4Compactness

IB Metric and Topological Spaces

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

X

is sequentially

compact if every sequence (

x

n

) in

X

has a convergent subsequence (that converges

to a point in X!).

Example. (0

,

1)

⊆ R

is not sequentially compact since no subsequence of (1

/n

)

converges to any x ∈ (0, 1).

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

first need a lemma.

Lemma. Let (

x

n

) be a sequence in a metric space (

X, d

) and

x ∈ X

. Then (

x

n

)

has a subsequence converging to

x

iff for every

ε >

0,

x

n

∈ B

ε

(

x

) for infinitely

many n (∗).

Proof.

If (

x

n

i

)

→ x

, then for every

ε

, we can find

I

such that

i > I

implies

x

n

i

∈ B

ε

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

Now suppose (

∗

) holds. We will construct a sequence

x

n

i

→ x

inductively.

Take n

0

= 0. Suppose we have defined x

n

0

, ··· , x

n

i−1

.

By hypothesis,

x

n

∈ B

1/i

(

x

) for infinitely many

n

. Take

n

i

to be smallest

such n with n

i

> n

i−1

.

Then d(x

n

i

, x) <

1

i

implies that x

n

i

→ 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

X

is sequentially compact.

Proof.

Suppose

x

n

is a sequence in

X

with no convergent subsequence. Then

for any

y ∈ X

, there is no subsequence converging to

y

. By lemma, there exists

ε > 0 such that x

n

∈ B

ε

(y) for only finitely many n.

Let

U

y

=

B

ε

(

y

). Now

V

=

{U

y

:

y ∈ X}

is an open cover of

X

. Since

X

is

compact, there is a finite subcover

{U

y

1

, ··· , U

y

m

}

. Then

x

n

∈

S

m

i=1

U

y

i

=

X

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

n

∈ X for all n!

So x

n

must have a convergent subsequence.

Example. Let

X

=

C

[0

,

1] with the topology induced

d

∞

(uniform norm). Let

f

n

(x) =

nx x ∈ [0, 1/n]

2 − nx x ∈ [1/n, 2/n]

0 x ∈ [2/n, 1]

x

y

Then

f

n

(

x

)

→

0 for all

x ∈

[0

,

1]. We now claim that

f

n

has no convergent

subsequence.

Suppose

f

n

i

→ f

. Then

f

n

i

(

x

)

→ f

(

x

) for all

x ∈

[0

,

1]. However, we know

that

f

n

i

(

x

)

→

0 for all

x ∈

[0

,

1]. So

f

(

x

) = 0. However,

d

∞

(

f

n

i

,

0) = 1. So

f

n

i

6→ 0.

It follows that

B

1

(0)

⊆ X

is not sequentially compact. So it is not compact.