Education: Closed Sets.

Metric Space: Closed Sets

Closure

Definition: The point

p

is called point of closure of a set

A

if there exists a sequence

a_n, \forall n, a_n \in A

, such that

a_n \rightarrow p

An equivalent definition using balls: The point

p

is called point of closure of a set

A

if for every open ball

B, p \in B

, we have

B \cap A \neq \emptyset

.
The proof is left as an exercise.

Intuitively, a point of closure is arbitrarily "close" to the set

A

. 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

[0,1) \in \mathbb{R}

, The point

p = 1

is not in

A

, but it is a point of closure: Let

a_n = 1 - \frac{1}{n}

a_n \in A

(

n > 0

, and therefore

a_n = 1 - \frac{1}{n} < 1

) and

a_n \rightarrow 1

(that's because

\frac{1}{n} \rightarrow 0

Definition: The closure of a set

A \subseteq X

({X},d)

, is the set of all points of closure. The closure of a set A is marked

\bar{A}

Cl(A)

Note that

A \subseteq \bar{A}

. a quick proof: For every

x \in A

, Let

(a_n = x)\forall n

Examples

For the metric space

\mathbb{R}

(the line), and let

a,b \in \mathbb{R}

we have:

$Cl([a,b]) = [a,b]$
$Cl((a,b]) = [a,b]$
$Cl([a,b)) = [a,b]$
$Cl((a,b)) = [a,b]$

Closed set

Definition: A set

A \subseteq X

is closed in

{X}\,

A = Cl(A)

.
Meaning: A set is closed, if it contains all its point of closure.

An equivalent definition is: A set

A \subseteq X

is closed in

{X}\,

If for every point

p \in A

, and for ever Ball

B, p \in B

, then

B \cap A \neq \emptyset

.
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

A,B

$A \subseteq Cl(A)$
$Cl(Cl(A))=Cl(A)$
$Cl(A \cup B) = Cl(A)\cup Cl(B)$
$A$ is closed iff $A = Cl(A)$
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 $A_n=\{[-1+\frac{1}{n},1-\frac{1}{n}]\}$ . We see that $\cup_{i=1}^\infty A_i=(-1,1)$ fails to contain its points of closure, $\pm 1.$

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

Cl(A)=\cap \{A\subseteq S|S\text{ is closed }\!\!\}\!\!\text{ }

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 $(X, d)$ , the set $X$ is both open and closed.

In any space with a discrete metric, every set is both open and closed.

In $\R$ , under the regular metric, the only sets that are both open and closed are $\R$ and $\emptyset$ . However, some sets are neither open nor closed. For example, a half-open range like $[0, 1)$ 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 $\pi$ ).

Complementary set

A Reminder/Definition: Let

A

be a set in the space

X

. We define the complement of

A

A^c

to be

X \setminus A

A Quick example: let

X = [0,1]; A = [0,\frac{1}{2}]

. Then

A^c = (\frac{1}{2},1]

The plot continues...

A very important Proposition: Let

A

be a set in the space

(X,d)

. Then, A is open iff

A^c

is closed.
Proof: (

\Rightarrow

) For the first part, we assume that A is an open set. We shall show that

A^c = Cl(A^c)

. It is enough to show that

Cl(A^c) \subseteq A^c

because of the properties of closure. Let

p \in Cl(A^c)

(we will show that

p \in A^c

).
for every ball

B, p\in B

we have, by definition that (*)

B \cap A^c \neq \emptyset

. If the point is not in

A^c

then

p \in A

A

is open and therefore, there is a ball

B

, such that:

p \in B \subseteq A

, that means that

B \cap A^c = \emptyset

, contradicting (*).
(

\Leftarrow

) On the other hand, Lets a assume that

A^c

is closed, and show that

A

is open. Let

p

be a point in

A

(we will show that

p \in int(A)

). If

p

is not in

int(A)

then for every ball

B, p \in B

we have that

B \nsubseteq A

. That means that

B \cap A^c \neq \emptyset

. And by definition of closure point

p

is a closure point of

A^c

so we can say that

p \in Cl(A^c)

A^c

is closed, and therefore

p \in A^c = Cl(A^c)

That contradicts the assumption that

p \in A

Note that, as mentioned earlier, a set can still be both open and closed!

On $\mathbb{R}^1$

The following is an important theorem characterizing open and closed sets on

\mathbb{R}^1

.
Theorem: An open set O in R¹ is a countable union of disjoint open intervals.
Proof: Let

x\in O

. Let

a=\sup\{t|t\notin O, t<x\}

and let

b=\inf\{t|t\notin O, t>x\}

. 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.

Examples of closed sets

In any metric space, a singleton $\{x\}$ is closed. To see why, consider the open set, $\{x\}^c$ . Let $y \in \{x\}^c$ . Then $y \neq x$ , so $d(y,x) > 0$ . Let $\epsilon = \frac{1}{2}d(y,x)$ . Then $B_\epsilon(y) \subseteq \{x\}^c$ . So $\{x\}^c$ is open, and hence $\{x\}$ is closed.
In any metric space, every finite set $T = \{x_1,x_2,...,x_n\}$ is closed. To see why, observe that $T^c = \Big[\bigcup\{x_i\}\Big]^c = \bigcap\{x_i\}^c$ is open, so $T$ is closed.
Closed intervals [a,b] are closed.
Cantor Set Consider the interval [0,1] and call it C₀. Let A₁ be equal {0, $\tfrac{2}{3}$ } and let d_n = $(\tfrac{1}{3})^{n}$ . Let A_n+1 be equal to the set A_n∪{x|x=a+2d_n, a∈A_n}. Let C_n be $\textstyle \bigcup_{a\in A_n}$ {[a,a+d_n]}, which is the finite union of closed sets, and is thus closed. Then the intersection $\textstyle \bigcap_{i=1}^\infty {C_i}$ is called the Cantor set and is closed.

24/10/2014

Closed Sets.