IB Metric and Topological Spaces (Full)

Part IB — Metric and Topological Spaces

Based on lectures by J. Rasmussen

Notes taken by Dexter Chua

Easter 2015

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Metrics

Definition and examples. Limits and continuity. Open sets and neighbourhoods.

Characterizing limits and continuity using neighbourhoods and open sets. [3]

Topology

Definition of a topology. Metric topologies. Further examples. Neighbourhoods, closed

sets, convergence and continuity. Hausdorff spaces. Homeomorphisms. Topologi-

cal and non-topological properties. Completeness. Subspace, quotient and product

top ologies. [3]

Connectedness

Definition using open sets and integer-valued functions. Examples, including inter-

vals. Components. The continuous image of a connected space is connected. Path-

connectedness. Path-connected spaces are connected but not conversely. Connected

op en sets in Euclidean space are path-connected. [3]

Compactness

Definition using open covers. Examples: finite sets and [0, 1]. Closed subsets of

compact spaces are compact. Compact subsets of a Hausdorff space must b e closed.

The compact subsets of the real line. Continuous images of compact sets are compact.

Quotient spaces. Continuous real-valued functions on a compact space are bounded

and attain their bounds. The product of two compact spaces is compact. The compact

subsets of Euclidean space. Sequential compactness. [3]

Contents

0 Introduction

1 Metric spaces

1.1 Definitions

1.2 Examples of metric spaces

1.3 Norms

1.4 Open and closed subsets

2 Top ological spaces

2.1 Definitions

2.2 Sequences

2.3 Closed sets

2.4 Closure and interior

2.4.1 Closure

2.4.2 Interior

2.5 New topologies from old

2.5.1 Subspace topology

2.5.2 Product topology

2.5.3 Quotient topology

3 Connectivity

3.1 Connectivity

3.2 Path connectivity

3.2.1 Higher connectivity*

3.3 Components

3.3.1 Path components

3.3.2 Connected components

4 Compactness

4.1 Compactness

4.2 Products and quotients

4.2.1 Products

4.2.2 Quotients

4.3 Sequential compactness

4.4 Completeness

0 Introduction

The course Metric and Topological Spaces is divided into two parts. The first is

on metric spaces, and the second is on topological spaces (duh).

In Analysis, we studied real numbers a lot. We defined many properties such

as convergence of sequences and continuity of functions. For example, if (

) is

a sequence in R, x

→ x means

(∀ε > 0)(∃N)(∀n > N) |x

− x| < ε.

Similarly, a function f is continuous at x

(∀ε > 0)(∃δ > 0)(∀x) |x − x

| < δ ⇒ |f(x) − f(x

)| < ε.

However, the definition of convergence doesn’t really rely on

being real

numbers, except when calculating values of

− x|

. But what does

− x|

really mean? It is the distance between

and

. To define convergence, we

don’t really need notions like subtraction and absolute values. We simply need a

(sensible) notion of distance between points.

Given a set

, we can define a metric (“distance function”)

X × X → R

where

(

x, y

) is the distance between the points

and

. Then we say a sequence

) in X converges to x if

(∀ε > 0)(∃N)(∀n > N) d(x

, x) < ε.

Similarly, a function f : X → X is continuous if

(∀ε > 0)(∃δ > 0)(∀x) d(x, x

) < δ ⇒ d(f(x), f(x

)) < ε.

Of course, we will need the metric

to satisfy certain conditions such as being

non-negative, and we will explore these technical details in the first part of the

course.

As people studied metric spaces, it soon became evident that metrics are not

really that useful a notion. Given a set

, it is possible to find many different

metrics that completely agree on which sequences converge and what functions

are continuous.

It turns out that it is not the metric that determines (most of) the properties

of the space. Instead, it is the open sets induced by the metric (intuitively, an

open set is a subset of

without “boundary”, like an open interval). Metrics

that induce the same open sets will have almost identical properties (apart from

the actual distance itself).

The idea of a topological space is to just keep the notion of open sets and

abandon metric spaces, and this turns out to be a really good idea. The second

part of the course is the study of these topological spaces and defining a lot of

interesting properties just in terms of open sets.

1 Metric spaces

1.1 Definitions

As mentioned in the introduction, given a set

, it is often helpful to have a

notion of distance between points. This distance function is known as the metric.

Definition (Metric space). A metric space is a pair (

X, d

) where

is a set

(the space) and

is a function

X × X → R

(the metric) such that for all

x, y, z,

– d

(x, y) ≥ 0 (non-negativity)

– d

(x, y) = 0 iff x = y (identity of indiscernibles)

– d

(x, y) = d

(y, x) (symmetry)

– d

(x, z) ≤ d

(x, y) + d

(y, z) (triangle inequality)

We will have two quick examples of metrics before going into other important

definitions. We will come to more examples afterwards.

Example.

– (Euclidean “usual” metric) Let X = R

. Let

d(v, w) = |v − w| =

i=1

− w

)

This is the usual notion of distance we have in the

vector space. It is

not difficult to show that this is indeed a metric (the fourth axiom follows

from the Cauchy-Schwarz inequality).

– (Discrete metric) Let X be a set, and

(x, y) =

(

1 x 6= y

0 x = y

To show this is indeed a metric, we have to show it satisfies all the axioms.

The first three axioms are trivially satisfied. How about the fourth? We

can prove this by exhaustion.

Since the distance function can only return 0 or 1,

(

x, z

) can be 0 or 1,

while

(

x, y

) +

(

y, z

) can be 0, 1 or 2. For the fourth axiom to fail, we

must have RHS

LHS. This can only happen if the right hand side is 0.

But for the right hand side to be 0, we must have

. So the left

hand side is also 0. So the fourth axiom is always satisfied.

Given a metric space (

X, d

), we can generate another metric space by picking

out a subset of X and reusing the same metric.

Definition (Metric subspace). Let (

X, d

) be a metric space, and

Y ⊆ X

Then (

Y, d

) is a metric space, where

(

a, b

) =

(

a, b

), and said to be a

subspace of X.

Example.

{

∈ R

n+1

= 1

}

, the

-dimensional sphere, is a subspace

of R

n+1

Finally, as promised, we come to the definition of convergent sequences and

continuous functions.

Definition (Convergent sequences). Let (

) be a sequence in a metric space

(

X, d

). We say (

) converges to

x ∈ X

, written

→ x

, if

(

, x

)

→

0 (as a

real sequence). Equivalently,

(∀ε > 0)(∃N )(∀n > N ) d(x

, x) < ε.

Example.

–

Let (v

) be a sequence in

with the Euclidean metric. Write v

(

, ··· , v

), and v = (

, ··· , v

)

∈ R

. Then v

→

v iff (

)

→ v

for

all i.

–

Let

have the discrete metric, and suppose

→ x

. Pick

Then there is some

such that

(

, x

)

whenever

n > N

. But if

(

, x

)

, we must have

(

, x

) = 0. So

. Hence if

→ x

then eventually all x

are equal to x.

Similar to what we did in Analysis, we can show that limits are unique (if

exist).

Proposition. If (

X, d

) is a metric space, (

) is a sequence in

such that

→ x, x

→ x

, then x = x

Proof.

For any

ε >

0, we know that there exists

such that

(

, x

)

< ε/

2 if

n > N . Similarly, there exists some N

such that d(x

, x

) < ε/2 if n > N

Hence if n > max(N, N

), then

0 ≤ d(x, x

)

≤ d(x, x

) + d(x

, x

)

= d(x

, x) + d(x

, x

)

≤ ε.

So 0 ≤ d(x, x

) ≤ ε for all ε > 0. So d(x, x

) = 0, and x = x

Note that to prove the above proposition, we used all of the four axioms. In

the first line, we used non-negativity to say 0

≤ d

(

x, x

). In the second line, we

used triangle inequality. In the third line, we used symmetry to swap

(

x, x

)

with

(

, x

). Finally, we used the identity of indiscernibles to conclude that

x = x

To define a continuous function, here, we opt to use the sequence definition.

We will later show that this is indeed equivalent to the

ε − δ

definition, as well

as a few other more useful definitions.

Definition (Continuous function). Let (

X, d

) and (

Y, d

) be metric spaces,

and

X → Y

. We say

is continuous if

(

)

→ f

(

) (in

) whenever

→ x (in X).

Example. Let

with the Euclidean metric. Let

with the discrete

metric. Then

X → Y

that maps

(

) =

is not continuous. This is since

1/n → 0 in the Euclidean metric, but not in the discrete metric.

On the other hand,

Y → X

(

) =

is continuous, since a sequence

in Y that converges is eventually constant.

1.2 Examples of metric spaces

In this section, we will give four different examples of metrics, where the first

two are metrics on

. There is an obvious generalization to

, but we will

look at R

specifically for the sake of simplicity.

Example (Manhattan metric). Let X = R

, and define the metric as

d(x, y) = d((x

, x

), (y

, y

)) = |x

− y

| + |x

− y

The first three axioms are again trivial. To prove the triangle inequality, we have

d(x, y) + d(y, z) = |x

− y

| + |x

− y

| + |y

− z

| + |y

− z

≥ |x

− z

| + |x

− z

= d(x, z),

using the triangle inequality for R.

This metric represents the distance you have to walk from one point to

another if you are only allowed to move horizontally and vertically (and not

diagonally).

Example (British railway metric). Let X = R

. We define

d(x, y) =

(

|x − y| if x = ky

|x| + |y| otherwise

To explain the name of this metric, think of Britain with London as the origin.

Since the railway system is

stupid

less than ideal, all trains go through London.

For example, if you want to go from Oxford to Cambridge (and obviously not

the other way round), you first go from Oxford to London, then London to

Cambridge. So the distance traveled is the distance from London to Oxford plus

the distance from London to Cambridge.

The exception is when the two destinations lie along the same line, in which

case, you can directly take the train from one to the other without going through

London, and hence the “if x = ky” clause.

