1549901369-Elements_of_Real_Analysis__Denlinger_

3.2 Closed Sets and Cluster Points 147

The closed set theorem claims that the intersection of any collection of

closed sets is closed, but only that the union of finitely many closed sets is

closed. The next example shows that the union of infinitely many closed sets

need not be closed. (Compare with Example 3.1.8.)

Example 3.2.5 (A Collection of Closed Sets Whose Union Is Not Closed)

u -CXl [1 , 1 - -1] = (0, 1).

n=2 n n

! II

[

0

1-k

] I

Figure 3. 7

Each set [~, 1 - ~] is closed, while the union (0, 1) is not closed. D

CLUSTER POINTS OF A SET

Definition 3.2.6 (Cluster Points) Suppose A<;;; JR and x ER Then x is a

cluster point of A if every neighborhood of x contains a point of A other than

x. [i.e., Ve> 0, NE(x) n (A-{x}) =j:. 0.]

Examples 3.2.7 (a) 0, 1, ~'and 0.789 are cluster points of the interval (0, 1).
(b) 0 is a cluster point of { ~ : n E N} but 160 and 9 ~ 6 are not. D

Theorem 3.2.8 A set is closed iff it contains all of its cluster points.

Proof. Part 1 (::::}): Suppose A is closed. Let x be a cluster point of A. We
must prove that x E A.
For contradiction, suppose x rJ. A. Then x E Ac. But Ac is open, so 3 e >
0 3 NE(x) <;;;Ac. Then NE(x ) is a neighborhood of x containing no point of A.
But x is a cluster point of A. Contradiction. Therefore, x E A.

Part 2 ( ~): Suppose A contains all its cluster points. We must prove that

A is closed.

For contradiction, suppose A is not closed. Then Ac is not open, so

3 x E Ac 3 Ve> 0, NE(x) Cl Ac,

so, NE(x) contains a point of A

(which can't be x, since x rJ. A)

so, NE(x) contains a point of A other than x.