2Topological spaces
IB Metric and Topological Spaces
2.4 Closure and interior
2.4.1 Closure
Given a subset
A ⊆ X
, if
A
is not closed, we would like to find the smallest
closed subset containing A. This is known as the closure of A.
Officially, we define the closure as follows:
Definition. Let X be a topological space and A ⊆ X. Define
C
A
= {C ⊆ X : A ⊆ C and C is closed in X}
Then the closure of A in X is
¯
A =
\
C∈C
A
C.
First we do a sanity check: since
¯
A
is defined as an intersection, we should
make sure we are not taking an intersection of no sets. This is easy: since
X
is closed in
X
(its complement
∅
is open),
C
A
6
=
∅
. So we can safely take the
intersection.
Since
¯
A
is an intersection of closed sets, it is closed in
X
. Also, if
C ∈ C
A
,
then A ⊆ C. So A ⊆
T
C∈C
A
C =
¯
A. In fact, we have
Proposition.
¯
A is the smallest closed subset of X which contains A.
Proof.
Let
K ⊆ X
be a closed set containing
A
. Then
K ∈ C
A
. So
¯
A
=
T
C∈C
A
C ⊆ K. So
¯
A ⊆ K.
We basically defined the closure such that it is the smallest closed subset of
X which contains A.
However, while this “clever” definition makes it easy to prove the above
property, it is rather difficult to directly use it to compute the closure.
To compute the closure, we define the limit point analogous to what we did
for metric spaces.
Definition (Limit point). A limit point of
A
is an
x ∈ X
such that there is a
sequence x
n
→ x with x
n
∈ A for all n.
In general, limit points are easier to compute, and can be a useful tool for
determining the closure of A.
Now let
L(A) = {x ∈ X : x is a limit point of A}.
We immediately get the following lemma.
Lemma. If C ⊆ X is closed, then L(C) = C.
Proof.
Exactly the same as that for metric spaces. We will also prove a more
general result very soon that implies this.
Recall that we proved the converse of this statement for metric spaces.
However, the converse is not true for topological spaces in general.
Example. Let
X
be an uncountable set (e.g.
R
), and define a topology on
X
by saying a set is open if it is empty or has countable complement. One can
check that this indeed defines a topology. We claim that the only sequences that
converge are those that are eventually constant.
Indeed, if x
n
is a sequence and x ∈ X, then consider the open set
U = (X \ {x
n
: n ∈ N}) ∪ {x}.
Then the only element in the sequence
x
n
that can possibly be contained in
U
is x itself. So if x
n
→ x, this implies that x
n
is eventually always x.
In particular, it follows that L(A) = A for all A ⊆ X.
However, we do have the following result:
Proposition. L(A) ⊆
¯
A.
Proof.
If
A ⊆ C
, then
L
(
A
)
⊆ L
(
C
). If
C
is closed, then
L
(
C
) =
C
. So
C ∈ C
A
⇒ L(A) ⊆ C. So L(A) ⊆
T
C∈C
A
C =
¯
A.
This in particular implies the previous lemma, since for any
A
, we have
A ⊆ L(A) ⊆
¯
A, and when A is closed, we have A =
¯
A.
Finally, we have the following corollary that can help us find the closure of
subsets:
Corollary. Given a subset
A ⊆ X
, if we can find some closed
C
such that
A ⊆ C ⊆ L(A), then we in fact have C =
¯
A.
Proof. C ⊆ L
(
A
)
⊆
¯
A ⊆ C
, where the last step is since
¯
A
is the smallest closed
set containing A. So C = L(A) =
¯
A.
Example.
– Let (a, b) ⊆ R. Then (a, b) = [a, b].
– Let Q ⊆ R. Then
¯
Q = R.
– R \ Q = R.
–
In
R
n
with the Euclidean metric,
B
r
(x)
=
¯
B
r
(
x
). In general,
B
r
(x) ⊆
¯
B
r
(
x
), since
¯
B
r
(
x
) is closed and
B
r
(
x
)
⊆
¯
B
r
(
x
), but these need not be
equal.
For example, if
X
has the discrete metric, then
B
1
(
x
) =
{x}
. Then
B
1
(x) = {x}, but
¯
B
1
(x) = X.
In the above example, we had
¯
Q
=
R
. In some sense, all points of
R
are
“surrounded” by points in Q. We say that Q is dense in R.
Definition (Dense subset). A ⊆ X is dense in X if
¯
A = X.
Example. Q and R \ Q are both dense in R with the usual topology.
2.4.2 Interior
We defined the closure of
A
to be the smallest closed subset containing
A
. We
can similarly define the interior of
A
to be the largest open subset contained in
A.
Definition (Interior). Let A ⊆ X, and let
O
A
= {U ⊆ X : U ⊆ A, U is open in X}.
The interior of A is
Int(A) =
[
U∈O
A
U.
Proposition. Int(A) is the largest open subset of X contained in A.
The proof is similar to proof for closure.
To find the closure, we could use limit points. What trick do we have to find
the interior?
Proposition. X \ Int(A) = X \ A
Proof. U ⊆ A ⇔
(
X \U
)
⊇
(
X \A
). Also,
U open in X ⇔ X \U is closed in X
.
So the complement of the largest open subset of
X
contained in
A
will be
the smallest closed subset containing X \ A.
Example. Int(Q) = Int(R \Q) = ∅.