Example (

-adic metric). Let

p ∈ Z

be a prime number. We first define the

norm

|n|

to be

−k

, where

is the highest power of

that divides

. If

= 0,

we let |n|

= 0. For example, |20|

= |2

· 5|

= 2

−2

Now take

, and let

(

a, b

) =

|a−b|

. The first three axioms are trivial,

and the triangle inequality can be proved by making some number-theoretical

arguments about divisibility.

This metric has rather interesting properties. With respect to

, we have

, ··· →

0, while 1

, ···

does not converge. We can also use it

to prove certain number-theoretical results, but we will not go into details here.

Example (Uniform metric). Let

1] be the set of all continuous

functions on [0, 1]. Then define

d(f, g) = max

x∈[0,1]

|f(x) − g(x)|.

The maximum always exists since continuous functions on [0

1] are bounded

and attain their bounds.

Now let

→ R

be defined by

(

) =

(

). Then this is continuous

with respect to the uniform metric on C[0, 1] and the usual metric on R:

Let

→ f

in the uniform metric. Then we have to show that

(

)

→ F

(

i.e. f

(

) → f (

). This is easy, since we have

0 ≤ |F (f

) − F (f)| = |f

(

) − f(

)| ≤ max |f

(x) − f(x)| → 0.

So |f

(

) − f(

)| → 0. So f

(

) → f (

1.3 Norms

There are several notions on vector spaces that are closely related to metrics.

We’ll look at norms and inner products of vector spaces, and show that they all

naturally induce a metric on the space.

First of all, we define the norm. This can be thought of as the “length” of a

vector in the vector space.

Definition (Norm). Let

be a real vector space. A norm on

is a function

k · k : V → R such that

– kvk ≥ 0 for all v ∈ V

– kvk = 0 if and only if v = 0.

– kλvk = |λ|kvk

– kv + wk ≤ kvk + kwk.

Example. Let

. There are several possible norms we can define on

kvk

i=1

kvk

i=1

kvk

∞

= max{|v

| : 1 ≤ i ≤ n}.

In general, we can define the norm

kvk

i=1

1/p

for any 1 ≤ p ≤ ∞, and kvk

∞

is the limit as p → ∞.

Proof that these are indeed norms is left as an exercise for the reader (in the

example sheets).

A norm naturally induces a metric on V :

Lemma. If k · k is a norm on V , then

d(v, w) = kv − wk

defines a metric on V .

Proof.

(i) d(v, w) = kv − wk ≥ 0 by the definition of the norm.

(ii) d(v, w) = 0 ⇔ kv − wk = 0 ⇔ v − w = 0 ⇔ v = w.

(iii) d(w, v) = kw − vk = k(−1)(v − w)k = | − 1|kv − wk = d(v, w).

(iv) d(u, v) + d(v, w) = ku − vk+ kv − w k ≥ ku − wk = d(u, w).

Example. We have the following norms on C[0, 1]:

kfk

|f(x)| dx

kfk

f(x)

kfk

∞

= max

x∈[0,1]

|f(x)|

The first two are known as the

and

norms. The last is called the uniform

norm, since it induces the uniform metric.

It is easy to show that these are indeed norms. The only slightly tricky part

is to show that kf k = 0 iff f = 0, which we obtain via the following lemma.

Lemma. Let

f ∈ C

1] satisfy

(

)

≥

0 for all

x ∈

1]. If

(

) is not

constantly 0, then

f(x) dx > 0.

Proof.

Pick

∈

1] with

(

) =

a >

0. Then since

is continuous, there is

such that

(

)

−f

(

)

| < a/

2 if

|x −x

| < δ

. So

(

)

| > a/

2 in this region.

Take

g(x) =

(

a/2 |x −x

| < δ

0 otherwise

Then f (x) ≥ g(x) for all x ∈ [0, 1]. So

f(x) dx ≥

g(x) dx =

· (2δ) > 0.

Example. Let X = C[0, 1], and let

(f, g) = kf − gk

|f(x) − g(x)| dx.

Define the sequence

(

1 − nx x ∈ [0,

]

0 x ≥

Then

kfk

· 1 =

→ 0

as n → ∞. So f

→ 0 in (X, d

) where 0(x) = 0.

On the other hand,

∞

= max

x∈[0,1]

kf(x)k = 1.

So f

6→ 0 in the uniform metric.

So the function (

, d

)

→

(

, d

∞

) that maps

f 7→ f

is not continu-

ous. This is similar to the case that the identity function from the usual metric

to the discrete metric of

is not continuous. However, the discrete metric

is a silly metric, but d

is a genuine useful metric here.

Using the same example, we can show that the function

: (

, d

)

→

(R, usual) with G(f ) = f(0) is not continuous.

We’ll now define the inner product of a real vector space. This is a general-

ization of the notion of the “dot product”.

Definition (Inner product). Let

be a real vector space. An inner product on

V is a function h·, ·i : V × V → R such that

(i) hv, vi ≥ 0 for all v ∈ V

(ii) hv, vi = 0 if and only if v = 0.

(iii) hv, wi = hw, vi.

(iv) hv

+ λv

, w) = hv

, wi + λhv

, wi.

Example.

(i) Let V = R

. Then

hv, wi =

i=1

is an inner product.

(ii) Let V = C[0, 1]. Then

hf, gi =

f(x)g(x) dx

is an inner product.

We just showed that norms induce metrics. The proof was completely trivial

as the definitions were almost the same. Now we want to prove that inner

products induce norms. However, this is slightly less trivial. To do so, we need

the Cauchy-Schwarz inequality.

Theorem (Cauchy-Schwarz inequality). If h·, ·i is an inner product, then

hv, wi

≤ hv, vihw, wi.

Proof. For any x, we have

hv + xw, v + xwi = hv, vi + 2xhv, wi+ x

hw, wi ≥ 0.

Seen as a quadratic in

, since it is always non-negative, it can have at most one

real root. So

(2hv, wi)

− 4hv, vihw, wi ≤ 0.

So the result follows.

With this, we can show that inner products induce norms (and hence metrics).

Lemma. If h·, ·i is an inner product on V , then

kvk =

hv, vi

is a norm.

Proof.

(i) kvk =

hv, vi ≥ 0.

(ii) kvk = 0 ⇔ hv, vi = 0 ⇔ v = 0.

(iii) kλvk =

hλv, λvi =

hv, vi = |λ|kvk.

(iv)

(kvk + kwk)

= kvk

+ 2kvkkwk+ kwk

≥ hv, vi + 2hv, wi + hw, wi

= kv + wk

1.4 Open and closed subsets

In this section, we will study open and closed subsets of metric spaces. We will

then proceed to prove certain key properties of open and closed subsets, and

show that they are all we need to define continuity. This will lead to the next

chapter (and the rest of the course), where we completely abandon the idea of

metrics and only talk about open and closed subsets.

To define open and closed subsets, we first need the notion of balls.

Definition (Open and closed balls). Let (

X, d

) be a metric space. For any

x ∈ X, r ∈ R,

(x) = {y ∈ X : d(y, x) < r}

is the open ball centered at x.

(x) = {y ∈ X : d(y, x) ≤ r}

is the closed ball centered at x.

Example.

(i) When X = R, B

(x) = (x − r, x + r).

(x) = [x − r, x + r].

(ii) When X = R

(a)

is the metric induced by the

, then an open ball

is a rotated square.

(b)

is the metric induced by the

+ v

, then an open ball

is an actual disk.

(c)

is the metric induced by the

∞

max{|v

|, |v

, then an

open ball is a square.

Definition (Open subset).

U ⊆ X

is an open subset if for every

x ∈ U

∃δ >

such that B

(x) ⊆ U .

C ⊆ X is a closed subset if X \ C ⊆ X is open.

As if we’ve not emphasized this enough, this is a very very important

definition.

We first prove that this is a sensible definition.

Lemma. The open ball

(

)

⊆ X

is an open subset, and the closed ball

(x) ⊆ X is a closed subset.

Proof. Given y ∈ B

(x), we must find δ > 0 with B

(y) ⊆ B

(x).

Since

y ∈ B

(

), we must have

(

y, x

)

< r

. Let

r − a >

0. Then if

z ∈ B

(y), then

d(z, x) ≤ d(z, y) + d(y, x) < (r − a) + a = r.

So z ∈ B

(x). So B

(y) ⊆ B

(x) as desired.

The second statement is equivalent to

X \

(

) =

{y ∈ X

(

y, x

)

> r}

open. The proof is very similar.

Note that openness is a property of a subset.

A ⊆ X

being open depends on

both

and

, not just

. For example, [0

) is not an open subset of

, but

is an open subset of [0

1] (since it is

(0)), both with the Euclidean metric.

However, we are often lazy and just say “open set” instead of “open subset”.

Example.

(i)

⊆ R

is open, while [0

⊆ R

is closed. [0

⊆ R

is neither closed

nor open.

(ii) Q ⊆ R

is neither open nor closed, since any open interval contains both

rational numbers and irrational numbers. So any open interval cannot be

a subset of Q or R \ Q.

(iii)

Let

= [

−

\ {

}

with the Euclidean metric. Let

= [

−

⊆ X

Then

is open since it is equal to

(

−

1).

is also closed since it is

equal to

(−

Definition (Open neighborhood). If

x ∈ X

, an open neighborhood of

is an

open U ⊆ X with x ∈ U.

This is not really an interesting definition, but is simply a convenient short-

hand for “open subset containing x”.

Lemma. If

is an open neighbourhood of

and

→ x

, then

∃N

such that

∈ U for all n > N.

Proof.

Since

is open, there exists some

δ >

0 such that

(

)

⊆ U

. Since

→ x

∃N

such that

(

, x

)

< δ

for all

n > N

. This implies that

∈ B

(

)

for all n > N. So x

∈ U for all n > N.

Definition (Limit point). Let

A ⊆ X

. Then

x ∈ X

is a limit point of

if there

is a sequence x

→ x such that x

∈ A for all n.

