• 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.]
  

• How to Read and Do Proofs by D.Solow
• The Keys to Advanced Mathematics by D. Solow
• How to Solve It by G. Polya

Space Filling Curves - YouTube

• 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?

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).`

• For a definition of a metric space see  http://mathworld.wolfram.com/MetricSpace.html

• 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

• 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. A similar argument leads to a contradiction by supposing there are `a,b in I` with `a<b, f(a)> f(b).

• Proofs about increasing/decreasing functions: If `f'(0)>0` on an interval, then `f` is an increasing function.
  
• Proof of The Fundamental Theorem of Calculus (second part) using Riemann Sums
  
• Concavity (convexity): If `f''(x) > 0` on an interval then `f` is concave up (convex) on the interval. See Spivak Ch 11 Appendix Theorem 2
• See Interactive real Analysis and  What is the point of the mean value theorem?

• Counter examples to the MVT:
  (1) without Continuity: f (x) = x  for x  in (0, 1] and f(0) = 1 on the interval [0,1]
  
  and
  (2) without differentiability:  f (x) = 1/2 -| x - 1/2|   for x  in [0, 1].
  Also  `f (x) = (1/2)^(1/3) -|( x - 1/2)|^(1/3)`   for x  in [0, 1].
  

• Intermediate Value Theorem
• Extreme Value Theorem   
  

• 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
  

• Fields

Week 3
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.

• Interesting: 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, Z_p, 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. This contradicts the fact that n was supposed to be the SMALLEST natural number for which n·1=0. IRMC.

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.

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`.
   

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|`.

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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

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2-4

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

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)`.

(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 

• Motivational Question I: What is a number?
• Motivational Question II: What does continuity mean?
  
• Motivational Question III: What does limit mean? for numbers? for function values? for functions?  for sets?

• Sequences of Numbers- Convergence
• Visualizing a sequence of real numbers:
  Def'n: A sequence is a function `a:N->S` where `S subset R`.
• See visualizations in Sensible Calculus:
  
• With a mapping diagram.
• As a graph.
• On a number line.

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

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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

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• Provisional Definition: limit  of  `a_n` is `a` means "after a while all `a_n` are close to `a`."
  2-9
  
• Def'n: We say `lim_{n->oo} a_n = a` if
  given any `epsilon >0` there is a natural number M so that
  if `n>M` then `| a_n - a | < epsilon`. 
  Comment: What kind of number is `epsilon`?
• Visualizing the definition of  `lim_{n->oo}a_n = a`
  

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  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.
• 
  

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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

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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`.

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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.

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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`.

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For a few more examples see: Convergence in the Reals

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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)`

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  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

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Theorems connected to algebra, inequalities and intervals.

•   Check this link for Proofs of  Properties of convergent sequences  (Algebra of sequences)
• Any convergent sequence is bounded (both above and below).
  For proof see Proofs of  Properties of convergent sequences



• If the sequence {a_n}→ a and {b_n}→ b then
  

  (i) (a_n+ b_n)→ 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-(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.

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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.

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

2-16
(iii) (a_nb_n)→ 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}`.
For 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) (^a_n/_bn)→ ^a/_b (provided b_n ≠ 0 and `b` ≠ 0).
Lemma: `1/{b_n} \to 1/b ` (provided b_n ≠ 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.

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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`.

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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
 

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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).

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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`. 

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  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.

•   Subsequences

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.

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2-22
Cauchy criterion for convergence:

Definition

A sequence ${\displaystyle \left\{a_n\right\}}$ is called a Cauchy sequence if the terms of the sequence eventually all become arbitrarily close to one another.
That is, given ${\displaystyle \epsilon }$ > 0 there exists N such that if m, n > N then |a_m- a_n| < ${\displaystyle \epsilon }$.
Proposition: If ${\displaystyle a_n}$ converges, then ${\displaystyle a_n}$ is a Cauchy sequence.
Proof: Suppose ${\displaystyle \underset{n\to \infty }{lim}{a}_n=L}$, then given ${\displaystyle \epsilon >0}$, there exists N such that if n > N then |a_n - L| < ${\displaystyle \frac{\epsilon }{2}}$. Thus if ${\displaystyle m,n>L}$ then .

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Theorem (CC): If ${\displaystyle {\mathrm{\{a}}_n}$} is a Cauchy sequence, then ${\displaystyle {\mathrm{\{a}}_n}$} converges.
Proof  Plan: 1 . Show ${\displaystyle {\mathrm{\{a}}_n}$} is bounded.
2. Use the B. W. Theorem to give a subsequence of ${\displaystyle {\mathrm{\{a}}_n}$} that converges to ${\displaystyle L}$.
3. Show ${\displaystyle \underset{n\to \infty }{lim}{a}_n=L}$.
   
