4ℝn as a normed space
IB Analysis II
4.2 Cauchy sequences and completeness
Definition (Cauchy sequence). Let (
V, ∥ · ∥
) be a normed space. A sequence
(x
(k)
) in V is a Cauchy sequence if
(∀ε)(∃N)(∀n, m ≥ N) ∥x
(n)
− x
(m)
∥ < ε.
Definition (Complete normed space). A normed space (
V, ∥ · ∥
) is complete if
every Cauchy sequence converges to an element in V .
We’ll start with some easy facts about Cauchy sequences and complete spaces.
Proposition. Any convergent sequence is Cauchy.
Proof. If x
k
→ x, then
∥x
k
− x
ℓ
∥ ≤ ∥x
k
− x∥ + ∥x
ℓ
− x∥ → 0 as k, ℓ → ∞.
Proposition. A Cauchy sequence is bounded.
Proof.
By definition, there is some
N
such that for all
n ≥ N
, we have
∥
x
N
−
x
n
∥ < 1. So ∥x
n
∥ < 1 + ∥x
N
∥ for n ≥ N. So, for all n,
∥x
n
∥ ≤ max{∥x
1
∥, ··· , ∥x
N−1
∥, 1 + ∥x
N
∥}.
Proposition. If a Cauchy sequence has a subsequence converging to an element
x, then the whole sequence converges to x.
Proof.
Suppose x
k
j
→
x. Since (x
k
) is Cauchy, given
ε >
0, we can choose an
N
such that
∥
x
n
−
x
m
∥ <
ε
2
for all
n, m ≥ N
. We can also choose
j
0
such that
k
j
0
≥ n and ∥x
k
j
0
− x∥ <
ε
2
. Then for any n ≥ N, we have
∥x
n
− x∥ ≤ ∥x
n
− x
k
j
0
∥ + ∥x − x
k
j
0
∥ < ε.
Proposition. If
∥ · ∥
′
is Lipschitz equivalent to
∥ · ∥
on
V
, then (x
k
) is Cauchy
with respect to
∥ · ∥
if and only if (x
k
) is Cauchy with respect to
∥ · ∥
′
. Also,
(V, ∥ · ∥) is complete if and only if (V, ∥ · ∥
′
) is complete.
Proof. This follows directly from definition.
Theorem. R
n
(with the Euclidean norm, say) is complete.
Proof.
The important thing is to know this is true for
n
= 1, which we have
proved from Analysis I.
If (x
k
) is Cauchy in
R
n
, then (
x
(k)
j
) is a Cauchy sequence of real numbers for
each
j ∈ {
1
, ··· , n}
. By the completeness of the reals, we know that
x
k
j
→ x
j
∈ R
for some
x
. So
x
k
→ x
= (
x
1
, ··· , x
n
) since convergence in
R
n
is equivalent to
componentwise convergence.
Note that the spaces
ℓ
1
, ℓ
2
, ℓ
∞
are all complete with respect to the standard
norms. Also,
C
([0
,
1]) is complete with respect to
∥ · ∥
∞
, since uniform Cauchy
convergence implies uniform convergence, and the uniform limit of continuous
functions is continuous. However,
C
([0
,
1]) with the
L
1
or
L
2
norms are not
complete (see example sheet).
The incompleteness of
L
1
tells us that
C
([0
,
1]) is not large enough to to be
complete under the
L
1
or
L
2
norm. In fact, the space of Riemann integrable
functions, say
R
([0
,
1]), is the natural space for the
L
1
norm, and of course
contains
C
([0
,
1]). As we have previously mentioned, this time
R
([0
,
1]) is too
large for
∥ · ∥
to be a norm, since
R
1
0
|f|
d
x
= 0 does not imply
f
= 0. This is a
problem we can solve. We just have to take the equivalence classes of Riemann
integrable functions, where
f
and
g
are equivalent if
R
1
0
|f − g|
d
x
= 0. But still,
L
1
is not complete on
R
([0
,
1])
/∼
. This is a serious problem in the Riemann
integral. This eventually lead to the Lebesgue integral, which generalizes the
Riemann integral, and gives a complete normed space.
Note that when we quotient our
R
([0
,
1]) by the equivalence relation
f ∼ g
if
R
1
0
|f − g|
d
x
= 0, we are not losing too much information about our functions.
