (Y,TY ) be a continuous map of topological spaces. IfA ⊆ X is compact then so is f(A) ⊆ Y .

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).

Corollary 5.10. Let f : (X,TX) → (Y,TY ) be a homeomorphism. Then X is compactif and only if Y is compact.

Theorem 5.11. Let (X,TX) and (Y,TY ) be topological spaces, and let A ⊆ X, B ⊆ Ybe compact subsets. Then, A×B is a compact subset of X × Y .

Proof. Before we prove the theorem, we show the following.

Step 1. Let a ∈ A and let N ⊆ X×Y be an open subset such that a×B ⊆ N . Thenthere exists some U ⊆ A such that a ×B ⊆ U ×B ⊆ N .

You can think of U ×B as a tube around the ray a ×B. To show the above point,let Wγγ∈Γ be an open cover of a × B, where, for each γ, the subset Wγ is anelement of the basis for the product topology such that Wγ ⊆ N . In other words, foreach γ ∈ Γ,

Wγ = Uγ × Vγ ⊆ N,

where Uγ ∈ TX and Vγ ∈ TY .

Since a × B ∼= B and B is compact, there exists a finite subcover W1, . . . ,Wn,convering a ×B. The following picture illustrates the situation.


A

B

W1

W2

W3

W4

W5W6

a

Recall that Wk = Uk × Vk for k = 1, . . . , n. Set

U = U1 ∩ . . . ∩ Un ∈ TX .

Then a ∈ U . Furthermore, if (x, y) ∈ U × B, then there exists some k such that(x, y) ∈ Uk × Vk. Indeed, since W1, . . . ,Wn covers a × B, there is some k suchthat (a, y) ∈ Wk = Uk × Vk, and then (x, y) ∈ Wk since U ⊆ Uk. Since Wγ ⊆ N for allγ ∈ Γ, it follows that U ×B ⊆ N .

Step 2. Proof of the theorem.

Let Wγγ∈Γ be an open cover of A×B. Given a ∈ A, the ray a×B is compact andmay be covered by finitely many elements Wa,1, . . . ,Wa,ma of the original cover. Theunion Na = Wa,1 ∪ . . . ∪Wa,ma is an open susbet containing a × B, and by Step 1we know that there exist a tube Ua ×B such that

a ×B ⊆ Ua ×B ⊆ Na.

Thus, for each a ∈ A we may choose some Ua ∈ TX containing the element a suchthat the tube Ua × B can be covered by finitely many elements of Wγγ∈Γ. Notethat Uaa∈A forms an open cover of A, and thus there exist a1, . . . , an ∈ A such thatUa1 , . . . , Uan is a finite subcover, since A is compact.

Note that A×B ⊆ Ua1 ×B ∪ . . . ∪ Uan ×B. Thus, the collection

Wa1,1 . . . ,Wa1,ma1,Wa2,1, . . . ,Wa2,ma2

, . . . ,Wan,1, . . . ,Wan,manis a finite subcover of A×B.

46 ALEX GONZALEZ

5.1. Compact subspaces of Rn. In this section we study compact subspaces of avery particular type. Be extremely careful with this section because most of theresults here apply only to Rn with the Euclidean metric or, in some cases, to metricspaces. We start with a definition.

Definition 5.12. Let (X, dX) be a metric space. A subset A ⊆ X is bouned if thereexist x ∈ X and ε > 0 such that A ⊆ Bx,ε.

Note that the above definition does not make much sense for topological spaces! Letus now characterize compact subsets of R.

Theorem 5.13 (Heine-Borel Theorem for R). Consider R with the Euclidean metric,and let 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 ∈ A\A. For each n ∈ N,let

Un =(Bx, 1

n

)c= R \Bx, 1

n.

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 \Bx, 1n

= R \⋂n∈N

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 ∈ A. Hence A = 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.

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 5.14. 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.

Theorem 5.15 (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. We already know from the homework that if A is compact then it is closed andbounded. Suppose then that A is closed and bounded. In particular, there is someinterval [m,M ] ⊆ R such that

A ⊆ I = [m,M ]× . . .× [m,M ] ⊆ Rn.

48 ALEX GONZALEZ

Now, the subset I ⊆ Rn is compact by Theorem 5.11, since I is the direct productof compact subsets of R, and then by Lemma 5.8 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 Rn iscompact.

Example 5.16. 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 5.17. 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.

5.2. Nets and Tychonoff’s Theorem. Tychonoff’s Theorem is one of the most im-portant results of this course. It states that the product of any arbitrary collection ofcompact spaces is again compact (with the product topology, of course). As we sawin previous sections, dealing with arbitrary (thus possibly infinite) products requiressubtlety. Thus we need a different approach than we used for the product of twospaces. Indeed, instead we will follow Chernoff’s proof using nets.

A directed set is a partially ordered set (A,≤) satisfying the following property: forall α, β ∈ A, there exists some γ ∈ A such that α, β ≤ γ. The concept of net in atopological space is a generalization of the notion of sequence in a metric space.

Definition 5.18. Let (X,T) be a topological space, and let (A,≤) be a directed set.A net in X is a function x : A → X. We use the following notation: for each α ∈ A,we set xα = x(α). This way, we can write the net x : A → X as xαα∈A, whichemphasizes the connection between nets and sequences (a sequence is just a netwhere A = N, the set of natural numbers).

The net xαα∈A is said to converge to a point p ∈ X if the following holds: for allU ∈ T such that p ∈ U , there is some α ∈ A such that xβ ∈ U for all β ∈ A withβ ≥ α. We then say that p is a limit point of the net x.

A point p ∈ X is a cluster point of the net xαα∈A if, given any U ∈ T such that p ∈ Uand any α ∈ A, there is some β ≥ α with xβ ∈ U .

In general, a net x : A → X does not converge to a unique point, but this is notsurprising, we already know that a sequence in a topological space may have severaldifferent limit points. As an exercise you can check that a subset V ⊆ X is closed ifand only if every limit point of every net in V is a point in V . Cluster points generalizethe idea of limits of subsequences of a given sequence. For example, in the sequencexn = (−1)nn∈N, the element −1 is a cluster point.


Definition 5.19. Let (A,≤) and (B,4) be directed sets. A map φ : B → A is cofinalif, for each α ∈ A, there exists some β ∈ B such that α ≤ φ(β′) whenever β 4 β′.Given a net x : A→ X and a cofinal map φ : B → A, the map x φ : B → X is againa net, and we say that x φ is a subnet of x.

Proposition 5.20. Let x : A→ X be a net. A point p ∈ X is a cluster point of xαα∈Aif and only if p is a limit point of some subnet of xαα∈A.

Proof. The proof is left as an exercise.

An interesting feature of nets is that compactness can be characterized in terms ofnets. This is in fact a generalization of the characterization of compactness for metricspaces in terms of sequences.

