3.2 Closed Sets and Cluster Points 151(b) (-¢=): Suppose :3 sequence {an} of points of A other than x, such that
an ----; x. Then every neighborhood of x contains infinitely many terms of the
sequence, and hence contains a point of A other than x. That is, x is a cluster
point of A. •
Theorem 3.2.19 (Sequential Criterion for Closed Sets) A set A is closed
iff V convergent sequences {an} of points of A, lim an E A.
n->oo
Proof. P art 1 ( =? ): Suppose A is a closed set. Suppose {an} is a convergent
sequence of points of A. Let L = lim an. For contradiction, suppose L i A.
n->oo
Then , Vn, an =J=. L. By Theorem 3.2.18, L is a cluster point of A, so by Theorem
3.2.8, L E A. Contradiction. Therefore, L E A.
Part 2 (-¢=): Suppose that V convergent sequence {an} of points of A,
lim an E A. Let x be a cluster point of A. Then, by Theorem 3.2.18, :3 se-
n->oo
quence {an} of points of A other than x, such that an ----; x. Then, by our
hypothesis, x E A. That is, A contains all its cluster points. Thus, by Theorem
3.2.8, A is closed. •
DENSE SUBSETS OF A SETDefinition 3.2.20 Suppose A, B ~ R We say that A is dense in B if B ~ A.
Equivalently, A is dense in B ¢} every member of B is either a member of A
or a cluster point of A. (Cf. Definition 1.5.6, Exercises 3.1.19- 20, and Exercises
3.2.27- 29.)
Theorem 3.2.21 (Sequential Criterion for Denseness) A set A is dense
in a set B iff Vb EB, :3 sequence {an} of points of A such that an----; b.
Proof. Exercise 30. •
Compare this result with Theorem 2.3.6.EXERCISE SET 3.2l. Prove Corollary 3.2.2.
- Prove Note 1, following Corollary 3.2.2.