For more details of the proof see Cauchy criterion for convergence

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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 ${\displaystyle f}$ is continuous on ${\displaystyle [a,b]}$ and then there is a number ${\displaystyle c\in (a,b)}$ with ${\displaystyle f\left(c\right)=0}$.
Proof: Plan. Create  sequences ${\displaystyle a_n}$ and ${\displaystyle b_n}$ where   and for any ${\displaystyle n}$, and ${\displaystyle |a_n-b_n|=\frac{b-a}{2^n}}$.
Then both sequences are monotonic and bounded and converge to the same number ${\displaystyle c\in (a,b)}$. Using continuity show that ${\displaystyle f\left(c\right)=0}$.

Here's how  to construct the ${\displaystyle a_n}$ and ${\displaystyle b_n}$. [The basic idea is bisection.]

Let ${\displaystyle a_0=a}$ and ${\displaystyle b_0=b}$.  Now let ${\displaystyle m_0=\frac{a_0+b_0}{2}.}$
If ${\displaystyle f\left(a_0\right)\cdot f\left(m_0\right)=0}$, we have found the desired ${\displaystyle c=m_0}$ for the theorem... stop here. :)

If ${\displaystyle f\left(a_0\right)\cdot f\left(m_0\right)>0}$, let ${\displaystyle a_1=m_0}$ and ${\displaystyle b_1=b_0}$.
If , let ${\displaystyle a_1=a_0}$ and ${\displaystyle b_1=m_0}$.
 Then and and  ${\displaystyle |a_1-b_1|=\frac{b-a}{2}}$.

Continue in the same way to construct ${\displaystyle a_{n+1}}$ and ${\displaystyle b_{n+1}}$ from ${\displaystyle a_n}$ and ${\displaystyle b_n}$:
 Now let ${\displaystyle m_n=\frac{a_n+b_n}{2}.}$
If ${\displaystyle f\left(a_n\right)\cdot f\left(m_n\right)=0}$, we have found the desired ${\displaystyle c=m_n}$ for the theorem... stop here. :)

If ${\displaystyle f\left(a_n\right)\cdot f\left(m_n\right)>0}$, let ${\displaystyle a_{n+1}=m_n}$ and ${\displaystyle b_{n+1}=b_n}$.
If , let ${\displaystyle a_{n+1}=a_n}$ and ${\displaystyle b_{n+1}=m_n}$.
 Then and ${\displaystyle |a_{n+1}-b_{n+1}|=\frac{b_n-a_n}{2}=\frac{1}{2}\frac{b-a}{2^n} =\frac{b-a}{2^{n+1}}}$.

Continuation of proof :
Now the sequences ${\displaystyle \left\{a_n\right\}}$ and ${\displaystyle \left\{b_n\right\}}$ are monotonic and bounded, so they each have a limit, call them ${\displaystyle c_a}$ and ${\displaystyle c_b}$. 
Noting that  ${\displaystyle |a_n-b_n|=\frac{b-a}{2^n}}$ it is an exercise for the reader to show that ${\displaystyle c_a=c_b}$.

We let ${\displaystyle c=c_a=c_b}$ and claim that ${\displaystyle f\left(c\right)=0.}$
Since ${\displaystyle f\left(a_n\right)\cdot f\left(a_{n+1}\right)>0}$ for all ${\displaystyle n}$, we can say the limit of  ${\displaystyle f\left(a_n\right)}$ is ${\displaystyle f\left(c_a\right)}$ by continuity , and  thus  ${\displaystyle f\left(a\right)\cdot f\left(c_a\right)\ge 0}$.
Likewise, since ${\displaystyle f\left(b_n\right)\cdot f\left(b_{n+1}\right)>0}$ for all ${\displaystyle n}$, we can say the limit of  ${\displaystyle f\left(b_n\right)}$  is ${\displaystyle f\left(c_b\right)}$ by continuity , and thus ${\displaystyle f\left(b\right)\cdot f\left(c_b\right)\ge 0}$.

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.

• Motivation from Sensible Calculus definitions of the definite integral.

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

• The Riemann Integral ( General Sums and upper and lower sums).

• Definition of the Riemann Integral (using upper and lower sums) with example of non- integrable function.

• 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 Function Constraint Result for Integral.
  

(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`.