Proposition 5.21. Let (X,T) be a topological space. Then X is compact if and only ifevery net in X has a cluster point (i.e., every net in X has a converging subnet).

Proof. Suppose first that X is compact, and let xαα∈A be a net in X. Suppose alsothat xαα∈A does not have any cluster point. Then, no element y ∈ X is the limitof any subnet, which means that there exists some Uy ∈ T and αy ∈ A such thaty ∈ Uy, and such that xβ /∈ Uy for all β ≥ αy.

Note that Uyy∈X is an open cover of X. Thus, since X is compact, there exists afinite subcover Uy1 , . . . , Uyn ⊆ Uyy∈X . Since A is a directed set, there exists someα ∈ A such that α ≥ αyi for all i = 1, . . . , n. Then, it follows that for all β ≥ α,xβ /∈ Uyi, for all i = 1, . . . , n, and thus xβ /∈ X =

⋃ni=1 Uyi, which is absurd. Thus

xαα∈A must have a cluster point.

Suppose now that every net in X has a cluster point, and let Uii∈I be an open coverof X. Suppose also that Uii∈I does not have any finite subcover. We can constructa net in X as follows. Let A be the set of all finite subsets of I, ordered by inclusion.That is, if α, β ∈ A, then α ≤ β if α ⊆ β as subsets of I. Now, for each α ∈ A, letWα =

⋃i∈α Ui ⊆ X. Since Uii∈I does not have any finite subcover, it follows that

Wα $ X for all α ∈ A, and we can choose xα ∈ X \Wα. This defined a net xαα∈A.

By assumption, the net xαα∈A must have a cluster point p ∈ X. Since Uii∈I isan open cover of X, there exists some Uj ∈ Uii∈I such that p ∈ Uj, and there existssome α ∈ A such that xβ ∈ Uj for all β ≥ α. Note that if j ∈ β, thenxβ /∈ Wβ. On theother hand Uj ⊆ Wβ by definition of Wβ, and this is a contradiction. Thus Uii∈Imust have a finite subcover.

Theorem 5.22 (Tychonoff’s Theorem). Let (Xγ,Tγ)γ∈Γ be a collection of topologicalspaces, and let (

∏γ∈ΓXγ,Tprod) be the product topological space. Then

∏γ∈ΓXγ is

compact if and only if Xγ is compact for all γ ∈ Γ.

50 ALEX GONZALEZ

Proof. Clearly, if∏

γ∈Γ Xγ is compact, then Xγ is compact for all γ ∈ Γ, since itis the image of

∏γ∈ΓXγ through the projection map πγ :

∏γ∈ΓXγ → Xγ, which is

continuous.

Suppose now that Xγ is compact for all γ ∈ Γ. We prove that∏

γ∈ΓXγ is compactby means of Proposition 5.21. We will also need Zorn’s Lemma: let Ω is a partiallyordered set, and suppose that every simply ordered subset has an upper bound; thenΩ has a maximal element.

Let xαα∈A be a net in X. We have to show that xαα∈A has a cluster point. Firstwe fix some notation.

• Let B be the usual basis for the product topology. Recall that U ∈ B is ofthe form U =

∏γ∈Γ Uγ such that Uγ = Xγ for all but finitely many indices γ.

Given U ∈ B, let F (U) ⊆ Γ be the subset of indices such that Uγ $ Xγ for allγ ∈ F (U).

• Let Γ′ ⊆ Γ, and let U ∈ B be such that F (U) ⊆ Γ′. Then, we can defineU ′ =

∏γ∈Γ′ Uγ, which is a element in the basis B′ for the product topology of∏

γ∈Γ′ Xγ.

• A partially defined element g of∏

γ∈ΓXγ is an element of∏

γ∈Γ′ Xγ, for someΓ′ ⊆ Γ.

• Given Γ′ ⊆ Γ, the net xαα∈A induces a net xΓ′α α∈A in

∏γ∈Γ′ Xγ, with

(xΓ′α )γ = (xα)γ for all γ ∈ Γ′ (basically, xΓ′

α is the restriction of xα to Γ′).

• A partially defined element g ∈∏

γ∈Γ′ Xγ is a partial cluster point of xαα∈A ifg is a cluster point of the net xΓ′

α α∈A.

Let P be the set of all partial cluster points of xαα∈A. To see that P is nonempty, letγ ∈ Γ, and set Γ′ = γ. Then, the restriction of xαα∈A to Γ′, namely x′αα∈A, hasa cluster point, since Xγ is compact, and this cluster point is a partial cluster pointof xαα∈A. We can define a partial order in P as follows. Let g, h ∈ P. By definition,g ∈

∏γ∈Γ′ Xγ and h ∈

∏γ∈Γ′′ Xγ, for some Γ′,Γ′′ ⊆ Γ. Then, g ≤ h if Γ′ ⊆ Γ′′ and

gγ = hγ for all γ ∈ Γ′.

We want to apply Zorn’s Lemma to P to show that it has maximal elements, but firstwe have to show that P satisfies the conditions in Zorn’s Lemma. Let L = gii∈Ibe a simply ordered subset of P. For each i ∈ I, we have gi ∈

∏γ∈Γi

Xγ, for someΓi ⊆ Γ, and if i ≤ j, then Γi ⊆ Γj. Define Γ′ =

⋃i∈I Γi ⊆ Γ, and let g0 ∈

∏γ∈Γ′ Xγ be

the element defined by (g0)γ = (gi)γ if γ ∈ Γi.

We have to check that g0 ∈ P, that is, g0 is a partial cluster point of xαα∈A. LetU ∈ B with F (U) ⊆ Γ′, and let U ′ ∈ B′ be the corresponding basis element. SinceF (U) is finite, there exists some i ∈ I such that F (U) ⊆ ΓI . Let x′αα∈A be therestriction of xαα∈A to ΓI , and let UI =

∏γ∈Γi

Uγ, which is an element in the basis


for the product topology of∏

γ∈ΓiXγ. Since gi is a cluster point of x′αα∈A and U ′

does not care about the coordinates away from F (U), it follows that g0 is a partialcluster point of the net induced by xαα∈A in

∏γ∈Γ′ Xγ. Thus g0 ∈ P. Clearly, g0 is

also an upper bound for L, and thus P satisfies the hypothesis of Zorn’s Lemma.

Let g ∈ P be a maximal element. By definition, g ∈∏

γ∈Γ′ Xγ, for some Γ′ ⊆ Γ, andwe claim that Γ′ = Γ (which means that g is a cluster point of the net xαα∈A).Suppose otherwise, and let ω ∈ Γ \ Γ′. Let x′αα∈A be the restriction of xαα∈A toΓ′. By definition, g is a cluster point of the net x′αα∈A. Let also x′′αα∈A be the netin Xω defined by x′′α = πω(xα). Since Xω is compact, the net x′′αα∈A has a clusterpoint p ∈ Xω.

