156 Chapter 3 • Topology of the Real Number SystemLet n be a fixed natural number. Then the sets N :!;. ( ~) = ( ~^1 , k~^1 ) for
k = 0, 1, 2, · · · , n form a collection of n + 1 sets in U. These sets collectively
cover A , since they overlap as follows:
N:!;. (~)(^0) - k-l ,-, k II -,-k+l ,
N:!;. (k:l)
(^0) Ii k k+l -,-, k+2 -,-,
Figure 3.10
Thus, U has a finite subcover of A. 0
Definition 3.3.4 A set A of real numbers is compact if every open cover of
A has a finite subcover of A.
The notion of compactness is not as complicated in the setting of the real
number system as this definition seems to imply. Examples, and the Heine-Borel
Theorem below, will make the concept more concrete.
Theorem 3.3.5 Every finite set is compact.
Proof. Exercise l. •
Theorem 3.3.6 Every compact set is bounded.
Proof. Exercise 2. •
Example 3.3.7 Is the open interval A= (0, 1) a compact set?
Discussion: The collection U = { N :!;. ( r) : r E IQ, n E N} is an open cover
of (0, 1). We have seen in Example 3.3.3 that Uhas a finite subcover. Does this
mean that (0, 1) is compact? No!
To show that (0, 1) is compact we would have to show that every open
cover of (0, 1) has a finite subcover of (0, 1). In fact, we shall now show that
(0, 1) is not compact.
Solution: Consider the collection U = { ( ~, 1 - ~) : n E N}. This is a col-
lection of open sets, and U U =(0, 1).
0! II I -! n
Figure 3.11