Week 1: 1-22: The first day of class was introductory.
The focus of the course will be to understand the foundations for
what is the substance of the first year of calculus.
Unlike Math 343 which is an introduction to contemporary abstract
algebra, where the concepts and results are new and eventually
abstract in their nature,
Math 316 covers the analysis of real numbers and real valued
functions of one real variable.
The subject matter of real anaysis is quite familiar. Real numbers
have been a topic of study since early courses in algebra and
functions of these numbers were the focus of preparations for
calculus as well as the main subject of calculus. Thus the basic
results of this course are fairly familiar, BUT the proofs for many
of the key theorems of calculus were not proven in the calculus
courses. These results appeared reasonable and so why bother to
spend precious time on proving "the obvious". Part of this course
will be to develop an appreciation for the need for rigor along with
the rigor itself.
One common approach to the study of real analysis starts with the
foundations: developing first what real numbers are in a rigorous
fashion from the natural numbers or the integers, then the rational
numbers, and finally the real numbers. This involves some vary
careful work in the algebra of equalities and inequalities for
arithmetic.
We will follow an alternative approach looking at some of the
key theorems of calculus and analyzing these results to see what is
needed to prove them. Eventually we will delve more deeply into the
nature of the subject arriving finally at a rigorous definition of
what a real number is.
Key theorems we will examine will involve continuity,
differentiability, and integrability. The first coherent and
somewhat successful presentation of calculus was by Cauchy in the
early 19th century- about 150 years after the works of Newton and
Leibniz. The final work on a careful definition of he real numbers
comes toward the end of the 19th century with separate work by
Weierstrass and Dedekind.
We will use a working definition for what a real number is: a real
number is any number that can be represented as a possibly infinite
decimal (positive or negative) or that can be thought of a the
measure of a length of a line segment either to one side or the
other of a specified point on on a given line an a speicic unit for
measurement.
We will visualize real valued functions from the real numbers with
both graphs and mapping figures. Mapping figures are discussed
extensively in the assigned readings from Sensible Calculus
materials on-line: SC
0.B1 Numbers [on-line] and SC
0.B2 Functions [on-line] . These materials as well as
other readings for the first week and the first assignment were
discussed briefly.
Details of the syllabus will be discussed further on Thursday.
1-24
This class was spent mainly on the organization of the course:
details like tests, homework, etc. as descibed in the main
course page.
In the discussion reference was made to Polya's :
How to Solve It... available on line:
https://notendur.hi.is/hei2/teaching/Polya_HowToSolveIt.pdf
Polya discribes 4 Phases of Problem Solving 1. Understand the problem. 2. See connections to devise a plan. 3. Carry out the plan. 4. Look back. Reflect on the process and results.
It is the first phase that is usually not recognized as being
essential. there is usually more to understand than is apparent.
Also mentioned were other texts in Real Analysis:
Analysis WebNotes by John Lindsay Orr
Interactive Real Analysis by Bert G. Wachsmuth
A First Analysis Course by John O'Connor
Basic Analysis: Introduction to Real Analysis by Jiri
Lebl
Introduction to Real Analysis by William Trench
Principles of Mathematical Analysis by Walter
Rudin (1953)
Calculus by Michael Spivak [A calculus book that
proved all the theorems! PDF (with some issues) now on
Moodle.]
Next class: We will watch a video "Space Filling Curves"
which were introduced in the late 19th Century and were the early
instances of what later came to be called fractals in the late
Twentieth Century. Fractals are now a subject of continuing
research.
Also you can see this in the library on shelf- please replace
accurately! VIDEO2577
Use of point-wise convergence for a sequence of approximating
curves.
Use of concept of continuity for curves.
Length of a curve.
How does a curve "fill" space?
Week 2: 1-28
Continuation on space filling curves:
To make sense of curves being close and sequence of curves having limits,
we introduced the concept of a metric space. - a set where
closeness can be measured using non-negative real numbers for the
measurements. A metric on a set X is a function m: '
XxX -> R_+ u {0}` where
For all `a,b in X, m(a,b)=m(b,a)`.
If `m(a,b) = 0` then `a = b`.
For all ` a,b,c in X, m(a,b) + m(b,c) >= m(ac).`
Elementary examples are the real numbers with `m(a,b) = |b-a|`, and
`R^2` and `R^3` with the usual distance between points
measurement.
Metrics for continuous functions on a closed interval `[a,b]` are
related to the issues of when functions are "close."
One such metric for continuous functions is `m(f,g) = max {
|f(x)-g(x)| , x in [a,b]}`.
Another is `m(f,g) = int_a^b ( f(x)- g(x))^2 dx`.
In Linear algebra, a real inner product on vectors,
`<f,g> ` gives rise to a norm, `||f|| =
sqrt{<f,f>}`, and then a metric, `m(f,g) =
||f-g|| `.
Mean
Value Theorem [proof based on Rolle's
theorem ]: If f is (i) continuous on [a, b] and (ii) differentiable on (a, b), then
there exists a number c in (a,
b)such that`f'(c)
= (f(b)-f(a))/(b-a)` .
Some theorems that rely on the Mean Value Theorem for
their proof.
If `f'(x) = 0` for an interval then there is a real
numberK where
`f(x) = K` for all`x` in the interval, I.
Two proofs were given: Direct: Pick `a in I`. It's enough to show for any `b in
I, a<b , then f(b) = f(a).` K will be `f(a)`.
Using differentiability implies continuity, `f` satisfies the
hypotheses of the MVT, so there is a number `c` where `c
in (a,b)` and `f'(c)= (f(b)-f(a))/(b-a)`.
But since `c in I` , `f'(c) = 0`, and thus `0 =
(f(b)-f(a))/(b-a)` and thus `f(b) = f(a)`. Indirect: Suppose there are `a,b in I` with `a<b,
f(a)< f(b)`. Using differentiability implies continuity, `f`
satisfies the hypotheses of the MVT, so there is a number
`c` where `c in (a,b)` and `f'(c)= (f(b)-f(a))/(b-a)`.
But since `c in I` , `f'(c) = 0`,thus `0=
(f(b)-f(a))/(b-a) > 0`. A contradiction.
1-29
More on the MVT;
Proofs about increasing/decreasing functions: If `f'(0)>0`
on an interval, then `f` is an increasing function.
Counter examples to the MVT:
(1) without Continuity: f (x) = x for
x in (0, 1] and f(0) = 1 on the interval
[0,1]
and
(2) without differentiability: f (x) = 1/2
-| x - 1/2| for x in [0,
1].
Also `f (x) = (1/2)^(1/3) -|( x - 1/2)|^(1/3)` for
x in [0, 1].
Theorem:
Rolle's Theorem [proof based on Extreme Value Theorem and critical
point analysis]: If f is (i) continuous on [a,
b] and (ii) differentiable on
(a, b), (iii) f
(a) = f (b)
then there exists a number c
in (a, b)
such that` f '(c) =0 = (f(b)-f(a))/(b-a)`.
Motivational Question I: What is a number?
Foundational Theorems for Continuity [also about existence].
See the following if you want to see a "simple" approach to
defining real numbers: [This is a main theme for the course!] Vector Calculus, Linear Algebra, and
Differential Forms: A Unified Approach by John
Hubbard and Barbara Burke Hubbard Appendix.A.1
Arithmetic of Real numbers.pdf
2-1
Example: algebraic numbers = {z: z is a complex number which is
the root of a polynomial with integer coefficients}
A few more properties of Fields:
Suppose F is a field.
Proposition: If a and b are
in F and ab = 0 then either a =0 or b=0.
Proof: Case 1. If a=0 then we are done.
Case 2. If a is not 0, then a has an inverse... c
where ca=1. Lemma: For any c in F, c*0 = 0.
Then b=1*b= (c*a)*b= c*(a*b)
= c*0 = 0. Thus either a =0 or b=0.
IRMC [I rest my case.]
Proof of Lemma: c*(0+0)=c*0, so c*0+c*0 = c*0.
Let g be the real number where g+c*0=0.
Then g+ (c*0 + c*0) = g+ c*0 = 0,
but g+ (c*0 + c*0=(g+ c*0) + c*0)= 0 + c*0 = c*0. Thus c*0 = 0.
If a, b and c are in F, a+b = a+c impliesb=c.