Let Γ′′ = Γ′ ∪ ω, and let h ∈∏

γ∈Γ′′ Xγ be the point with coordinates hγ = gγ, forγ ∈ Γ′, and hω = p. Then h is a partial cluster point of xαα∈A, and g ≤ h aselements of P. This contradicts the maximality of g in P, and thus we must haveΓ′ = Γ, so g is a cluster point of xαα∈A. This finishes the proof.

52 ALEX GONZALEZ

6. Countability and separation axioms

In general, it is a rather difficult question to show if a certain topology corresponds tothe collection of open subsets with respect to some metric. In some sense, if a giventopology does not satisfy this property, then it is a strange topology. The countabilityaxioms and the separation axioms are designed to somehow measure how far is atopology from being associated to a metric in the sense above, or in other words tomeasure how strange the space is.

Eventually, we will combine the results in this section to prove a general result abouttopologies that are associated to metrics in the sense above.

6.1. The countability axioms. Roughly speaking, the countability axioms are con-ditions about how many subsets of a set are necessary in order to generate a giventopology: the lesser the better. Recall the definition of basis of a topology in 3.21.

Definition 6.1. Let (X,TX) be a topological space. X has a countable basis at x ifthere is a countable collection B of open subsets satisfying the following:

(i) x ∈ U for each U ∈ B; and

(ii) if V ∈ TX is such that x ∈ V , then there exists some U ∈ B such that U ⊆ V .

If X has a countable basis at every element x ∈ X, then X satisfies the first count-ability axiom. We also say that X is first-countable.

If (X,T) is first-countable, then we can form a basis B for T as follows. For eachx ∈ X, let Bx ⊆ T be a countable basis at x, and let B =

⋃x∈X Bx. This turns out

to be a basis for T (check out the conditions!), although not necessarily a countablebasis.

Theorem 6.2. Let (X, dX) be a metric space, and let TX be the topology associated todX . Then (X,TX) is first-countable.

Proof. For each x ∈ X, the collection B = Bx, 1/n satisfies conditions (i) and (ii) inthe definition above.

Definition 6.3. A topological space (X,TX) satisfies the second countability axiom ifit has a countable basis for TX . We also say that X is second-countable.

Remark 6.4. If (X,TX) is second-countable then it is first-countable, but the con-verse is not true in general.

Example 6.5. The obvious example of a second-countable space is the Euclideanline (R,TEucl). Consider the collection

B = (x, y) |x, y ∈ Q, x < y ∪ ∅.This forms a basis of TEucl, and it is easily checked to be a countable basis.


Although the following example may bot be very interesting, it shows that metricspaces need not be second-countable.

Example 6.6. Let X be an uncountable set (for example X = R), and let d be thediscrete metric on X. Then, (X, d) is not second-countable.

To see a more interesting example, we need a lemma first.

Lemma 6.7. Let (X,TX) be a second-countable space. Then, every discrete subspaceA ⊆ X is countable.

Proof. Let B ⊆ TX be a countable basis. Since A is discrete, every singleton a ⊆ Ais an open subset of A, which means that there exists some U ∈ TX such thatU ∩ A = a. Hence, for each a ∈ A we may choose a basis element Ua ∈ B suchthat Ua ∩ A = a. This produces an injective map A → B, where a 7→ Ua (if a 6= bthen Ua ∩ Ub = ∅, so this map is indeed injective), and thus A is countable.

Example 6.8. Let Y = R, and let dY be the Euclidean metric (i.e. absolute value).Let also d : R× R→ [0,∞) be the map defined by

d(x, y) = min|x− y|, 1.As an exercise, we can show that this is a metric on Y . Now, consider X = Πγ∈ΓR, aninfinite (possibly uncountable) product of copies of R. Define also dX : X×X → [0,∞)by

dX = supdY (xγ, yγ) | γ ∈ Γ.This defines a metric on X, called the uniform metric. Again, details that this is ametric are left as an exercise. Since (X, dX) is a metric space, it is first-countable byTheorem 6.2. We claim that it is not second-countable. Let A ⊆ X be the subset ofelements (xγ)γ∈Γ such that xγ ∈ 0, 1 for each γ ∈ Γ. Then A is a discrete subspace,because

dX((xγ)Γ, (yγ)Γ) = 1

if (xγ)Γ 6= (yγ)Γ. By Lemma 6.7, if (X,TX) was second-countable then A should becountable, but A is not countable.

Proposition 6.9. The following holds.

(i) A subset of a first-countable space is again first-countable.

(ii) A countable product of first-countable spaces is again first-countable.

(iii) A subset of a second-countable space is again second-countable.

(iv) A countable product of second-countable spaces is again second-countable.

Proof. The proof is left as an exercise.

Definition 6.10. Let (X,TX) be a topological space. A subset A ⊆ X is dense ifA = X.

54 ALEX GONZALEZ

Theorem 6.11. Let (X,TX) be second-countable, and let A ⊆ X. Then the followingholds.

(i) Every open cover of A contains a countable subcover.

(ii) There exists a countable subset of X that is dense in X.

Proof. Let B be a countable basis forX, and let BA = W∩A |W ∈ B be a countablebasis for the subspace topology on A. Since BA is a countable basis, we may write itas W1,W2, . . ..

Part (i). Let Uγγ∈Γ be an open covering of A. For each n ∈ N, let Uγn ∈ Uγ besuch that Wn ⊆ Uγn or, if no such Uγn exists, set Uγn = ∅. Let also

Ω = Uγn |n ∈ N and Uγn 6= ∅ ⊆ Uγγ∈Γ.

The collection Ω is countable, and also covers A. Indeed, for each a ∈ A there issome Uα ∈ Uγγ∈Γ such that a ∈ Uα, and thus Uα contains some element of BA bydefinition of basis.

Part (ii). From each nonempty basis element Wn ∈ BA, choose some xn ∈ Wn, andlet D = xn |Wn ∈ BA , Wn 6= ∅. Then D is dense in A: for each a ∈ A, every basiselement in BA that contains a must intersect D, so x ∈ D.

6.2. The separation axioms. The idea behind the separation axioms is to answerthe following question: can we separate points in a given space? If a topology isassociated to a metric, then it is not difficult to check that the answer is yes always,but this is not the case in general for topological spaces. In fact, when it comes totopological spaces, a whole spectrum of axioms is available to describe increasinglyrestrictive situations, which in a sense measure how close is our topology to beassociated to a metric.

Definition 6.12. Let (X,T) be a topological space.

