Metric Space: Closed Sets
Closure
Definition: The point
is called point of closure of a set
if there exists a sequence
, such that 
An equivalent definition using balls: The point
is called point of closure of a set
if for every open ball
, we have
.
The proof is left as an exercise.
The proof is left as an exercise.
Intuitively, a point of closure is arbitrarily "close" to the set
. It is so close, that we can find a sequence in the set that converges to any point of closure of the set.
Example: Let A be the segment
, The point
is not in
, but it is a point of closure: Let
.
(
, and therefore
) and
(that's because
).
Definition: The closure of a set
, is the set of all points of closure. The closure of a set A is marked
or
.
Note that
. a quick proof: For every
, Let
.
Examples
For the metric space
(the line), and let
we have:
Closed set
Definition: A set
is closed in
if
.
Meaning: A set is closed, if it contains all its point of closure.
Meaning: A set is closed, if it contains all its point of closure.
An equivalent definition is: A set
The proof of this definition comes directly from the former definition and the definition of convergence.
Properties
Some basic properties of Cl (For any sets
):
is closed iff
- While the above implies that the union of finitely many closed sets is also a closed set, the same does not necessarily hold true for the union of infinitely many closed sets. To see an example on the real line, let
. We see that
fails to contain its points of closure,
This union can therefore not be a closed subset of the real numbers.
The proofs are left to the reader as exercises. Hint for number 5: recall that
.
Open vs Closed
That is, an open set approaches its boundary but does not include it; whereas a closed set includes every point it approaches. These two properties may seem mutually exclusive, but they are not:
- In any metric space
, the set
is both open and closed.
- In any space with a discrete metric, every set is both open and closed.
- In
, under the regular metric, the only sets that are both open and closed are
and
. However, some sets are neither open nor closed. For example, a half-open range like
is neither open nor closed. As another example, the set of rationals is not open because an open ball around a rational number contains irrationals; and it is not closed because there are sequences of rational numbers that converge to irrational numbers (such as the various infinite series that converge to
).
Complementary set
A Reminder/Definition: Let
be a set in the space
. We define the complement of
,
to be
.
A Quick example: let
. Then
.
The plot continues...
A very important Proposition: Let
be a set in the space
. Then, A is open iff
is closed.
Proof: (
) For the first part, we assume that A is an open set. We shall show that
. It is enough to show that
because of the properties of closure. Let
(we will show that
).
for every ball
we have, by definition that (*)
. If the point is not in
then
.
is open and therefore, there is a ball
, such that:
, that means that
, contradicting (*).
(
) On the other hand, Lets a assume that
is closed, and show that
is open. Let
be a point in
(we will show that
). If
is not in
then for every ball
we have that
. That means that
. And by definition of closure point
is a closure point of
so we can say that
.
is closed, and therefore
That contradicts the assumption that 
Proof: (
for every ball
(
Note that, as mentioned earlier, a set can still be both open and closed!
On 
The following is an important theorem characterizing open and closed sets on
.
Theorem: An open set O in R1 is a countable union of disjoint open intervals.
Proof: Let
. Let
and let
. There exists an open ball (x-ε,x+ε) such that (x-ε,x+ε) ⊆ O because O is open. Thus, a≤x-ε and b≥x+ε. Thus, x ∈(a,b). The set O contains all elements of (a,b) since if a number is greater than a, and less than x but is not within O, then a would not be the supremum of {t|t∉O, t<x}. Similarly, if there is a number is less than b and greater than x, but is not within O, then b would not be the infimum of {t|t∉O, t>x}. Thus, O also contains (a,x) and (x,b) and so O contains (a,b). If y≠x and y∈(a,b), then the interval constructed from this element as above would be the same. If y<a, then inf{t|t∉O, t>y} would also be less than a because there is a number between y and a which is not within O. Similarly if y>b, then sup{t|t∉O, t<y} would also be greater than b because there is a number between y and b which is not within O. Thus, all possible open intervals constructed from the above process are disjoint. The union of all such open intervals constructed from an element x is thus O, and so O is a union of disjoint open intervals. Because the rational numbers is dense in R, there is a rational number within each open interval, and since the rational numbers is countable, the open intervals themselves are also countable.
Theorem: An open set O in R1 is a countable union of disjoint open intervals.
Proof: Let
Examples of closed sets
- In any metric space, a singleton
is closed. To see why, consider the open set,
. Let
. Then
, so
. Let
. Then
. So
is open, and hence
is closed.
- In any metric space, every finite set
is closed. To see why, observe that
is open, so
is closed.
- Closed intervals [a,b] are closed.
- Cantor Set Consider the interval [0,1] and call it C0. Let A1 be equal {0,
} and let dn =
. Let An+1 be equal to the set An∪{x|x=a+2dn, a∈An}. Let Cn be
{[a,a+dn]}, which is the finite union of closed sets, and is thus closed. Then the intersection
is called the Cantor set and is closed.