Martin Flashman's Courses
MATH 316 Real Analysis I Spring, 2016
Class Notes and Summaries
[Based on notes from 2013]


Monday
Tuesday
Thursday
Friday
Week1:
Beginnings.
1-18
MLK day
1-19
1-21
1-22
Week 2 1-25
1-26
1-28
1-29
Week 3 2-1
2-2
2-4
2-5
Week 4
2-8
2-9
2-11
2-12
Week 5
2-15
2-16
2-18
2-19
Week 6
2-22
2-23
2-25
2-26
Week 7
2-29
3-1
3-3
3-4
Week 8
3-7
3-8
3-10
3-11
Week 9
Spring Break
Week 10
3-21
3-22
3-24
3-25
Week 11
3-28
3-29
3-31
CC Day
4-1
Week 12
4-4
4-5
4-7
4-8
Week 13
4-11
4-12
4-14
4-15
Week 14
4-18
4-19
4-21
4-22
week 15
4-25
4-26
4-28
4-29
week 16
5-2
5-3
5-5
5-6


Week 1:
1-19: Introduction.

The focus of the course will be to understand 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 analysis 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.
We will look at some of the key theorems of calculus and analyze 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.
History: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. This work formed a foundation for the new empirical approach to science as a whole ( in opposition to religion). a major criticism of the calculus was found in the work of  Bishop Berkeley (The Analyst) . The efforts of the next hundred plus years arrives at a careful definition of the real numbers comes toward the end of the 19th century with separate works by Weierstrass and Dedekind. At times I will suggest reading from original sources to illuminate some of the difficulties and key issues in the development of a rigorous analysis for the calculus.

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 specific unit for measurement.

We will visualize real valued functions from the real numbers with both graphs and mapping diagrams.
Mapping diagrams 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]  as well as in Mapping Diagrams from A(lgebra) B(asics) to C(alculus) and D(ifferential) E(quation)s. A Reference and Resource Book on Function Visualizations Using Mapping Diagrams.
Details of the syllabus will be discussed  further on Thursday.

1-21
  Organization of the course: details like tests, homework, etc. as described in the main course page.

Reference to Polya's : How to Solve It... available on line: https://notendur.hi.is/hei2/teaching/Polya_HowToSolveIt.pdf
Polya describes  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.

Next class: Discuss 5 theorems at the heart of calculus.
 Mention other texts in Real Analysis:
On Proofs:   Watch video  "Space Filling Curves"Space Filling Curves - YouTube 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.
1-22

Space Filling Curves - YouTube

Also you can see this in the library on shelf- please replace accurately! VIDEO2577
Week 2:
1-25
Continuation on space filling curves:
To make sense of curves being close and sequence of curves having limits, we consider 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:  ` X xx X -> R_+ uu {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 euclidean 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`.

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

Some theorems that rely on the Mean Value Theorem for their proof.
1-29
More on the MVT;
1-31

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].

Real numbers:
Sensible Calculus Chapter 0 on numbers



Other field results: (-1)*a = -a;   -(-a)=a;  (-a)*b= -(a*b);  (-a)*(-b)= a*b

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 many simple algebraic or order properties of R.

Examples

  1. 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.

  2.   If a > 0 and b > c then a b > a c.
    Proof
    b > c if and only if b - c > 0
    Hence a (b- c) > 0 and so ab - ac > 0 and we have ab > ac.
  3. For and real numbers `a` and `b` with `a<b`, `a < (a+b)/2 < b`
    Proof:
    `a+a < a + b < b + b` so `a = 1/2 (a+a) < (a+b)/2 < 1/2 (b+b) = b`.


2-2

Remarks on Assignment 2.
    Absolute value inequalities: Proofs were best 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|.`


Using inequalities with decimals to prove the IVT:
Outline:

Assume  `f(a) < P < f(b)`.
Consider `a < (a+b)/2 < b` and apply the trichotomy property to `f((a+b)/2)` and P.
This may give that`f((a+b)/2)=P` and we have the conclusion needed to prove the theorem. 
Otherwise consider a smaller interval depending on whether
`f((a+b)/2)>P` or `f((a+b)/2)<P.`
Continuing with this process on smaller (and smaller) intervals will lead to a sequence of  intervals of lengths `(b-a)/{2^n}` and eventually, the estimation of a decimal number `c` will arise.
Because of  the continuity of `f`,
it must be that `f(c) = P.`
Note: We interpret continuity here informally as "when `x  approx c` then `f(x) approx f(c)`  " and use that and the trichotomy law  of inequality to understand that `f(c)` must equal P. we will clarify this as we evolve a more precise definition of continuity.

An example created with GeoGebra illustrating this "bisection" argument with a graph and mapping diagram for the function `f(x) = x^2 -2` on the interval `[0,2]`.


Bisection and Proof Outline of IVT vizualized with GEOGEBRA.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com



2-4

Axioms for the Real numbers- from A first Analysis course by John O'Connor


What distinguishes `R` from `Q` (and from other subfields of `R`) is the Completeness Axiom.

Definitions:
    An upper bound of a non-empty subset A of R is an element `b in R` with `b ge a` for all `a in A`.
    An element `M in 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 ge M`.