(T0) We say that (X,T) is T0 or Kolmogorov if, given two points x, y ∈ X, x 6= y,there exists U ∈ T such that either x ∈ U 63 y or x /∈ U 3 y.

(T1) We say that (X,T) is T1 if, given two points x, y ∈ X, x 6= y, there existU, V ∈ T such that x ∈ U 63 y and x /∈ V 3 y.

(T2) We say that (X,T) is T2 or Hausdorff if, given two points x, y ∈ X, x 6= y,there exist U, V ∈ T such that x ∈ U 63 y and x /∈ V 3 y, and such thatU ∩ V = ∅.

(T3) We say that (X,T) is T3 or regular if it is T1 and, given a point x ∈ X and aclosed subset B ⊆ X with x /∈ B, there exist U, V ∈ T such that x ∈ U + Band x /∈ V ⊇ B, and U ∩ V = ∅.


(T4) We say that (X,T) is T4 or normal if it is T1 and, given closed subsetsA,B ⊆ Xwith A ∩ B = ∅, there exist U, V ∈ T such that A ⊆ U + B and A 6⊆ V ⊇ B,and U ∩ V = ∅.

Lemma 6.13. Let (X,TX) be a T1 space. Then every singleton x ⊆ X is a closedsubset.

Proof. Fix x ∈ X and let y 6= x be another element of X. By definition there existU, V ∈ TX such that x ∈ U 63 y and x /∈ V 3 y. This implies that V ∩ x = ∅, andthus y does not belong to x.

Theorem 6.14. Let (X,T) be a topological space. For each i = 0, . . . , 4, if (X,T) is Tithen it is Ti−1.

Proof. The statement is clear for i = 0, 1, 2. Suppose now that (X,T) is T3. By Lemma6.13 we know that singletons are closed subsets in X. Let x, y ∈ X, and let B = y.Since X is T3, there exist U, V ∈ T such that x ∈ U + B and x /∈ V ⊇ B, andU ∩ V = ∅. Hence, (X,T) is T2.

Finally, suppose that (X,T) is T4, and we have to show that it is also T3. By definitionof T4 space, (X,T) is T1. Let x ∈ X be a point and let B ⊆ X bet a closed subset.Let also A = x ⊆ X, which is a closed subset by Lemma 6.13. Then, there existU, V ∈ T such that A ⊆ U + B and A 6⊆ V ⊇ B, and U ∩ V = ∅. Hence (X,T) isT3.

Remark 6.15. We will usually talk about Hausdorff, regular or normal spaces, ratherthan T2, T3 or T4 spaces.

The above result shows how each axiom Ti is stronger (i.e. more restrictive) than theprevious one. In general, being normal is considered to be a good property, althoughusually too restrictive, whereas not even being T0 is usually very bad. Often, whenworking on some topological problem, it is asked that the spaces considered are atleast Hausdorff, as a good balance between normality and T0-ness. Next we showthat in fact being a metric space is an even more restrictive condition.

Theorem 6.16. Let (X, dX) be a metric space, and let TX be the topology associatedto dX . Then (X,TX) is normal (T4).

Proof. First we have to show that (X,TX) is T1. Let x, y ∈ X, with x 6= y, and letε = d(x, y)/2. Then, x ∈ Bx,ε 63 y and x /∈ By,ε 3 y. In fact, it is not hard to seeusing the Triangle Inequality that Bx,ε ∩ By,ε = ∅, so (X,TX) is not only T1 but alsoHausdorff (T2).

Let now A,B ⊆ X be closed subsets with A ∩B = ∅. Since both A and B are closedand disjoint, we have A = A and B = B, and A ∩B = ∅. Thus,

(1) for each a ∈ A there is some ε(a) such that Ba,ε(a) ∩B = ∅; and

56 ALEX GONZALEZ

(2) for each b ∈ B there is some ε(b) such that Bb,ε(b) ∩ A = ∅.

Let U =⋃a∈ABa,ε(a)/2 and V =

⋃b∈B Bb,ε(b)/2. Then clearly A ⊆ U and B ⊆ V , and

we have to show that U ∩ V = ∅. Suppose there exists some x ∈ U ∩ B. Then thereexist a ∈ A and b ∈ B such that x ∈ Ba,ε(a)/2 ∩ Bb,ε(b)/2. Suppose for simplicity thatε(a) ≤ ε(b). Then,

dX(a, b) ≤ dX(a, x) + dX(b, x) < ε(a)/2 + ε(b)/2 ≤ ε(b).

Thus a ∈ Bb,ε(b), which contradicts our choice of ε(b). Thus U ∩ V = ∅.

Theorem 6.17. Let (X,TX) be a Hausdorff space, and let xnn∈N be a sequence. Ifxnn∈N is convergent then its limit is unique.

Proof. Suppose that xnn∈N converges to x, and let y 6= x be a different element ofX. By definition there exist U, V ∈ TX such that x ∈ U , y ∈ V and U ∩ V = ∅. Sincethe sequence converges to x, the open subset U contains all but finitely many termsof the sequence, which implies that V contains at most finitely many terms of thesequence. Thus xnn∈N cannot converge to y.

Next we present several examples of topological spaces exhibiting some Ti spacesthat are not Ti+1. Try to find other examples on your own!

Example 6.18. Let X be a set with at least two elements, and let T be the indiscretetopology on X. Then (X,T) is not a T0 space.

Example 6.19. A space (X,T) which is T0 but not T1. This example is called theSierpińsky space. Let X = x, y, with T = ∅, x, X. Then, U = x ∈ T satisfiesx ∈ U 63 y, and thus (X,T) is T0. However, there does not exist U, V ∈ T such thatx ∈ U 63 y and x /∈ V 3 y, so (X,T) is not T1.

Example 6.20. This example is very important in algebraic geometry, and it is anexample of a space that is T0 but not T1. Let A be a commutative ring, and let X bethe set of prime ideals of A (recall that an ideal P A is prime if the following holds:whenever ab ∈ P for a, b ∈ A, either a ∈ P or b ∈ P ).

For each a ∈ A, let Xa be the set of prime ideals P A such that a /∈ P . Note thatXa ∩ Xb = Xab for all a, b ∈ A. Also, X0 = ∅ (because 0 ∈ P for all P ∈ X), andX1 = X. Thus, B = Xa | a ∈ A is a basis for a topology on X! This topology iscalled the Zariski topology. The topological space (X,T) is called the prime spectrumof the ring A, and it is sometimes denoted by Spec(A). As an exercise, find a proofthat the space (X,T) is T0 but not T1.

Example 6.21. A space (X,T) which is T1 but not T2. Let X = Z, and let

T = ∅, X ∪ U ⊆ X | the complement U c is a finite subset of X.Note that if U, V ∈ T are nonempty subsets, then U ∩V 6= ∅. This implies that (X,T)cannot be Hausdorff. Let’s check that it is T1. Let x, y ∈ X, with x 6= y, and let


U = X \ y = yc and V = X \ x = xc. Then it is clear that x ∈ U 63 y andx /∈ V 3 y, so (X,T) is T1.

