Advanced book on Mathematics Olympiad

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116 3 Real Analysis


347.Letkbe an integer greater than 1. Supposea 0 >0, and define


an+ 1 =an+

1

√kan

forn>0. Evaluate

nlim→∞

ank+^1
nk

.

We conclude the discussion about limits of sequences with the theorem of Georg
Cantor.


Cantor’s nested intervals theorem.Given a decreasing sequence of closed intervals
I 1 ⊃I 2 ⊃ ··· ⊃In⊃ ···with lengths converging to zero, the intersection∩∞n= 1 In
consists of exactly one point.


This theorem is true in general if the intervals are replaced by closed and bounded
subsets ofRnwith diameters converging to zero. As an application of this theorem we
prove the compactness of a closed bounded interval. A set of real numbers is called
compact if from every family of open intervals that cover the set one can choose finitely
many that still cover it.


The Heine–Borel Theorem.A closed and bounded interval of real numbers is compact.


Proof.Let the interval be[a, b]and assume that for some family of open intervals(Iα)α
that covers[a, b]one cannot choose finitely many that still cover it. We apply the
dichotomic (division into two parts) method. Cut the interval[a, b]in half. One of the
two intervals thus obtained cannot be covered by finitely manyIα’s. Call this intervalJ 1.
CutJ 1 in half. One of the newly obtained intervals will again not be covered by finitely
manyIα’s. Call itJ 2. Repeat the construction to obtain a decreasing sequence of intervals
J 1 ⊃J 2 ⊃J 3 ⊃ ···,with the length ofJkequal tob− 2 kaand such that none of these
intervals can be covered by finitely manyIα’s. By Cantor’s nested intervals theorem, the
intersection of the intervalsJk,k≥1, is some pointx. This point belongs to an open
intervalIα 0 , and so an entire-neighborhood ofxis inIα 0. But thenJk⊂Iα 0 forklarge
enough, a contradiction. Hence our assumption was false, and a finite subcover always
exists. 


Recall that the same dichotomic method can be applied to show that any sequence
in a closed and bounded interval (and more generally in a compact metric space) has a
converent subsequence. And if you find the following problems demanding, remember
Charlie Chaplin’s words: “Failure is unimportant. It takes courage to make a fool of
yourself.’’

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