Math 316 Notes and Summaries This page requires Mozilla/Firefox/Netscape 7+ or Internet Explorer 6+ MathPlayer .

Martin Flashman's Courses
MATH 316 Real Analysis I Spring, 2013
Class Notes and Summaries
[Based on notes from 2008]
 Monday Tuesday Thursday Friday Week1: Beginnings. 1-21 MLK day 1-22 1-24 1-25 Week 2 1-28 1-29 1-31 2-1 Week 3 2-4 2-5 2-7 2-8 Week 4 2-11 2-12 2-14 2-15 Week 5 2-18 2-19 2-21 2-22 Week 6 2-25 2-26 2-28 3-1 Week 7 3-4 3-5 3-7 3-8 Week 8 3-11 3-12 3-14 3-15 Week 9 Spring Break Week 10 3-25 3-26 3-28 3-29 Week 11 4-1 CC day 4-2 4-4 4-5 Week 12 4-8 4-9 4-11 4-12 Week 13 4-15 4-16 4-18 4-19 Week 14 4-22 4-23 4-25 4-26 week 15 4-29 4-30 5-2 5-3 week 16 5-6 5-7 5-9 5-10

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.]
• stanford.courses math M115 summary.pdf
On Proofs:
• How to Read and Do Proofs by D.Solow
• The Keys to Advanced Mathematics by D. Solow
• How to Solve It by G. Polya
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.

1-25

### Space Filling Curves - YouTube

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
1. For all a,b in X, m(a,b)=m(b,a).
2. If  m(a,b) = 0 then a = b.
3. 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) .

### See the MVT on http://www.cut-the-knot.org

Some theorems that rely on the Mean Value Theorem for their proof.
• If f'(x) = 0 for an interval then there is a real number K 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;
1-31
• Counter examples to the MVT:
(1) without Continuity: f (x) = for in (0, 1] and f(0) = 1 on the interval [0,1]

and
(2) without differentiability:  f (x) = 1/2 -| x - 1/2|   for in [0, 1].
Also  f (x) = (1/2)^(1/3) -|( x - 1/2)|^(1/3)   for 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].
• Intermediate Value Theorem
• Extreme Value Theorem

Real numbers:
Sensible Calculus Chapter 0 on numbers
• 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 implies b=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 implies b=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 case n·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".
The order axioms from http://www-groups.mcs.st-andrews.ac.uk/~john/analysis/Lectures/L5.html

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 thatf((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|.

• ## Axioms for the Real numbers-

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 b R with b a for all a A.
An element M R 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 b M.

A lower bound of a non-empty subset A of R is an element d R with d a for all a A.
An element m R 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 m d.
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.

• ## 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}.

## Rethinking the proof of The intermediate value theorem (IVT) using the LUB property. Opportunity: Student presentations on applications of the IVT?

• 2-8
Motivation Question(s) II  What does  limit mean? for numbers? for function values? for functions?  for sets? What does continuity mean?

• Sequences of Numbers- Convergence
• 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.
Convergence in the Reals

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.

Cauchy criterion for convergence:

Definition

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.

For more details of the proof see Cauchy criterion for convergence.

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.

Begin with proof of the extreme value theorem.
2-21
The boundedness theorems
.

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).

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, solim_{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.

For alternative views see  http://en.wikipedia.org/wiki/Mathematical_analysis#History.
and http://en.wikipedia.org/wiki/Timeline_of_calculus_and_mathematical_analysis

Beginnings of Integration: An interval [a,b] with a < b.
A partition of the interval: P = {a=x_0 < x_1 < x_2< ... < x_{n-1} < x_n=b}.
The partition P determines closed intervals [x_k, x_{k+1}] where k = 0, 1, 2, ... , n-1.
These intervals give rise to numbers:Deltax_k = x_k - x_{k-1} for k = 1,2,3,..., n.

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.

• Various definitions and an example of a function that is not integrable.
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.

See these on-line references:
Continuity of Real functions
Images of intervals

Limits of functions

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

See the following link for a similar presentation of the argument.
The epsilon-delta definition : Equivalence of definition with sequence definition.