Intuitively, a limit point is a point we can get arbitrarily close to.

Example.

(i)

a ∈ A

, then

is a limit point of

, by taking the sequence

a, a, a, a, ···

(ii) If A = (0, 1) ⊆ R, then 0 is a limit point of A, e.g. take x

(iii) Every x ∈ R is a limit point of Q.

It is possible to characterize closed subsets by limit points. This is often a

convenient way of proving that sets are closed.

Proposition.

C ⊆ X

is a closed subset if and only if every limit point of

an element of C.

Proof.

(

⇒

) Suppose

is closed and

→ x

∈ C

. We have to show that

x ∈ C.

Since

is closed,

X \ C ⊆ X

is open. Suppose the contrary that

x 6∈ C

Then

x ∈ A

. Hence

is an open neighbourhood of

. Then by our previous

lemma, we know that there is some

such that

∈ A

for all

n ≥ N

. So

∈ A

. But we know that

∈ C

by assumption. This is a contradiction. So

we must have x ∈ C.

(⇐) Suppose that C is not closed. We have to find a limit point not in C.

Since

is not closed,

is not open. So

∃x ∈ A

such that

(

)

6⊆ A

for all

δ > 0. This means that B

(x) ∩ C 6= ∅ for all δ > 0.

So pick

∈ B

(

)

∩C

for each

n >

0. Then

∈ C

(

, x

) =

→

0. So

→ x. So x is a limit point of C which is not in C.

Finally, we get to the Really Important Result

that tells us metrics are

useless.

Proposition (Characterization of continuity). Let (

X, d

) and (

Y, d

) be metric

spaces, and f : X → Y . The following conditions are equivalent:

(i) f is continuous

(ii) If x

→ x, then f (x

) → f (x) (which is the definition of continuity)

(iii) For any closed subset C ⊆ Y , f

−1

(iv) For any open subset U ⊆ Y , f

−1

(U) is open in X.

(v)

For any

x ∈ X

and

ε >

∃δ >

0 such that

(

))

⊆ B

(

)).

Alternatively, d

(x, z) < δ ⇒ d

(f(x), f(z)) < ε.

Proof.

– 1 ⇔ 2: by definition

–

⇒

3: Suppose

C ⊆ Y

is closed. We want to show that

−1

(

) is closed.

So let x

→ x, where x

∈ f

−1

(C).

We know that

(

)

→ f

(

) by (2) and

(

)

∈ C

. So

(

) is a limit

point of

. Since

is closed,

(

)

∈ C

. So

x ∈ f

−1

(

). So every limit

point of f

−1

(C). So f

−1

–

⇒

4: If

U ⊆ Y

is open, then

Y \ U

is closed in Y. So

−1

(

Y \ U

) =

X \ f

−1

(U) is closed in X. So f

−1

(U) ⊆ X is open.

–

⇒

5: Given

x ∈ X, ε >

(

)) is open in

. By (4), we know

−1

(

))) =

is open in

. Since

x ∈ A

∃δ >

0 with

(

)

⊆ A

f(B

(x)) ⊆ f (A) = f(f

−1

(f(x)))) = B

(f(x))

–

⇒

2: Suppose

→ x

. Given

ε >

∃δ >

0 such that

(

))

⊆

(

)). Since

→ x

∃N

such that

∈ B

(

) for all

n > N

. Then

f(x

) ∈ f (B

(x)) ⊆ B

(f(x)) for all n > N . So f(x

) → f (x).

The third and fourth condition can allow us to immediately decide if a subset

is open or closed in some cases.

Example. Let f : R

→ R be defined as

f(x

, x

) = x

+ x

Then this is continuous. So

{

∈ R

(x)

≤

}

−1

((

−∞,

1]) is closed in

Before we end, we prove some key properties of open subsets. These will be

used as the defining properties of open subsets in the next chapter.

Lemma.

(i) ∅ and X are open subsets of X.

(ii) Suppose V

⊆ X is open for all α ∈ A. Then U =

[

α∈A

is open in X.

(iii) If V

, ··· , V

⊆ X are open, then so is V =

i=1

Proof.

(i) ∅

satisfies the definition of an open subset vacuously.

is open since for

any x, B

(x) ⊆ X.

(ii)

x ∈ U

, then

x ∈ V

for some

. Since

is open, there exists

δ >

such that B

(x) ⊆ V

. So B

(x) ⊆

[

α∈A

= U. So U is open.

(iii)

x ∈ V

, then

x ∈ V

for all

= 1

, ··· , n

. So

∃δ

0 with

(

)

⊆ V

Take

min{δ

, ··· , δ

}

. So

(

)

⊆ V

for all

. So

(

)

⊆ V

. So

open.

Note that we can take infinite unions or finite intersection, but not infinite

intersections. For example, the intersection of all (

−

) is

{

}

, which is not

open.

2 Topological spaces

2.1 Definitions

We have previously shown that a function

is continuous iff

−1

(

) is open

whenever

is open. Convergence can also be characterized using open sets only.

This suggests that we can dump the metric and just focus on the open sets.

What we will do is to define the topology of a space

to be the set of open

sets of

. Then this topology would define most of the structure or geometry of

X, and we don’t need to care about metrics.

Definition (Topological space). A topological space is a set

(the space)

together with a set U ⊆ P(X) (the topology) such that:

(i) ∅, X ∈ U

(ii) If V

∈ U for all α ∈ A, then

[

α∈A

∈ U.

(iii) If V

, ··· , V

∈ U, then

i=1

∈ U.

The elements of

are the points, and the elements of

are the open subsets of

Definition (Induced topology). Let (

X, d

) be a metric space. Then the topology

induced by d is the set of all open sets of X under d.

Example. Let

and consider the two metrics

y) =

−

and

∞

(x, y) = kx − yk

∞

. We will show that they induce the same topology.

Recall that the metrics are defined by

kvk =

i=1

|, kvk

∞

= max

1≤i≤n

This implies that

kvk

∞

≤ kvk

≤ nkvk

∞

This in turn implies that

∞

(x) ⊇ B

∞

r/n

(x).

∞

To show that the metrics induce the same topology, suppose that

is open with

respect to

, and we show that it is open with respect to

∞

. Let

x ∈ U

. Since

is open with respect to

, there exists some

δ >

0 such that

(

)

⊆ U

. So

∞

δ/n

(x) ⊆ B

(x) ⊆ U . So U is open with respect to d

∞

The other direction is similar.

Example. Let

1]. Let

(

f, g

) =

kf − gk

and

∞

(

f, g

) =

kf − gk

∞

Then they do not induce the same topology, since (

X, d

)

→

(

X, d

∞

) by

f 7→ f

is not continuous.

It is possible to have some other topologies that are not induced by metrics.

Example.

(i) Let X be any set.

(a) U = {∅, X} is the coarse topology on X.

(b) U

(

) is the discrete topology on

, since it is induced by the

discrete metric.

{A ⊆ X

X \A is finite or A

∅}

is the cofinite topology on

(ii) Let X = R, and U = {(a, ∞) : a ∈ R} is the right order topology on R.

Now we can define continuous functions in terms of the topology only.

Definition (Continuous function). Let

X → Y

be a map of topological

spaces. Then f is continuous if f

−1

(U) is open in X whenever U is open in Y .

When the topologies are induced by metrics, the topological and metric

notions of continuous functions coincide, as we showed in the previous chapter.

Example.

(i) Any function f : X → Y is continuous if X has the discrete topology.

(ii) Any function f : X → Y is continuous if Y has the coarse topology.

(iii)

and

both have cofinite topology, then

X → Y

is continuous iff

−1

({y}) is finite for every y ∈ Y .

Lemma. If

X → Y

and

Y → Z

are continuous, then so is

g ◦ f

X → Z

Proof.

U ⊆ Z

is open,

is continuous, then

−1

(

) is open in

. Since

also continuous, f

−1

(U)) = (g ◦ f)

−1

(U) is open in X.

In group theory, we had the notion of isomorphisms between groups. Isomor-

phic groups are equal-up-to-renaming-of-elements, and are considered to be the

same for most purposes.

Similarly, we will define homeomorphisms between topological spaces, and

homeomorphic topological spaces will be considered to be the same (notice the

“e” in homeomorphism).

Definition (Homeomorphism). f : X → Y is a homeomorphism if

(i) f is a bijection

(ii) Both f and f

−1

are continuous

Equivalently, f is a bijection and U ⊆ X is open iff f (U) ⊆ Y is open.

Two spaces are homeomorphic if there exists a homeomorphism between

them, and we write X ' Y .

Note that we specifically require

and

−1

to be both continuous. In

group theory, if

is a bijective homomorphism, then

−1

is automatically a

homomorphism as well. However, this is not true for topological spaces.

being continuous does not imply

−1

is continuous, as illustrated by the example

below.

Example. Let

1] with the topology induced by

k · k

and

with the topology induced by

k · k

∞

. Then

Y → X

f 7→ f

is continuous

but F

−1

is not.

Example. Let

= [0

) and

{z ∈ C

|z|

= 1

}

. Then

X → Y

given by f(x) = e

is continuous but its inverse is not.

Similar to isomorphisms, we can show that homeomorphism is an equivalence

relation.

Lemma. Homeomorphism is an equivalence relation.

Proof.

(i) The identity map I

: X → X is always a homeomorphism. So X ' X.

(ii)

X → Y

is a homeomorphism, then so is

−1

Y → X

. So

X ' Y ⇒ Y ' X.

(iii) If f : X → Y and g : Y → Z are homeomorphisms, then g ◦ f : X → Z is

a homeomorphism. So X ' Y and Y ' Z implies X ' Z.

Example.

(i)

Under the usual topology, the open intervals (0

(

a, b

) for all

a, b ∈ R

using the homeomorphism x 7→ a + (b − a)x.

Similarly, [0, 1] ' [a, b]

(ii) (−1, 1) ' R by x 7→ tan(

x).

