5Metric spaces
IB Analysis II
5.3 Cauchy sequences and completeness
Definition (Cauchy sequence). Let (X, d) be a metric space. A sequence (x
n
)
in X is Cauchy if
(∀ε)(∃N)(∀n, m ≥ N) d(x
n
, x
m
) < ε.
Proposition. Let (X, d) be a metric space. Then
(i) Any convergent sequence is Cauchy.
(ii)
If a Cauchy sequence has a convergent subsequence, then the original
sequence converges to the same limit.
Proof.
(i) If x
k
→ x, then
d(x
m
, x
n
) ≤ d(x
m
, x) + d(x
n
, x) → 0
as m, n → ∞.
(ii)
Suppose
x
k
j
→ x
. Since (
x
k
) is Cauchy, given
ε >
0, we can choose an
N
such that
d
(
x
n
, x
m
)
<
ε
2
for all
n, m ≥ N
. We can also choose
j
0
such
that k
j
0
≥ n and d(x
k
j
0
, x) <
ε
2
. Then for any n ≥ N , we have
d(x
n
, x) ≤ d(x
n
, x
k
j
0
) + d(x, x
k
j
0
) < ε.
Definition (Complete metric space). A metric space (
X, d
) is complete if all
Cauchy sequences converge to a point in X.
Example. Let X = R
n
with the Euclidean metric. Then X is complete.
It is easy to produce incomplete metric spaces. Since arbitrary subsets of
metric spaces are subspaces, we can just remove some random elements to make
it incomplete.
Example. Let
X
= (0
,
1)
⊆ R
with the Euclidean metric. Then this is
incomplete, since
1
k
is Cauchy but has no limit in X.
Similarly,
X
=
R \ {
0
}
is incomplete. Note, however, that it is possible to
construct a metric
d
′
on
X
=
R \{
0
}
such that
d
′
induces the same topology on
X
, but makes
X
complete. This shows that completeness is not a topological
property. The actual construction is left as an exercise on the example sheet.
Example. We can create an easy example of an incomplete metric on
R
n
. We
start by defining h : R
n
→ R
n
by
h(x) =
x
1 + ∥x∥
,
where
∥ · ∥
is the Euclidean norm. We can check that this is injective: if
h(x) = h(y), taking the norm gives
∥x∥
1 + ∥x∥
=
∥y∥
1 + ∥y∥
.
So we must have ∥x∥ = ∥y∥, i.e. x = y. So h(x) = h(y) implies x = y.
Now we define
d(x, y) = ∥h(x) − h(y)∥.
It is an easy check that this is a metric on R
n
.
In fact, we can show that
h
:
R
n
→ B
1
(0), and
h
is a homeomorphism (i.e.
continuous bijection with continuous inverse) between
R
n
and the unit ball
B
1
(0), both with the Euclidean metric.
To show that this metric is incomplete, we can consider the sequence
x
k
=
(
k −
1)
e
1
, where
e
1
= (1
,
0
,
0
, ··· ,
0) is the usual basis vector. Then (
x
k
) is
Cauchy in (R
n
, d). To show this, first note that
h(x
k
) =
1 −
1
k
e
1
.
Hence we have
d(x
n
, x
m
) = ∥h(x
n
) − h(x
m
)∥ =
1
n
−
1
m
→ 0.
So it is Cauchy. To show it does not converge in (
R
n
, d
), suppose
d
(
x
k
, x
)
→
0
for some x. Then since
d(x
k
, x) = ∥h(x
k
) − h(x)∥ ≥
∥h(x
k
)∥ − ∥h(x)∥
,
We must have
∥h(x)∥ = lim
k→∞
∥h(x
k
)∥ = 1.
However, there is no element with ∥h(x)∥ = 1.
What is happening in this example, is that we are pulling in the whole
R
n
in
to the unit ball. Then under this norm, a sequence that “goes to infinity” in the
usual norm will be Cauchy in this norm, but we have nothing at infinity for it
to converge to.
Suppose we have a complete metric space (
X, d
). We know that we can
form arbitrary subspaces by taking subsets of
X
. When will this be complete?
Clearly it has to be closed, since it has to include all its limit points. It turns it
closedness is a sufficient condition.
Theorem. Let (X, d) be a metric space, Y ⊆ X any subset. Then
(i) If (Y, d|
Y ×Y
) is complete, then Y is closed in X.
(ii)
If (
X, d
) is complete, then (
Y, d|
Y ×Y
) is complete if and only if it is closed.
Proof.
(i)
Let
x ∈ X
be a limit point of
Y
. Then there is some sequence
x
k
→ x
,
where each
x
k
∈ Y
. Since (
x
k
) is convergent, it is a Cauchy sequence.
Hence it is Cauchy in
Y
. By completeness of
Y
, (
x
k
) has to converge to
some point in
Y
. By uniqueness of limits, this limit must be
x
. So
x ∈ Y
.
So Y contains all its limit points.
(ii)
We have just showed that if
Y
is complete, then it is closed. Now suppose
Y
is closed. Let (
x
k
) be a Cauchy sequence in
Y
. Then (
x
k
) is Cauchy in
X
. Since
X
is complete,
x
k
→ x
for some
x ∈ X
. Since
x
is a limit point
of Y , we must have x ∈ Y . So x
k
converges in Y .