4-5
Open, closed, bounded. Openness and continuity.
Open and closed sets in metric spaces (consider the  real numbers with the metric given by the absolute value of the difference).[Topology]

Notation: N(L, epsilon) = ( L - epsilon, L+ epsilon) = {x: |x-L| < epsilon}.

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.

Continuity and open sets

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 xinI.
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.

EOP.

More on Connected and disconnected sets

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.

Continuous functions and connected sets

Compact sets(topological definition)
• 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

• 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.

Another application of Compactness:
4-18
Uniform Continuity:
Demonstration on wolfram.com of UniformContinuity

Theorem: Any continuous function on [a,b] is Darboux (or Riemann) integrable.
Proof: [An application of uniform continuity.]

4-19
Theorem: A continuous function on a compact set of real numbers was uniformly continuous.
Proof: ...http://pirate.shu.edu/~wachsmut/ira/cont/proofs/ctunifct.html
Heine-Cantor_Theorem

Week 14

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:
•  the divergence test,
• geometric series,
• convergence of positive series,
• the comparison test,
• the integral test,
• the harmonic series,
• p- series and
• absolute convergence implies convergence.

The Ratio and Root Tests and other stuff.
Rearrangements of series and conditional convergence
Power series
Series and Power Series

5-6
Taylor's theorem: approximating functions with partial sums of power series
Taylor Series (with remainders - Integral & Lagrange)
Proof of Taylor Theorem (Integral remainder)
Proof of Taylor Theorem(Lagrange Remainder)

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  $g(t)=f(t)+f'(t)(b-t)+f''(t)2(b-t)2+...+f(n)(t)n!(b-t)n+Rn(b)b(n+1)(b-t)(n+1)$ where we assume $b$ is not $0$ and $t$ is in the interval I.

Then $g(0)=Pn(b,f)+Rn(b)=f(b)$ and $g(b)=f(b)$ as well. Thus, [using the Mean Value Theorem or Rolle's Theorem] for some $c$ between $0$ and $b$ we have $g'(c)=0$. But, after extensive use of the product rule, we find that $+[f(n)(t)n!n(b-t)(n-1)(-1)+f(n+1)(t)n!(b-t)n]+Rn(b)bn+1(n+1)(b-t)n(-1)$

so $g'(t)=f(n+1)(t)n!(b-t)n-Rn(b)bn+1(n+1)(b-t)n$.  Now using $t=c$ and $g'(c)=0$ we can conclude that $f(n+1)(c)n!(b-c)n=Rn(b)bn+1(n+1)(b-c)n$ or   $Rn(b)=f(n+1)(c)(n+1)!b(n+1)$.   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

Uniform limits of continuous functions are continuous

Uniform limits of Riemann integrable functions

The Weierstrass M-test

Convergence of power series

A nowhere differentiable function
Metric spaces: definition and examples
Convergence in a metric space
Convergence in infinite dimensional spaces
Properties of uniform convergence
The Weierstrass approximation theorem

# 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:
•  the divergence test,
• geometric series,
• convergence of positive series,
• the comparison test,
• the integral test,
• the harmonic series,
• p- series and
• how absolute convergence implies convergence.

The Ratio and Root Tests
and other stuff.
Rearrangements of series and conditional convergence
Power series
Series and Power Series

Taylor's theorem: approximating functions with partial sums of power series
Taylor Series (with remainders - Integral & Lagrange)

Proof of Taylor Theorem (Integral remainder- Inductive based on Integration by parts)
Proof of Taylor Theorem(Lagrange Remainder base on integral remainder)

Definition of uniform convergence

Uniform limits of continuous functions are continuous

Uniform limits of Riemann integrable functions

The Weierstrauss M-test

Convergence of power series

A nowhere differentiable function

Metric spaces: definition and examples
Convergence in a metric space
Convergence in infinite dimensional spaces
Properties of uniform convergence
The Weierstrass approximation theorem

More on the definition of the real numbers:
Review:
Axioms for the Real numbers-

Construction of the real numbers- revisited on line-
including the proof of the least upper bound property.
["Close to what we outlined last class.]

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