(iii) R ' (0, ∞) by x 7→ e

(iv) (a, ∞) ' (b, ∞) by x 7→ x + (b − a).

The fact that

is an equivalence relation implies that any 2 open intervals in

are homeomorphic.

It is relatively easy to show that two spaces are homeomorphic. We just have

to write down a homeomorphism. However, it is rather difficult to prove that

two spaces are not homeomorphic.

For example, is (0

1) homeomorphic to [0

1]? No, but we aren’t able

to immediately prove it now. How about

and

? Again they are not

homeomorphic, but to show this we will need some tools that we’ll develop in

the next few lectures.

How about

and

in general? They are not homeomorphic, but we

won’t be able to prove this rigorously in the course. To properly prove this, we

will need tools from algebraic topology.

So how can we prove that two spaces are not homeomorphic? In group theory,

we could prove that two groups are not isomorphic by, say, showing that they

have different orders. Similarly, to distinguish between topological spaces, we

have to define certain topological properties. Then if two spaces have different

topological properties, we can show that they are not homeomorphic.

But before that, we will first define many many useful definitions in topological

spaces, including sequences, subspaces, products and quotients. The remainder

of the chapter will be mostly definitions that we will use later.

2.2 Sequences

To define the convergence of a sequence using open sets, we again need the

concept of open neighbourhoods.

Definition (Open neighbourhood). An open neighbourhood of

x ∈ X

is an open

set U ⊆ X with x ∈ U .

Now we can use this to define convergence of sequences.

Definition (Convergent sequence). A sequence

→ x

if for every open

neighbourhood U of x, ∃N such that x

∈ U for all n > N.

Example.

(i)

has the coarse topology, then any sequence

converges to every

x ∈ X, since there is only one open neighbourhood of x.

(ii)

has the cofinite topology, no two

s are the same, then

→ x

for every

x ∈ X

, since every open set can only have finitely many

not

inside it.

This looks weird. This is definitely not what we used to think of sequences.

At least, we would want to have unique limits.

Fortunately, there is a particular class of spaces where sequences are well-

behaved and have at most one limit.

Definition (Hausdorff space). A topological space

is Hausdorff if for all

, x

∈ X

with

, there exist open neighbourhoods

such that U

∩ U

= ∅.

Lemma. If

is Hausdorff,

is a sequence in

with

→ x

and

→ x

then x = x

, i.e. limits are unique.

Proof.

Suppose the contrary that

x 6

. Then by definition of Hausdorff, there

exist open neighbourhoods U, U

of x, x

respectively with U ∩ U

= ∅.

Since

→ x

and

is a neighbourhood of

, by definition, there is some

such that whenever

n > N

, we have

∈ U

. Similarly, since

→ x

, there is

some N

such that whenever n > N

, we have x

∈ U

This means that whenever

n > max

(

N, N

), we have

∈ U

and

∈ U

So x

∈ U ∩ U

. This contradicts the fact that U ∩ U

= ∅.

Hence we must have x = x

Example.

(i)

has more than 1 element, then the coarse topology on

is not

Hausdorff.

(ii)

has infinitely many elements, the cofinite topology on

is not

Hausdorff.

(iii) The discrete topology is always Hausdorff.

(iv)

If (

X, d

) is a metric space, the topology induced by

is Hausdorff: for

, let

(

, x

)

0. Then take

r/2

(

). Then

∩U

∅

2.3 Closed sets

We will define closed sets similarly to what we did for metric spaces.

Definition (Closed sets). C ⊆ X is closed if X \ C is an open subset of X.

Lemma.

(i) If C

is a closed subset of X for all α ∈ A, then

α∈A

is closed in X.

(ii) If C

, ··· , C

are closed in X, then so is

i=1

Proof.

(i)

Since

is closed in

X \ C

is open in

. So

α∈A

(

X \ C

) =

X \

α∈A

is open. So

α∈A

is closed.

(ii)

is closed in

, then

X \ C

is open. So

i=1

(

X \ C

) =

X \

i=1

is open. So

i=1

is closed.

This time we can take infinite intersections and finite unions, which is the

opposite of what we have for open sets.

Note that it is entirely possible to define the topology to be the collection of

all closed sets instead of open sets, but people seem to like open sets more.

Corollary. If X is Hausdorff and x ∈ X, then {x} is closed in X.

Proof.

For all

y ∈ X

, there exist open subsets

, V

with

y ∈ U

, x ∈ V

∩ V

= ∅.

Let

X \ U

. Then

is closed,

y 6∈ C

x ∈ C

. So

{x}

y6=x

closed since it is an intersection of closed subsets.

2.4 Closure and interior

2.4.1 Closure

Given a subset

A ⊆ X

, if

is not closed, we would like to find the smallest

closed subset containing A. This is known as the closure of A.

Officially, we define the closure as follows:

Definition. Let X be a topological space and A ⊆ X. Define

= {C ⊆ X : A ⊆ C and C is closed in X}

Then the closure of A in X is

A =

C∈C

First we do a sanity check: since

is defined as an intersection, we should

make sure we are not taking an intersection of no sets. This is easy: since

is closed in

(its complement

∅

is open),

∅

. So we can safely take the

intersection.

Since

is an intersection of closed sets, it is closed in

. Also, if

C ∈ C

then A ⊆ C. So A ⊆

C∈C

C =

A. In fact, we have

Proposition.

A is the smallest closed subset of X which contains A.

Proof.

Let

K ⊆ X

be a closed set containing

. Then

K ∈ C

. So

C∈C

C ⊆ K. So

A ⊆ K.

We basically defined the closure such that it is the smallest closed subset of

X which contains A.

However, while this “clever” definition makes it easy to prove the above

property, it is rather difficult to directly use it to compute the closure.

To compute the closure, we define the limit point analogous to what we did

for metric spaces.

Definition (Limit point). A limit point of

is an

x ∈ X

such that there is a

sequence x

→ x with x

∈ A for all n.

In general, limit points are easier to compute, and can be a useful tool for

determining the closure of A.

Now let

L(A) = {x ∈ X : x is a limit point of A}.

We immediately get the following lemma.

Lemma. If C ⊆ X is closed, then L(C) = C.

Proof.

Exactly the same as that for metric spaces. We will also prove a more

general result very soon that implies this.

Recall that we proved the converse of this statement for metric spaces.

However, the converse is not true for topological spaces in general.

Example. Let

be an uncountable set (e.g.

), and define a topology on

by saying a set is open if it is empty or has countable complement. One can

check that this indeed defines a topology. We claim that the only sequences that

converge are those that are eventually constant.

Indeed, if x

is a sequence and x ∈ X, then consider the open set

U = (X \ {x

: n ∈ N}) ∪ {x}.

Then the only element in the sequence

that can possibly be contained in

is x itself. So if x

→ x, this implies that x

is eventually always x.

In particular, it follows that L(A) = A for all A ⊆ X.

However, we do have the following result:

Proposition. L(A) ⊆

Proof.

A ⊆ C

, then

(

)

⊆ L

(

). If

is closed, then

(

) =

. So

C ∈ C

⇒ L(A) ⊆ C. So L(A) ⊆

C∈C

C =

This in particular implies the previous lemma, since for any

, we have

A ⊆ L(A) ⊆

A, and when A is closed, we have A =

Finally, we have the following corollary that can help us find the closure of

subsets:

Corollary. Given a subset

A ⊆ X

, if we can find some closed

such that

A ⊆ C ⊆ L(A), then we in fact have C =

Proof. C ⊆ L

(

)

⊆

A ⊆ C

, where the last step is since

is the smallest closed

set containing A. So C = L(A) =

Example.

– Let (a, b) ⊆ R. Then (a, b) = [a, b].

– Let Q ⊆ R. Then

Q = R.

– R \ Q = R.

–

with the Euclidean metric,

(x)

(

). In general,

(x) ⊆

(

), since

(

) is closed and

(

)

⊆

(

), but these need not be

equal.

For example, if

has the discrete metric, then

(

) =

{x}

. Then

(x) = {x}, but

(x) = X.

In the above example, we had

. In some sense, all points of

are

“surrounded” by points in Q. We say that Q is dense in R.

Definition (Dense subset). A ⊆ X is dense in X if

A = X.

Example. Q and R \ Q are both dense in R with the usual topology.

2.4.2 Interior

We defined the closure of

to be the smallest closed subset containing

. We

can similarly define the interior of

to be the largest open subset contained in

Definition (Interior). Let A ⊆ X, and let

= {U ⊆ X : U ⊆ A, U is open in X}.

The interior of A is

Int(A) =

[

U∈O

Proposition. Int(A) is the largest open subset of X contained in A.

The proof is similar to proof for closure.

To find the closure, we could use limit points. What trick do we have to find

the interior?

Proposition. X \ Int(A) = X \ A

Proof. U ⊆ A ⇔

(

X \U

)

⊇

(

X \A

). Also,

U open in X ⇔ X \U is closed in X

So the complement of the largest open subset of

contained in

will be

the smallest closed subset containing X \ A.

Example. Int(Q) = Int(R \ Q) = ∅.

2.5 New topologies from old

In group theory, we had the notions of subgroups, products and quotients. There

are exact analogies for topological spaces. In this section, we will study the

subspace topology, product topology and quotient topology.

2.5.1 Subspace topology

Definition (Subspace topology). Let

be a topological space and

Y ⊆ X

The subspace topology on

is given by:

is an open subset of

if there is

some U open in X such that V = Y ∩ U.

If we simply write

Y ⊆ X

and don’t specify a topology, the subspace topology

is assumed. For example, when we write Q ⊆ R, we are thinking of Q with the

subspace topology inherited from R.

Example. If (

X, d

) is a metric space and

Y ⊆ X

, then the metric topology on

(Y, d) is the subspace topology, since B

(y) = Y ∩ B

(y).

To show that this is indeed a topology, we need the following set theory facts:

Y ∩

[

α∈A

[

α∈A

(Y ∩ V

)

Y ∩

α∈A

(Y ∩ V

)

Proposition. The subspace topology is a topology.

