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5.1
Definition (Sequence.)
Let
![$A$](img50.gif)
be a set. A
sequence in ![$A$](img50.gif)
is a function
![$f\colon \mbox{{\bf N}}\to A$](img1259.gif)
. I
sometimes denote the
sequence
![$f$](img38.gif)
by
![$\{f(n)\}$](img1260.gif)
or
![$\{f(0),f(1),f(2),\cdots\}$](img1261.gif)
. For
example, if
![$f\colon\mbox{{\bf N}}\to\mbox{{\bf Q}}$](img1262.gif)
is defined by
![$\displaystyle {f(n)={1\over {n+1}}}$](img1263.gif)
, I might write
5.2
Warning.
The notation
![$\{f(0),f(1),f(2),\cdots\}$](img1261.gif)
is always ambiguous.
For example,
might denote
![$\{2^n\}$](img1266.gif)
. It might also denote
![$\{\phi(n)\}$](img1267.gif)
where
![$\phi(n)$](img1268.gif)
is the
number of regions into which a circle
is divided when all the segments joining the
vertices of an inscribed regular
![$(n+1)$](img1269.gif)
-gon are drawn.
5.3
Entertainment.
Show that
![$\phi(4)=2^4$](img1271.gif)
, but that it is not true that
![$\phi(n)=2^n$](img1272.gif)
for all
![$n\in\mbox{{\bf N}}$](img359.gif)
.
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
![$k\in\mbox{{\bf Z}}$](img1016.gif)
, then
Thus,
![$\mbox{{\bf Z}}_{\geq 0}=\mbox{{\bf N}}$](img1018.gif)
. Occasionally I will want to consider sequences whose
domain is
![$\mbox{{\bf Z}}_{\geq k}$](img1015.gif)
where
![$k\neq 0$](img1276.gif)
. I will denote such a sequence by
Hence, if
then
![$f(n)=n+1$](img1279.gif)
for all
![$n\in\mbox{{\bf N}}$](img359.gif)
, and if
then
![$g(n)=n$](img1281.gif)
for all
![$n\in\mbox{{\bf Z}}_{\geq 1}$](img1085.gif)
.
5.6
Remark.
Most of the results we prove for sequences
![$\{f(n)\}$](img1260.gif)
have obvious analogues for
sequences
![$\{f(n)\}_{n\geq k}$](img1282.gif)
, and I will assume these analogues without
explanation.
5.7
Examples.
![$\{i^n\}=\{1,i,-1,-i,1,i,\cdots\}$](img1284.gif)
is a sequence in
![$\mbox{{\bf C}}_{\mathbf{Q}}$](img1224.gif)
.
is a sequence of intervals in an ordered field
![$F$](img406.gif)
.
5.8
Definition (Open and closed intervals.)
An interval
![$J$](img727.gif)
in an ordered field is
closed if it contains all of its
endpoints.
![$J$](img727.gif)
is
open if it contains none of its endpoints. Thus,
5.9
Definition (Binary search sequence.)
Let
![$F$](img406.gif)
be an ordered field. A
binary search sequence ![$\{[a_n,b_n]\}$](img1289.gif)
in
![$F$](img406.gif)
is a sequence of closed intervals with end points
![$a_n,b_n$](img1290.gif)
in
![$F$](img406.gif)
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
![$F$](img406.gif)
be an ordered field, let
![$\{[a_n,b_n]\}$](img1289.gif)
be a binary search sequence in
![$F$](img406.gif)
,
and let
![$x\in F$](img489.gif)
. We say
converges to
![$x$](img85.gif)
and
write
![$\{[a_n,b_n]\}\to x$](img1294.gif)
if
![$x\in[a_n,b_n]$](img1295.gif)
for all
![$n\in\mbox{{\bf N}}$](img359.gif)
.
We say
converges, if there is some
![$x\in F$](img489.gif)
such that
![$\{[a_n,b_n]\}\to x$](img1294.gif)
. We say
diverges if there is no such
![$x$](img85.gif)
.
5.12
Example.
Let
![$F$](img406.gif)
be an ordered field. Then
![$\displaystyle { \left\{\left[0,{1\over
{2^n}}\right]\right\}}$](img1296.gif)
is a binary search sequence and
![$\displaystyle {\left\{\left[0,{1\over
{2^n}}\right]\right\}\to 0}$](img1297.gif)
.
5.13
Exercise.
Let
![$F$](img406.gif)
be an ordered field, let
![$a,b\in F$](img476.gif)
with
![$a<b$](img1298.gif)
. Let
![$\displaystyle {m={{a+b}\over 2}}$](img1299.gif)
. Show that
- 1)
- 2)
-
(Conditions 1) and 2) say that
![$m$](img1109.gif)
is the midpoint
of
![$a$](img24.gif)
and
![$b$](img42.gif)
.)
5.14
Exercise.
Let
![$F$](img406.gif)
be an ordered field and let
![$a,b\in F$](img476.gif)
with
![$a\leq b$](img705.gif)
and let
![$c,d$](img1302.gif)
be
points in
![$[a,b]$](img1303.gif)
. 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
![$2^n\geq n$](img1305.gif)
for all
![$n\in\mbox{{\bf N}}$](img359.gif)
.
5.16
Example (A divergent binary search sequence.)
Define a binary search
sequence
![$\{[a_n,b_n]\}$](img1289.gif)
in
![$\mbox{{\bf Q}}$](img243.gif)
by the rules
Thus,
Since
is the midpoint of
, we have
and
![\begin{displaymath}
b_{n+1}-a_{n+1}={1\over 2}(b_n-a_n)
\end{displaymath}](img1313.gif) |
(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
![\begin{displaymath}
a_n^2<2<b_n^2 \mbox{ for all } n\in\mbox{{\bf N}}.
\end{displaymath}](img1324.gif) |
(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
![\begin{displaymath}
\vert x^2-2\vert\leq b_n^2-a_n^2=(b_n-a_n)(b_n+a_n)
\leq\left({b_0-a_0 \over 2^n}\right)(b_0+b_0)={4\over
{2^n}}
\end{displaymath}](img1327.gif) |
(5.19) |
for all
. Since
is not a square in
,
. Write
, where
. Then
so
By exercise 5.15A, for all
,
![\begin{displaymath}
n\leq 2^n\leq 4q.
\end{displaymath}](img1333.gif) |
(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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