5Metric spaces
IB Analysis II
5.4 Compactness
Definition ((Sequential) compactness). A metric space (
X, d
) is (sequentially)
compact if every sequence in X has a convergent subsequence.
A subset
K ⊆ X
is said to be compact if (
K, d|
K×K
) is compact. In other
words,
K
is compact if every sequence in
K
has a subsequence that converges to
some point in K.
Note that when we say every sequence has a convergent subsequence, we
do not require it to be bounded. This is unlike the statement of the Bolzano-
Weierstrass theorem. In particular, R is not compact.
It follows from definition that compactness is a topological property, since it
is defined in terms of convergence, and convergence is defined in terms of open
sets.
The following theorem relates completeness with compactness.
Theorem. All compact spaces are complete and bounded.
Note that
X
is bounded iff
X ⊆ B
r
(
x
0
) for some
r ∈ R, x
0
∈ X
(or
X
is
empty).
Proof.
Let (
X, d
) be a compact metric space. Let (
x
k
) be Cauchy in
X
. By
compactness, it has some convergent subsequence, say
x
k
j
→ x
. So
x
k
→ x
. So
it is complete.
If (
X, d
) is not bounded, by definition, for any
x
0
, there is a sequence (
x
k
)
such that
d
(
x
k
, x
0
)
> k
for every
k
. But then (
x
k
) cannot have a convergent
subsequence. Otherwise, if x
k
j
→ x, then
d(x
k
j
, x
0
) ≤ d(x
k
j
, x) + d(x, x
0
)
and is bounded, which is a contradiction.
This implies that if (
X, d
) is a metric space and
E ⊆ X
, and
E
is compact,
then
E
is bounded, i.e.
E ⊆ B
R
(
x
0
) for some
x
0
∈ X, R >
0, and
E
with the
subspace metric is complete. Hence E is closed as a subset of X.
The converse is not true. For example, recall if we have an infinite-dimensional
normed vector space, then the closed unit sphere is complete and bounded, but
not compact. Alternatively, we can take
X
=
R
with the metric
d
(
x, y
) =
min{
1
, |x − y|}
. This is clearly bounded (by 1), and it is easy to check that this
is complete. However, this is not compact since the sequence
x
k
=
k
has no
convergent subsequence.
However, we can strengthen the condition of boundedness to total bound-
edness, and get the equivalence between “completeness and total boundedness”
and compactness.
Definition (Totally bounded*). A metric space (
X, d
) is said to be totally
bounded if for all
ε >
0, there is an integer
N ∈ N
and points
x
1
, ··· , x
N
∈ X
such that
X =
N
[
i=1
B
ε
(x
i
).
It is easy to check that being totally bounded implies being bounded. We
then have the following strengthening of the previous theorem.
Theorem. (non-examinable) Let (
X, d
) be a metric space. Then
X
is compact
if and only if X is complete and totally bounded.
Proof.
(
⇐
) Let
X
be complete and totally bounded, (
y
i
)
∈ X
. For every
j ∈ N
,
there exists a finite set of points
E
j
such that every point is within
1
j
of one of
these points.
Now since
E
1
is finite, there is some
x
1
∈ E
1
such that there are infinitely
many y
i
’s in B(x
1
, 1). Pick the first y
i
in B(x
1
, 1) and call it y
i
1
.
Now there is some
x
2
∈ E
2
such that there are infinitely many
y
i
’s in
B
(
x
1
,
1)
∩ B
(
x
2
,
1
2
). Pick the one with smallest value of
i > i
1
, and call this
y
i
2
.
Continue till infinity.
This procedure gives a sequence x
i
∈ E
i
and subsequence (y
i
k
), and also
y
i
n
∈
n
\
j=1
B
x
j
,
1
j
.
It is easy to see that (
y
i
n
) is Cauchy since if
m > n
, then
d
(
y
i
m
, y
i
n
)
<
2
n
. By
completeness of X, this subsequence converges.
(
⇒
) Compactness implying completeness is proved above. Suppose
X
is not
totally bounded. We show it is not compact by constructing a sequence with no
Cauchy subsequence.
Suppose ε is such that there is no finite set of points x
1
, ··· , x
N
with
X =
N
[
i=1
B
ε
(x
i
).
We will construct our sequence iteratively.
Start by picking an arbitrary
y
1
. Pick
y
2
such that
d
(
y
1
, y
2
)
≥ ε
. This exists
or else B
ε
(y
1
) covers all of X.
Now given
y
1
, ··· , y
n
such that
d
(
y
i
, y
j
)
≥ ε
for all
i, j
= 1
, ··· , n
,
i
=
j
, we
pick
y
n+1
such that
d
(
y
n+1
, y
j
)
≥ ε
for all
j
= 1
, ··· , n
. Again, this exists, or
else
S
n
i=1
B
ε
(
y
i
) covers
X
. Then clearly the sequence (
y
n
) is not Cauchy. So
done.
In IID Linear Analysis, we will prove the Arzel`a-Ascoli theorem that charac-
terizes the compact subsets of the space
C
([
a.b
]) in a very concrete way, which
is in some sense a strengthening of this result.