Example 6.22. A space (X,T) which is Hausdorff but not regular. Let X = R, andlet K = 1/n |n ∈ N. Let also

B = (a, b) | a, b ∈ R ∪ (a, b) \K | a, b ∈ R.This forms a basis for a topology on X, namely T. The space (X,T) is easily seento be Hausdorff. However, we show next that it is not regular: the set K is a closedsubset of X, but it cannot be separated from 0.

To see that K is closed, let A = (−∞, 0)∪ (1,∞) and let B = (−1, 1)\K. Both A andB are open subsets of X, and thus A∪B is also open. Note that A∪B = Kc = R\K,and thus K is closed.

Suppose now that (X,T) is regular, and let U, V ∈ T be disjoint subsets such that0 ∈ U and K ⊆ V . Let also W ∈ B be such that 0 ∈ W ⊆ U . Then W must be of theform W = (a, b) \K (if W = (a, b) then W ∩K 6= ∅). Choose n ∈ N large enough sothat 1/n ∈ (a, b), and let Z ∈ B be such that 1/n ∈ Z ⊆ V . Notice that Z = (c, d) forsome c, d ∈ R. Finally, choose z ∈ R so that

z < 1/n and z > maxc, 1/(n+ 1).Then z ∈ U ∩ V , contradicting the hypothesis that they are disjoint.

In order to give an example of an space that is regular but not normal, we need toprove some results first.

Lemma 6.23. Let (X,TX) be a topological space whose singletons are closed subsets.Then the following holds.

(i) X is regular if and only if given a point x ∈ X and U ∈ TX containing x, there isa some V ∈ TX such that x ∈ V ⊆ V ⊆ X.

(ii) X is normal if and only if given a closed subset A ⊆ X and U ∈ TX containing A,there is a some V ∈ TX such that A ⊆ V ⊆ V ⊆ X.

Proof. We only prove part (i), part (ii) being an easy exercise. Suppose X is regular,and let x ∈ X, U ∈ TX be given. Let B = U c, which is a closed subset not containingx. Since X is regular, there exist disjoint subsets V,W ∈ TX such that x ∈ V andB ⊆ W . We claim that V ∩ B = ∅: if y ∈ B, then W ∈ TX is such that y ∈ W andW ∩ V = ∅. Thus, V ⊆ U as desired.

To prove the converse, let x ∈ X and let B ⊆ X be a closed subset not containing x.We have to show that there are disjoint open subsets separating x and B. Let U =Bc ∈ TX . Then x ∈ U , and there exists some V ∈ TX be such that x ∈ V ⊆ V ⊆ U .Let W = V

c. Then x ∈ V , B ⊆ W and V ∩W = ∅. Thus X is regular.

Theorem 6.24. The following holds.

58 ALEX GONZALEZ

(i) A subset of a Hausdorff space is Hausdorff.

(ii) A product of Hausdorff spaces is Hausdorff.

(iii) A subset of a regular space is regular.

(iv) A product of regular spaces is regular.

Proof. Point (i). Let (X,TX) be a Hausdorff topological space, let A ⊆ X, and letx, y ∈ A. Since X is Hausdorff, there exist U, V ∈ TX such that x ∈ U , y ∈ V andU ∩B = ∅. Then U ∩A, V ∩A ∈ TA, x ∈ U ∩A, y ∈ V ∩A and (U ∩A)∩ (V ∩A) = ∅.

Point (ii). Let Xγγ∈Γ be a family of Hausdorff spaces, letX = Πγ∈ΓXγ be the productspace, and let x = (xγ), y = (yγ) be two distinct elements of X. Since x 6= y, thereexists some γ ∈ Γ such that xγ 6= yγ. Choose W,Z ∈ TXγ to be disjoint subsetsseparating xγ and yγ, and let U = π−1

γ (W ) and V = π−1γ (Z), where πγ : X → Xγ is

the projection map. Then U and V are disjoint open subsets of X that separate xand y.

Point (iii). Let (X,TX) be a regular space, and let A ⊆ X. By (i), A is Hausdorff, andthus singletons are closed subsets. Let also x ∈ A be a point and B ⊆ A be a closedsubset (B is closed in A), x /∈ B. Let C ⊆ X be the closure of B in X, so B = C ∩A.Then x /∈ C, and the regularity of X implies that there exist U, V ∈ TX such thatx ∈ U , C ⊆ V and U ∩ V = ∅. Thus U ∩ A and V ∩ A are disjoint open subsets of Aseparating x and B.

Point (iv). Let Xγγ∈Γ be a family of regular spaces, and let X = Πγ∈ΓXγ be theproduct space. By (ii), X is Hausdorff, and thus singletons are closed subsets. Letx = (xγ) ∈ X be a point, and let U ⊆ X be an open subset containing x. Choose abasis open subset W = ΠγUγ such that x ∈ W ⊆ U . Since each Xγ is regular, wemay choose for each γ an open subset Vγ such that x ∈ Vγ ⊆ Vγ ⊆ Uγ. Furthermore,whenever Uγ = Xγ, we choose Vγ = Xγ. This way, V = ΠγVγ satisfies

x ∈ V ⊆ V = ΠγVγ ⊆ ΠγUγ ⊆ U.

By Lemma 6.23 (i) it follows that X is regular.

Example 6.25. A space (X,T) which is regular but not normal. This example isknown as the Sorgenfrey plane. First, consider the set of real numbers R, and let

B = ∅ ∪ [a, b) | a, b ∈ R.This forms a basis for a topology on R, namely TS, and (R,TS) is called the Sorgenfreyline. We start by showing that (R,TS) is regular. Indeed, let x ∈ R be a point, andlet B ⊆ R be a closed subset. Then there exists some U = [a, b) ∈ B such that x ∈ Uand U ∩ B = ∅. Let also V = (∞, a) ∪ [b,∞). Then V can be described as a unionof elements in the basis B, and thus V ∈ TS. Clearly, B ⊆ V , and U ∩ V = ∅, andthus (R,TS) is a regular space. As an exercise, show that in fact (R,TS) is a normalspace.


Let now (X,T) be the product space of two copies of the Sorgenfrey line, to be calledfrom now on the Sorgenfrey plane. By the above discussion, together with Theorem6.24 (iv), the Sorgenfrey plane is a regular space, and we show now that it is notnormal.

Suppose otherwise that (X,T) is normal, and let L = (x,−x) |x ∈ R. Then thefollowing properties hold (the proof of each property is left as an exercise):

(1) L is closed in X; and

(2) L has the discrete topology as a subspace of (X,T).