    A  lower bound of a non-empty subset A of R is an element `b in R` with `b le a` for all `a in A`.
    An element `m in R` is a greatest lower bound or infimum of A if
    `m` is an lower bound of `A` and if `b` is a lower bound of `A` then `b le m`.
   


The Completeness Axiom

    If a non-empty set A has an upper bound, it has a least upper bound.

Look at Exercises 3 from A first Analysis course by John O'Connor


Motivation for believing Completeness Axiom:
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. ...

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.

Credibility without proof for the Completeness:
We can use decimals, representing real numbers, to help give credibility to the belief that the Completeness Property is true for the real numbers.

Discussion: Since S is nonempty, there is an `x_0 in S`. Suppose B is an upper bound for S, so `B>=x_0`. If `B = x_0` then 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`.  [Why?]

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. If `m_1 in S ` then `m_1` is the least upper bound of S.  Otherwise 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_1*0.1` where `0< k_1 le 10` is an integer and `n_1+k_1*0.1` is an upper bound for S but `n_1+ (k_1-1)*0.1` is not an upper bound for S.

Let `m_2 = n_1 +k_1*0.1`
. If `m_2 in S ` then `m_2` is the least upper bound of S.
Otherwise,  let `n_2 = n_1 +(k_1-1)*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_2*0.01` where `0< k_2 le 10` is an integer and `n_2+k_2*0.01` is an upper bound for S but `n_2+ (k_2-1)*0.01` is not an upper bound for S.
Let `m_3 = n_2 +k_2*0.01`. If `m_3 in S ` then `m_3` is the least upper bound of S.
Otherwise,  let  `n_3 = n_2 +(k_2-1)*0.01` and now consider the interval `[n_3,m_3]`.

We can continue in this fashion. After each step that continues 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 indirect 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
here as this argument is given only to suggest the truth of the completeness property for real numbers.

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 that is an upper bound for N, and thus there is 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`.


2-5

Another important order property of the real ( and rational) numbers

The Archimedean Property of the real (and rational) numbers:
Proposition: If `r ` is a positive real ( or rational) number, then there is natural number `n' > 0` where `0 < 1/{n'} < r`.
Proof:  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.
Alternative Proof: [A decimal approach.] 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`.


Greatest Lower Bound Theorem:

If a non-empty set B has a lower bound, it has a greatest lower bound.
Proof: Suppose `B` is a non-empty set and `L` is a lower bound for `B`.
Let `A =` { `x : x= -y ` for some `y in B`}.
Then `A` is non-empty and `-L` is an upper bound for `A`, so by the completeness axiom, there is a number `M` where ` M` is the least upper bound of `A`. Then `-M` is the greatest lower bound of `B`. [Justify this last statement using the fact that if `a le b` then `-a ge -b`.]






Rethink proof of The intermediate value theorem (IVT) using the LUB property. 


Assume  that `f` is continuous on the interval `[a,b]` and `f(a) < 0 < f(b)`. The theorem concludes that there is a number `c in (a,b)` with `f(c) = 0`.
For the purpose of this "proof" we interpret "continuous at ` t` " as meaning that when `x ~~ t`
then `f(x) ~~ f(t)`.

The Plan: Use the completeness property for a subset of `[a,b]` to find a candidate for `c` and then show that `f(c) =0` by showing that `f(c)` cannot be positive or negative, leaving `f(c) = 0` as the only possibility.

The Set: Let `X = \{ x in [a,b]:` for all `y in [a,x], f(y) le 0 \}`. `X` is not empty since `f(a) <0`, and thus there is some small interval `[a, a+epsilon]` where
by continuity `f` is negative for all `a le x le a +epsilon`.
`X` is bounded above by `b`, so by  the completeness of the real numbers, it has a least upper bound, which we denote `c`.

From the remark that showed that `X` is not empty, it also follows that `a<c`.
Because `f(b) >0` there is some small interval `[b - epsilon, b]` where by continuity `f` is positive for all `b -epsilon le x le b`. Thus `c<b`

We now proceed to examine `f(c)` showing that `f(c) < 0 ` and `f(c) > 0` are both impossible, leading to the conclusion that `f(c) = 0`.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

(i) Suppose `f(c) < 0`. Then by continuity, there is an interval, `I`, containing `c` where `f(y)< 0` for all `y in I`.
In `I` there is some `s` with `s<c`, so `s` is not an upper bound for `X`. Thus there is some `x`* in X with `s<x`*`<c`.
Notice that there is some `r` with `r in I` and `c< r`.
Considering `[a,x`*`]` and `[x`*`,r]` we have that for all `y in [a,r], f(y) le 0`. Thus `r in X`. This contradicts that `c` is an upper bound for `X`, so  `f(c) < 0` is false.

(ii) Suppose `f(c)>0`. Then by continuity, there is an interval, `J`, containing `c` where `f(y)> 0` for all `y in J`.
But then there is some `s` with `s<c`,  with  `f(s) > 0`. So if  `x in X` then `x < s` and thus ` s` is an upper bound for `X` smaller than `c`. This contradicts that `c` is the least upper bound of `X`, so `f(c) >0` is false

Since (i) and (ii) have eliminated `f(c) < 0` and `f(c) > 0` as possibilities, it must be that ``f(c) =0`.    EOP 


2-8


GeoGebra Applet for visualizing  sequences with Mapping Diagrams and Graphs for a sequence defined by `a_n = f(n)`. Examples.
   
