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# 5.1 Sequences and Search Sequences

5.1   Definition (Sequence.) Let be a set. A sequence in is a function . I sometimes denote the sequence by or . For example, if is defined by , I might write

5.2   Warning. The notation is always ambiguous. For example,

might denote . It might also denote where is the number of regions into which a circle is divided when all the segments joining the vertices of an inscribed regular -gon are drawn.

5.3   Entertainment. Show that , but that it is not true that for all .

5.4   Warning. The notation for a sequence and a set are the same, but a sequence is not a set. For example, as sets,

But as sequences,

5.5   Notation ( ) Recall from section 3.65, that If , then

Thus, . Occasionally I will want to consider sequences whose domain is where . I will denote such a sequence by

Hence, if

then for all , and if

then for all .

5.6   Remark. Most of the results we prove for sequences have obvious analogues for sequences , and I will assume these analogues without explanation.

5.7   Examples. is a sequence in .

is a sequence of intervals in an ordered field .

5.8   Definition (Open and closed intervals.) An interval in an ordered field is closed if it contains all of its endpoints. is open if it contains none of its endpoints. Thus,

5.9   Definition (Binary search sequence.) Let be an ordered field. A binary search sequence in is a sequence of closed intervals with end points in such that
1)
for all , and
2)
for all .
Condition 1) is equivalent to

5.10   Warning. Note that the intervals in a binary search sequence are closed. This will be important later.

5.11   Definition (Convergence of search sequence.) Let be an ordered field, let be a binary search sequence in , and let . We say converges to and write if for all . We say converges, if there is some such that . We say diverges if there is no such .

5.12   Example. Let be an ordered field. Then is a binary search sequence and .

5.13   Exercise. Let be an ordered field, let with . Let . Show that
1)
2)
(Conditions 1) and 2) say that is the midpoint of and .)

5.14   Exercise. Let be an ordered field and let with and let be points in . Show that

i.e., if two points lie in an interval then the distance between the points is less than or equal to the length of the interval.

5.15   Exercise. A Show that for all .

5.16   Example (A divergent binary search sequence.) Define a binary search sequence in by the rules

Thus,

Since is the midpoint of , we have

and
 (5.17)

It follows from (5.17) that

Hence is a binary search sequence. For each , let be the proposition

Then says , so is true. Let . If , then

If , then

Hence, in all cases, , and by induction, for all . Since for all , we have
 (5.18)

I now will show that diverges. Suppose, in order to get a contradiction, that for some , . Then

so

Combining this with (5.18), we get
 (5.19)

for all . Since is not a square in , . Write , where . Then

so

By exercise 5.15A, for all ,
 (5.20)

Statement (5.20) is false when , and hence our assumption that was false.

Next: 5.2 Completeness Up: 5. Real Numbers Previous: 5. Real Numbers   Index