Let A ⊆ be a subset. Since L has the discrete topology, it follows that A is closed,and since L is closed in X it follows that A is also closed as a subset of X. Thus,for each A ⊆ X, we have that both A,L \ A are closed in X, and the normalitycondition applies: there exist disjoint open subsets UA and VA such that A ⊆ UA andL \ A ⊆ VA. For each A ⊆ L, fix UA and VA as above.

Let also D ⊆ X be the subset of points in X = R × R with rational coordinates,which is a dense subset of X, and define a map θ : subsets of L → subsets of Das follows:

θ(A) =

∅, A = ∅D, A = LD ∩ UA, otherwise.

Here, UA is the open subset of X fixed in the previous paragraph. We claim that themap θ is injective. Let A ⊆ L be a proper subset. Since D is dense in X, it followsthat D ∩ UA, D ∩ VA 6= ∅, and thus

θ(A) = D ∩ UA 6= ∅ and θ(A) = D ∩ UA 6= D.

Now, if A,B ⊆ L are proper, distinct subsets, then one of them contains an elementwhich is not contained in the other. For simplicity suppose that (x,−x) ∈ A and(x,−x) /∈ B. Then, (x,−x) ∈ L \B, and thus

(x,−x) ∈ UA ∩ VB.Note that UA ∩ VB is open and nonempty, and thus K = D ∩ (UA ∩ VB) 6= ∅, since Dis dense. By definition, the elements of K are elements of D∩UA, but not of D∩UB,and thus θ(A) 6= θ(B).

Next we construct an injective map φ : subsets of D → L. Since D is countableand L has the cardinality of R, it is enough to construct a map from subsets of Ninto R. For each S ⊆ N, define

φ(S) =∞∑k=1

ak10k

,

where ak = 0 if k ∈ S and ak = 1 if k /∈ S. This is clearly injective. Combining themap φ with the map θ, we obtain an injective map subsets of L → L, but no suchmap exists. Hence a contradiction.

60 ALEX GONZALEZ

Theorem 6.26. Let (X,TX) be a second-countable, regular space. Then X is normal.

Proof. Let B be a countable basis for TX , and let A,B ⊆ X be disjoint closed subsets.Then for each point a ∈ A, the set X \B is open and contains x. Since X is regular,we may choose Va ∈ B such that a ∈ Va ⊆ Va ⊆ X \ B. Since B is regular, we mayreindex and rename the collection Uaa∈A (by removing repetitions) as Uii∈N. Notethat A ⊆

⋃i∈N Ui, and B ∩ Ui = ∅ for all i. In a similar way, reversing the roles of A

and B above, we produce a set Vii∈N of basis elements such that B ⊆⋃i∈N Vi, and

A ∩ Vi = ∅ for all i.

For each i ∈ N, define the sets Yi and Zi by

Yi = Ui \ (i⋃

k=1

Vk) Zi = Vi \ (i⋃

k=1

Ui).

By definition, Yi and Zi are open subsets. Let U =⋃i∈N Yi and V =

⋃i∈N Zi, which

again are open sets. We claim that A ⊆ U , B ⊆ V , and U ∩ V = ∅.

To show that U ∩ V = ∅, suppose otherwise that there exists some x ∈ U ∩ V . Thenthere are i, j ∈ N such that x ∈ Yi ∩Zj. We may as well suppose that i ≥ j, in whichcase x ∈ Yi = Ui \ (∪ik=1Vi) ⊆ Ui \ Vj, and x ∈ Zj ⊆ Vj, which is a contradiction.Clearly, A ⊆ U , since A ⊆

⋃i∈N Ui, and A ∩ Vj = ∅. Similarly, B ⊆ V . This finishes

the proof.

Theorem 6.27. Let (X,TX) be a compact Hausdorff space. Then X is normal.

Proof. Let A,B ⊆ X be disjoint closed subsets. Fix a ∈ A. For each b ∈ B thereexist disjoint subsets Ub, Vb ∈ TX containing a and b respectively. Note that Vbb∈Bis an open cover of B, and B is compact because it is a closed subset of a compactspace. Thus there exist b1, . . . , bn ∈ B such that Ub1 , . . . , Ubn is a finite subcover.Let U =

⋂ni=1 Ubi and V =

⋃ni=1 Vbi. Then a ∈ U , B ⊆ V and U ∩V = ∅. In particular,

this shows that (X,TX) is regular.

Since (X,TX) is regular, for each a ∈ A, there exist disjoint subsets Wa, Za ∈ TXsuch that a ∈ Wa and B ⊆ Za. The collection Waa∈A is an open cover of A, andA is compact because it is a closed subset of a compact space, and thus there exista1, . . . , am ∈ A such that Wa1 , . . . ,Wam is a finite subcover. Set

W =m⋃j=1

Waj and Z =m⋂j=1

Zaj .

Then A ⊆ W , B ⊆ Z, and W ∩ Z = ∅. Thus X is normal.


7. Urysohn metrization theorem

We now combine results from previous chapters to give conditions for a given topologyto be associated to a metric. First we need the following result.

Theorem 7.1 (Urysohn Lemma). Let (X,T) be a normal topological space, and letA,B ⊆ X be disjoint, closed subsets. Let also [a, b] be a closed interval in the realEuclidean line. Then, there exists a continuous map f : X → [a, b] such that f(x) = afor all x ∈ A and f(x) = b for all x ∈ B.

Proof. For simplicity we may assume that [a, b] = [0, 1]. The proof is divided intosmaller steps.

Step 1. Let Q ⊆ [0, 1]. The goal of this step is to construct a family Uqq∈Q of opensubsets in X such that Uq ⊆ Uq′ whenever q < q′.

Start with U1 = Bc = X \ B. Note that A ∩ B = ∅, which implies that A ⊆ U1.Now, since (X,T) is normal and A ⊆ X is closed, there exists U0 ∈ T such thatA ⊆ U0 ⊆ U0 ⊆ U1. Fix U0 and U1 as above.

To define Uq in general, arrange the elements of Q in a list, with 1 and 0 as its twofirst terms. For each n ∈ N, let Qn ⊆ Q be the list of the first n elements in Q. Then,the set Qn+1 \ Qn contains only one element. Now, suppose that for some n ∈ N wehave already defined Uq for all q ∈ Qn, subject to the condition

q < q′ ⇐⇒ Uq ⊆ Uq′ .

Let r ∈ Qn+1 \ Qn. Since Qn+1 contains finitely many elements of [0, 1], and sincer 6= 0, 1, it follows that there exists some r1, r2 ∈ Qn such that r1 < r < r2. Inparticular, we have already defined Ur1 and Ur2, and they satisfy that Ur1 ⊆ Ur2.Thus, since X is normal, there exists Ur ∈ T such that

Ur1 ⊆ Ur ⊆ Ur ⊆ Ur2 .

