3.4 *The Cantor Set 169Since
3: < c, either an E Nc;(x) or bn E Nc;(x). Recall that both an and
bn are in the Cantor set. Hence, Ve: > 0, Nc;(x) contains a point of the Cantor
set other than x. Therefore, x is a cluster point of C ; i.e., x EC'.
Therefore, C ~ C'. •Definition 3.4.16 A set A of real numbers is nowhere dense if A contains
no nonempty open intervals.Theorem 3.4.17 The Cantor set is nowhere dense.Proof. Exercise 10. •The following t heorem is an interesting application of compactness as char-
acterized by Theorem 3.3.17.Theorem 3.4.18 Every nonempty perfect set is uncountable.Proof. Let A be a perfect set. Since A has cluster points it must be
infinite, by Corollary 3.2.12. For contradiction, suppose A is countable, say
A= {a 1 ,a2, ···,an,· ··}. We construct the following sequence {Nk} of neigh-
borhoods in A:
Let N 1 = Nc;(a 1 ), for some c > 0.
Since a 1 is a cluster point of A, N 1 contains a point a~ of A other than
a 1. Since a~ must be a cluster point of A, it has a neighborhood N 2 such that
N2 ~ N 1, a1 tj. N2, and N2 n A-=/:-0.
Since a~ is a cluster point of A, N 2 contains a point a~ of A other than
a 2. Since a~ must be a cluster point of A, it has a neighborhood N 3 such that
N3 ~ N2, az tJ. N3, and N3 n A-=/:-0.
Continuing inductively, we obtain a sequence { Nk} of neighborhoods (open
intervals) such that
(1) N1 2 N2 2 .. · 2 Nk 2 .. ·;
(2) 't:/k EN, ak tj. Nk+li
(3) Each Nk contains a point of A; i.e., Nk n A-=/:-0.
Now, 't:/k EN, define Ck= NknA. Then each Ck is compact. The collection
{Ck : k E N} is a collection of compact sets with the "finite intersection prop-
erty" 4 since n~= 1 Ck; =Cm, where m = max{k1, kz, · · · , kn}· So, by Theorem
3.3.17, n%: 1 Ck -=/:-0. But 't:/k EN, ak tj. Ck+l, soak tj. n%: 1 Ck. Therefore,
n%: 1 Ck= 0, since n%: 1 Ck ~A. Contradiction.
Therefore, A cannot be countable. •
- See Definition 3.3.15.