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com


GeoGebra Applet for visualizing  sequences with Mapping Diagrams and Graphs for a sequence defined by `a_n = f(n)`.
        


This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com



This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com


2-11
Example:
Show that `lim_{n->oo}n/{n+1}` is not `1.1`.
Solution: Supose `lim_{n->oo}n/{n+1}=1.1` . Then let `epsilon = 0.05` and consider any natural number `n`. Then `n/{n+1} lt 1 lt 1.1` and `|n/{n+1}-1.1| gt |1-1.1| =0.1 > 0.05`. So for any natural number `M`, if  `n >M` then `|n/{n+1}-1.1|  > 0.05` which contradicts the definition of limit which says that for any `epsilon >0` there is a natural number M so that if `n>M` then `| a_n - a | = |n/{n+1}-1.1| < epsilon`.

Example: Show that `lim_{n->oo}(-1)^n` does not exist.
Solution: Supose there is a number `L` where `lim_{n->oo}(-1)^n = L` . Then let `epsilon = 0.5`.  So if  `lim_{n->oo}(-1)^n = L` then there is a natural number `M`, where if  `n >M` then `|(-1)^n - L| < 0.5` or `L-0.5 < (-1) ^n < L +0.5`. Consider- if n is even then `(-1)^n = 1` so `L-0.5 < 1 < L +0.5` and `0.5 <L < 1.5`. While if n is odd the `(-1)^n = -1` and `L-0.5 < -1 < L +0.5` so `-1.5<L < -.5`. But then  `L` is both positive and negative - which is a contradiction.

Proposition: [ "Limits are unique."] If  `lim_{n->oo} a_n = A` and `lim_{n->oo} a_n = B` then `A = B`.
Proof: Suppose
`lim_{n->oo} a_n = A` and `lim_{n->oo} a_n = B`.
Let ` epsilon = {|A-B|}/3`. 
By the definition there is a natural number `M_A` where  if `n>M_A` then `| a_n - A | < epsilon`. 
Likewise there is a natural number `M_B` where  if `n>M_B` then `| a_n - B | < epsilon`.
Now let `n = M_B + M_A +1` so that `n > M_A ` and `n > M_B`.
Then `| a_n - A | < epsilon=  {|A-B|}/3` or  `A - {|A-B|}/3 < a_n <A +  {|A-B|}/3`.
Similiarly
`B - {|A-B|}/3 < a_n <B +  {|A-B|}/3`.

If  `B<A` then  these inequalities say `{2 A +B}/3 < a_n< {4A - B}/3` while `{4B - A}/3 < a_n < {2B + A}/3 `.
But this means `{2 A +B}/3 < a_n < {2B + A}/3 ` which implies `2 A +B <  2B + A ` or `A <B`.  A contradiction!
Similarly `B>A` also leads to a contradiction. So it must be that `A = B`.

For a few more examples see: Convergence in the Reals

2-12
Motivation:
Preview of definition for limit of a function and continuity based on sequence limits.
Suppose `a in `domain of  `f`.
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)` or
`lim_{x->a} f(x) = f(a)`

  This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Theorems connected to algebra, inequalities and intervals.
Proof: Suppose {`a_n`}→ `a` and {`b_n`}→ `b` and  we are given `epsilon > 0`.
[Plan:We need to find a natural number `M` so that if `n>M`, we can show that `|a_n+b_n-(a+b)| < epsilon`. We will use the convergence of `a_n` and `b_n` to find `M`.]
For the sequence `\{a_n\}` there is a natural number `M_a` where if `n>M_a` we have that `|a_n-a|< epsilon/3` or `a-epsilon/3 < a_n < a + epsilon/3` .

Similarly for the sequence `\{b_n\}` there is a natural number `M_b` where if `n>M_b` we have that `|b_n-b|< epsilon/3` or `b-epsilon/3 < b_n < b + epsilon/3`.
Now let `M = M_a +M_b +1`.

Then for `n > M`, `n > M_a` and `n> M_b` so
`a-epsilon/3 < a_n < a + epsilon/3` and `b-epsilon/3 < b_n < b + epsilon/3`.

Adding these inequalities gives
`a+b- 2epsilon/3 < a_n +b_n <a + b + 2epsilon/3` or `| a_n +b_n - (a+b) | < 2 epsilon/3 < epsilon.` EOP.

2-15
Class started with discussion of Problem 3 of assignment due Feb. 8.
Work on understanding the problem led to specifying the function as `f(x) = x^2` and intervals `[-1/2, 1/3]` and `[1,2]` and the set `{3,5,sqrt 2}`. Discussion considered the graph and mapping diagram and how the extreme value theorem and intermediate vlue theorems were used to find intervals and number to describe `f(D)`.

Reference was made to the following proof for (i).
Alternative Proof: Choose `M = max \{M_a,M_b\}`.
Then for `n > M`, `n > M_a` and `n> M_b` so
`|a_n + b_n - (a+b)| = |(a_n - a)  + (b_n - b)| le |a_n - a|  + |b_n - b| < epsilon/3 + epsilon/3= 2 epsilon/3 < epsilon.` EOP.

