24/10/2014

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} or 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}\, if 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 AA^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 AA 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

  1. 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.
  2. 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.
  3. Closed intervals [a,b] are closed.
  4. Cantor Set Consider the interval [0,1] and call it C0. Let A1 be equal {0, \tfrac{2}{3}} and let dn = (\tfrac{1}{3})^{n}. Let An+1 be equal to the set An∪{x|x=a+2dn, a∈An}. Let Cn be \textstyle \bigcup_{a\in A_n}{[a,a+dn]}, 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.