Proof: Let k be the element of the field where k
+ a = 0 Then b = 0 + b = (k +a)+b =k
+(a+b)= k +(a+c) =(k +a)+c
= 0 + c = c.
If a, b and c are in F with a
not equal to 0, ab = ac impliesb=c.
Proof: Since a is not equal to 0, let k be the
element of the field where k * a = 1 Then
b = 1* b = (k *a)*b =k(a*b)= k *(a*c) =(k *a)*c = 1* c = c.
Other field results discussed: (-1)*a = -a; -(-a)=a;
(-a)*b= -(a*b); (-a)*(-b)= a*b
Interesting but NOT Discussed: Let n·1 stand
for 1 + 1 + 1 + ... + 1 with n summands. Either (i) for
all n, n·1 is not 0 in which case we say the field has
"characteristic 0" , or (ii) for some n, n·1=0. In the casen·1=0, there is a smallest n for which n·1=0, in which case
we say the field has "characteristic n". For example: R, Q,
and C all have characteristic 0, while Z2, Zp,
where p is a prime number, and any finite field such as
F4, all have a non zero characteristic.
If the characteristic of F is not zero, then it is a
prime number.
Proof: If n is not a prime, n = rs with 1<r,s<n. Let a
= 1+1+...+1 r times and b= 1+1+...+1
s times. Then ab=n·1=0, so either a = 0
or b = 0, contradicting the fact the n was supposed to
be the SMALLEST natural number for which n·1=0. IRMC.
Order for the Real Numbers: At the heart of "close".
There is a relation > on R.
(That is, given any pair a, b then a
> b is either true or false).
This can be based on establishing a subset of
the Real Numbers R+ - the "positive real
numbers" with properties:
i) for any number a, either a = 0, `a in
R_+` , or `-a in R_+`, but not any two of these are true.
ii) if `a,b in R_+` then `a+b in R_+` and `a*b in R_+`.
This allows a definition of "<" by saying
`a<b` is true if (and only if) `b-a in R_+`.
It follow that `0<a` if and only if `a in
R_+`.
The following properties hold for the
relation "<": [Proofs are based on algebra and the
properties of `R_+`.]
a) Trichotomy: For any a, b `in` R
exactly one of a > b, a = b,
b > a is true.
b) If a, b > 0 then a
+ b > 0 and a.b > 0
c) If a > b then a
+ c > b + c for any c
Something satisfying the field and order
axioms is called an ordered field.
The field and order axioms for the real numbers can be used to
deduce any simple algebraic or order properties of R.
Example
The ordering > on R is transitive.
That is, if a > b and b > c
then a > c.
Proof a > b if and only if a - b
> 0 by definition. b > c if and only if b - c
> 0
Hence (a - b) + (b- c) > 0
and so a - c > 0 and we have a >
c.
Using inequalities with decimals to prove the IVT: Outline-
Assume `f(a) < P < f(b)`. Then consider `a <
(a+b)/2 < b` and apply the trichotomy property to
`f((a+b)/2)` and P. This will either give that`f((a+b)/2)=P`
and the IVT is proven, or will allow further
investigations on a smaller interval depending on
whether `f((a+b)/2)>P` or `f((a+b)/2)<P.`
Continuing with this process will lead to smaller intervals
and eventually because of continuity the estimation of a
decimal number `c` will arise that satisfies `f(c) = P.`
Week 3
2-4
Remarks on Assignment 1.
Absolute value inequalities: Proofs were besat
that did not proceed by cases.
Examples: For any `x -|x|<=x<=|x|`
so ` -|a|<=a<=|a|` and ` -|b|<=b<=|b|` so
` -|a| -|b|<=a+b<=|a| +|b|`. Thus `|a+b|<=|a|+|b|`.
Or `(|a+b|)^2 = (a+b)^2 = a^2 +2ab+b^2 <= |a|^2 +
2|a||b| + |b|^2 = (|a| + |b|)^2.`
since `0< |a+b|` and `0< |a| + |b| `,
`|a+b|<|a|+|b|.`
The thing which distinguishes R from Q (and
from other subfields) is the Completeness Axiom.
Definitions
An upper bound of a non-empty subset A
of R is an element bR with ba for all aA.
An element MR
is a least upper bound or supremum of A
if M is an upper bound of A
and if b is an upper bound of A
then bM.
A lower bound of a non-empty subset A of R
is an element dR
with da for all
aA.
An element mR
is a greatest lower bound or infimum of A if m is a lower bound of A and if d is an
lower bound of A then md.
2-5
Motivation: Using decimals to find decimal places of
`sqrt{ 3}`:
Start with `1^1 = 1` and `2^2=4`, so the units decimal for
`sqrt{ 3}` is 1.
Now to find the tenths decimal digit consider
`1^2, 1.1^2 , 1.2^2, 1.3^2, ...,``1.7^2 = 2.89 , 1.8^2 =
3.24` `, 1.9^2, 2^2`.
So the tenth digit is 7.
To find the hundredth digit consider
`1.70^2, 1.71^2 , 1.72^2, 1.73^2 = 2.9929, 1.74^2= 3.0276, ... ,
1.78^2, 1.79^2, 1.8^2`.
So the hundredth digit is 3.
Continue in this fashion, for each decimal place, examine the
eleven numbers that might have the correct digit, and choose the
digit where the squared value changes from being smaller than 3
to being larger than 3.
Now prove the infinite decimal that results must have its square
equal to 3.
.... The Least Upper Bound
(Completeness) Property of the Real Numbers:
If S is a nonempty set of real numbers that has an
upper bound, then there is a least upper bound for S.
Fact: If L and L' are least upper bounds for a set S, then L =
L'.
Proof: Since L' is an upper bound for S and L is a least upper
bound for S, L `<=` L'. Likewise, L' `<=` L. So L = L'.
(trichotomy property) EOP.
We can use decimals, representing real numbers, to help give
credibility to the belief that the least upper bound property
is true.
Discussion: Since S is nonempty, there is `x_0 in S`. Suppose
B is an upper bound for S, so `B>=x_0`. If `B = x_0` then B
it is not hard to show that B is the least upper bound
of S.
So,we suppose `B>x_0`. Now there is an integer `m_0` with
`m_0 > B` and an integer `n_0` with `n_0 < x_0`.
In the interval `[n_0,m_0]` there is an integer `m_1` where
`m_1` is an upper bound for S but `m_1 -1` is not an upper
bound for S. We let `n_1 = m_1-1` and now consider the
interval `[n_1,m_1]`.
In the interval `[n_1,m_1]` there is a decimal `n_1 +k*0.1`
where `n_1+k*0.1` is an upper bound for S but `n_1+ (k-1)*0.1`
is not an upper bound for S.
We let `m_2 = n_1 +k*0.1`, `n_2 = n_1 +(k-1)*0.1` and
now consider the interval `[n_2,m_2]`.
In the interval `[n_2,m_2]` there is a decimal `n_2 +k*0.01`
where `n_2+k*0.01` is an upper bound for S but `n_2+
(k-1)*0.01` is not an upper bound for S.
We let `m_3 = n_2 +k*0.01`, `n_3 = n_2 +(k-1)*0.01` and
now consider the interval `[n_3,m_3]`.
We can continue in this fashion. After each step we obtain a
decreasing sequence of terminating decimals `m_j` which are
upper bounds for S and an increasing sequence of terminating
decimals `n_j` each of which is just one digit smaller in the
last decimal place than `m_j` and each of which is not an
upper bound for S .
These decimals allow us to determine a number to any decimal
precision which approximates the least upper bound L. L
will be the limit of the two sequences `n_j` and `m_j`.
The proof that L is the least upper bound of S would involve
an argument assuming either that L was not an upper bound for
S or that there was an upper bound of S that was smaller
than L and consideration of how the two sequences were
determined. This argument is omitted as the presentation here
is given only to suggest the truth of the completeness
property.
Using the Least Upper Bound (Completeness) Property:
Proposition: There is a real number `r` where `r^2 =
3`. [ `r = sqrt{ 3} `]
Proof: Let `S= { x in R : 0< x^2 < 3}`. Since `1 in
S`, we have that `S` is not empty.
Since `2^2 = 4`, `2 notin S`. More important, for any
`x in S` , `x^2 < 3 < 4`, so `4 -x^2 > 0`.
