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.