Proof.

(i) Since ∅ is open in X, ∅ = Y ∩ ∅ is open in Y .

Since X is open in X, Y = Y ∩ X is open in Y .

(ii) If V

is open in Y , then V

= Y ∩ U

for some U

open in X. Then

[

α∈A

[

α∈A

(Y ∩ U

) = Y ∩

[

α∈U

Since

is open in X, so

is open in Y .

(iii) If V

is open in Y , then V

= Y ∩ U

for some open U

⊆ X. Then

i=1

(Y ∩ U

) = Y ∩

i=1

Since

is open,

is open.

Recall that if

Y ⊆ X

, there is an inclusion function

Y → X

that sends

y 7→ y

. We can use this to obtain the following defining property of a subspace.

Proposition. If

has the subspace topology,

Z → Y

is continuous iff

ι ◦ f : Z → X is continuous.

Proof.

(

⇒

) If

U ⊆ X

is open, then

−1

(

) =

Y ∩ U

is open in

. So

continuous. So if f is continuous, so is ι ◦ f .

(

⇐

) Suppose we know that

ι ◦ f

is continuous. Given

V ⊆ Y

is open, we

know that

Y ∩ U

−1

(

). So

−1

(

) =

−1

(

−1

(

))) = (

ι ◦ f

)

−1

(

) is

open since ι ◦ f is continuous. So f is continuous.

This property is “defining” in the sense that it can be used to define a

subspace:

is a subspace of

if there exists some function

Y → X

such

that for any f , f is continuous iff ι ◦ f is continuous.

Example.

{

∈ R

| ≤

}

is the

-dimensional closed unit disk.

n−1

= {v ∈ R

: |v| = 1} is the n − 1-dimensional sphere.

We have

Int(D

) = {v ∈ R

: |v| < 1} = B

(0).

This is, in fact, homeomorphic to

. To show this, we can first pick our favorite

homeomorphism

: [0

7→

, ∞

). Then v

7→ f

(

)v is a homeomorphism

Int(D

) → R

2.5.2 Product topology

If X and Y are sets, the product is defined as

X × Y = {(x, y) : x ∈ X, y ∈ Y }

We have the projection functions π

: X × Y → X, π

: X × Y → Y given by

(x, y) = x, π

(x, y) = y.

If A ⊆ X, B ⊆ Y , then we have A × B ⊆ X × Y .

Given topological spaces

, we can define a topology on

X ×Y

as follows:

Definition (Product topology). Let

and

be topological spaces. The product

topology on X × Y is given by:

U ⊆ X × Y

is open if: for every (

x, y

)

∈ U

, there exist

⊆ X, W

⊆ Y

open neighbourhoods of x and y such that V

× W

⊆ U .

Example.

–

V ⊆ X

and

W ⊆ Y

are open, then

V × W ⊆ X × Y

is open (take

= V , W

= W ).

–

The product topology on

R × R

is same as the topology induced by the

k · k

∞

, hence is also the same as the topology induced by

k · k

Similarly, the product topology on

n−1

×R

is also the same as that

induced by k · k

∞

– (0, 1) × (0, 1) × ··· × (0, 1) ⊆ R

is the open n-dimensional cube in R

Since (0, 1) ' R, we have (0, 1)

' R

' Int(D

–

× S

' {

∈ R

n+1

: 1

≤ |

| ≤

}

, where the last

homeomorphism is given by (

7→ t

w with inverse v

7→

(

). This is

a thickened sphere.

–

Let

A ⊆ {

(

r, z

) :

r >

} ⊆ R

(

) be the set obtained by rotating

around the z axis. Then R(A) ' S × A by

(x, y, z) = (v, z) 7→ (

v, (|v|, z)).

In particular, if

is a circle, then

(

)

' S

× S

is the two-

dimensional torus.

The defining property is that

Z → X × Y

is continuous iff

◦ f

and

◦ f are continuous.

Note that our definition of the product topology is rather similar to the

definition of open sets for metrics. We have a special class of subsets of the

form

V × W

, and a subset

is open iff every point

x ∈ U

is contained in some

V × W ⊆ U . In some sense, these subsets “generate” the open sets.

Alternatively, if U ⊆ X × Y is open, then

U =

[

(x,y)∈U

× W

U ⊆ X ×Y

is open if and only if it is a union of members of our special class

of subsets.

We call this special class the basis.

Definition (Basis). Let

be a topology on

. A subset

B ⊆ U

is a basis if

“U ∈ U iff U is a union of sets in B”.

Example.

– {V ×W

V ⊆ X, W ⊆ Y are open}

is a basis for the product topology for

X × Y .

– If (X, d) is a metric space, then

1/n

(x) : n ∈ N

, x ∈ X}

is a basis for the topology induced by d.

2.5.3 Quotient topology

is a set and

∼

is an equivalence relation on

, then the quotient

X/∼

is the

set of equivalence classes. The projection

X → X/∼

is defined as

(

) = [

the equivalence class containing x.

Definition (Quotient topology). If

is a topological space, the quotient topology

on X/∼ is given by: U is open in X/∼ if π

−1

(U) is open in X.

We can think of the quotient as “gluing” the points identified by

∼

together.

The defining property is

X/∼ → Y

is continuous iff

f ◦ π

X → Y

continuous.

Example.

–

Let

x ∼ y

iff

x − y ∈ Z

. Then

X/∼

R/Z ' S

, given by

[x] 7→ (cos 2πx, sin 2πx).

–

Let

. Then v

∼

w iff v

−

∈ Z

. Then

X/∼

(

R/Z

)

(

R/Z

)

' S

×S

. Similarly,

×S

×···×S

–

A ⊆ X

, define

∼

x ∼ y

iff

x, y ∈ A

. This glues everything

in A together and leaves everything else alone.

We often write this as

X/A

. Note that this is not consistent with the

notation we just used above!

◦

Let

= [0

1] and

{

}

, then

X/A ' S

by, say,

t 7→

(

cos

πt, sin

πt

). Intuitively, the equivalence relation says that the

two end points of [0

1] are “the same”. So we join the ends together

to get a circle.

◦

Let

and

n−1

. Then

X/A ' S

. This can be pictured

as pulling the boundary of the disk together to a point to create a

closed surface

–

Let

= [0

1] with

∼

given by (0

, y

)

∼

, y

) and (

∼

(

1),

then X/∼ ' S

× S

= T

, by, say

(x, y) 7→



(cos 2πx, sin 2πx), (cos 2πy, sin 2πy)



Similarly, T

= [0, 1]

/∼, where the equivalence is analogous to above.

Note that even if

is Hausdorff,

X/∼

may not be! For example,

R/Q

is not

Hausdorff.

3 Connectivity

Finally we can get to something more interesting. In this chapter, we will study

the connectivity of spaces. Intuitively, we would want to say that a space is

“connected” if it is one-piece. For example,

is connected, while

R \ {

}

not. We will come up with two different definitions of connectivity - normal

connectivity and path connectivity, where the latter implies the former, but not

the other way round.

3.1 Connectivity

We will first look at normal connectivity.

Definition (Connected space). A topological space

is disconnected if

can

be written as

A ∪ B

, where

and

are disjoint, non-empty open subsets of

We say A and B disconnect X.

A space is connected if it is not disconnected.

Note that being connected is a property of a space, not a subset. When we

say “

is a connected subset of

”, it means

is connected with the subspace

topology inherited from X.

Being (dis)connected is a topological property, i.e. if

is (dis)connected,

and

X ' Y

, then

is (dis)connected. To show this, let

X → Y

be the

homeomorphism. By definition,

is open in

iff

(

) is open in

. So

and

B disconnect X iff f(A) and f (B) disconnect Y .

Example.

– If X has the coarse topology, it is connected.

– If X has the discrete topology and at least 2 elements, it is disconnected.

–

Let

X ⊆ R

. If there is some

α ∈ R \ X

such that there is some

a, b ∈ X

with

a < α < b

, then

is disconnected. In particular,

X ∩

(

−∞, α

) and

X ∩ (α, ∞) disconnect X.

For example, (0, 1) ∪ (1, 2) is disconnected (α = 1).

We can also characterize connectivity in terms of continuous functions:

Proposition.

is disconnected iff there exists a continuous surjective

X →

{0, 1} with the discrete topology.

Alternatively,

is connected iff any continuous map

X → {

}

constant.

Proof. (⇒) If A and B disconnect X, define

f(x) =

(

0 x ∈ A

1 x ∈ B

Then

−1

(

∅

) =

∅

−1

(

{

}

) =

−1

(

{

}

) =

and

−1

(

{

}

) =

are all

open. So f is continuous. Also, since A, B are non-empty, f is surjective.

(

⇐

) Given

X 7→ {

}

surjective and continuous, define

−1

(

{

}

B = f

−1

({1}). Then A and B disconnect X.

Theorem. [0, 1] is connected.

Note that

Q ∩

1] is disconnected, since we can pick our favorite irrational

number

, and then

x < a}

and

x > a}

disconnect the interval. So we

better use something special about [0, 1].

The key property of R is that every non-empty A ⊆ [0, 1] has a supremum.

Proof.

Suppose

and

disconnect [0

1]. wlog, assume 1

∈ B

. Since

non-empty, α = sup A exists. Then either

– α ∈ A

. Then

α <

1, since 1

∈ B

. Since

is open,

∃ε >

0 with

(

)

⊆ A

So α +

∈ A, contradicting supremality of α; or

– α 6∈ A

. Then

α ∈ B

. Since

is open,

∃ε >

0 such that

(

)

⊆ B

. Then

a ≤ α −ε

for all

a ∈ A

. This contradicts

being the least upper bound of

Either option gives a contradiction. So

and

cannot exist and [0

1] is

connected.

To conclude the section, we will prove the intermediate value property. The

key proposition needed is the following.