Factoring gives ` (2 -x)(2+x) > 0`. Since for any `x in
S` , `x >0`, so `2 + x >0` and thus `2 -x > 0` so
`2 > x`. Thus 2 is an upper bound for S. Now we can apply
the LUB property to say that there is a LUB for the set S-
call it `r`. Now we will show that `r^2 = 3` which
will complete the proof.
To show `r^2 = 3` we show that it cannot be otherwise
by using the Trichotomy property and showing `r^2 > 3`
and `r^2 < 3` are not possible if `r` is the LUB of `S`.
So suppose that `r^2 > 3`. Look at `(r -1/n)^2 =
r^2 - {2r}/n + 1 /n^2 > r^2 - {2r}/n`.
Question: Is there an n for which `(r-1/n)^2 > 3
`?
Answer: `r^2 - {2r}/n > 3` happens if and only if `r^2 -
3 > {2r}/n` or `1/n < (r^2 -3)/{2r}`. Since `(r^2-
3)/{2r} >0 ` we can choose an n* by the Archimedean
property with `1/{n*} < (r^2 -3)/{2r}`. Thus `(r -
1/{n*})^2 >3` and `r-1/{n*}` is an upper bound for
`S`, contradicting the assumption that `r` was the least
upper bound.
Similarly, if `r^2 < 3` then `(r + 1/n)^2 = r^2 +
{2r}/n + 1/n^2 < r^2 + {2r}/n + 1/n = r^2 +{2r+1}/n`.
Question: Is there an n for which `(r + 1/n)^2< 3`
?
Answer: `r^2 + {2r+1}/n < 3` happens if and only if
`3 - r^2 > (2 r+1}/n` or `1/n < (3 - r^2)/{2r+1}`.
Since `(3 -r^2)/{2r+1} >0 ` we can choose an n* by the
Archimedean property with `1/{n*} <
(3-r^2)/{2r+1}`. Then `r+1/{n*} in S` which leads to
the conclusion that `r` would not be an upper bound, a
contradiction.
Note on the Archimedean Property of the real numbers:
Proposition: If `r ` is a positive real number, then there is
natural number `n' > 0` where `0 < 1/{n'} < r`.
Lemma: The set of natural numbers, N, is not a bounded set.
Proof of Lemma: (Indirect) If N is
bounded, there is a real number B so that an upper bound,
and thus a least upper bound L for the set N.
Then L-1 is not an upper bound for N, so there is a
natural number `m in N` where `L-1 < m < L`. But then
`L < m+1 < L+1 ` which contradicts L as an upper bound
for N since `m+1 in N`.
Proof of Proposition: Consider `1/r`: `1/r
> 0`, and by the lemma, there is a natural number
`n'> 0` where `n'> 1/r > 0 `. Then `0 < 1/{n'}
< r`. EOP.
A decimal approach to the proof: Express r as a decimal.
If `r>=1` use `n' = 2`. If `r<1`, express `r` as a
decimal. Suppose the first nonzero digit of `r` is in the
kth decimal place. Let `n' = 10^{k+1}`. Then
`0<1/{n'}< r`.
2-7 Time
spent reviewing material from last class on `sqrt{ 3}`.
Visualizing a
sequence of real numbers:
Def'n: A sequence is a function
`a:N->S` that is onto where `S subset R`.
With a
mapping figure.
As a graph.
On a number
line.
Provisional Definition: limit
of `a_n` is L means "after a while
all `a_n` are close to L."
Visualizing
the definition of
`lim_{n->oo}a_n = L`
Def'n: We say `lim_{n->oo} a_n = L` if
given any `epsilon >0` there is a natural number M so
that
if `n>M` then `| a_n - L | < epsilon`.
Example: `lim_{n->oo}
1/n = 0.`
Proof: Suppose `epsilon >0.` Then by the Archimedean
Property there is a natural number M where `1/M
< epsilon`. To show this M works, suppose `n > M`.
Then we have `1/n < 1/M` [Prove this---] . Thus `|1/n
- 0| = 1/n < 1/M < epsilon`. EOP.
Week 4 2-11 Motivation:
Preview of definition for limit of a function and continuity
based on sequence limits.
We say `lim_{x->a} f(x) = L` if for any sequence
`a_n` in the domain of `f -{a}`
with `lim_{n->oo}a_n = a`, `lim_{n->oo} f(a_n) =
L`.
We say `f ` is continuous on its domain D if
for any `a in D` and for any sequence `a_n` in the
domain of `f` where `lim_{n->oo}a_n = a`,
`lim_{n->oo} f(a_n) = f(a)`.
Theorems connected to algebra, inequalities
and intervals. 2-12
Properties of convergent sequences
(Algebra of sequences) 2-14
Monotonic sequences (A bounded
monotonic increasing sequence is convergent.) 2-15
Subsequences
We covered definitions and proofs of some results
about limits of sequences and continuity. For example: If `f(x)
= x^2` then `f ` is continuous on `(-oo,oo)`.
The Squeeze lemma was proven: If `a_n <= c_n <= b_n ` for
all `n` and
`lim_{n->oo}a_n = L` and `lim_{n->oo}b_n
= L`, then `lim_{n->oo}c_n = L`.
The Bolzano Weierstass Theorem:Every
bounded sequence has a convergent subsequence.
The proof of BWT was covered using Spivak's
lemma:
Every bounded sequence has a monotone subsequence.
Week 5 2-18
warm up: Composition of continuous is continuous.
Prop: Suppose `f : D -> R` and ` g: E -> R` with `f(D)
subset E` and `f` and `g` are continuous on the domains, and `g
circ f (x) = g(f(x)) ` for all ` x in D`, then `g circ f` is
continuous on D.
Proved using definition of continuity with sequences.
A sequence `{a_n}` is called a Cauchy sequence
if the terms of the sequence eventually all become arbitrarily
close to one another.
That is, given `epsilon`
> 0 there exists N such that if m, n
> N then |am- an|
< `epsilon`.
Proposition: If `a_n` converges, then `a_n` is a Cauchy
sequence.
Proof: Suppose `lim_{n->oo}a_n = L`, then given `epsilon
>0`, there exists N
such that if n > N then |an
- L| < `epsilon /2`. Thus if ` m, n > L` then `|a_n -
a_m| = |a_n - L + L - a_m| < |a_n - L | + |L- a_m| <
epsilon/2 + epsilon/2 = epsilon`.
Theorem:If `a_n` is a Cauchy
sequence, then `a_n` converges.
Proof Plan: 1 . Show `a_n` is bounded.
2. Use the B. W. Theorem to give a subsequence of `a_n` that
converges to `L`.
3. Show `lim_{n->oo}a_n = L`.
2-19 The intermediate value theorem (again!). I V Theorem [0 version]: If `f ` is continuous on `[a,b]`
and `f(a) * f(b) < 0` then there is a number `c in (a,b) `
with `f(c) = 0`. Proof: Outline [latest revision 2-21]. Create
sequences `a_n` and `b_n` where `a_0 = a, b_0 = b , a_n<b_n,
a_n <=a_{n+1}, b_{n+1}<=b_n` and for any `n`, `
f(a_n)* f(a_{n+1}) >0 , f(b_n)* f(b_{n+1}) >0, [so] f(a_n)
* f(b_n) < 0` and `|a_n - b_n| = (b-a)/ {2^{n}}`.
Then both sequences are monotonic and bounded and converge to
the same number `c in (a,b)`. Using continuity show that `f(c) =
0`.
Here's how to construct the `a_n` and `b_n`. [The
basic idea is bisection.]
Let `a_0=a` and `b_0 = b`. Now let `m_0 = {a_0+b_0}/2.`
If `f(a_0)*f(m_0) =0 `, we have found the desired `c=m_0` for
the theorem... stop here. :)
If `f(a_0)*f(m_0) > 0 `, let `a_1 = m_0` and `b_1=b_0`.
If `f(a_0)*f(m_0) < 0 `, let `a_1 = a_0` and `b_1=m_0`.
Then `a_1<b_1, a_0 <=a_{1}, b_{1}<=b_0` and `
f(a_0)* f(a_{1}) >0 , f(b_0)* f(b_{1}) >0, [so] f(a_1) *
f(b_1) < 0` and `|a_1 - b_1| = (b-a)/ {2}`.
