the
Real Analysis appendices of
Why Slopes
and
More Math
Volume 3
Printed in Canada
ISBN 0-9697564-3-7
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These Real Analysis appendices
continue the decimal
viewpoint of limits, continuity and convergence in chapter 14. and
this further lesson
A. What's Next
B. Pigeon Hole Principle
B. Bolzano-Weierstrass
C1. Triangle Inequality
C2. Triangle Inequality
C. More T.Inequality
D. Sets & Sequences
D. Monotone Sequences
E. Limits, Properties
E Limits & Error Control
F. Continuous Functions
F. Closed Range Thm
F. Intermediate Val. Thm
F. Compactness Thm
F. Equicontinuity Thm
F Extreme Value Thm
G. Rolle's Theorem etc
G. Mean Val. Thm.
G. Constant Difference Thm
G. Lipschitz Continuity I
PS: One Sided Range Theorems
G. Velocity Revisited
G. Sufficient Conditions
H. Riemann Sums Conv
H. Lipschitz Continuity II
Proofs of one-sided theorems could be of interest in the
study of 2D topology.
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Limits of Monotonic Sequences
Vocabulary. A sequence qk
with the property qk ³
qk+1 for every whole number k,
is said to be a decreasing sequence. A sequence qk
with the property qk £
qk+1 for every whole number k,
is said to be an increasing sequence. A sequence qk
is said to be monotonic if it is an increasing sequence or a
decreasing sequence.
Theorem D.2. [On Decreasing Sequences]
If the infinite sequence q1,q2,q3¼
has the property that for each whole numbers k, qk
³ qk+1 ³
Q for some real number Q, a lower bound, then
the sequence converges to a limit A ³
Q. The limit A is the greatest lower bound of
the set of points in the sequence.
| Proof:
First, by mathematical induction, n < j
implies qn ³
qj.
Second, if there exist a whole number K
> 0 such that k ³
K implies qk = qK
then the limit is given by A = qK
and the sequence takes on at most K distinct
values q1,¼,qK.
Otherwise for every K > 0, there is a k
> K such that qk ¹
qK. In this case, the range
of values assumed by the sequence is an infinite
set. The case where the range of values is an
infinite set is considered next.
An small assertion: observe A > qm
for some m implies that A > qm
³ qn
for all n ³ m.
This in turn implies A is not a limit point.
The interval of length qm-A
centered at A contains at most m
elements of the sequence. That A > qm
for some m implies that A is not a
limit point of the sequence qj
³ qj+1.
According to the Weierstrass theorem, the
infinite set formed by the sequence qj
has a limit point A. The contrapositive of
the above assertion now implies that the limit point
A £ qp
for all whole numbers p. But now for each
whole number k, the interval A+[1/2]10-k
to A (with A included) contains
infinitely many elements of the sequence. Therefore
there is some sequence element qn
with the property that A+[1/2]10-k
³ qn
³ A. Therefore p
³ n implies A+[1/2]10-k
³ qn
³ qp
³ A since qp
³ A and qn
³ qp
whenever p > n. Finally, since k
is arbitrary, we conclude include the sequence qn
converges to a limit A. Moreover, we can say
there are no points in the sequence qn
to the left of A. |
The proof of the following theorem is similar.
Theorem D.3 [On Increasing Sequences]
If the infinite sequence q1,q2,q3¼
has the property that for each whole numbers k, qk
£ qk+1 £
Q for some real number Q, an upper bound, then
the sequence converges to a limit B £
Q. The limit B is the least upper bound of the
set of points in the sequence.
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