Proposition. If

X → Y

is continuous and

is connected, then

im f

is also

connected.

Proof.

Suppose

and

disconnect

im f

. We will show that

−1

(

) and

−1

(

)

disconnect X.

Since

A, B ⊆ im f

are open, we know that

im f ∩ A

and

im f ∩ B

for some

, B

open in

. Then

−1

(

) =

−1

(

) and

−1

(

) =

−1

(

) are

open in X.

Since

A, B

are non-empty,

−1

(

) and

−1

(

) are non-empty. Also,

−1

(

)

∩

−1

(

) =

−1

(

A ∩ B

) =

−1

(

∅

) =

∅

. Finally,

A ∪ B

im f

. So

−1

(

)

∪

−1

(B) = f

−1

(A ∪ B) = X.

−1

(

) and

−1

(

) disconnect

, contradicting our hypothesis. So

im f

is connected.

Alternatively, if

im f

is not connected, let

im f → {

}

be continuous

surjective. Then g ◦ f : X → {0, 1} is continuous surjective. Contradiction.

Theorem (Intermediate value theorem). Suppose

X → R

is continuous

and

is connected. If

∃x

, x

such that

(

)

< f

(

), then

∃x ∈ X

with

f(x) = 0.

Proof.

Suppose no such

exists. Then 0

6∈ im f

while 0

> f

(

)

∈ im f

< f

(

)

∈ im f

. Then

im f

is disconnected (from our previous example),

contradicting X being connected.

Alternatively, if

(

)

= 0 for all

, then

f(x)

|f(x)|

is a continuous surjection from

X to {−1, +1}, which is a contradiction.

Corollary. If

: [0

→ R

is continuous with

(0)

< f

(1), then

∃x ∈

with f (x) = 0.

Is the converse of the intermediate value theorem true? If

is disconnected,

can we find a function g that doesn’t satisfy the intermediate value property?

The answer is yes. Since

is disconnected, let

X → {

}

be continu-

ous. Then let

(

) =

(

)

−

. Then

is continuous but doesn’t satisfy the

intermediate value property.

3.2 Path connectivity

The other notion of connectivity is path connectivity. A space is path connected

if we can join any two points with a path. First, we need a definition of a path.

Definition (Path). Let

be a topological space, and

, x

∈ X

. Then a

path from

is a continuous function

: [0

7→ X

such that

(0) =

γ(1) = x

Definition (Path connectivity). A topological space

is path connected if for

all points x

, x

∈ X, there is a path from x

to x

Example.

(i)

(

a, b

)

[

a, b

)

(

a, b

]

, R

are all path connected (using paths given by linear

functions).

(ii) R

is path connected (e.g. γ(t) = tx

+ (1 − t)x

(iii) R

}

is path-connected for

n >

1 (the paths are either line segments or

bent line segments to get around the hole).

Path connectivity is a stronger condition than connectivity, in the sense that

Proposition. If X is path connected, then X is connected.

Proof.

Let

be path connected, and let

X → {

}

be a continuous function.

We want to show that f is constant.

Let

x, y ∈ X

. By path connectedness, there is a map

: [0

→ X

such that

(0) =

and

(1) =

. Composing with

gives a map

f ◦ γ

: [0

→ {

}

Since [0

1] is connected, this must be constant. In particular,

(

(0)) =

(

(1)),

i.e. f (x) = f(y). Since x, y were arbitrary, we know f is constant.

We can use connectivity to distinguish spaces. Apart from the obvious “

is connected while

is not”, we can also try to remove points and see what

happens. We will need the following lemma:

Lemma. Suppose

X → Y

is a homeomorphism and

A ⊆ X

, then

A →

f(A) is a homeomorphism.

Proof.

Since

is a bijection,

is a bijection. If

U ⊆ f

(

) is open, then

(

)

∩ U

for some

open in

. So

−1

(

) =

−1

(

)

∩ A

is open in

So f |

is continuous. Similarly, we can show that (f|

)

−1

is continuous.

Example. [0

1). Suppose it were. Let

: [0

→

1) be a homeomor-

phism. Let

= (0

1]. Then

: (0

→

\ {f

(0)

}

is a homeomorphism.

But (0, 1] is connected while (0, 1) \ {f (0)} is disconnected. Contradiction.

Similarly, [0

1] and [0

1). Also,

6' R

for

n >

1, and

not homeomorphic to any subset of R.

3.2.1 Higher connectivity*

We were able to use path connectivity to determine that

is not homeomorphic

for

n >

1. If we want to distinguish general

and

, we will need to

generalize the idea of path connectivity to higher connectivity.

To do so, we have to formulate path connectivity in a different way. Recall

that

{−

} ' {

} ⊆ R

, while

= [

−

⊆ R

. Then we

can formulate path connectivity as:

is path-connected iff any continuous

f : S

→ X extends to a continuous γ : D

→ X with γ|

= f.

This is much more easily generalized:

Definition (

-connectedness).

-connected if for any

k ≤ n

, any continuous

f : S

→ X extends to a continuous F : D

k+1

→ X such that F |

= f.

For any point

p ∈ R

\{p}

-connected iff

m ≤ n −

2. So

\{p} 6'

\ {q} unless n = m. So R

6' R

Unfortunately, we have not yet proven that

6' R

. We casually stated

that

\ {p}

-connected iff

m ≤ n −

2. However, while this intuitively

makes sense, it is in fact very difficult to prove. To actually prove this, we will

need tools from algebraic topology.

3.3 Components

If a space is disconnected, we could divide the space into different components,

each of which is (maximally) connected. This is what we will investigate in this

section. Of course, we would have a different notion of “components” for each

type of connectivity. We will first start with path connectivity.

3.3.1 Path components

Defining path components (i.e. components with respect to path connectedness)

is relatively easy. To do so, we need the following lemma.

Lemma. Define

x ∼ y

if there is a path from

. Then

∼

is an

equivalence relation.

Proof.

(i)

For any

x ∈ X

, let

: [0

→ X

(

) =

, the constant path. Then

this is a path from x to x. So x ∼ x.

(ii)

: [0

→ X

is a path from

, then

¯γ

: [0

→ X

t 7→ γ

− t

)

is a path from y to x. So x ∼ y ⇒ y ∼ x.

(iii)

is a path from

and

is a path from

, then

∗γ

defined

t 7→

(

(2t) t ∈ [0, 1/2]

(2t − 1) t ∈ [1/2, 1]

is a path from x to z. So x ∼ y, y ∼ z ⇒ x ∼ z.

With this lemma, we can immediately define the path components:

Definition (Path components). Equivalence classes of the relation “

x ∼ y

there is a path from x to y” are path components of X.

3.3.2 Connected components

Defining connected components (i.e. components with respect to regular connec-

tivity) is slightly more involved. The key proposition we need is as follows:

Proposition. Suppose

⊆ X

is connected for all

α ∈ T

and that

α∈T

∅

Then Y =

α∈T

is connected.

Proof.

Suppose the contrary that

and

disconnect

. Then

and

are

open in

. So

Y ∩ A

and

Y ∩ B

, where

, B

are open in

. For

any fixed α, let

= Y

∩ A = Y

∩ A

, B

= Y

∩ B = Y

∩ B

Then they are open in Y

. Since Y = A ∪ B, we have

= Y ∩ Y

= (A ∪ B) ∩ Y

= A

∪ B

Since A ∩ B = ∅, we have

∩ B

= Y

∩ (A ∩ B) = ∅.

, B

are disjoint. So

is connected but is the disjoint union of open

subsets A

, B

By definition of connectivity, this can only happen if A

= ∅ or B

= ∅.

However, by assumption,

α∈T

∅

. So pick

y ∈

α∈T

. Since

y ∈ Y

either

y ∈ A

y ∈ B

. wlog, assume

y ∈ A

. Then

y ∈ Y

for all

implies that

y ∈ A

for all

. So

is non-empty for all

. So

is empty for all

. So

B = ∅.

So A and B did not disconnect Y after all. Contradiction.

Using this lemma, we can define connected components:

Definition (Connected component). If x ∈ X, define

C(x) = {A ⊆ X : x ∈ A and A is connected}.

Then C(x) =

[

A∈C(x)

A is the connected component of x.

(

) is the largest connected subset of

containing

. To show this, we

first note that

{x} ∈ C

(

). So

x ∈ C

(

). To show that it is connected, just note

that

x ∈

A∈C(x)

. So

A∈C(x)

is non-empty. By our previous proposition,

this implies that C(x) is connected.

Lemma. If y ∈ C(x), then C(y) = C(x).

Proof.

Since

y ∈ C

(

) and

(

) is connected,

(

)

⊆ C

(

). So

x ∈ C

(

). Then

C(y) ⊆ C(x). So C(x) = C(y).

It follows that

x ∼ y

x ∈ C

(

) is an equivalence relation and the connected

components of X are the equivalence classes.

Example.

–

Let

= (

−∞,

∪

, ∞

)

⊆ R

. Then the connected components are

(−∞, 0) and (0, ∞), which are also the path components.

–

Let

Q ⊆ R

. Then

(

) =

{x}

for all

x ∈ X

. In this case, we say

is totally disconnected.

Note that

(

) and

X \ C

(

) need not disconnect

, even though it is the

case in our first example. For this, we must need

(

) and

X \C

(

) to be open

as well. For example, in Example 2, C(x) = {x} is not open.

It is also important to note that path components need not be equal to the

connected components, as illustrated by the following example. However, since

path connected spaces are connected, the path component containing

must be

a subset of C(x).

Example. Let Y = {(0, y) : y ∈ R} ⊆ R

be the y axis.

Let Z = {(x,

sin

) : x ∈ (0, ∞)}.

Let

Y ∪ Z ⊆ R

. We claim that

and

are the path components of

Since

and

are individually path connected, it suffices to show that there is

no continuous γ : [0, 1] → X with γ(0) = (0, 0), γ(1) = (1, sin 1).