Continue in the same way to construct `a_{n+1}` and `b_{n+1}`
from `a_n` and `b_n`:
Now let `m_n = {a_n+b_n}/2.`
If `f(a_n)*f(m_n) =0 `, we have found the desired `c=m_n` for
the theorem... stop here. :)
If `f(a_n)*f(m_n) > 0 `, let `a_{n+1} = m_n` and
`b_{n+1}=b_n`.
If `f(a_n)*f(m_n) < 0 `, let `a_{n+1} = a_n` and
`b_{n+1}=m_n`.
Then ` f(a_n)* f(a_{n+1}) >0 , f(b_n)* f(b_{n+1})
>0, [so] f(a_{n+1}) * f(b_{n+1}) < 0` and `|a_{n+1} -
b_{n+1}| = (b_n-a_n)/ {2} = 1/2 (b - a)/{2^n} =
(b-a)/{2^{n+1}}`.
Continuation of proof :
Now the sequences `{a_n} ` and `{b_n}` are monotonic and
bounded, so they each have a limit, call them `c_a` and
`c_b`.
Noting that `|a_n - b_n| = (b-a)/ {2^{n}}` it is an
exercise for the reader to show that `c_a = c_b`.
We let `c = c_a = c_b` and claim that `f(c) = 0.`
Since `f(a_n)*f(a_{n+1}) >0` for all `n`, we can say the
limit of `f(a_n)` is `f(c_a)` by continuity , and
thus `f(a)*f(c_a)>=0`.
Likewise, since `f(b_n)*f(b_{n+1}) >0` for all `n`, we can
say the limit of `f(b_n)` is `f(c_b) ` by continuity
, and thus `f(b)*f(c_b)>=0`.
But since `c_a = c_b = c`,
we have `f(a)*f(c)>=0` and
`f(b)*f(c)>=0`,
and thus by multiplication,
`f(a)*f(c)*f(b)*f(c)>=0` (`\square`)
But from the hypothesis, `f(a)* f(b) <0`,
and `f(c)*f(c) >=0`. If `f(c)*f(c) >0` then `f(a)*f(c)*f(b)*f(c)<
0`, a contradiction of (`\square`)
Thus `f(c)*f(c) = 0` and `f(c)= 0`. EOP.
Cor. I V Theorem [general version]: If `f ` is continuous
on `[a,b]` and `v` is between `f(a) ` and ` f(b)` then
there is a number `c in (a,b) ` with `f(c) = v` Proof: Consider `h(x) = f(x) - v ` for all `x in
[a,b]`. Then `h ` is continuous on `[a,b]` and ` h(a)*h(b) =
(f(a) -v)*(f(b)-v)<0 `, so `h` satisfies the hypotheses of IV
T[0 -version], so there is a number `c in (a,b) ` where `h(c) =
f(c) -v = 0.` Thus ` f(c) = v`. EOP.
2-22 Review of Rational
Numbers `Q` and Irrational Numbers, `R-Q`: Def'n. A number `x` is a rational number
if there exist integers `n` and `m` with `m \ne 0` where `x =
n/m`.
A number `x` is a d-rational number if there
exists an integer `q` and a sequence `{d_i}_{i=1,2,...}` where
`d_i in N`, `0<=d_i<=9`
with `x=q +.d_1d_2 ... = q + \sum_{i=1}^{oo) d_i*10^{-i}`,
and there exist `M` and `r` `in N` where if `i>=M` then
`d_{i+r}=d_i.`
Examples: `3/5` and `{-11}/7` are rational numbers.
`sqrt{3}` is not a rational number.
`57.4368686868...` is a d-rational number, `q =57, d_1=4,
d_2 = 3, d_3=6, d_4=8, d_{3+2j}=6, d_{4+2j}=8` for ` j in
N`. `M=3, r=2`. Theorem: `x`
is a rational number if and only if `x` is a d-rational number. Proof outline:
`=>`: Suppose `x` is a rational number, ` x= n/m` with
`m,n in Z` and `m > 0`. Do the "long division". By the
division algorithm for natural numbers, the remainders `R in N`
are always `0<= R < m`, so eventually the long division
will give repeating blocks of digits and this shows `x` is an d-
rational number.
`\Leftarrow`: Suppose `x` is a d- rational number, `x=q
+.d_1d_2 ... = q + \sum_{i=1}^{oo) d_i*10^{-i}`. Then consider
`10^r x - x = (10^r -1)*x ` a terminating decimal, which shows
that `x` can be recognized as a rational number by some
simple arithmetic. "Density" Proposition: If `a` and `b` are rational numbers with `a
< b`, then there is an irrational number `c` with `a < c
< b`. Proof: Using the linear function `f(x) = a + (b-a)(x-1)`
we see that `f(1) = a, f(2) = b`, and if `1<x<2`
then `a<f(x) <b`.
Since `1< sqrt{ 3}< 2`, we have `c = f(sqrt{ 3}) in
(a,b).`
But if c is a rational number then `c = a +
(b-a)(sqrt{ 3} -1) ` is a rational number.
Solving for `sqrt{ 3}` we have `sqrt{ 3} = {c-a}/{b-a} + 1`, a
rational number. This contradicts the fact that `sqrt{ 3}` is
not a rational number, so `c` is not a rational number, i.e.,
`c` is an irrational number.
EOP. Proposition: If `a` and `b` are irrational numbers with
`a < b`, then there is an rational number `c` with `a < c
< b`.
Proof: Suppose `a` and `b` are irrational numbers with `a<b`.
Treating these as infinite decimals `a= n_a + .a_1a_2a_3... `
and `b= n_b + .b_1b_2b_3...`. there is some decimal position
where these two numbers differ. If `n_a<n_b` then since `b`
is not a rational number `a<n_b<b` and we are done. Id
`n_a= n_b` let k be the smallest natural number where
`a_k<b_k`. Then let `c = n_a+ .a_1a_2...b_k00000...`.
Then `a< c<b`. EOP.
Remark: In fact-- If `a` and `b` are real numbers with `a <
b`, then there is an rational number `c_0` with `a < c_0
< b` and there is an irrational number `c_1` with `a <
c_1 < b`.
Examples: 1. Let `f(x) = 1` if `x` is rational and `f(x) =
0` if `x` is irrational.
Then f is not continuous at any real number.
2. Let `f(x) = x` if `x` is rational and `f(x) = 0`
if `x` is irrational.
Then f is continuous at `x=0` but not continuous at any other
real number.
Week 6 2-25
More examples of some horrible
functions for limits and
continuity, including the mathematician's sine.
Examples: i. If `f(x) =sin(1/x), x \ne 0` then there is no
number `L` where `lim_{x->0} f(x) = L`.
Explanation: Consider the sequence `a_n= 1/{n pi}`.
`lim_{n->oo} a_n = 0` and `lim_{n->oo}f(a_n) =
lim_{n->oo}sin (n pi) =0 `.
While for the sequence `b_n= 1/{pi/2 + 2n pi }`.
`lim_{n->oo} b_n = 0` and `lim_{n->oo}f(b_n) =
lim_{n->oo}sin (pi/2+2n pi) =1 `.
ii. If `g(x) =xsin(1/x), x \ne 0` and
`g(0)=0` then `g` is continuous.
Proof for `x = 0`: Note that `|sin(t)| <=1` for all `t`, so
`|g(x)| =|xsin(1/x)| <= |x| ` for all `x \ne 0`. So
`lim_{x->0} g(x) = 0`.
2-26
Review of definition for limit of a function and continuity
based on sequence limits.
We say `lim_{x->a} f(x) = L` if for any sequence `a_n` in the
domain of `f -{a}` with `lim_{n->oo}a_n = a`,
`lim_{n->oo} f(a_n) = L`.
We say `f ` is continuous on its domain D if for any `a in D`
and for any sequence `a_n` in the domain of `f` where
`lim_{n->oo}a_n = a`, `lim_{n->oo} f(a_n) = f(a)`.
[We will return to continuity and limits later in the course.]
But now a look at the rest of the first year of calculus-
Derivatives and Integrals! Definition of derivative and differentiable
functions. [REVISED* !]
Suppose `f'` is defined on an interval open I and `a in I`. We
say that `f` is differentiable at `a` if there is a
number `L` so that `lim_{x->a} {f(x)-f(a)}/{x-a} = L` or
`lim_{x->a} {f(x) -f(a) - L*(x-a)}/{x-a} = 0`*.
Comment: If `f` is differentiable at `a` then the number
`L` in the definition is unique and is denoted f '(a)
or `Df (a)`.
