14. Prove that every Cauchy sequence is bounded (Theorem 1.4)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Transcribed Image Text:14. Prove that every Cauchy sequence is bounded (Theorem 1.4).
Transcribed Image Text:1.4 THEOREM Every Cauchy sequence is bounded.
Proof The proof is left as Exercise 14.
Consider again a Cauchy sequence (an)-1. The terms of this sequence all lie in
an interval (C, D), since the sequence is bounded; moreover, the terms of the sequence
get closer and closer together as one goes out farther and farther in the sequence. It
seems reasonable to suspect that there is a real number A at which the terms of the
sequence "pile up." We must find this number A and prove that the sequence converges
to A. The existence of such a real number will be proved as a corollary to a theorem
that we shall use later. First, the notion of a "pile-up" must be formulated in precise
terminology.
DEFINITION Let S be a set of real numbers. A real number A is an accumulation
point of S iff every neighborhood of A contains infinitely many points of S.
A few remarks concerning this definition are in order. First, if A is an accumulation
point of S, then every neighborhood of A contains at least one point of S that is different
from A. (Indeed, A might not belong to S.) On the other hand, if A is not an accumulation
point of S, then some neighborhood of A contains only a finite number of members of
S. In this case it is possible to find a smaller neighborhood of A that excludes all points
of S different from A. Thus, A is an accumulation point of S iff every neighborhood of
A contains a member of S that is different from A. We state this result as follows.
LEMMA Let S be a set of real numbers. Then A is an accumulation point of S
iff each neighborhood of A contains a member of S different from A.
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