140 Chapter 3 • Topology of the Real Number System(b) Let C be any collection of open sets. To prove that UC is open, let
x E UC. Then 3A EC 3 x EA. Since A is open, :le> 0 3 Nc:(x) ~A. But
A<;.;;; UC. Thus, Ne(x) <;.;;;UC. Therefore, 't/x E UC, :L:: > 0 3 Ne(x) <;.;;;UC. That
is , UC is open.
n
(c) Let {A 1 ,A 2 , · · · , An} be a finite collection of open sets. To prove n Ai
i=l
n
is open, let x E n Ai· Now for each i = 1, 2, ... 'n, Ai is open, so 3 Ci >
i=l
0 3 Ne, (x) <;.;;; Ai· Let e = min{e1,e2,··· ,en}· Then e > 0 and for each
i = 1, 2 , · · · , n, e ::; ei, so
Ne(x) <;.;;; Ne, (x) ~Ai, so
n
Ne(x) <;.;;; n Ai.
i=l
n
Therefore, by Definition 3.1.3, n Ai is open. •
i=l
Theorem 3.1. 7 A nonempty open set must be an infinite set. That is, a
nonempty set with only finitely many elements cannot be open.Proof. Let A= {a 1 ,a 2 ,· ··,an} be a finite set. Let x EA. Now, 'Ve> 0,
the set Ne(x) is an infinite set, hence it cannot be a subset of A, since A is a
finite set. Thus, there is no c > 0 3 Ne(x) <;.;;; A. That is, a finite set A cannot
be open. Therefore, if A is a nonempty open set, it must be infinite. •The open set theorem (Theorem 3.1.6) claims that the union of any collec-
tion of open sets is open, but only that the intersection of finitely many open
sets is open. The next example shows that the intersection of infinitely many
open sets need not be open.Example 3.1.8 (A Collection of Open Sets Whose Intersection Is Not Open)(^00) n ( --, (^1) - 1) = {O}.
n=l n n
! 11
0
Figure 3.4
Each set (-~, ~) is open, while the intersection {O} is not open. 0
The open set idea is extremely powerful. We shall see in the remainder of
this section that it enables us to define other intrinsically interesting and useful
concepts.