Examples: (i) If `f(x) = mx + b` then `f '(x) = m`.
(ii)If `f(x) = x^2`, then `f '(3) = 6`.
(iii) If `f(x) = |x|`, then ` f` is not differentiable at
`0`.
:)
"Differentiability implies continuity." Theorem (DIC): If `f ` is defined on an interval and `f`
is differentiable at `a`, then `f ` is continuous at `a`.
Proof: By hypothesis, there is a number `L` so that
`lim_{x->a} {f(x)-f(a)}/{x-a} = L`. Then `lim_{x->a}
f(x) = lim_{x->a} f(a) + {f(x)-f(a)}/{x-a}*(x-a) =
f(a)`.
[Or since `lim_{x->a} {f(x)
-f(a) - L*(x-a)}/{x-a} = 0`, we have
`lim_{x->a} f(x) -f(a) - L*(x-a) = 0`, so`lim_{x->a} f(x)
= lim_{x->a}f(a) - L*(x-a) = f(a)`.]
Thus `f` is continuous at `a`. EOP.
2-28
Proof of linearity, product and chain rules.
These proofs use DIC:
Theorem: Suppose `f ` and `g` are differentiable at `a`, then
`alpha*f +g` and `f*g` are differentiable at `a`.
In fact: `(alpha *f +g)'(a) = (alpha*f ' + g')(a) = alpha*f '(a)
+g'(a)` [Linearity] and
`(f*g)'(a) = (f '* g + g' * f)(a) = f '(a)* g(a) + g'(a) *
f(a).` [Leibniz Rule].
Proof:
See Differentiable
Functions 6.5
Note on Linearity: We can consider the set `F(R,R)= { f:
R -> R }` as a real vector space using function addition and
scalar multiplication. It is a real linear algebra using
function value multiplication.
The subset `DIFF(R,R) = { f : R -> R` where `f ` is
differentiable for all `x in R }` is a vector subspace of
`F(R,R)`
and the transformation `D: DIFF(R,R) -> F(R,R) ` is linear
and in the language of linear algebras is described as a
derivation
because it satisfies the Leibniz Rule.
3-1
Theorem ( Chain Rule): .....
Critical Point Theorem. [This completes first proof of MVT!]
Week 7
3-4 The Flashman version of the
brief history of integration... from Euclid, Archimedes, and
Aristotle to Newton, Leibniz, Cauchy,
and Euler.
Key concepts: Measurement of curves and planar and spatial
regions. Estimation
with finite approximations, use of proportions and trichotomy
law, the "method of exhaustion".
Archimedes physical balancing with infinitesmals or
indivisibles.
Introduction of decimals, logarithms, Descartes "analytic
geometry" allows interpretation of higher powers of numbers as
lengths, general solution of the area problems for most
algebraic curves - except hyperbolae.
Galileo connects position and motion to areas in solving the
problem of motion with constant acceleration. The development of
logarithms as tools for solving trigonometric proportions and
the creation of the hyperbolic/natural logarithm from an area
problem.
Barrow's theorem relates area and tangent problems.
Newton considers curves as related to motion and velocity
determines "slope of tangent line".
Leibniz uses infinitesmals to find area of regions determined by
variables that related as functions.
Inability to find simple arithmetic formulae to determine
"integrals" leads to desire for more precision in estimation and
more accurate definitions returning to precision of Euclid and
the method of Exhaustion to justify and clarify the key concepts
used in the calculus of Euler
and Cauchy.
Note: Euler used a partition where `Deltax_j =Deltax_k ` for all
`j` and `k` , giving
`Sigma_{k=1}^nDeltax_k = Sigma_{k=1}^nDeltax_1 = nDeltax_1 = b -
a`. So, for Euler `Deltax = {b-a}/n`. 3-5
Continuation: How to define the definite integral:
Suppose `f `is a function, `f: [a,b] -> R`. We can interpret
`f` as the height of a curve above the "`x`" axis or as the
velocity of an object moving on a straight line at time `x`.
We can define various sums based on an Euler partition of the
interval with `n` equally sized sub-intervals:
`L_n : Sigma_{k=1}^{n} f(x_{k-1}) Delta x` [A Left Hand Sum]
`R_n : Sigma_{k=1}^{n} f(x_k) Delta x` [A Right Hand Sum]
If we let `m_k = {x_{k-1} + x_k}/2` for `k = 1,2,... ,n` then
`M_n : Sigma_{k=0}^{n-1} f(m_k) Delta x` [A Midpoint Sum].
With these Euler sums we have three sequences determined by the
function `f` and the interval `[a,b]`.
First provisional attempt to define an integral based on
experience using these sums in Calculus numerical integrals:
If there is a number `I` where `lim_{n->oo} L_n =
lim_{n->oo} R_n =lim_{n->oo} M_n = I`
then we say that `f` is integrable over `[a,b]` and we denote `I
= \int_a^b f`.
Comment: In calculus courses it is usually asserted that these
approximations all work for continuous function, as well as
other related estimates.
`T_n = {L_n + R_n}/2` Trapezoidal estimate.
`S_n = 2/3 M_n + 1/3 T_n` Simpson's (Parabolic) estimate.
It is not hard to see also that `lim_{n->oo} T_n =
lim_{n->oo} S_n = I`
If we try to use this definition we would want to basic
properties to hold true:
1. If `f(x) >= 0` for all ` x in [a,b]` then `\int_a^b f
>= 0`.
2. If `c in (a,b)`, then `\int_a^b f = \int_a^c
f + \int_c^b f .`
Example: Consider `f(x) = 1` if `x` is
rational and `f(x) = 0` if `x` is irrational.
On `[0,1]` we have `L_n=R_n=M_n = 1` for all `n`, so `\int_0^1 f
= 1`.
However on `[1, sqrt{3}]` we have `L_n= 1/n; R_n=M_n = 0`
for all `n`, so `\int_1^sqrt{3} f = 0`.
And on `[0, sqrt{3}]` we have `L_n= {sqrt{3}}/n; R_n=M_n =
0` for all `n`, so `\int_0^sqrt{3} f = 0`.
But then if this definition of integration also satisfies
property 2 we have
`\0 = int_0^sqrt{3} f = \int_0^1 f + \int_1^sqrt{3}
f = 1+0=1.`
If property 2 is going to hold then this function leads to
a contradiction. This is an example of why we need to be
more careful in defining the definite integral.
General Euler Integral: Choose a set `C` with `n` points
`C = {c_1, c_2, ... , c_n}` where `c_k in [x_{k-1},x_k]` [An
Euler set].
The define `S_C = Sigma_{k=1}^{n} f(c_k) Delta x` The general
Euler sum that depends on `C`.
We say that ` f` is Euler integrable over `[a,b]` if there
is a number `I` so that as `n -> oo`, the sums `S_C -> I`.
If `I` exists it is called the Euler integral of `f` over `[a,b]
`.
Since the size of the intervals `Delta x = {b-a}/n -> 0` we
can make this more precise:
We say `lim_{n->oo} S_C = I` if given any `epsilon
>0`, there is a natural number `M` so that if `n > M` and
`C` is an Euler set with `n` points, then `|S_C - I | <
epsilon.`
Example revisited: Consider `f(x) = 1` if `x` is
rational and `f(x) = 0` if `x` is irrational.
On the interval [0,1] for any `n` use `c_k = x_k`. Then
for any `n`, `S_C =1.`
But for any `n`, choose `r_k in [x_{k-1},x_k] ` where `r_k` is
irrational. Then for `C = {r_1,r_2, ..., r_n}` we have `S_C =
0`.
Thus the function `f` is not "Euler" Integrable. 3-7 and 3-8
Discussion of integration and connection to continuity.
Week 8 3-11
Discussion of connection between Riemann Integrable functions
and Darboux Integrable functions for bounded functions.
Theorem: Suppose `f ` is a bounded function on `[a,b]`. `f` is
Riemann integrable if and only if `f` is Darboux integrable.
When either of these conditions hold, the Riemann integral of
f equals the Darboux integral of `f`.
Work was started on the proof of this result.