(ii) (an- bn)→ a - b
Proof: Leave till after (iii). Then apply i) and iii) to
(an+ (-1) bn) → a +(-1) b = a - b.

2-16
(iii)
(anbn)→ a b

Proof: Suppose {`a_n`}→ `a` and {`b_n`}→ `b` and  we are given `epsilon > 0`.
[Plan:We need to find a natural number `M` so that if `n>M`, we can show that `|a_n b_n-(ab)| < epsilon`. We will use the convergence of `a_n` and `b_n`  and the fact that convergent sequences are bounded to find `M`. Also we will  consider these products as areas of two rectangular regions: one with sides `a` and `b` and and the other with sides `a_n`and `b_n`.]
This visualizes the algebra that `|a_n b_n - ab| = |(a_n - a)b_n + a(b_n - b)|`.
Let `B>0` with `B> |b_n|` for all `n`. [Explain why this is possible.]
Let `A >0` so that `A>|a|`  so `0 le |a|/A < 1`. 

For the sequence `\{a_n\}` there is a natural number `M_a` where if `n>M_a` we have that `|a_n-a|< epsilon/{3B}`.
F
or the sequence `\{b_n\}` there is a natural number `M_b` where if `n>M_b` we have that `|b_n-b|< epsilon/{3A}`.
Choose `M = max \{M_a,M_b\}`.
Then for `n > M`, `n > M_a` and `n> M_b` so
`|a_n b_n - ab| = |(a_n - a)b_n + a(b_n - b)| le |a_n - a||b_n|  + |a| |b_n - b| < epsilon/{3B}B + |a| epsilon/{3A}< 2 epsilon/3 < epsilon.` EOP.

(iv) (an/bn)a/b (provided bn ≠ 0 and `b` ≠ 0).
Lemma:
`1/{b_n} \to 1/b ` (provided bn ≠ 0 and `b` ≠ 0).
Proof of lemma:
Suppose {`b_n`}→ `b ne 0` and  we are given `epsilon > 0`.
[Plan:We need to find a natural number `M` so that if `n>M`, we can show that `|1/ {b_n}-1/b| < epsilon`. We will use the convergence of  `b_n`, `b_n ne 0`, `b ne 0`  and the fact that convergent sequences are bounded to find `M`.]
Consider that
by (iii) the sequence {`b_n b`} converges to `b^2 ne 0`. 
From this it can be shown (an exercise for the reader) that there is a number `B`
so that `|b b_n|> B>0` and `B/{b b_n} < 1` for all `n`.
For the sequence `\{b_n\}` there is a natural number `M_b` where if `n>M_b` we have that `|b_n-b|< B*epsilon/2`. Then for `n > M_b` we have

`|1/{b_n} - 1/b| = |{b- b_n}/{b b_n}|  =  {|b- b_n|}/{|b b_n|} <  epsilon B/{2 |b b_n|}<  epsilon/2 < epsilon.` EOP.


Application of arithmetic for sequences to limits and continuity of functions:
Example: Let `f(x) = x^2`. Prove that `lim_{x \to 3} f(x)= 9 `[ which `= f(3)`].
Demonstration: Referring to the limit definition for a function we suppose that we have a sequence  {`a_n`} with `a_n ne 3` and `lim_{n \to oo} a_n = 3`. Then we consider the sequence {`f(a_n) = a_n^2`} and apply (iii)  so that 
`lim_{n \to oo} a_n^2 = lim_{n \to oo} a_n cdot a_n = lim_{n \to oo} a_n cdot  lim_{n \to oo} a_n = 3 cdot 3 = 9.`
Notes: 1. This example shows the function `f(x) = x^2` is continuous at `x = 3`.
It is not difficult to show then that this function
`f(x) = x^2` is continuous for any `a`, and so is continuous for the domain `(-oo, oo)`.
2.
Let `g(x) = x^2` when `x ne 3` and `g(3) = 7`. Prove that `lim_{x \to 3} g(x) = 9` [ which ` ne f(3)`].
Demonstration: Referring to the limit definition for a function we suppose that we have a sequence  {`a_n`} with `a_n ne 3` and `lim_{n \to oo} a_n = 3`. Then we consider the sequence {`g(a_n) = a_n^2`} and apply (iii)  so that 
`lim_{n \to oo} a_n^2 = 9.` Here the limit exists as `x` approaches 3, but the limit is not equal to `g(3) =7`, so `g` is not continuous at `x=3`.
3. Similar arguments can show that if `f(x) = b_0 +b_1x+b_2x^2 + ...+b_nx^n` and  `g(x) = c_0 +c_1x+c_2x^2 + ...+c_mx^m`
then as long as `g(a) ne 0`, the function `h(x) = {f(x)}/{g(x)}` is continuous at `x=a`.

