5Metric spaces

IB Analysis II


5.2 Topology of metric spaces
We will define open subsets of a metric space in exactly the same way as we did
for normed spaces.
Definition (Open subset). Let (
X, d
) be a metric space. A subset
U ⊆ X
is
open if for every y ∈ U, there is some r > 0 such that B
r
(y) ⊆ U.
This means we can write any open U as a union of open balls:
U =
[
y∈U
B
r(y)
(y)
for appropriate choices of r(y) for every y.
It is easy to check that every open ball
B
r
(
y
) is an open set. The proof is
exactly the same as what we had for normed spaces.
Note that two different metrics
d, d
′
on the same set
X
may give rise to the
same collection of open subsets.
Example. Lipschitz equivalent metrics give rise to the same collection of open
sets, i.e. if
d, d
′
are Lipschitz equivalent, then a subset
U ⊆ X
is open with
respect to
d
if and only if it is open with respect to
d
′
. Proof is left as an easy
exercise.
The converse, however, is not necessarily true.
Example. Let
X
=
R
,
d
(
x, y
) =
|x − y|
and
d
′
(
x, y
) =
min{
1
, |x − y|}
. It is
easy to check that these are not Lipschitz equivalent, but they induce the same
set collection of open subsets.
Definition (Topology). Let (
X, d
) be a metric space. The topology on (
X, d
) is
the collection of open subsets of
X
. We say it is the topology induced by the
metric.
Definition (Topological notion). A notion or property is said to be a topological
notion or property if it only depends on the topology, and not the metric.
We will introduce a useful terminology before we go on:
Definition (Neighbourhood). Given a metric space
X
and a point
x ∈ X
, a
neighbourhood of x is an open set containing x.
Some people do not require the set to be open. Instead, it requires a
neighbourhood to be a set that contains an open subset that contains
x
, but
this is too complicated, and we could as well work with open subsets directly.
Clearly, being a neighbourhood is a topological property.
Proposition. Let (
X, d
) be a metric space. Then
x
k
→ x
if and only if for every
neighbourhood
V
of
x
, there exists some
K
such that
x
k
∈ V
for all
k ≥ K
.
Hence convergence is a topological notion.
Proof.
(
⇒
) Suppose
x
k
→ X
, and let
V
be any neighbourhood of
x
. Since
V
is open, by definition, there exists some
ε
such that
B
ε
(
x
)
⊆ V
. By definition
of convergence, there is some
K
such that
x
k
∈ B
ε
(
x
) for
k ≥ K
. So
x
k
∈ V
whenever k ≥ K.
(
⇒
) Since every open ball is a neighbourhood, this direction follows directly
from definition.
Theorem. Let (X, d) be a metric space. Then
(i) The union of any collection of open sets is open
(ii) The intersection of finitely many open sets is open.
(iii) ∅ and X are open.
Proof.
(i)
Let
U
=
S
α
V
α
, where each
V
α
is open. If
x ∈ U
, then
x ∈ V
α
for
some
α
. Since
V
α
is open, there exists
δ >
0 such that
B
δ
(
x
)
⊆ V
α
. So
B
δ
(x) ⊆
S
α
V
α
= U. So U is open.
(ii)
Let
U
=
T
n
i=1
V
α
, where each
V
α
is open. If
x ∈ V
, then
x ∈ V
i
for all
i
= 1
, ··· , n
. So
∃δ
i
>
0 with
B
δ
i
(
x
)
⊆ V
i
. Take
δ
=
min{δ
1
, ··· , δ
n
}
. So
B
δ
(x) ⊆ V
i
for all i. So B
δ
(x) ⊆ V . So V is open.
(iii) ∅
satisfies the definition of an open subset vacuously.
X
is open since for
any x, B
1
(x) ⊆ X.
This theorem is not important in this course. However, this will be a key
defining property we will use when we define topological spaces in IB Metric
and Topological Spaces.
We can now define closed subsets and characterize them using open subsets,
in exactly the same way as for normed spaces.
Definition (Limit point). Let (
X, d
) be a metric space and
E ⊆ X
. A point
y ∈ X
is a limit point of
E
if there exists a sequence
x
k
∈ E
,
x
k
=
y
such that
x
k
→ y.
Definition (Closed subset). A subset
E ⊆ X
is closed if
E
contains all its limit
points.
Proposition. A subset is closed if and only if its complement is open.
Proof.
Exactly the same as that of normed spaces. It is useful to observe that
y ∈ X
is a limit point of
E
if and only if (
B
r
(
y
)
\{y}
)
∩E
=
∅
for all
r >
0.
We can write down an analogous theorem for closed sets:
Theorem. Let (X, d) be a metric space. Then
(i) The intersection of any collection of closed sets is closed
(ii) The union of finitely many closed sets is closed.
(iii) ∅ and X are closed.
Proof. By taking complements of the result for open subsets.
Proposition. Let (
X, d
) be a metric space and
x ∈ X
. Then the singleton
{x}
is a closed subset, and hence any finite subset is closed.
Proof.
Let
y ∈ X \{x}
. So
d
(
x, y
)
>
0. Then
B
d(y,x)
(
x
)
⊆ X \{x}
. So
X \{x}
is open. So {x} is closed.
Alternatively, since
{x}
has no limit points, it contains all its limit points.
So it is closed.