2.6 Subsequences and Cluster Points 113Theorem 2.6.17 A bounded sequence converges {::} it has one and only one
cluster point.Proof. Part 1 (::::}): Exercise 11.
Part 2 (~):Suppose {xn} is a bounded sequence with one and only one
cluster point, say L. We shall prove that Xn __, L. For contradiction, suppose
Xn ft L. Then
:3 c > 0 3 Vno E N, :3 n 2'. no 3 lxn - LI 2'. €.
Thus, there exists a strictly increasing sequence {an} of natural numbers
such that
(12)
Since { x°'n} is a subsequence of { Xn} it is also bounded, so by the Bolzano-
Weierstrass Theorem, it has a convergent subsequence {x,en}; say x,en __, M.
Then Mis a cluster point of {xn} since {x,en} is a subsequence of {xn}· But
{xn} has only one cluster point, so M = L. That is, x,en __, L. Then,Ve > 0, :3 no E N 3 n 2'. no ::::} lx,en - LI < €.
This contradicts (12), since {x,en} is a subsequence of {xan}. Therefore,
Xn __, L. •*Theorem 2.6.18 Every convergent sequence has a monotone subsequence
(converging to the same limit). In fact, if a convergent sequence is not eventu-
ally constant, then it has a strictly monotone subsequence.*Proof. Let {xn} be a convergent sequence, Xn , L. If {xn} is eventually
constant, it has a constant tail, which serves as a monotone subsequence of { xn}·
Henceforth, we assume that { xn} is not eventually constant. Then infinitely
many terms of { Xn} are different from L. So, one of the two intervals (-oo, L)
or (L, +oo) must contain infinitely many terms of {xn}·
Case 1, (-oo, L) contains infinitely many terms of {xn}: Let€> 0. Since
Xn , L, all but finitely many terms of { Xn} are in ( L - c, L + €). Thus, by our
Case 1 hypothesis, infinitely many terms of {xn} must be in (L -c, L). Thus,
Ve> 0, infinitely many terms of {xn} are in (L-c , L). (13)
Let c 1 = l. Then, by (13), :3n 1 EN3 Xn, E (L-c 1 ,L). This defines the
first term Xn, of our subsequence.
Let c 2 = L - Xn,. Then c 2 > 0, so by (1), :3 infinitely many n EN 3 Xn E
(L - €2, L ). Hence, :3 n2 > ni 3 Xn 2 E (L - c2, L ). Note that Xn 2 > L - c2 =
L - ( L - Xn,) = Xn,. That is , Xn 1 < Xn 2 • This defines the second term, Xn 2 •
We proceed by the principle of mathematical induction. In the general step,
we assume that we have found nk EN (k 2'. 2) 3