Why Read This Book? - Index of

6.3 The Nested Interval Property 199
Proof. By Exercises 6.3.1 and 6.3.2, both 〈an〉 and 〈bn〉 converge, and by Corollary
6.2.14, they have the same limit L. We show that ∩ ∞ n=1 [an,bn] ={L} ....
EXERCISE 6.3.4 Finish the proof of Theorem 6.3.3.
6.3.2 The NIP Applied to Subsequences
In Section 6.4, we will need some results based on taking a given sequence and
creating from it a new sequence by perhaps deleting some of its terms and preserving
the order of the kept terms. The new sequence is called a subsequence, and
we take the opportunity now to introduce the term and note what the NIP has to
say about certain sequences and subsequences.
Creating a subsequence of a given sequence can make for a little notational
mess when it comes to indexing the subsequence, so let’s look at an example.
Given a sequence 〈an〉, suppose we want to consider the following subsequence.
〈a2,a8,a10,a40,a44,a66,...〉 (6.20)
The notational dilemma can be resolved by noting that the indices
form a strictly increasing sequence 〈nk〉 ∞ k=1
〈2, 8, 10, 40, 44, 66,...〉 (6.21)
of positive integers. If we denote
n1 = 2, n2 = 8, n3 = 10, and so on, then the sequence 〈n1,n2,n3,...〉 is a way
of indexing our subsequence. The subsequence can then be denoted by
〈an1 ,an2 ,an3 ,an4 ,...〉=〈ank 〉∞ k=1
where k is now the indexing variable. This brings us to a definition.
(6.22)
Definition 6.3.5 Suppose 〈an〉 ∞ n=1 is a sequence of real numbers, and suppose
〈nk〉 ∞ k=1 is a strictly increasing sequence of positive integers. Then the sequence
is called a subsequence of the sequence 〈an〉.
〈ank 〉∞ k=1
Suppose you love to play darts. You play every day from now to eternity,
and you’re pretty good at it. At least you’re good enough so that you always hit
somewhere on the target. If every dart you throw leaves a tiny hole behind, what
must eventually happen to the set of pinholes on the target as time goes on? Well,
it is certainly possible that your darts could always land on one of a finite number
of spots on the target, in which case you would hit at least one spot infinitely many
times. But if you do not hit any one spot infinitely many times, then your target
will have infinitely many holes in it. If so, then the finite size of the target will
imply that the pinholes are going to cluster in one or more places. Of course, if