2-18
For Friday class discussion:
Theorem: Suppose `f` and `g` are functions with domains `(-oo,oo)`, `h(x) = g(f(x))` for all `x`,  `f ` is continuous at `x=a` and `g` is continuous at `b = f(a)`. Prove `h` is continuous at `x=a`.
Here is a GeoGebra mapping diagram to help you visualize the theorem.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
 

The Squeeze Lemma (for sequences): If `a_n <= c_n <= b_n ` for all `n` and `lim_{n->oo}a_n = c` and `lim_{n->oo}b_n = c`, then `lim_{n->oo}c_n = c`.
Proof: Suppose the hypotheses and that we are given `epsilon > 0`.
[Plan: We need to find an `M` so that if `n > M`, then `|c_n - c| < epsilon`.  Connect `|a_n-c|` and  `|b_n -c|` with
`a_n <= c_n <= b_n ` and the limit assumptions `|c_n-c|`]
Since  `a_n <= c_n <= b_n `,  `a_n -c <= c_n -c <= b_n -c ` and this  `0 <= |c_n -c| <= max \{|b_n -c|, |a_n -c| \}  `
For the sequence `\{a_n\}` there is a natural number `M_a` where if `n>M_a` we have that `|a_n-c|< epsilon`.
For the sequence `\{b_n\}` there is a natural number `M_b` where if `n>M_b` we have that `|b_n-c|< epsilon`.
Let `M= max\{ M_a,M_b\}` and then ... [Finish the argument.]
Class Exercise: State and prove the squeeze lemma (for functions).

2-19
Theorem: Suppose `f` and `g` are functions with domains `(-oo,oo)`, `h(x) = g(f(x))` for all `x`,  `f ` is continuous at `x=a` and `g` is continuous at `b = f(a)`. Prove `h` is continuous at `x=a`.
Proof: Suppose {`a_n`} with `lim_{n to oo} a_n = a`. By continuity of `f` at `x=a` we have that  `lim_{n to oo} f(a_n) = f(a)= b`.  By continuity of `g` at `x=b` we have that  `lim_{n to oo} g(f(a_n)) = g(f(a))= g(b) = h(a)`.  Thus  we have that  `lim_{n to oo} h(a_n) = h(a)`, showing that `h` is continuous at `x=a`. 


  Monotonic sequences
Theorem BMC:
A bounded monotonic (increasing) sequence is convergent.
Proof: Suppose {`a_n`}  is a sequence with ` a_n le a_{n+1}`  and `a_n < B` for all `n`. 
By the completeness of the real numbers let L be the least upper bound of  {`a_n`}. Claim : `lim_{n to oo} a_n = L`. Here's why: Suppose `epsilon > 0`. Then `L- epsilon < L` and hence `L - epsilon` is not an upper bound for  {`a_n`}. So there is a natural number `M` where `L - epsilon < a_M < L` and so for `n > M` we have `L - epsilon < a_M < a_n < L or `| a_n - L | < epsilon`.
Comment: This result gives a sufficient condition for a sequence to have a limit without knowing in advance what that limit is. The proof finds the limit by using the completeness of the real numbers.
The Bolzano Weierstass Theorem: Every bounded sequence has a convergent subsequence.
The proof of BWT can be developed using Spivak's lemma:
Lemma: Every (bounded) sequence has a monotone subsequence.
Proof: (Spivak) ... Two cases:
Case i) There are an infinite number of `n` with the property that if `m>n` then `a_m < a_n`. Put these  `n`'s in order, so `n_1< n_2<...` and `a_{n_1} > a_{n_2}> ...` and we have found a subsequence that is monotonic decreasing.
Case ii) There is a largest number `N` with the property that if `m>N` then `a_m < a_N`.  Then let `n*_0 = N+1> N`. Then there is a number `n*_1 > n*_0` where `a_{n*_1} ge a_{n*_0}`. We can continue to create sequence of  `n*`'s where `a_{n*_{k+1}} ge a_{n*_k}` and we have found a subsequence that is monotonic increasing.
EOP.

Proof of BWT: Suppose {`a_n`} is a bounded sequence. Then it has a monotonic subsequence which is still bounded. This subsequence is convergent by the BMC theorem.


2-22
Cauchy criterion for convergence:

Definition

A sequence {an} is called a Cauchy sequence if the terms of the sequence eventually all become arbitrarily close to one another.
That is, given
ε
> 0 there exists N such that if m, n > N then |am- an| < ε.
Proposition: If an converges, then an is a Cauchy sequence.
Proof: Suppose limnan=L, then given ε>0
, there exists N such that if n > N then |an - L| < ε2. Thus if m,n>L then |an-am|=|an-L+L-am|<|an-L|+|L-am|<ε2+ε2=ε.

Theorem (CC): If {an} is a Cauchy sequence, then {an} converges.
Proof  Plan: 1 . Show {an} is bounded.
2. Use the B. W. Theorem to give a subsequence of {an} that converges to L.
3. Show limnan=L.
 
 
For more details of the proof see Cauchy criterion for convergence

2-23
Some "monster" examples of functions and continuity
Note: Between any two rational numbers there is an irrational number.

Between any two irrational numbers there is a rational number.
Discussed extensively in class using the decimal estimation of a irrational numbers and `sqrt 2` for finding an irrational number between two rational numbers.