Suppose

existed. Then the function

◦ γ

: [0

→ R

projecting the

path to the

direction is continuous. So it is bounded. Let

be such that

◦ γ

(

)

≤ M

for all

t ∈

1]. Let

X ∩

(

R ×

(

−∞, M

]) be the part of

that lies below y = M. Then im γ ⊆ W .

However,

is disconnected: pick

with

sin

> M

. Then

W ∩

((

−∞, t

)

× R

) and

W ∩

((

, ∞

)

× R

) disconnect

. This is a contradiction,

since γ is continuous and [0, 1] is connected.

We also claim that

is connected: suppose

and

disconnect

. Then

since

and

are connected, either

Y ⊆ A

Y ⊆ B

;

Z ⊆ A

Z ⊆ B

. If both

Y ⊆ A, Z ⊆ A, then B = ∅, which is not possible.

So wlog assume

. This is also impossible, since

is not open

as it is not a union of balls (any open ball containing a point in

will also

contain a point in Z). Hence X must be connected.

Finally, recall that we showed that path-connected subsets are connected.

While the converse is not true in general, there are special cases where it is true.

Proposition. If U ⊆ R

is open and connected, then it is path-connected.

Proof. Let A be a path component of U. We first show that A is open.

Let

a ∈ A

. Since

is open,

∃ε >

0 such that

(

)

⊆ U

. We know that

(

)

' Int

(

) is path-connected (e.g. use line segments connecting the points).

Since

is a path component and

a ∈ A

, we must have

(

)

⊆ A

. So

is an

open subset of U.

Now suppose b ∈ U \ A. Then since U is open, ∃ε > 0 such that B

(b) ⊆ U .

Since

(

) is path-connected, so if

(

)

∩ A 6

∅

, then

(

)

⊆ A

. But this

implies

b ∈ A

, which is a contradiction. So

(

)

∩ A

∅

. So

(

)

⊆ U \ A

Then U \ A is open.

A, U \ A

are disjoint open subsets of

. Since

is connected, we must

have U \ A empty (since A is not). So U = A is path-connected.

4 Compactness

Compactness is an important concept in topology. It can be viewed as a

generalization of being “closed and bounded” in

. Alternatively, it can also be

viewed as a generalization of being finite. Compact sets tend to have a lot of

really nice properties. For example, if

is compact and

X → R

is continuous,

then f is bounded and attains its bound.

There are two different definitions of compactness - one based on open covers

(which we will come to shortly), and the other based on sequences. In metric

spaces, these two definitions are equal. However, in general topological spaces,

these notions can be different. The first is just known as “compactness” and the

second is known as “sequential compactness”.

The actual definition of compactness is rather weird and unintuitive, since it

is an idea we haven’t seen before. To quote Qiaochu Yuan’s math.stackexchange

answer (http://math.stackexchange.com/a/371949),

The following story may or may not be helpful. Suppose you live

in a world where there are two types of animals: Foos, which are

red and short, and Bars, which are blue and tall. Naturally, in your

language, the word for Foo over time has come to refer to things

which are red and short, and the word for Bar over time has come to

refer to things which are blue and tall. (Your language doesn’t have

separate words for red, short, blue, and tall.)

One day a friend of yours tells you excitedly that he has discovered

a new animal. “What is it like?” you ask him. He says, “well, it’s

sort of Foo, but. . . ”

The reason he says it’s sort of Foo is that it’s short. However,

it’s not red. But your language doesn’t yet have a word for “short,”

so he has to introduce a new word — maybe “compact”. . .

The situation with compactness is sort of like the above. It

turns out that finiteness, which you think of as one concept (in the

same way that you think of “Foo” as one concept above), is really

two concepts: discreteness and compactness. You’ve never seen

these concepts separated before, though. When people say that

compactness is like finiteness, they mean that compactness captures

part of what it means to be finite in the same way that shortness

captures part of what it means to be Foo.

But in some sense you’ve never encountered the notion of com-

pactness by itself before, isolated from the notion of discreteness (in

the same way that above you’ve never encountered the notion of

shortness by itself before, isolated from the notion of redness). This

is just a new concept and you will to some extent just have to deal

with it on its own terms until you get comfortable with it.

4.1 Compactness

Definition (Open cover). Let

U ⊆ P

(

) be a topology on

. An open cover of

X is a subset V ⊆ U such that

[

V ∈V

V = X.

We say V covers X.

If V

⊆ V, and V

covers X, then we say V

is a subcover of V.

Definition (Compact space). A topological space

is compact if every open

cover V of X has a finite subcover V

= {V

, ··· , V

} ⊆ V.

Note that some people (especially algebraic geometers) call this notion “quasi-

compact”, and reserve the name “compact” for “quasi-compact and Hausdorff”.

We will not adapt this notion.

Example.

(i)

is finite, then

(

) is finite. So any open cover of

is finite. So

is compact.

(ii)

Let

and

{

(

−R, R

) :

R ∈ R, R >

}

. Then this is an open cover

with no finite subcover. So

is not compact. Hence all open intervals are

not compact since they are homeomorphic to R.

(iii) Let X = [0, 1] ∩ Q. Let

= X \ (α − 1/n, α + 1/n).

for some irrational α in (0, 1) (e.g. α = 1/

√

2).

Then

n>0

since

is irrational. Then

n ∈ Z >

}

an open cover of X. Since this has no finite subcover, X is not compact.

Theorem. [0, 1] is compact.

Again, since this is not true for [0

∩ Q

, we must use a special property of

the reals.

Proof. Suppose V is an open cover of [0, 1]. Let

A = {a ∈ [0, 1] : [0, a] has a finite subcover of V}.

First show that

is non-empty. Since

covers [0

1], in particular, there is some

that contains 0. So {0} has a finite subcover V

. So 0 ∈ A.

Next we note that by definition, if 0 ≤ b ≤ a and a ∈ A, then b ∈ A.

Now let α = sup A. Suppose α < 1. Then α ∈ [0, 1].

Since

covers

, let

α ∈ V

. Since

is open, there is some

such that

(

)

⊆ V

. By definition of

, we must have

α − ε/

∈ A

. So [0

, α − ε/

has a finite subcover. Add

to that subcover to get a finite subcover of

, α

ε/

2]. Contradiction (technically, it will be a finite subcover of [0

, η

] for

η = min(α + ε/2, 1), in case α + ε/2 gets too large).

So we must have α = sup A = 1.

Now we argue as before:

∃V

∈ V

such that 1

∈ V

and

∃ε >

0 with

− ε,

⊆ V

. Since 1

− ε ∈ A

, there exists a finite

⊆ V

which covers

[0, 1 − ε/2]. Then W = V

∪ {V

} is a finite subcover of V.

We mentioned that compactness is a generalization of “closed and bounded”.

We will now show that compactness is indeed in some ways related to closedness.

Proposition. If

is compact and

is a closed subset of

, then

is also

compact.

Proof.

To prove this, given an open cover of

, we need to find a finite subcover.

To do so, we need to first convert it into an open cover of

. We can do so by

adding

X \C

, which is open since

is closed. Then since

is compact, we can

find a finite subcover of this, which we can convert back to a finite subcover of

Formally, suppose

is an open cover of

. Say

α ∈ T }

. For

each

, since

is open in

C ∩ V

for some

open in

. Also, since

α∈T

= C, we have

α∈T

⊇ C.

Since

is closed,

X \ C

is open in

. So

α ∈ T } ∪ {U}

is an open cover of

. Since

is compact,

has a finite subcover

, ··· , V

, U}

(

may or may not be in there, but it doesn’t matter). Now

U ∩ C = ∅. So {V

, ··· , V

} is a finite subcover of C.

The converse is not always true, but holds for Hausdorff spaces.

Proposition. Let

be a Hausdorff space. If

C ⊆ X

is compact, then

closed in X.

Proof. Let U = X \ C. We will show that U is open.

For any

, we will find a

such that

⊆ U

and

x ∈ U

. Then

x∈U

will be open since it is a union of open sets.

To construct

, fix

x ∈ U

. Since

is Hausdorff, for each

y ∈ C

∃U

, W

open neighbourhoods of x and y respectively with U

∩ W

= ∅.

Then

∩ C

y ∈ C}

is an open cover of

. Since

is compact, there

exists a finite subcover W

= {W

∩ C, ··· , W

∩ C}.

Let

i=1

. Then

is open since it is a finite intersection of open

sets. To show

⊆ U

, note that

i=1

⊇ C

since

∩ C}

is an

open cover. We also have W

∩ U

= ∅. So U

⊆ U . So done.

After relating compactness to closedness, we will relate it to boundedness.

First we need to define boundedness for general metric spaces.

Definition (Bounded metric space). A metric space (X, d) is bounded if there

exists M ∈ R such that d(x, y) ≤ M for all x, y ∈ X.

Example. A ⊆ R is bounded iff A ⊆ [−N, N] for some N ∈ R.

Note that being bounded is not a topological property. For example, (0

but (0

1) is bounded while

is not. It depends on the metric

, not just the

topology it induces.

Proposition. A compact metric space (X, d) is bounded.

Proof.

Pick

x ∈ X

. Then

(

) :

r ∈ R

}

is an open cover of

. Since

X is compact, there is a finite subcover {B

(x), ··· , B

(x)}.

Let

max{r

, ··· , r

}

. Then

(

x, y

)

< R

for all

y ∈ X

. So for all

y, z ∈ X,

d(y, z) ≤ d(y, x) + d(x, z) < 2R

So X is bounded.

Theorem (Heine-Borel). C ⊆ R is compact iff C is closed and bounded.

Proof. Since R is a metric space (hence Hausdorff), C is also a metric space.

So if

is compact,

is closed in

, and

is bounded, by our previous two

propositions.

Conversely, if

is closed and bounded, then

C ⊆

[

−N, N

] for some

N ∈ R

Since [

−N, N

]

1] is compact, and

C ∩

[

−N, N

] is closed in [

−N, N

is compact.

