4Compactness
IB Metric and Topological Spaces
4.4 Completeness
The course ends with a small discussion about completeness. This really should
belong to the chapter on metric spaces, since this is all about metric spaces. How-
ever, we put it here because the (only) proposition we have is about compactness,
which was just defined recently.
Similar to what we did in Analysis, we can define Cauchy sequences.
Definition (Cauchy sequence). Let (X, d) be a metric space. A sequence (x
n
)
in
X
is Cauchy if for every
ε >
0,
∃N
such that
d
(
x
n
, x
m
)
< ε
for all
n, m ≥ N
.
Example.
(i) x
n
=
P
n
k=1
1/k is not Cauchy.
(ii)
Let
X
= (0
,
1)
⊆ R
with
x
n
=
1
n
. Then this is Cauchy but does not
converge.
(iii)
If
x
n
→ x ∈ X
, then
x
n
is Cauchy. The proof is the same as that in
Analysis I.
(iv)
Let
X
=
Q ⊆ R
. Then the sequence (2
,
2
.
7
,
2
.
71
,
2
.
718
, ···
) is Cauchy but
does not converge in Q.
Exactly analogous to what we did in Analysis, we can also define a complete
space.
Definition (Complete space). A metric space (
X, d
) is complete if every Cauchy
sequence in X converges to a limit in X.
Example. (0, 1) and Q are not complete.
Proposition. If X is a compact metric space, then X is complete.
Proof.
Let
x
n
be a Cauchy sequence in
X
. Since
X
is sequentially compact,
there is a convergent subsequence x
n
i
→ x. We will show that x
n
→ x.
Given
ε >
0, pick
N
such that
d
(
x
n
, x
m
)
< ε/
2 for
n, m ≥ N
. Pick
I
such that
n
I
≥ N
and
d
(
x
n
i
, x
)
< ε/
2 for all
i > I
. Then for
n ≥ n
I
,
d(x
n
, x) ≤ d(x
n
, x
n
I
) + d(x
n
I
, x) < ε. So x
n
→ x.
Corollary. R
n
is complete.
Proof.
If (
x
n
)
⊆ R
n
is Cauchy, then (
x
n
)
⊆
¯
B
R
(0) for some
R
, and
¯
B
R
(0) is
compact. So it converges.
Note that completeness is not a topological property.
R '
(0
,
1) but
R
is
complete while (0, 1) is not.