"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{ 2}< 2`, we have `c = f(sqrt{ 2}) in (a,b).`
But if c is a rational number  then   `c = a + (b-a)(sqrt{ 2} -1) ` is a rational number.
Solving for `sqrt{ 2}` we have `sqrt{ 2} = {c-a}/{b-a} + 1`, a rational number. This contradicts the fact that `sqrt{ 2}` 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. `f(x) = 1` for `x le 5`, `f(x) = 0` for `x> 5`.
    `f` is continuous for all `x ne 5`. `f` is not continuous at ` x = 5`. Consider `a_n = 5+1/n; n = 1,2,...`.` and note that `lim_{n to oo} f(a_n) = 0 ne 1 = f(5)`. 
  2. `f(x) = 1` for `x` a rational number, `f(x) = 0` for `x` an irrational number.
    `f` is not continuous at any ` x = a`.
    Consider when `a` is irrational `a_n` the `n`th truncated estimate of `a` as a decimal; n = 1,2,...`.` and note that `lim_{n to oo} f(a_n) = 1 ne 0 = f(a)`. 
    Consider when `a` is rational `a_n = a+ {sqrt 2}/n`; n = 1,2,...`.` and note that `lim_{n to oo} f(a_n) = 0 ne 1 = f(a)`. 
  3. `f(x) = 1/q` for `x = p/q` where `gcd(p,q) = 1`, `f(x) = 0` for `x` an irrational number.
  4.  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.

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(a,b) with f(c)=0.
Proof: Plan. Create  sequences an and bn where a0=a,b0=b,an<bn,anan+1,bn+1bn  and for any n, f(an)f(an+1)>0,f(bn)f(bn+1)>0,[so]f(an)f(bn)<0 and |an-bn|=b-a2n.
Then both sequences are monotonic and bounded and converge to the same number c(a,b). Using continuity show that f(c)=0.

Here's how  to construct the an and bn. [The basic idea is bisection.]

Let a0=a and b0=b.  Now let m0=a0+b02.
If f(a0)f(m0)=0, we have found the desired c=m0 for the theorem... stop here. :)

If f(a0)f(m0)>0, let a1=m0 and b1=b0.
If f(a0)f(m0)<0, let a1=a0 and b1=m0.
 Then a1<b1,a0a1,b1b0 and f(a0)f(a1)>0,f(b0)f(b1)>0,[so]f(a1)f(b1)<0 and  |a1-b1|=b-a2.

Continue in the same way to construct an+1 and bn+1 from an and bn:
 Now let mn=an+bn2.
If f(an)f(mn)=0, we have found the desired c=mn for the theorem... stop here. :)

If f(an)f(mn)>0, let an+1=mn and bn+1=bn.
If f(an)f(mn)<0, let an+1=an and bn+1=mn.
 Then f(an)f(an+1)>0,f(bn)f(bn+1)>0,[so]f(an+1)f(bn+1)<0 and |an+1-bn+1|=bn-an2=12b-a2n =b-a2n+1.

Continuation of proof :
Now the sequences {an} and {bn} are monotonic and bounded, so they each have a limit, call them ca and cb
Noting that  |an-bn|=b-a2n it is an exercise for the reader to show that ca=cb.

We let c=ca=cb and claim that f(c)=0.
Since f(an)f(an+1)>0 for all n, we can say the limit of  f(an) is f(ca) by continuity , and  thus  f(a)f(ca)0.
Likewise, since f(bn)f(bn+1)>0 for all n, we can say the limit of  f(bn)  is f(cb) by continuity , and thus f(b)f(cb)0.

But since ca=cb=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 ()

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 ()
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(a,b) with f(c)=v
Proof: Consider  h(x)=f(x)-v for all x[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(a,b) where h(c)=f(c)-v=0. Thus f(c)=v. EOP.

2-25
Begin with proof of the extreme value theorem.
The boundedness theorems.

2-26
More on horrible functions from previous class:

  1. `f(x) = 1/q` for `x = p/q` where `gcd(p,q) = 1`, `f(x) = 0` for `x` an irrational number.
    Result: `f` is not continuous at any rational number, `x = p/q`. Let `a_n =x + sqrt 2` then `lim_{n to oo} a_n =x` and `f(a_n) =0` so
    `lim_{n to oo} f(a_n) = 0 ne 1/q` for `x = p/q, gcd(p.q)=1`.
    `f` is continuous for all `x`, where `x` is irrational.

  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.

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.



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`

2-29
Theorem: If `f` is a continuous function `f:[a,b] to R` then there exist `c,d in R` where `R_f = \{ f(x): x in [a.b]\} = [c,d]`.
Proof: Plan. Apply the extreme value theorem to find `c` and `d` in `R_f` so that `c le f(x) le d` for all `x in [a,b]`, so `R_f sub [c,d]`.
The use the intermediate value throrem to show that if `c le z le d` then `z in R_f` and so `R_f =[c,d]`.

