148 Chapter 3 • Topology of the Real Number System
Hence, :3 x E Ac 3 x is a cluster point of A. That is, there is a cluster point of
A that does not belong to A. This contradicts our hypothesis that A contains
all its cluster points. Therefore, A is closed. •
Lemma 3.2.9 (Cluster Points vs. Interior Points and Boundary Points)
(a) Every interior point of A is a cluster point of A.
(b) If x is a boundary point of A and x tJ. A , then x is a cluster point of A.(c) If x is a cluster point of A and x tJ. A, then x is a boundary point of A.Proof. Exercise 9. •Corollary 3.2.10 A set is closed iff it contains all of its boundary points.
Proof. Exercise 10. •Theorem 3.2.11 Suppose A ~ JR and x E JR. Then x is a cluster point of A
iff every neighborhood of x contains infinitely many points of A.
Proof. Part 1 (::::}): Suppose x is a cluster point of A. For contradiction,
suppose :3 neighborhood Nc(x) containing only finitely many points of A. Only
finitely many of these points will be different from x, say a 1 , a 2 , · · · , an. Let
O = min{lx-a1 I, lx-a2I, · · · , lx-anl}. Then O > 0 and N8(x) is a neighborhood
of x containing no point of A different from x. This is a contradiction, since x
is a cluster point of A. Therefore, every neighborhood of x contains infinitely
many points of A.
Part 2 ( {=): Suppose that every neighborhood of x contains infinitely many
points of A. Then, every neighborhood of x contains a point of A other than
x, so x is a cluster point of A. •
Corollary 3.2.12 (Finite Sets)
(a) Finite sets have no cluster points.(b) All finite sets are closed.Proof. Exercise 11. •BOLZANO-WEIERSTRASS THEOREM (AGAIN)We first encountered the Bolzano-Weierstrass theorem as a result about
sequences: every bounded sequence has a convergent subsequence. This theorem
has an analogue for bounded sets. Like the sequential version, it is a theorem
of considerable power.