If `P` is a partition of `[a,b]` then any Riemann sum for this
partition `S(P,C)` will satisfy `L(P)<= S(P,C)<=U(P)`. 3-12 Review for exam:
Discussion of assignment due Friday March 8 problem 6:
Suppose f is continuous on `[a,b]` and for all `x,y in [a,b]`
if `x<y` then `f(x) <f(y)` .
a. f([a,b]) = [f(a),f(b)].
b. There is a unique function `g: [f(a),f(b)] -> [a,b]` where
`g(f(t)) = t` for all `t in [a,b]` and `f(g(s)) = s ` for all `s in
[f(a),f(b)]`.
c. `g` is continuous.
Part a - Use definitions and assumptions to prove the set equality.
[IVT used.]
Part b. Show f is one to one.... then review of "inverse
functions" from Math 240. Proof of c.
Consider a sequence `{c_n}` where `lim c_n = c in [f(a),f(b)]`. Then
let `r_n = g(c_n) in [a,b] ` so `f(r_n) = c_n`. Since `{r_n}` is a
bounded sequence, using BW Theorem we have a convergent subsequence
`{s_m}` where `s_m = r_n` for some `n>m` and `lim s_m = s.` Thus
`f(s_m) = f(r_n) =c_n` and by continuity of `f`, `lim f(s_m) = f(s)
= lim c_n = c.` Thus `g(c)= s`. It suffices to show that `lim r_n =
s.`
Suppose not. Then there is an `epsilon >0` where for any natural
number `M` there is a number `j > M` where `|r_j - s| >
epsilon `. Then for all `j`, `s>z = s- epsilon >
r_j` or `s< w = s+epsilon< r_j`. Thus `c= f(s) > f(z)
> f(r_j)=c_j ` or `c = f(s)< f(w)< f(r_j) = c_j`. But this
contradicts `lim c_n = c`. Thus `lim r_n = lim g(c_n) = s =
g(c)` and g is continuous.
3-14. Discussed Monotonicity. Linearity.
3-15. FTofC and inverse function theorem connected to Ln and `e^X` ,
Arctan and tangent functions, Arcsin and sin functions can be
defined through integration and Inverse Function Theorem.
Bounded Constraint Result for Integral.
Week 10
3-25 Listed Main properties of the definite integral that we will
prove:
Included: Monotonicity. Linearity. Additivity. Bounded Constraint.
Continuity of Integral Function for integrable functions. Continuous
Functions are integrable. Fundamental Theorem (Derivative form) for
Continuous Functions. FTofC (Evaluation form) for Continuous
Functions. Mean Value Theorem for Integrals for Continuous
Functions.
3-26. Additivity. Continuity of Integral Function for Integrable
Functions. Continuous Functions are Integrable.
3-28. FTof Calc I and II, MVT for Integrals. Alternative proof of
FTofC using MVT for Integrals.
3-29 Discussion of Improper Integral for Discontinuities on bounded
intervals. Sets of Measure Zero introduced. Countable Sets have
measure zero. Continuity, integrability, and measure zero sets
related. The Cantor Set: has measure 0 but is uncountable.
These lectures and notes are found on a separate link:
Integration Notes from Week 10
Week 11 4-2
Some further discussion of the cantor set.
Let C = the cantor set = ` cap C_j ` for `j in N`. [See Integration
Notes from Week 10 ]
If `f(x) = 1` for `x in C` and `f(x)=0` for `x in [0,1]-C`, then
`f` is Darboux integrable and `\int_0^1f =0`.
Proof: If `P_n ={ 0, 1/{3^n}, 2/{3^n}, ... ,1}` then `U(P_n) =
(2/3)^n ` and `L(P_n) = 0`.
Non-sequence based limits and continuity.
Limits of sequences with intervals:
Note: `lim_{n ->oo} a_n = L` if for any `epsilon >0` there
is a number `M in N` where if `n>M` then `|a_n -L| <
epsilon.`
Rephrased:..... `a_n in ( L-epsilon, L+epsilon)`.
Definition of an open set. Suppose O is a set of
real numbers. O is open if (and only if) for any `L in O` there
is a `epsilon > 0` where `( L-epsilon, L+epsilon) sub `
O.
Example: The interval `(1,3)` is an open set.
Proof: Choose ` L in (1,3)`. Let ` epsilon = 1/2 min{L-1, 3-L}`.
Suppose `x in ( L-epsilon, L+epsilon)`. Then `1=
L-(L-1)\le L- min{L-1,3-L}< L-epsilon< x < L+epsilon
< L + min{L-1,3-L} \le L+ (3-L)=3`. Thus `x in (1,3)` , `(
L-epsilon, L+epsilon) sub (1,3)` and `(1,3)` is
open. EOP
Remark: It can be proven similarly that for any `a < b
`, `(a,b) ` is an open set.
Note continued:
`lim_{n ->oo} a_n = L` if for any open set,`O`, with `L in
O`, there is a number `M in N` where if `n>M` then `a_n in
O.`
Continuity: `f ` is continuous on its domain D
if for any `a in D` and for any sequence `a_n` in the
domain of `f` where `lim_{n->oo}a_n = a`,
`lim_{n->oo} f(a_n) = f(a)`.
With open sets: If `a_n`
is a sequence in
the domain of `f` where `lim_{n->oo}a_n = a` then for any
open set` O` with `f(a) in O`, there is an number `M in N` where
if `n>M` then `f(a_n) in O.`
`epsilon -delta ` Definition of Continuity. `f ` is
continuous on its domain D if for any `a in D` and `epsilon
>0`, there is a real number `delta>0` where if `x in D`
and `|x-a|< delta` , then `|f(x)-f(a)|<epsilon`.
4-4
Example: Let `f(x) = C`, a constant function. Then `f` is
continuous for the set of real numbers.
The justification of this statement is left as an exercise for
the reader.
Example: Let `f(x) = 3x+5`, a linear function. Then `f` is
continuous for the set of real numbers.
Proof: Suppose `a in R` and `epsilon > 0`.
[ We worked with inequalities to see that a likely effective
choice for `delta` was `epsilon /3`.]
Let `delta = epsilon/3`. Then suppose `x in R` and `|x-a|
< delta = epsilon/3`.
Then `|f(x)-f(a)| = |(3x+5) - (3a+5) | = |3(x-a)| = 3|x-a`|.
But we assumed `|x-a| < epsilon/3` so `|f(x)-f(a)| = 3|x-a|
< 3 epsilon/3 = epsilon`.
Thus `f` is continuous for every real number. EOP
Compare the previous proof with the sequence based proof:
Suppose `a_n` and `lim_{n->oo} a_n =a`. Then by applying the
results we have about limits of sequences,`lim_{n->oo} f(a_n)
= lim_{n->oo} 3a_n+ 5 = 3a +5 = f(a)`.
Theorem: Suppose `f` is a function with domain D. `f` is `epsilon-delta` continuous on D if and only
if `f` is sequence continuous on D. Proof: (1) ` = > ` : Suppose `f` is `epsilon-delta`
continuous on D. Assume `a in D` and `a_n in D` with
`lim_{n->oo} a_n =a`. Given `epsilon > 0`, there is a
`delta` where if `x in D` and `|x-a| < delta` then `|f(x) -
f(a)| < epsilon`. Since `lim_{n->oo} a_n =a` there
is a number `M in N` where if `n > M` then `|a_n -a| <
delta` and hence `|f(a_n)-f(a)| < epsilon`. Thus
`lim_{n->oo} f(a_n) = f(a)` and `f ` is sequence continuous
on D.
(2) ` < =` : Assume `f` is sequence continuous
on D. This will be indirect - so we assume `f` is not
`epsilon-delta` continuous on D. Then there is an element
of D, `a`* and a positive real number, `epsilon`*` > 0` where
for any `delta > 0` there is an `x in D` with `|x-a`*`| <
delta` while `|f(x) -f(a`*`)| \ge epsilon`*.
Now use `delta = 1/2` and let `x_1 in D` be so that
`|x_1-a`*`| < 1/2` while `|f(x_1) -f(a`*`)| \ge epsilon`*.
Next use `delta = 1/4` and let `x_2 in D` be so
that `|x_2-a`*`| < 1/4` while `|f(x_2) -f(a`*`)|
\ge epsilon`*.
Continuing use `delta = 1/2^n` and let `x_n in D` be so
that `|x_n-a`*`| < 1/2^n` while `|f(x_n)
-f(a`*`)| \ge epsilon`*.
