1Metric spaces
IB Metric and Topological Spaces
1.4 Open and closed subsets
In this section, we will study open and closed subsets of metric spaces. We will
then proceed to prove certain key properties of open and closed subsets, and
show that they are all we need to define continuity. This will lead to the next
chapter (and the rest of the course), where we completely abandon the idea of
metrics and only talk about open and closed subsets.
To define open and closed subsets, we first need the notion of balls.
Definition (Open and closed balls). Let (
X, d
) be a metric space. For any
x ∈ X, r ∈ R,
B
r
(x) = {y ∈ X : d(y, x) < r}
is the open ball centered at x.
¯
B
r
(x) = {y ∈ X : d(y, x) ≤ r}
is the closed ball centered at x.
Example.
(i) When X = R, B
r
(x) = (x − r, x + r).
¯
B
r
(x) = [x − r, x + r].
(ii) When X = R
2
,
(a)
If
d
is the metric induced by the
k
v
k
1
=
v
1

+
v
2

, then an open ball
is a rotated square.
v
1
v
2
(b)
If
d
is the metric induced by the
k
v
k
2
=
p
v
2
1
+ v
2
2
, then an open ball
is an actual disk.
v
1
v
2
(c)
If
d
is the metric induced by the
k
v
k
∞
=
max{v
1
, v
2
}
, then an
open ball is a square.
v
1
v
2
Definition (Open subset).
U ⊆ X
is an open subset if for every
x ∈ U
,
∃δ >
0
such that B
δ
(x) ⊆ U.
C ⊆ X is a closed subset if X \ C ⊆ X is open.
As if we’ve not emphasized this enough, this is a very very important
definition.
We first prove that this is a sensible definition.
Lemma. The open ball
B
r
(
x
)
⊆ X
is an open subset, and the closed ball
¯
B
r
(x) ⊆ X is a closed subset.
Proof. Given y ∈ B
r
(x), we must find δ > 0 with B
δ
(y) ⊆ B
r
(x).
v
1
v
2
r
y
Since
y ∈ B
r
(
x
), we must have
a
=
d
(
y, x
)
< r
. Let
δ
=
r − a >
0. Then if
z ∈ B
δ
(y), then
d(z, x) ≤ d(z, y) + d(y, x) < (r − a) + a = r.
So z ∈ B
r
(x). So B
δ
(y) ⊆ B
r
(x) as desired.
The second statement is equivalent to
X \
¯
B
r
(
x
) =
{y ∈ X
:
d
(
y, x
)
> r}
is
open. The proof is very similar.
Note that openness is a property of a subset.
A ⊆ X
being open depends on
both
A
and
X
, not just
A
. For example, [0
,
1
2
) is not an open subset of
R
, but
is an open subset of [0
,
1] (since it is
B
1
2
(0)), both with the Euclidean metric.
However, we are often lazy and just say “open set” instead of “open subset”.
Example.
(i)
(0
,
1)
⊆ R
is open, while [0
,
1]
⊆ R
is closed. [0
,
1)
⊆ R
is neither closed
nor open.
(ii) Q ⊆ R
is neither open nor closed, since any open interval contains both
rational numbers and irrational numbers. So any open interval cannot be
a subset of Q or R \ Q.
(iii)
Let
X
= [
−
1
,
1]
\ {
0
}
with the Euclidean metric. Let
A
= [
−
1
,
0)
⊆ X
.
Then
A
is open since it is equal to
B
1
(
−
1).
A
is also closed since it is
equal to
¯
B
1
2
(−
1
2
).
Definition (Open neighborhood). If
x ∈ X
, an open neighborhood of
x
is an
open U ⊆ X with x ∈ U.
This is not really an interesting definition, but is simply a convenient short
hand for “open subset containing x”.
Lemma. If
U
is an open neighbourhood of
x
and
x
n
→ x
, then
∃N
such that
x
n
∈ U for all n > N .
Proof.
Since
U
is open, there exists some
δ >
0 such that
B
δ
(
x
)
⊆ U
. Since
x
n
→ x
,
∃N
such that
d
(
x
n
, x
)
< δ
for all
n > N
. This implies that
x
n
∈ B
δ
(
x
)
for all n > N. So x
n
∈ U for all n > N .
Definition (Limit point). Let
A ⊆ X
. Then
x ∈ X
is a limit point of
A
if there
is a sequence x
n
→ x such that x
n
∈ A for all n.
Intuitively, a limit point is a point we can get arbitrarily close to.
Example.
(i)
If
a ∈ A
, then
a
is a limit point of
A
, by taking the sequence
a, a, a, a, ···
.
(ii) If A = (0, 1) ⊆ R, then 0 is a limit point of A, e.g. take x
n
=
1
n
.
(iii) Every x ∈ R is a limit point of Q.
It is possible to characterize closed subsets by limit points. This is often a
convenient way of proving that sets are closed.
Proposition.
C ⊆ X
is a closed subset if and only if every limit point of
C
is
an element of C.
