4Compactness

IB Metric and Topological Spaces

4.1 Compactness

Definition (Open cover). Let

U ⊆ P

(

X

) be a topology on

X

. An open cover of

X is a subset V ⊆ U such that

[

V ∈V

V = X.

We say V covers X.

If V

0

⊆ V, and V

0

covers X, then we say V

0

is a subcover of V.

Definition (Compact space). A topological space

X

is compact if every open

cover V of X has a finite subcover V

0

= {V

1

, ··· , V

n

} ⊆ 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)

If

X

is finite, then

P

(

X

) is finite. So any open cover of

X

is finite. So

X

is compact.

(ii)

Let

X

=

R

and

V

=

{

(

−R, R

) :

R ∈ R, R >

0

}

. Then this is an open cover

with no finite subcover. So

R

is not compact. Hence all open intervals are

not compact since they are homeomorphic to R.

(iii) Let X = [0, 1] ∩ Q. Let

U

n

= X \ (α − 1/n, α + 1/n).

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

√

2).

Then

S

n>0

U

n

=

X

since

α

is irrational. Then

V

=

{U

n

:

n ∈ Z >

0

}

is

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

,

1]

∩ 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

A

is non-empty. Since

V

covers [0

,

1], in particular, there is some

V

0

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

0

. 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

V

covers

X

, let

α ∈ V

α

. Since

V

α

is open, there is some

ε

such that

B

ε

(

α

)

⊆ V

α

. By definition of

α

, we must have

α − ε/

2

∈ A

. So [0

, α − ε/

2]

has a finite subcover. Add

V

α

to that subcover to get a finite subcover of

[0

, 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

1

∈ V

such that 1

∈ V

1

and

∃ε >

0 with

(1

− ε,

1]

⊆ V

1

. Since 1

− ε ∈ A

, there exists a finite

V

0

⊆ V

which covers

[0, 1 − ε/2]. Then W = V

0

∪ {V

1

} 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

X

is compact and

C

is a closed subset of

X

, then

C

is also

compact.

Proof.

To prove this, given an open cover of

C

, we need to find a finite subcover.

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

X

. We can do so by

adding

X \C

, which is open since

C

is closed. Then since

X

is compact, we can

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

C.

Formally, suppose

V

is an open cover of

C

. Say

V

=

{V

α

:

α ∈ T }

. For

each

α

, since

V

α

is open in

C

,

V

α

=

C ∩ V

0

α

for some

V

0

α

open in

X

. Also, since

S

α∈T

V

a

= C, we have

S

α∈T

V

0

α

⊇ C.

Since

C

is closed,

U

=

X \ C

is open in

X

. So

W

=

{V

0

α

:

α ∈ T } ∪ {U}

is an open cover of

X

. Since

X

is compact,

W

has a finite subcover

W

0

=

{V

0

α

1

, ··· , V

0

α

n

, U }

(

U

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

U ∩ C = ∅. So {V

α

1

, ··· , V

α

n

} is a finite subcover of C.

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

Proposition. Let

X

be a Hausdorff space. If

C ⊆ X

is compact, then

C

is

closed in X.

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

For any

x

, we will find a

U

x

such that

U

x

⊆ U

and

x ∈ U

x

. Then

U

=

S

x∈U

U

x

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

To construct

U

x

, fix

x ∈ U

. Since

X

is Hausdorff, for each

y ∈ C

,

∃U

xy

, W

xy

open neighbourhoods of x and y respectively with U

xy

∩ W

xy

= ∅.

W

xy

U

xy

C

x

y

Then

W

=

{W

xy

∩ C

:

y ∈ C}

is an open cover of

C

. Since

C

is compact, there

exists a finite subcover W

0

= {W

xy

1

∩ C, ··· , W

xy

n

∩ C}.

Let

U

x

=

T

n

i=1

U

xy

i

. Then

U

x

is open since it is a finite intersection of open

sets. To show

U

x

⊆ U

, note that

W

x

=

S

n

i=1

W

xy

i

⊇ C

since

{W

xy

i

∩ C}

is an

open cover. We also have W

x

∩ U

x

= ∅. So U

x

⊆ U . So done.

W

x

U

x

C

x

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

,

1)

'

R

but (0

,

1) is bounded while

R

is not. It depends on the metric

d

, not just the

topology it induces.

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

Proof.

Pick

x ∈ X

. Then

V

=

{B

r

(

x

) :

r ∈ R

+

}

is an open cover of

X

. Since

X is compact, there is a finite subcover {B

r

1

(x), ··· , B

r

n

(x)}.

Let

R

=

max{r

1

, ··· , r

n

}

. Then

d

(

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

C

is compact,

C

is closed in

R

, and

C

is bounded, by our previous two

propositions.

Conversely, if

C

is closed and bounded, then

C ⊆

[

−N, N

] for some

N ∈ R

.

Since [

−N, N

]

'

[0

,

1] is compact, and

C

=

C ∩

[

−N, N

] is closed in [

−N, N

],

C

is compact.

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

Proof.

Since

A

is compact, it is bounded. Let

α

=

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

A

is compact, it is closed in

R

. So

R \A

is open. So

∃ε >

0 such that

B

ε

(

α

)

⊆ R \A

, which implies that

a ≤ α −ε

for all

a ∈ A

. This contradicts the assumption that

α

=

sup A

. So we can

conclude α ∈ A.

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

We have previously proved that if

X

is connected and

f

:

X → Y

, then

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

Proposition. If

f

:

X → Y

is continuous and

X

is compact, then

im f ⊆ Y

is

also compact.

Proof.

Suppose

V

=

{V

α

:

α ∈ T }

is an open cover of

im f

. Since

V

α

is open in

im f , we have V

α

= im f ∩V

0

α

, where V

0

α

is open in Y . Then

W

α

= f

−1

(V

α

) = f

−1

(V

0

α

)

is open in

X

. If

x ∈ X

then

f

(

x

) is in

V

α

for some

α

, so

x ∈ W

α

. Thus

W = {W

α

: α ∈ T } is an open cover of X.

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

α

1

, ··· , W

α

n

} of W.

Since V

α

⊆ im f, f(W

α

) = f(f

−1

(V

α

)) = V

α

. So

{V

α

1

, ··· , V

α

n

}

is a finite subcover of V.

Theorem (Maximum value theorem). If

f

:

X → R

is continuous and

X

is

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

Proof.

Since

X

is compact,

im f

is compact. Let

α

=

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

f

: [0

,

1]

→ R

is continuous, then

∃x ∈

[0

,

1] such that

f

(

x

)

≥ f

(

y

)

for all y ∈ [0, 1]

Proof. [0, 1] is compact.