The sequence `x_n in D` has `lim_{n->oo} x_n =a`* but
`lim_{n->oo} f(x_n) ne f(a`*`)`. This contradicts the
assumption that `f ` is sequence continuous on D. EOP
Recall: Definition of an open set. O is open if for
any `L in O` there is a `epsilon > 0` where `N(L, epsilon
)sub` O.
Review Definitions: Given f : D `->` R, and `A sub
D`
The image of A under f , `f (A) = { y in R : y = f
(x)` for some `x in A}`and
the preimage of A under f, `f ^{-1}(A) = {x in R :
f (x) in A}`.
Example: `f(x) = x^2`.
`f ^{-1}((-1,1))= (-1,1)` ; `f ^{-1}((1,4))= (1,2) cup
(-2,-1)`.
Continuity and open sets Theorem: Given `f
: R ->R` ,
`f` is `epsilon- delta` continuous if and only
if whenever `O` is an open subset of R, `f ^{-1}(O)`
is also an open set.
["A function is continuous if and only if the inverse image of
an open set is open."]
Proof: (1) => : Suppose `f` is
`epsilon-delta` continuous on D.
Suppose `O` is an open set and that `a in f ^{-1}(O)`. Then
`f(a) in O`. Since `O` is an open set, there is an `epsilon
>0` where `N(f(a),epsilon) sub O`. Since `f` is `epsilon-
delta` continuous, there is a `delta >0` where if `x in
N(a,delta)` then `f(x) in N(a,epsilon)`. Thus `N(a,delta) sub
f ^{-1}(O)` and `f ^{-1}(O)` is an open set.
(2) <= : Suppose that whenever `O` is an
open subset of R, `f ^{-1}(O)` is also an open set.
Suppose `epsilon >0` and `a in R`. Then `N(f(a),epsilon)`
is an open set with `a in f ^{-1}(N(f(a),epsilon))`. By the
hypothesis `f ^{-1}(N(f(a),epsilon))` is an open set, so there
is a number `delta>0` where `N(a,delta) sub f
^{-1}(N(f(a),epsilon))`. Then if `x in N(a,delta)` then
`f(x) in N(a,epsilon)` and `f` is `epsilon-delta` continuous. EOP. Week 12 4-8
Prop: (i) `O/` and
`R` are open sets.
(ii) If U and V are open sets, then `U nnn V` is an open
set.
(iii) If O is a family of open sets, then `uuu O`
= { `x in R : x in U` for some `U in O`} is an
open set.
Fact:If `a in R` then `(a, oo )` and `(- oo, a)`
are open sets. Fact: {a} is not an open set.
Cor. : If `a < b` , then `(a,b)` is an open set.
The family of all open sets of R is sometimes called the
topology of R.
Sidenotes: A topological space is a set X together with a family
of subsets `O` that satisfies the three properties:
(i) `O/ in O` and `X in O`. (ii)
If `U , V in O` , then `U nnn V in O`.[Closure
under "finite intersection".]
(iii) If {`U_{alpha}`} is a family of sets with
each `U_{alpha} in O`, then `uuu U_{alpha} = {x in X : x in
U_{alpha}` for some `U_{alpha}` in the family}`in O`.
[Closure under "arbitrary union".] 4-9 Definition of a closed set. A set C is closed if (and
only if) `R - C` is open.
Prop: (i)
`O/` and `R` are closed sets. (ii) If
`a in R` then `[a, oo )` and `(- oo, a]` are closed
sets.
Fact: {a} is a closed set.
Prop: (i) If U and V are closed sets, then `U uuu V` is
a closed set.
(ii) If K is a family of closed sets, then `nnn K`
= { `x in R : x in C` for some `C in K`} is a
closed set.
Cor. : If `a < b` , then `[a,b]` is a closed set.
Connectedness , Continuity, and Topological
Proof of the Intermediate Value Theorem. Def'n: A subset I of the R is an interval: If
a, b `in` I and a < x< b, then x`in`I. Def'n: A subset S of (a metric space) R is disconnected
if there are open sets U and V where
(i) U `uuu` V `sup` S; (ii) U `nnn` S `!=` `O/`
, V `nnn` S `!=` `O/`
, AND `V nnn U = O/`. Def'n: A
subset S of (a metric space) R is connected if it is
not disconnected.
Thus If S is connected with open
sets U and V where
(i) U `uuu` V `sup` S; and
(ii) U `nnn` S `!=`
`O/`
, V `nnn` S `!=` `O/`,
then `V nnn U !=` `O/`.
Theorem: A subset `S` of R is connected if
and only if it is an interval. [Skip to IVT] 4-11
Proof: =>: [Indirect-
contrapositive] Suppose `S sub R` and `S` is not an
interval. Then there exist `a in S` and `b in S` with
`a<b` and `c` where `a<c<b` and `c` not in `S`. Let
`U = (-oo,c)` and `V=(c,oo)`. Then `U uuu V sup S`; `a in U
nnn S ` so `U cap C!= O/` and `b in V nnn S ` so `V cap C!=
O/` and certainly `V nnn U =O/`, so `S` is not connected.
<=: Suppose S is an interval. To show `S` is connected,
suppose open sets U and V where
(i) U `uuu` V `sup` S; and (ii) U `nnn` S `!=` `O/` , V
`nnn` S `!=` `O/` . [We need to show that `V
nnn U != O/`.] Also suppose `V cap U = O/`.
Choose `a in U cap S` and `b in V cap S`. For convenience
assume `a<b`.
Let `R = {t : [a,t] sub U}`.
Since ` a in R`, `R !=O/`. Since `b in V` , if `x in R` then
`x < b` and thus `R` is nonempty and bounded above, so by
the least upper bound property of the real numbers we can
let `z = lub(R)`. Then `a le z le b` and since `S ` is an
interval, `z in S`. So either `z in U` or `z in V`. [We will
show that neither of these alternatives are possible.]
(a) Suppose `z in U`.
Since `U` is open, there is an `epsilon_U>0`
where `N(z,epsilon_U) sub U`. Then `z-epsilon_U <z` is
not an upper bound for `R` and there is `w in R` where
`z-epsilon_U < w le z`. But then `[a,w]sub U` and
`[a,z+{epsilon_U}/2] = [a,w]cup [w,z+{epsilon_U}/2] sub U`.
Thus `z+{epsilon_U}/2 in R` and z is not an upper
bound for R. So `z!in U`.
(b) Suppose `z in V`.
Since `V` is open, there is an `epsilon_V>0`
where `N(z,epsilon_V) sub V`. Then `z-epsilon_V <z` is
not an upper bound for `R` and there is `w in R` where
`z-epsilon_V < w le z`. But then `[a,w]sub U`. But then
`w in U` while `w in N(z,epsilon_V)` so `w in V`. This
contradicts `U cap V=O/` so `z!in V`.
IVT. Theorem: If `f : R ->
R` is continuous and `a<b` and `v` is between
`f(a)` and `f(b)` then there is a number `c in [a,b]` so
that `f(c) = v`.
Proof: Suppose not. Then let `U= (-oo,v)`
and `V= (v,oo)` and
(i) `f([a,b]) sub U uuu V` ; and (ii) `U nnn f([a,b]) !=
O/` , `V nnn f([a,b]) != O/`, AND `V nnnU = O/` . We will show
this implies that `[a,b]` is not a connected set.
Since f is continuous for `R`, `hat U =f
^{-1}(U)` and `hatV =f ^{-1}(V)` are open sets.
Now (i) `hatU uuu hatV sup [a,b]` ; and (ii) `hatU nnn [a,b]!=
O/` , `hatV nnn [a,b] != O/` , AND `hatU nnn hatV = O/`.
[why?] But this contradicts the fact that `[a,b]` is a
connected set. EOP.
Cor: If `a<b`
and `f : [a,b]->
R` is continuous and `v` is between `f(a)` and `f(b)` then
there is a number `c in [a,b]` so that `f(c) = v`.
Proof: Extend the function `f` by letting
`f(x) = f(a)` for `x < a` and `f(x) =f(b)` for `x>b`.
Then `f: R->R` and `f` is continuous so the previous
theorem can be applied. EOP.
Definitions: Suppose `F` is a family of sets
and `A sub R`. We say `F`
is a cover of `A` if
`cup F = {x: x in U` for some `U in F} sup A`.
If `F` is a cover of `A` and every `U in F` is an open
set, then we say `F` is an open cover of `A`. If `hatF` is a subfamily of `F`, i.e., `hatF
sub F`, and `hatF` is a cover of `A`, we say that `hatF`
is a subcover of `F`.