We have to show that Uq ⊆ Uq′ for all q < q′ in Qn+1:

• if q, q′ ∈ Qn, then Uq ⊆ Uq′ by induction;

• if q ∈ Qn and q′ = r, then q ≤ r1 as above, and thus we have

Uq ⊆ Ur1 ⊆ Ur1 ⊆ Ur;

• if q = r and q′ ∈ Qn, then r2 ≤ q′, and it follows that

Ur ⊆ Ur2 ⊆ Ur2 ⊆ Uq′ .

Inductively, we construct the family Uqq∈Q such that Uq ⊆ Uq′ whenever q < q′.

Step 2. We now define the function f : X → R. We will see in the next step that thisis the function we are looking for.

62 ALEX GONZALEZ

First, we extend the family Uqq∈Q to a family Uqq∈Q, and such that Uq ⊆ Uq′ forall q < q′ in Q. For each q ∈ Q, define

Uq =

∅, q < 0Uq, q ∈ QX, q > 1

Clearly, if q < q′, then Uq ⊆ Uq′.

Given x ∈ X, set also Q(x) = q ∈ Q |x ∈ Uq. Note that, for all x ∈ X, the followingholds

q < 0⇐⇒ q /∈ Q(x) and q > 1⇐⇒ q ∈ Q(x).

For each x ∈ X, define f(x) = infQ(x) = infq ∈ Q |x ∈ Uq. Note that f(x) ∈ [0, 1]by the properties above.

Step 3. The function f : X → [0, 1] in Step 2 is continuous and satisfies f(A) = 0and f(B) = 1.

Let x ∈ X. If x ∈ A, then x ∈ Uq for each q ≥ 0 by definition (since A ⊆ U0), and thusf(x) = 0. On the other hand, if x ∈ B, then x /∈ Uq for any q ≤ 1 (since U1 = Bc),and thus f(x) = 1.

It remains to show that f is a continuous function. We fist prove the following.

(1) If x ∈ Uq, then f(x) ≤ q.

(2) If x /∈ Uq, then f(x) ≥ q.

If x ∈ Uq, then x ∈ Ur for all r > q. Therefore, Q(x) contains all rational numbersgreater than q, and f(x) = infQ(x) ≤ r. Similarly, if x /∈ Uq, then x is not inUr for all r < q, so Q(x) does not contain any element smaller than q, and f(x) =infQ(x) ≥ q.

Fix now x ∈ X, and let (c, d) ⊆ R be an open set that contains f(x). Choose q, q′ ∈ Qsuch that c < q < f(x) < q′ < d (all the inequalities are strict!). We claim thatWx = Uq′ \ Uq satisfies

x ∈ Wx ⊆ f−1(c, d).

To show that x ∈ U , note that f(x) < q′, so x ∈ Uq′. On the other hand, we havex /∈ Uq since f(x) > q. We have to check that Wx ⊆ f−1(c, d). Let y ∈ Wx. Bydefinition, y ∈ Uq′ ⊆ Uq′, and thus f(y) ≤ q′ by (1). Also, y /∈ Uq, so y /∈ Uq, and thusf(y) ≥ q′ by (2). Hence f(y) ∈ [q, q′] ⊆ (c, d), and the claim follows.

Set W = f−1(c, d). Then, by the above discussion, we may write W =⋃x∈W Wx. It

follows that W is open. Since the intervals (c, d) form a basis for the topology of R,it follows also that f is continuous. Finally, since f(x) ∈ [0, 1] for all x ∈ X, we mayrestrict f to f : X → [0, 1], which is a continuous map by the universal property ofthe subspace topology.


Example 7.2. Let (R, dEucl) be the usual Euclidean metric space, where dEucl(x, y) =|x− y|, and let

d : R× R −→ [0, 1]

be defined by d(x, y) = min|x − y|, 1. We have already seen in the homework thatd is a metric on R. Let now X =

∏n∈N R. Denote the elements of X as (xn)n∈N as

usual, and define D : X ×X −→ [0, 1] by the formula

D((xn)n∈N, (yn)n∈N) = supd(xn, yn)

n|n ∈ N.

Then, as we saw in the homework, D is a metric on X, and the product topology(from the Euclidean topology on each copy of R in X) corresponds to the collectionof open subsets of X with respect to D.

Let (X,T) be a topological space. We say that (X,T) is metrizable if T is the collectionof open subsets with respect to some metric d on X.

Theorem 7.3 (Urysohn Metrization Theorem). Every normal, second-countable space(X,T) is metrizable.

Proof. The idea is to prove that (X,T) is homeomorphic to a subspace of a metrizablespace.

Step 1. There exists a (countable) set of functions fn : X → [0, 1]n∈N satisfying thefollowing condition: for each x0 ∈ X and each U ∈ T such that x0 ∈ U , there is n ∈ Nsuch that fn(x0) = 1 and fn(U c) = 0.

To prove the above, we will use Urysohn’s Lemma 7.1. Let B = Bn be a countablebasis for T. Set M = (n,m) ∈ N2 |Bn ⊆ Bm. For each pair (n,m) ∈ M , we mayapply Urysohn’s Lemma 7.1: there exists a continuous function

gn,m : X → [0, 1]

such that gn,m(Bn) = 1 and gn,m((Bm)c) = 0. We claim that the collection gn,m(n,m)∈Msatisfies the conditions required above. Clearly, M is a countable set. Moreover,given U ∈ T with x0 ∈ U , there exists some Bm ∈ B such that x0 ∈ Bm ⊆ U (becauseB is a basis for T). Since X is a normal space, by Lemma 6.23 we know that thereexists some V ∈ T such that x0 ∈ V ⊆ V ⊆ U . Again, using the fact that B is a basisfor T, there is some Bn ∈ B such that x0 ∈ Bn ⊆ V , and we deduce that

x0 ∈ Bn ⊆ Bn ⊆ Bm.

Thus, the pair (n,m) is an element of M , and the function gn,m satisfies gn,m(x0) = 1and gn,m((Bn)c) = 0. For simplicity, rename the set of functions to fnn∈N for therest of the proof.

Step 2. Set Y =∏

n∈N R, and let Tprod be the product topology. Recall from Example7.2 that Tprod is the topology associated to a certain metric on Y . Define a function

64 ALEX GONZALEZ

F : X → Y by the following formula

F (x) = (fn(x))n∈N,

where fnn∈N is a collection of functions satisfying the properties in Step 1. Weclaim that the restriction of F to its image, F : X → F (X), is a homeomorphism.

• F is continuous.

We only have to apply the universal property of the product topology: for each n ∈ N,the composition of F with the projection onto the n-th factor, πn : Y → R, is acontinuous function, and it follows that F is continuous.