Proof.
(
⇒
) Suppose
C
is closed and
x
n
→ x
,
x
n
∈ C
. We have to show that
x ∈ C.
Since
C
is closed,
A
=
X \C ⊆ X
is open. Suppose the contrary that
x 6∈ C
.
Then
x ∈ A
. Hence
A
is an open neighbourhood of
x
. Then by our previous
lemma, we know that there is some
N
such that
x
n
∈ A
for all
n ≥ N
. So
x
N
∈ A
. But we know that
x
N
∈ C
by assumption. This is a contradiction. So
we must have x ∈ C.
(⇐) Suppose that C is not closed. We have to find a limit point not in C.
Since
C
is not closed,
A
is not open. So
∃x ∈ A
such that
B
δ
(
x
)
6⊆ A
for all
δ > 0. This means that B
δ
(x) ∩ C 6= ∅ for all δ > 0.
So pick
x
n
∈ B
1
n
(
x
)
∩C
for each
n >
0. Then
x
n
∈ C
,
d
(
x
n
, x
) =
1
n
→
0. So
x
n
→ x. So x is a limit point of C which is not in C.
Finally, we get to the Really Important Result
TM
that tells us metrics are
useless.
Proposition (Characterization of continuity). Let (
X, d
x
) and (
Y, d
y
) be metric
spaces, and f : X → Y . The following conditions are equivalent:
(i) f is continuous
(ii) If x
n
→ x, then f(x
n
) → f(x) (which is the definition of continuity)
(iii) For any closed subset C ⊆ Y , f
−1
(C) is closed in X.
(iv) For any open subset U ⊆ Y , f
−1
(U) is open in X.
(v)
For any
x ∈ X
and
ε >
0,
∃δ >
0 such that
f
(
B
δ
(
x
))
⊆ B
ε
(
f
(
x
)).
Alternatively, d
x
(x, z) < δ ⇒ d
y
(f(x), f(z)) < ε.
Proof.
– 1 ⇔ 2: by definition
–
2
⇒
3: Suppose
C ⊆ Y
is closed. We want to show that
f
−1
(
C
) is closed.
So let x
n
→ x, where x
n
∈ f
−1
(C).
We know that
f
(
x
n
)
→ f
(
x
) by (2) and
f
(
x
n
)
∈ C
. So
f
(
x
) is a limit
point of
C
. Since
C
is closed,
f
(
x
)
∈ C
. So
x ∈ f
−1
(
C
). So every limit
point of f
−1
(C) is in f
−1
(C). So f
−1
(C) is closed.
–
3
⇒
4: If
U ⊆ Y
is open, then
Y \ U
is closed in Y. So
f
−1
(
Y \ U
) =
X \ f
−1
(U) is closed in X. So f
−1
(U) ⊆ X is open.
–
4
⇒
5: Given
x ∈ X, ε >
0,
B
ε
(
f
(
x
)) is open in
Y
. By (4), we know
f
−1
(
B
ε
(
f
(
x
))) =
A
is open in
X
. Since
x ∈ A
,
∃δ >
0 with
B
δ
(
x
)
⊆ A
.
So
f(B
δ
(x)) ⊆ f(A) = f(f
−1
(B
ε
(f(x)))) = B
ε
(f(x))
–
5
⇒
2: Suppose
x
n
→ x
. Given
ε >
0,
∃δ >
0 such that
f
(
B
δ
(
x
))
⊆
B
ε
(
f
(
x
)). Since
x
n
→ x
,
∃N
such that
x
n
∈ B
δ
(
x
) for all
n > N
. Then
f(x
n
) ∈ f(B
δ
(x)) ⊆ B
ε
(f(x)) for all n > N . So f (x
n
) → f(x).
The third and fourth condition can allow us to immediately decide if a subset
is open or closed in some cases.
Example. Let f : R
3
→ R be defined as
f(x
1
, x
2
, x
3
) = x
2
1
+ x
4
2
x
6
3
+ x
8
1
x
2
3
.
Then this is continuous. So
{
x
∈ R
3
:
f
(x)
≤
1
}
=
f
−1
((
−∞,
1]) is closed in
R
3
.
Before we end, we prove some key properties of open subsets. These will be
used as the defining properties of open subsets in the next chapter.
Lemma.
(i) ∅ and X are open subsets of X.
(ii) Suppose V
α
⊆ X is open for all α ∈ A. Then U =
[
α∈A
V
α
is open in X.
(iii) If V
1
, ··· , V
n
⊆ X are open, then so is V =
n
\
i=1
V
i
.
Proof.
(i) ∅
satisfies the definition of an open subset vacuously.
X
is open since for
any x, B
1
(x) ⊆ X.
(ii)
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) ⊆
[
α∈A
V
α
= U. So U is open.
(iii)
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.
Note that we can take infinite unions or finite intersection, but not infinite
intersections. For example, the intersection of all (
−
1
n
,
1
n
) is
{
0
}
, which is not
open.