We know that for the integral to be zero,
f − g
cannot be non-zero at a point of
continuity. Hence they agree on all points of continuities. We also know that
by Lebesgue’s theorem, the set of points of discontinuity has Lebesgue measure
zero. So they disagree on at most a set of Lebesgue measure zero.
Example. Let
V = {(x
n
) ∈ R
N
: x
j
= 0 for all but finitely many j}.
Take the supremum norm
∥ · ∥
∞
on
V
. This is a subspace of
ℓ
∞
(and is
sometimes denoted
ℓ
0
). Then (
V, ∥ · ∥
∞
) is not complete. We define
x
(k)
=
(1,
1
2
,
1
3
, ··· ,
1
k
, 0, 0, ···) for k = 1, 2, 3, ···. Then this is Cauchy, since
∥x
(k)
− x
(ℓ)
∥ =
1
min{ℓ, k} + 1
→ 0,
but it is not convergent in
V
. If it actually converged to some
x
, then
x
(k)
j
→ x
j
.
So we must have x
j
=
1
j
, but this sequence not in V .
We will later show that this is because
V
is not closed, after we define what
it means to be closed.
Definition (Open set). Let (
V, ∥ · ∥
) be a normed space. A subspace
E ⊆ V
is
open in V if for any y ∈ E, there is some r > 0 such that
B
r
(y) = {x ∈ V : ∥x − y∥ < r} ⊆ E.
We first check that the open ball is open.
Proposition. B
r
(y) ⊆ V is an open subset for all r > 0, y ∈ V .
Proof. Let x ∈ B
r
(y). Let ρ = r − ∥x − y∥ > 0. Then B
ρ
(x) ⊆ B
r
(y).
x
y
Definition (Limit point). Let (
V, ∥ · ∥
) be a normed space,
E ⊆ V
. A point
y
∈ V
is a limit point of
E
if there is a sequence (x
k
) in
E
with x
k
= y for all
k
and x
k
→ y.
(Some people allow x
k
= y, but we will use this definition in this course)
Example. Let
V
=
R
,
E
= (0
,
1). Then 0
,
1 are limit points of
E
. The set of
all limit points is [0, 1].
If E
′
= (0, 1) ∪ {2}. Then the set of limit points of E
′
is still [0, 1].
There is a nice result characterizing whether a set contains all its limit points.
Proposition. Let
E ⊆ V
. Then
E
contains all of its limit points if and only if
V \ E is open in V .
Using this proposition, we define the following:
Definition (Closed set). Let (
V, ∥ · ∥
) be a normed space. Then
E ⊆ V
is
closed if V \ E is open, i.e. E contains all its limit points.
Note that sets can be both closed or open; or neither closed nor open.
Before we prove the proposition, we first have a lemma:
Lemma. Let (
V, ∥ · ∥
) be a normed space,
E
any subset of
V
. Then a point
y ∈ V is a limit point of E if and only if
(B
r
(y) \ {y}) ∩ E = ∅
for every r.
Proof.
(
⇒
) If y is a limit point of
E
, then there exists a sequence (x
k
)
∈ E
with x
k
= y for all
k
and x
k
→
y. Then for every
r
, for sufficiently large
k
,
x
k
∈ B
r
(y). Since x
k
= {y} and x
k
∈ E, the result follows.
(
⇐
) For each
k
, let
r
=
1
k
. By assumption, we have some x
k
∈
(
B
1
k
(y)
\
{y}) ∩ E. Then x
k
→ y, x
k
= y and x
k
∈ E. So y is a limit point of E.
Now we can prove our proposition.
Proposition. Let
E ⊆ V
. Then
E
contains all of its limit points if and only if
V \ E is open in V .
Proof.
(
⇒
) Suppose
E
contains all its limit points. To show
V \ E
is open,
we let y
∈ V \ E
. So y is not a limit point of
E
. So for some
r
, we have
(B
r
(y) \ {y}) ∩ E = ∅. Hence it follows that B
r
(y) ⊆ V \ E (since y ∈ E).
(
⇐
) Suppose
V \ E
is open. Let y
∈ V \ E
. Since
V \ E
is open, there is
some
r
such that
B
r
(y)
⊆ V \E
. By the lemma, y is not a limit point of
E
. So
all limit points of E are in E.