• As a map between sets, F is injective.

Let x, y ∈ X, x 6= y. Since X is normal, there exists Bm ∈ B such that x ∈ Bm andy /∈ Bm. By Step 1, there exists some n ∈ N such that fn(x) = 1 and fn(y) = 0 (sincey ∈ (Bm)c). Thus F (x) 6= F (y).

• F is a homeomorphism (when restricted to its image).

Let Z = F (X) ⊆ Y , and let TZ be the subspace topology on Z. To prove thatF : X → Z is a homeomorphism, it is enough to check that F (U) ∈ TZ for all U ∈ T.Let U ∈ T, and let z ∈ F (U). We have to show that there exists Wz ∈ TZ such that

z ∈ Wz ⊆ F (U).

This way, we can write F (U) =⋃z∈F (U) Wz, which implies that F (U) is open in Z.

Let x ∈ X be the (unique) element such that F (x) = z. Choose N ∈ N such that

fN(x) = 1 and fN(U c) = 0,

and consider the projection map πN : Y → R that sends every point in Y to its N-thcoordinate. Then, π−1

N (0,∞) ∈ Tprod, which implies that Wzdef= Z ∩ π−1

N (0,∞) is openin Z. Note also that z ∈ Wz, since

πN(z) = πN(F (x)) = fN(x) = 1 ∈ (0,∞).

To show that Wz ⊆ F (U), let z′ ∈ Wz. Then, z′ = F (x′) for some x′ ∈ X such that

πN(z′) = πN(F (x′)) = fN(x′) ∈ (0,∞).

Since fN(U c) = 0, this implies that x′ ∈ U , and thus F (x′) ∈ F (U).

The above shows that F : X → Z = F (X) ⊆ Y is a homeomorphism. Since (Y,Tprod)is metrizable by Example 7.2, it follows that (Z,TZ) is also metrizable (just restrictthe metric on Y to a metric on Z), and hence so is (X,T).


8. Topological manifolds

We now introduce one of the most important classes of topological spaces: (topo-logical) manifolds. Roughly speaking, a (topological) manifold is a space that locally(i.e. in a small open set around each point) looks like an Euclidean space. Manifoldsappear in many areas of mathematics. For example, surfaces like a sphere or a torusare manifolds. They appear as solution sets of systems of equations, as graphs offunctions, and so on.

For the moment, we will not talk about manifolds with boundary. This is an inter-esting generalization of manifolds (for example they arise as solution sets of systemsof equations involving inequalities). This, however, adds to the level of complication,so let’s leave this aside for now.

Definition 8.1. Let (X,T) be a topological space, and let x ∈ X. A neighborhood ofx is any open subset U ∈ T such that x ∈ U .

Definition 8.2. Letm ∈ N. A topologicalm-manifold is a Hausdorff, second countablespace (M,T) satisfying the following property.

(M) For each x ∈ M there exists some neighborhood U ∈ T which is homeomor-phic to Rm.

We also say that (M,T) is a manifold of dimension m.

A (very difficult) problem in topology (in fact still an open question) is how to distin-guish any two (compact) manifolds of the same dimension. For dimensions 1 and 2the problem is completely solved, but things get wilder when we move to dimensionof higher dimension.

Remark 8.3. The following condition is equivalent to (M).

(M’) For each x ∈ M there exists some neighborhood U ∈ T which is homeomor-phic to some open subset of Rm.

Clearly, if condition (M) is satisfied then so is (M’). Here is a sketch of how to prove theequivalence (details are left to the reader). Let x ∈ M , and let U ∈ T be a neighbor-hood satisfying condition (M’). Let V ⊆ Rm be an open subset such that U ∼= V . Wemay first restrict to the connected component of U that contains x, and restrict thehomeomorphism U ∼= V to a homeomorphisms between the corresponding connectedcomponents, namely U0 and V0. Then V0 must be a product of either open intervalsor copies of R (why? how does a connected open subset of R look like?). Since everyopen interval in R is homeomorphic to R, it follows that V0 is homeomorphic to Rm,and then so is U0. Thus condition (M) follows.

Example 8.4. The circle S1 is a topological 1-manifold, and more generally the sphereSn is a topological n-manifold. The cylinder C = S1 × [0, 1] is not a manifold (whathappens to point of the form (z, 0) and (z, 1), for z ∈ S1?). However, the infinite

66 ALEX GONZALEZ

cylinder C = S1 × R is a topological 2-manifold. The m-fold product S1 × . . . × S1

is usually called the m-torus, and is and m-manifold. In fact, we show below that aproduct of two manifolds is again a manifold.

Theorem 8.5. Let (M,TM) be anm-manifold, and let (N,TN) be an n-manifold. Then,the product space (M ×N,Tprod) is an (m+ n)-manifold.

Proof. Let (x, y) ∈ M × N . By definition, there exist neighborhoods U ∈ TM andV ∈ TN , of x and y respectively, such that U ∼= Rm and V ∼= Rn. Then, U ×V ∈ Tprodis a neighborhood of (x, y), and U×V ∼= Rm×Rn = Rm+n. This finishes the proof.

Theorem 8.6. Let (M,T) be a topological m-manifold, for some m ∈ N. Then (M,T)is metrizable.

Proof. The proof is left as an exercise (homework).

8.1. Compact manifolds. For practical reasons, one usually works with compactmanifolds, instead of just manifolds in general. We have already seen that everymanifold is metrizable. However, the metric that one obtains in general is not veryuseful (see for example the metric in 7.2). The goal of this section is to refine thismetrization result for compact manifolds: we will show that every compact manifoldis homeomorphic to a subspace of a (finite dimensional) Euclidean space. First weneed to introduce some tools and results.

Definition 8.7. Let φ : (X,T)→ (R,TEucl) be a continuous map. The support of φ isthe closure of φ−1(R \ 0):

Sφdef= (φ−1(R \ 0)) ⊆ X.

Remark 8.8. Suppose x /∈ Sφ, and let U = (Sφ)c, which is open in X since Sφ isclosed by definition. Then φ(y) = 0 for all y ∈ U , and U contains x. Thus we deducethe following: if x /∈ Sφ, then there exists a neighborhood U ∈ T such that φ(U) = 0.

Definition 8.9. Let Ω = U1, . . . , Un be a finite open cover of (X,T). A partition ofunity dominated by Ω is a family of functions φi : X → [0, 1]i=1,...,n, satisfying thefollowing properties:

(i) Sφi ⊆ Ui for each i = 1, . . . , n; and

(ii)∑n

i=1 φi(x) = 1 for all x ∈ X.

Theorem 8.10. Let (X,T) be a normal space, and let Ω = U1, . . . , Un be a finiteopen cover of X. Then, there exists a partition of unity dominated by Ω.


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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