4. Where in the proof of Theorem 2.3.3 did we use the following?
(a) The least upper bound axiom
(b) The assumption that {x } is bounded
(c) The assumption that {x } is increasing
5. Prove that a bounded decreasing sequence {xx } converges by
9(a) using the result proved in the text for increasing sequences;
(b) using the amp; – no definition of convergence and the set A = {xn | n EN}.
Theorem 2.3.3. A bounded monotonic sequence converges.
Proof. We prove the theorem for an increasing sequence; the decreasing sequence
case is left for the exercises (Problem 5).
Assume that {In } is bounded and increasing. To show that { n } is convergent, we
use the amp; – no definition of convergence, and to use this definition, we need to know
the limit of the sequence. We determine the limit using the set A = {xn | n EN},
the set of points in R consisting of the terms of the sequence {x}. Because {x} is
bounded, there is an M gt; 0 such that |X, | lt; M for all n, and this M is an upper bound
for A. Hence A is a bounded nonempty set of real numbers and so has a least upper
bound. Let a =. lubA. This number a is the limit of {x}, which we now show.
Take any amp; gt; 0. Since a = lubA, there is an element a E A greater than a – amp;;
that is, there is an integer no such that Xno = a gt; a – E. But {x } is increasing, so
Xno lt; Xn for all n gt; no, and a an upper bound of A implies that In _ a for all n.
Therefore, for n gt; no, a – amp; lt; xx lt; a, so |xn – a| lt;E.
Remark Note that, from the above proof, it follows immediately that, if a is the limit
of an increasing sequence {Xn}, then *n lt; a for all n E N. (Similarly, we have that
the limit of a convergent decreasing sequence is a lower bound for the terms of the
decreasing sequence.)Math