We say that a set`K` is (topologically)
compact if any open cover `F`
of `K` has a subcover `hatF` that is a finite set,
i.e., for any open cover `F` of `K`, there exist
`U_1,U_2, ..., U_n in F` where `U_1 cup U_2 cup
... cup U_n sup K`.
Examples: i. A finite set is compact.
ii. `{1,1/2,1/3,...}={x:x=1/n` for `n>0, n in N}` is
not compact. [This set is not closed!]
iii. `{0,1,1/2,1/3,...}={0}cup{x:x=1/n` for `n>0, n in
N}` is compact.[This set is closed.]
iv. `{1,2,3,...}={x:x=n`
for `n>0, n in N}` is not compact. [This set is closed,
but not bounded.]
Heine Borel Theorem (simplest version):
`[a,b]` is a compact subset of R. Proof: (outline). Suppose O is a family of open
sets where `uu O sup [a,b]`. Let G= {x `in` `[a,b]`
where `[a,x]` is covered by a finite subset of the members
of O.}. Claim ( "easy"): G `!=`
`O/` and G is bounded
by `b`. Then let `alpha` = lub of G.
Show `alpha in ` G and then `alpha = b`
See also : Closed_Bounded_Subset_of_Real_Numbers_is_Compact
Week 13 4-15
Theorem: A closed subset of a compact set is
compact.
Proof: Supposed
`C` is a closed set and `K` is a compact set with `C sub
K`. Now suppose O
is a family of open sets where `uuu O sup C`. Since `C` is
closed, let `U=C^c` and `hatO = O cup {U}`. Then `hatO`
covers `K` and by compactness a finite subcover of `hatO`
covers `K sup C`. But then if needed that finite subcover
less `U` will be a finite subcover of `O` that covers `C`.
So `C` is compact. EOP.
Corollary: If K is a closed and bounded
subset of R, then K is compact.
Proof: K is bounded- so K `sub [a,b]` for some `a` and
`b`. But by HB, `[a,b]` is compact, so K is a closed
subset of a compact set thus K is compact. EOP
Theorem: If `K` is a compact subset of `R`,
then `K` is closed and bounded. [HB in R] Proof:
(i) `K` is bounded: Consider the family of open
sets, `O = {N(0,n) for n = 1,2,3.... }`. Certainly this
family covers any subset of `R`, so it covers `K`. But a
finite subcover will be bounded by `n+1` for the largest
`n` of the sets in that finite subcover. Thus `K` is
bounded.
(ii) `K` is closed. Let `U` = the complement
of `K = K^c`. We show that `U` is open.
Suppose `p in U` and `x in K`.
Then let `r(x) = |x-p|/3` so `r(x) > 0` and `N(p,r(x))
nnn N(x,r(x)) = O/`.
Let `O` be the family of open sets `O= {N(x, r(x)) : x in
K}`. Then `O` covers `K` and because `K` is compact
there a finite number of points `x_1, ... , x_m` where the
family of open sets `{N(x_k, r(x_k)) : k = 1,2,..., m}`
covers `K`.
Then let `r = min{r(x_k) : k = 1,2,..., m}` and we have
that `N(p,r) nnn N(x_k,r(x_k)) = O/` for all k.
Hence `N(p,r) nnn K = O/`
Thus `N(p,r) sub U` and `U` is an open set. EOP.
4-16 Compactness and
continuity. Theorem:If
`f: R -> R` is continuous and `K` is a compact
then `f(K)` is compact.
Proof: To show
`f(K)` is compact, suppose `O` is an open cover of `f(K)`.
Because `f` is continuous, for each `U in O`, there is an open
set `hatU` where `hatU= f ^{-1}(U)`.
Let `hatO = { hatU : U in O}`. Note that `hatO` is open
cover of `K` , and since `K` is compact, `hatO` has a finite
subcover of `K`. For these (finite) open sets we have
corresponding `U` so that `hatU=
f ^{-1}(U)`. Thus ....
these `U` are a finite subcover of the family `O`. Topological proof of the
Extreme Value Theorem.
Sequences and series: Review some of
the properties for convergence related primarily to numbers.
Note how geometric series could be thought of as a sequence
of functions.
Review and prove:
Key: `1/{n!} \int_c^x (x-t)^n f^{n+1}(t)dt = 1/{n!} f^{n+1}(r)
\int_c^x (x-t)^n dt = 1/{(n+1)!} f^{n+1}(r) (x-c)^{n+1}`
Proof of Taylor Theorem (Lagrange Remainder
based on MVT)
Proof (for `a=0` from Sensible Calculus ): Let
where we assume is not
and is in
the interval I.
Then
and
as well. Thus, [using the Mean Value Theorem or Rolle's Theorem]
for some between
and we have
.
But, after extensive use of the product rule, we find that
so .
Now using
and
we can conclude that
or .
EOP
Given the integers as an
ordered integral domain ("commutative ring with unity and no
zero divisors") consider P= {(a,b): a,b integers and b not
0} and the equivalence relation on P where (a,b) ~ (c,d) [in
P] if and only if ad= bc.
As a relation on P, ~ is symmetric, reflexive, and
transitive.
Let [a,b] = { (r,s) in P : (a,b) ~ (r,s) } and show this
partitions P, i.e., (i) every (a,b) in P is in [a,b], and
(ii) if [a,b] amd [c,d] have some element in common, then
[a,b]= [c,d] - as sets.
The rational numbers Q as a set = {[a,b]: a,b are integers
and b is not 0.}
Define operations on Q for addition and
multiplication-
Proposition: These operations are "well defined" and
with them Q is a field.
Finally: Characterize the positive rational numbers
and from this define an order relation on Q.
The result makes Q an ordered field with a 1:1
function from Z to Q defined by `a ->[a,1]` that
preserves the order and algebraic structures of Z.
Defining the real numbers:
Consider the ways we describe real numbers- a
discussion of the decimal notation as a sequence of rational
numbers.
Note its connection to infinite series using powers of
10 which converge by comparison with geometric series.
This led to a discussion of other sequences that
characterize real numbers- in particular e as the
limit of the sums from the Taylor series `sum 1/(n!)` and
the limit of the powers - `(1+1/n)^n`.
To define real numbers we need to eliminate the assumptions
of a limit existing for these sequences of rational numbers.
We use the theory that characterizes convergence with the
Cauchy condition. This is the start of a more general
definition of a real number using "cauchy sequences" of
rational numbers.
Let QC = { S={`a_n`} : where `a_n` is a rational number for
each n and S is a Cauchy sequence }
Define an equivalence relation on QC , namely S~T [T =
{`b_n`} in R] if for any positive rational number `epsilon`
there is a natural number M where for any k >M, `|a_k -
b_k| < epsilon`.
Proposition: ~ is a reflexive, symmetric, and transitive
relation.
This leads to the partition of QC into equivalence classes
[S]= {T in R: S~T}. `R` = {[S]: S is in R} can then be
used to define operations and an order relation [
inherited from Q] so that `R` is an ordered field which
satisfies the least upper bound property. Thus `R` is a complete
ordered field- "the real numbers." wikipedia.org
on the Construction_of_the_real_numbers Review:Axioms for the Real numbers-
Construction
of the real numbers- revisit on line- including
the proof of the least upper bound property.Definition
of uniform convergence
Below this line are notes from Math 415 Fall,
2008
Functions:
Discussion of the intermediate value theorem, its proofs,
and counterexamples related to its hypotheses;
Comment: the proof of IV Theorem broke into three cases-
one of which actually found the number c = (a+b)/2. This
case actually did construct the desired number - and
demonstrated it was a number with the required properties
using the fact that the real numbers are a field with an
order relation! This was connected to the first Motivational
question on the nature of the real numbers.
Alternative proofs of the MVT [ Direct and geometric]
(ii) If `a in R` then `(a, oo )` and `(- oo,
a)` are open sets.
Fact: {a} is not an open set.
Prop:
Discuss:
An investigation of functions and comparing them to numbers-
in particular we introduce two distinct norms on functions:
the sup norm and the L2 norm.
These will be discussed later.
We turned our attention to sequences and series
reviewing some of the properties for convergence related
primarily to numbers. We noted how geometric series could be
thought of as a sequence of functions.
We reviewed and proved:
R with b 
a for all a 
a for all
a