Derivatives and differentiable functions.
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.
Suppose `f ` is defined on an
open interval, `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`*.
3-1
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`.
`lim_{h->0} {f(a+h) -f(a) - m*h}/h = lim_{h->0} {m(a+h)+b -(ma +b) - m*h}/h = lim_{h->0} 0/h = 0`.
(ii)If  `f(x) = x^2`, then `f '(3) = 6`.
 `lim_{h->0} {(3+h)^2 -3^2 - 6*h}/h = lim_{h->0} {9+6h +h^2 -9 - 6*h}/h = lim_{h->0} {h^2}/h = 0`.
(iii) If `f(x) = |x|`, then  ` f` is not differentiable at `0`.
Consider  `h_n={(-1)^n}/n` so `lim_{n to oo} h_n = 0`.
Now for any `m`, ` {f(0+h_n) -f(0) - m*h_n}/{h_n} = {|h_n|- m*h_n}/{h_n}={1/n-(-1)^nm/n}/{(-1)^n/n}={(-1)^n -m} = 1-m` when `n` is even and `-1-m` when `n` is odd, so there can be no limit.



"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.

3-3
Proof of linearity, product and chain rules.
These proofs use DIC:
Theorem: Suppose `f` and `g` are differentiable at `a`, then `α⋅f+g` and `f⋅g` are differentiable at `a`.
In fact: `(α⋅f+g)'(a)=(α⋅f'+g')(a)=α⋅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∈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.


Theorem ( Chain Rule): [ Proof in Sensible Calculus.]

 
Critical Point Theorem. [This completes first proof of MVT!]

Theorem. (The Critical Point Theorem) Suppose that `c` is a point in the open interval `(a,b)` where `f(c)` is an extreme value for `f` , then
     either i) `f` is not differentiable at `c`
     or ii) `f` is differentiable at `c`, and in this case `f '(c) = 0`.


Proof:
If `f` is not differentiable at `c` then the first alternative is satisfied and the theorem is true.

We may assume then that we are considering a situation where `f(c)` is an extreme value for `f` and `f` is differentiable at `c`.
Let's assume that `f(c)` is actually the minimum value of `f` and leave the case when it is the maximum value as an exercise. 
Since `f(c)` is the minimum value for `f(x)`, `f(x) \ge f(c)` for any `x`, so `f(x) - f(c)\ge 0` for any `x`. Consider `x_n > c` and `x_n \to c`. [See Figure].

 
In this case `x_n - c > 0` so `\frac{f(x_n) - f(c)} {x-c} \ge 0`. But this means that `f '(c)` cannot be negative because there are quotients as close as we want to `f '(c)` that are non negative.
When we consider  `x_n<c` and `x_n \to c` then `x_n - c<0` so `\frac{f(x) - f(c)}{x-c} \le 0`. But this means that `f '(c)` cannot be positive. [See Figure.]
So the only possibility for `f '(c)` is that `f '(c) = 0`, because we have assumed that `f '(c)` exists.                 
EOP.

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". See Euclid Book XII  Proposition 2 Circles are to one another as the squares on their diameters.
Archimedes physical balancing with infinitesmals or indivisibles.    The Method of Archimedes
The Lost Book of Archimedes : Documentary on the Lost Manuscript of the Mathematical Genius [youtube (48:22)]

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
Vignettes of Ancient Mathematics by Henry Mendell, Cal. State U., L.A. Special Topics
 http://en.wikipedia.org/wiki/Mathematical_analysis#History.
and http://en.wikipedia.org/wiki/Timeline_of_calculus_and_mathematical_analysis

The Mechanical Universe -Ep7- Integration - YouTube



3-5 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`.

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.

3-8
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].
Then 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.

 
Discussion of integration and connection to continuity.
3-10&11


Week 10
3-21 through 3- 25
These lectures and notes are found on a separate link: Integration Notes from Week 10
Discussion of assignment due March 8 problem 5:
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.


Main properties of the definite integral that we will prove:
Including: 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.

  Additivity. Continuity of Integral Function for Integrable Functions. Continuous Functions are Integrable.
  FTof Calc I and II, MVT for Integrals. Alternative proof of FTofC using MVT for Integrals.
  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.


Week 11
  3-28

Non-sequence based limits and continuity.

Limits of sequences with intervals:
Note: `lim_{n ->oo} a_n = a` if for any `epsilon >0` there is a number `M in N` where if `n>M` then `|a_n -a| < epsilon.`
Rephrase:..... `a_n in ( a-epsilon, a+epsilon)`.

Definition of an open set.
Suppose O  is a set of real numbers. O is open if (and only if) for any `a in O` there is a `epsilon > 0` where `( a-epsilon, a+epsilon)  sub ` O.

Example: The interval `(1,3)` is an open set.
Proof: Choose ` a in (1,3)`. Let ` epsilon = 1/2 min{a-1, 3-a}`. Suppose `x in ( a-epsilon, a+epsilon)`.  Then `1= a-(a-1)\le a- min{a-1,3-a}< a-epsilon< x < a+epsilon < a + min{a-1,3-a} \le a+ (3-a)=3`. Thus `x in (1,3)` , `( a-epsilon, a+epsilon)  sub (1,3)`  and `(1,3)` is open. EOP

Example: The interval `(1,3]` is not an open set.
Proof:  Consider  `a=3 in (1,3]`. Choose any `epsilon >0`. Then `3 +epsilon/2 in  ( a-epsilon, a+epsilon)` but `3 +epsilon/2` is not an element of `(1,3]` so for any `epsilon >0`, `( 3-epsilon, 3+epsilon)` is not a subset of `(1,3]`  and thus `(1,3]` is not an open set. 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 = a`: if for any open set,`O`, with `a in O`, there is a number `M in N` where if `n>M` then `a_n in O.`


Continuity: With sequences:   `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 sequences and 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 Limits and Continuity. 
The limit of `f ` as `x -> a` for `a in D` is `L` if for any `epsilon >0`, there is a real number `delta>0` where if `x in D` and `0<|x-a|< delta` , then `|f(x)-L|<epsilon`. This is written as `lim_{x ->a} f(x) = L`.
 `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`. That is, for any  `a in D` , `lim_{x ->a} f(x) = f(a)`.

