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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
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