Corollary. If A ⊆ R is compact, ∃α ∈ A such that α ≥ a for all a ∈ A.

Proof.

Since

is compact, it is bounded. Let

sup A

. Then by definition,

α ≥ a for all a ∈ A. So it is enough to show that α ∈ A.

Suppose

α 6∈ A

. Then

α ∈ R \ A

. Since

is compact, it is closed in

. So

R \A

is open. So

∃ε >

0 such that

(

)

⊆ R \ A

, which implies that

a ≤ α − ε

for all

a ∈ A

. This contradicts the assumption that

sup A

. So we can

conclude α ∈ A.

We call α = max A the maximum element of A.

We have previously proved that if

is connected and

X → Y

, then

im f ⊆ Y is connected. The same statement is true for compactness.

Proposition. If

X → Y

is continuous and

is compact, then

im f ⊆ Y

also compact.

Proof.

Suppose

α ∈ T }

is an open cover of

im f

. Since

is open in

im f, we have V

= im f ∩ V

, where V

is open in Y . Then

= f

−1

) = f

−1

)

is open in

. If

x ∈ X

then

(

) is in

for some

, so

x ∈ W

. Thus

W = {W

: α ∈ T } is an open cover of X.

Since X is compact, there’s a finite subcover {W

, ··· , W

} of W.

Since V

⊆ im f , f(W

) = f(f

−1

)) = V

. So

, ··· , V

}

is a finite subcover of V.

Theorem (Maximum value theorem). If

X → R

is continuous and

compact, then ∃x ∈ X such that f(x) ≥ f(y) for all y ∈ X.

Proof.

Since

is compact,

im f

is compact. Let

max{im f}

. Then

α ∈ im f

So ∃x ∈ X with f(x) = α. Then by definition f(x) ≥ f (y) for all y ∈ X.

Corollary. If

: [0

→ R

is continuous, then

∃x ∈

1] such that

(

)

≥ f

(

)

for all y ∈ [0, 1]

Proof. [0, 1] is compact.

4.2 Products and quotients

4.2.1 Products

Recall the product topology on

X × Y

U ⊆ X × Y

is open if it is a union of

sets of the form V × W such that V ⊆ X, W ⊆ Y are open.

The major takeaway of this section is the following theorem:

Theorem. If X and Y are compact, then so is X × Y .

Proof.

First consider the special type of open cover

X × Y

such that every

U ∈ V has the form U = V × W , where V ⊆ X and W ⊆ Y are open.

For every (x, y) ∈ X × Y , there is U

∈ V with (x, y) ∈ U

. Write

= V

× W

where V

⊆ X, W

⊆ Y are open, x ∈ V

, y ∈ W

Fix

x ∈ X

. Then

y ∈ Y }

is an open cover of

. Since

compact, there is a finite subcover {W

, ··· , W

Then

i=1

is a finite intersection of open sets. So

is open in

Moreover, V

= {U

, ··· , U

} covers V

× Y .

× Y

Now

x ∈ X}

is an open cover of

. Since

is compact, there is

a finite subcover

, ··· , V

}

. Then

i=1

is a finite subset of

which covers all of X × Y .

In the general, case, suppose

is an open cover of

X × Y

. For each (

x, y

)

∈

X × Y

∃U

∈ V

with (

x, y

)

∈ U

. Since

is open,

∃V

⊆ X, W

⊆ Y

open with V

× W

⊆ U

and x ∈ V

, y ∈ W

Then

× W

: (

x, y

)

∈

(

X, Y

)

}

is an open cover of

X × Y

the type we already considered above. So it has a finite subcover

, ··· , V

×W

}

. Now

×W

⊆ U

. So

, ··· , U

}

is a finite subcover of X × Y .

Example. The unit cube [0, 1]

= [0, 1] × [0, 1] × ··· × [0, 1] is compact.

Corollary (Heine-Borel in

C ⊆ R

is compact iff

is closed and bounded.

Proof.

is bounded,

C ⊆

[

−N, N

]

for some

N ∈ R

, which is compact. The

rest of the proof is exactly the same as for n = 1.

4.2.2 Quotients

It is easy to show that the quotient of a compact space is compact, since every

open subset in the quotient space can be projected back to an open subset the

original space. Hence we can project an open cover from the quotient space to

the original space, and get a finite subcover. The details are easy to fill in.

Instead of proving the above, in this section, we will prove that compact

quotients have some nice properties. We start with a handy proposition.

Proposition. Suppose

X → Y

is a continuous bijection. If

is compact

and Y is Hausdorff, then f is a homeomorphism.

Proof.

We show that

−1

is continuous. To do this, it suffices to show (

−1

)

−1

(

)

is closed in

whenever

is closed in

. By hypothesis,

is a bijection . So

−1

)

−1

Supposed

is closed in

. Since

is compact,

is compact. Since

continuous,

(

) = (

im f|

) is compact. Since

is Hausdorff and

(

)

⊆ Y

compact, f (C) is closed.

We will apply this to quotients.

Recall that if

∼

is an equivalence relation on

X → X/∼

is continuous

iff f ◦ π : X → Y is continuous.

Corollary. Suppose

X/∼ → Y

is a bijection,

is compact,

is Hausdorff,

and f ◦ π is continuous, then f is a homeomorphism.

Proof.

Since

is compact and

X 7→ X/∼

is continuous,

im π ⊆ X/∼

compact. Since

f ◦ π

is continuous,

is continuous. So we can apply the

proposition.

Example. Let

and

⊆ X

. Then

X/A 7→ S

by (

r, θ

)

7→

(1, πr, θ) in spherical coordinates is a homeomorphism.

We can check that

is a continuous bijection and

is compact. So

X/A ' S

4.3 Sequential compactness

The other definition of compactness is sequential compactness. We will not

do much with it, but only prove that it is the same as compactness for metric

spaces.

Definition (Sequential compactness). A topological space

is sequentially

compact if every sequence (

) in

has a convergent subsequence (that converges

to a point in X!).

Example. (0

⊆ R

is not sequentially compact since no subsequence of (1

)

converges to any x ∈ (0, 1).

To show that sequential compactness is the same as compactness, we will

first need a lemma.

Lemma. Let (

) be a sequence in a metric space (

X, d

) and

x ∈ X

. Then (

)

has a subsequence converging to

iff for every

ε >

∈ B

(

) for infinitely

many n (∗).

Proof.

If (

)

→ x

, then for every

, we can find

such that

i > I

implies

∈ B

(x) by definition of convergence. So (∗) holds.

Now suppose (

∗

) holds. We will construct a sequence

→ x

inductively.

Take n

= 0. Suppose we have defined x

, ··· , x

i−1

By hypothesis,

∈ B

1/i

(

) for infinitely many

. Take

to be smallest

such n with n

> n

i−1

Then d(x

, x) <

implies that x

→ x.

Here we will only prove that compactness implies sequential compactness,

and the other direction is left as an exercise for the reader.

Theorem. If (

X, d

) is a compact metric space, then

is sequentially compact.

Proof.

Suppose

is a sequence in

with no convergent subsequence. Then

for any

y ∈ X

, there is no subsequence converging to

. By lemma, there exists

ε > 0 such that x

∈ B

(y) for only finitely many n.

Let

(

). Now

y ∈ X}

is an open cover of

. Since

compact, there is a finite subcover

, ··· , U

}

. Then

∈

i=1

for only finitely many n. This is nonsense, since x

∈ X for all n!

So x

must have a convergent subsequence.

Example. Let

1] with the topology induced

∞

(uniform norm). Let

(x) =











nx x ∈ [0, 1/n]

2 − nx x ∈ [1/n, 2/n]

0 x ∈ [2/n, 1]

Then

(

)

→

0 for all

x ∈

1]. We now claim that

has no convergent

subsequence.

Suppose

→ f

. Then

(

)

→ f

(

) for all

x ∈

1]. However, we know

that

(

)

→

0 for all

x ∈

1]. So

(

) = 0. However,

∞

(

0) = 1. So

6→ 0.

It follows that

(0)

⊆ X

is not sequentially compact. So it is not compact.

4.4 Completeness

The course ends with a small discussion about completeness. This really should

belong to the chapter on metric spaces, since this is all about metric spaces. How-

ever, we put it here because the (only) proposition we have is about compactness,

which was just defined recently.

Similar to what we did in Analysis, we can define Cauchy sequences.

Definition (Cauchy sequence). Let (X, d) be a metric space. A sequence (x

)

is Cauchy if for every

ε >

∃N

such that

(

, x

)

< ε

for all

n, m ≥ N

Example.

(i) x

k=1

1/k is not Cauchy.

(ii)

Let

= (0

⊆ R

with

. Then this is Cauchy but does not

converge.

(iii)

→ x ∈ X

, then

is Cauchy. The proof is the same as that in

Analysis I.

(iv)

Let

Q ⊆ R

. Then the sequence (2

718

, ···

) is Cauchy but

does not converge in Q.

Exactly analogous to what we did in Analysis, we can also define a complete

space.

Definition (Complete space). A metric space (

X, d

) is complete if every Cauchy

sequence in X converges to a limit in X.

Example. (0, 1) and Q are not complete.

Proposition. If X is a compact metric space, then X is complete.

Proof.

Let

be a Cauchy sequence in

. Since

is sequentially compact,

there is a convergent subsequence x

→ x. We will show that x

→ x.

Given

ε >

0, pick

such that

(

, x

)

< ε/

2 for

n, m ≥ N

. Pick

such that

≥ N

and

(

, x

)

< ε/

2 for all

i > I

. Then for

n ≥ n

d(x

, x) ≤ d(x

, x

) + d(x

, x) < ε. So x

→ x.

Corollary. R

is complete.

Proof.

If (

)

⊆ R

is Cauchy, then (

)

⊆

(0) for some

, and

(0) is

compact. So it converges.

Note that completeness is not a topological property.

R '

1) but

complete while (0, 1) is not.