See these on-line references:
GeoGebra Mapping Diagram for limits: http://ggbtu.be/mNLaWHLas
Continuity of Real functions
Images of intervals

Limits of functions


3-29

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

For a visualization of `epsilon-delta` proof of continuity for linear functions see this GeoGebra worksheet:

http://ggbtu.be/mfvWrWyBu 

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) "only if" :  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) "if" :  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.


4-1 and 4-4
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}` `\(\[0\,1\)\)= (-1,1)` ;
`f ^{-1}` `\(\(0\,1\)\)= (-1,0) cup (0,1)`;
`f ^{-1}` `\(\(1,4\)\)= (1,2) cup (-2,-1)`.


Continuity and open sets

4-5
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.


Prop: (i) `O/` and `R` are open sets.
(ii) If U and V are open sets,  then `U nn 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.

4-7
Proof of (ii)
Suppose `a in U nn V`. Then  `a in U` and `a in V`.  Since `U` is an open set, there are numbers `epsilon_U` where `N(a, epsilon_U) sub U` and  `epsilon_V` where `N(a, epsilon_V) sub V`. Let `epsilon = min(epsilon_U, epsilon_V)`. Then it is not hard to show that `N(a,epsilon) sub U nn V` and thus `U nn V` is an open set. EOP

Proof of (iii)
Suppose `a in uuu O`. Then  `a in U` for some `U in O`.  Since `U` is an open set, there is a number `epsilon_U` where `N(a, epsilon_U) sub U`. Let `epsilon = epsilon_U`  Then it is not hard to show that `N(a,epsilon) sub uuuO` and thus `uuu O` is an open set.
EOP

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.
Example: Let `U_n = (-1/n,1/n)`. Then `uuu{U_n} = (-1,1)` which is open. BUT `nnn{U_n} = { 0}` which is not open.

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 nn 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".]


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. [ `R-{a} = (-oo,a)uu(a,oo)` which is an open set.]
4-8
Prop: (i) If U and V are closed sets,  then `U uu V` is a closed set.
(ii) If   K is a family of closed sets, then `nnn K` = { `x in R : x in C` for all  `C in  K`} is a closed set.
Cor. : If `a < b` , then `[a,b]` is a closed set.
Proof: `[a,b]= [a,oo) nn (-oo,b]`.
 


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 `uu` V `sup` S; (ii) 
U `nn` S `!=` `O/` , V `nn` S `!=` `O/` , AND `V nn 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 `uu` V `sup` S;
and (ii) U `nn` S `!=` `O/` , V `nn` S `!=` `O/`, then `V nn U !=` `O/`.



The following result leads immediately to the IVT.
Theorem:
A subset `S` of R is connected if and only if it is an interval. [Skip to IVT]
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 uu V sup S`; `a in U nn S ` so `U cap C!= O/` and `b in V nn S ` so `V cap C!= O/` and certainly `V nn U =O/`, so `S` is not connected.

<=:  Deferred till 4-11
Suppose S is an interval. To show `S` is connected, suppose open sets U and V where
(i) U `uu` V `sup` S; and (ii) U `nn` S `!=` `O/` , V `nn` S `!=` `O/` .
[We need to show that `V nn 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.

Theorem: If `f : R -> R`  is continuous and `C` is a connected set, the `f(C)` is a connected set.
Proof: Suppose not. Then let `U` and `V`  be open sets and
(i) `f(C) sub U uu V` ; and (ii)  `U nn f(C) != O/` , `V nn f(C) != O/`, AND `V nnU = O/` . We will show this implies that `R` 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 uu hatV sup R` ; and (ii) `hatU nn R!= O/` , `hatV nn R != O/` , AND `hatU nn hatV = O/`. [why?]
But this contradicts the fact that `R` is a connected set. 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 uu V` ; and (ii)  `U nn f([a,b]) != O/` , `V nn f([a,b]) != O/`, AND `V nnU = O/` .  So `f([a,b]` is not connected. but `[a,b]` is a connected set, so `f([a,b])` is connected.
A contradiction. 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.

4-11 Deferred from  4-8
Theorem: A subset `S` of R is connected if and only if it is an interval.
Proof <=:Suppose S is an interval. To show `S` is connected, suppose open sets U and V where
(i) U `uu` V `sup` S; and (ii) U `nn` S `!=` `O/` , V `nn` S `!=` `O/` . [We need to show that `V nn 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`.

Continuous functions and connected sets

A topological analysis for the Extreme Value Theorem:
 
Compact sets(topological definition)