IB Metric and Topological Spaces

4Compactness

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.