2.6 Subsequences and Cluster Points 111Proof. Exercise 7. •Theorem 2.6.13 L et {xn} be a sequence. Then(a) {xn} diverges to + oo ¢=> '\/M > 0, {xn} is eventually in (M, +oo).
(b) {xn} has a subsequence diverging to +oo ¢=> '\/M > 0, {xn} is frequently
in (M, +oo).(c) {xn} diverges to - oo ¢=> '\/M > 0, {xn} is eventually in (-oo, -M).
(d) {xn} has a subsequence diverging to - oo ¢=> '\/M > 0, {xn} is frequently
in (-oo, -M).Proof. Exercises 8 and 9. •CLUSTER POINTS
Definition 2.6.14 A real number xis a cluster point of a sequence {xn}
if the sequence { xn} has a subsequence converging to x. [Equivalently, \le: > 0,
Xn E (x -c:, x + c:) for infinitely many n.]
We say that +oo is a cluster point of { Xn} if { Xn} has a subsequence
diverging to +oo. [Equivalently, '\/M > 0, Xn > M for infinitely many n .]
We say that -oo is a cluster point of { Xn} if { xn} has a subsequence
diverging to -oo. [Equivalently, '\/M > 0, Xn < -M for infinitely many n.]
Examples 2.6.15 (a) Find all the cluster points of the sequence {sin :7f}·
Solution. Writing out this sequence, we see that{
n1f } { 1 1 1 1 1 }
sin 4 = v'2' 1, v'2' 0, - v'2' -1, - v'2' 0, v'2' 1, ·...1 1
Thus, this sequence has five cluster points: 0, 1, -1, v'2' and -v'2"(b) Find all the cluster points of the sequence { Xn}, where{n if n is odd
Xn = 1 + -1 n/2
(
2) if n is even.Solution. Writing out this sequence, we have{xn} = {l,0,3, l ,5,0,7,l,9,0,ll, 1,13, 0, · ··}