• Definitions: Suppose ${\displaystyle F}$ is a family of sets and ${\displaystyle A\subset R}$. We say ${\displaystyle F}$ is a cover of ${\displaystyle A}$ if `uuu`${\displaystyle }$F={x:x∈U for some ${\displaystyle U\in F\}\supset A}$.
  If ${\displaystyle F}$ is a cover of ${\displaystyle A}$ and every ${\displaystyle U\in F}$ is an open set, then we say ${\displaystyle F}$ is an open cover of ${\displaystyle A}$.
  If ${\displaystyle \hat{F}}$ is a subfamily of ${\displaystyle F}$, i.e., ${\displaystyle \hat{F}\subset F}$, and ${\displaystyle \hat{F}}$ is a cover of ${\displaystyle A}$, we say that ${\displaystyle \hat{F}}$ is a subcover of ${\displaystyle F}$.
• We say that a set ${\displaystyle K}$ is (topologically) compact if any open cover ${\displaystyle F}$ of ${\displaystyle K}$ has a subcover ${\displaystyle \hat{F}}$ that is a finite set, i.e., for any open cover ${\displaystyle F}$ of ${\displaystyle K}$, there exist ${\displaystyle U_1,U_2,...,U_n\in F}$ where  ${\displaystyle U_1\cup U_2\cup ...\cup U_n\supset K}$.
• Examples: i. A finite set is compact.
  ii. ${\displaystyle \{1,\frac{1}{2},\frac{1}{3},...\}=\{x:x=\frac{1}{n}}$ for ${\displaystyle n>0,n\in N\}}$ is not compact. [This set is not closed!]
  iii. ${\displaystyle \{0,1,\frac{1}{2},\frac{1}{3},...\}=\left\{0\right\}\cup \{x:x=\frac{1}{n}}$ for ${\displaystyle n>0,n\in N\}}$ is compact.[This set is closed.]
  iv. ${\displaystyle \{1,2,3,...\}=\{x:x=n}$ for ${\displaystyle n>0,n\in N\}}$ is not compact. [This set is closed, but not bounded.]
  

•  Outline:
• Heine Borel Theorem (simplest version): ${\displaystyle [a,b]}$ is a compact subset of R.
• Theorem: A closed subset of a compact set is compact.
• Theorem: If ${\displaystyle K}$ is a compact subset of ${\displaystyle R}$, then ${\displaystyle K}$ is closed and bounded. [HB in R]
• Theorem: If  ${\displaystyle f:R\to R}$ is continuous and ${\displaystyle K}$ is a compact  then  ${\displaystyle f\left(K\right)}$ is compact.

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• Extreme Value Theorem (Topological Proof) Suppose `f: [a,b] -> R` is continuous. Then there exist`c_M` and `c_m` in `[a,b]` where `f(c_m) le f(x) le f(c_M)` for all `x in [a,b]`.
  Proof: We can suppose that `f(x) = f(b)` for `x > b` and `f(x) = f(a)` for `x<a`, so `f: R -> R` is continuous.
  Since `[a,b]` is compact, `f([a,b])` is compact.
  Since `[a,b]` is also an interval, `f([a,b])` is a closed and bounded interval.
  Thus `f([a,b]) = [m,M]` where for all `x in [a,b]`, `m le f(x) le M`.
  But since `m, M in f([a,b])`, there exist `c_m` and `c_M` in `[a,b]` where `f(c_m) = m` and `f(c_M) = M.`  EOP!

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  4-14 Proofs remaining from outline.
  
  Heine Borel Theorem (simplest version): ${\displaystyle [a,b]}$ is a compact subset of R.
  Proof: (outline). Suppose O is a family of open sets where ${\displaystyle `uuu`O\supset [a,b]}$.
  Let G= {x ${\displaystyle \in }$ ${\displaystyle [a,b]}$ where ${\displaystyle [a,x]}$ is covered by a finite subset of the members of `O`.}. 
  Claims: `a in G` because `[a,a] = {a}` and since `[a,b] subset uuu O`,  for some `U in O`, ` a in U`.
  Thus `G` ${\displaystyle \ne }$ ${\displaystyle \varnothing }$ and by its definition G is bounded by ${\displaystyle b}$.
  Then let ${\displaystyle \alpha }$ = lub of G.
  Plan: Show `alpha = b` and ${\displaystyle \alpha \in }$`G`${\displaystyle \mathrm{.<br>}}$
  `alpha = b`: Suppose `alpha < b`. Then `a le alpha < b`. Since `O` is an open covervof `[a,b]`, there is an open set `U_{alpha}` where `alpha in U_{alpha} in O` and `epsilon_{alpha} > 0` where `N(alpha, epsilon_{alpha}) sub U_{alpha} nn [a,b]`.
  
  
  
  
  Considering `alpha - epsilon_{alpha} < alpha` is not an upper bound for `G`  so there is a number `c in G` with `alpha-epsilon_{alpha} < c<alpha`.
  
  
  
  Since `c in G`, the interval `[a,c]` is covered by a finite subset of the members of `O`. 
  But then add `U_{alpha}` to this family of members of `O` and we see that `alpha + epsilon_{alpha}/2 < b` is also in `G` which contradicts that `alpha` is an upper bound for `G`.
  
  
  
  So `alpha =b`.
   α∈`G`: Using `U_{alpha}` from above, we have `epsilon> 0` where  `N(alpha, epsilon) sub U_{alpha}`. Following an argument similar to the last one, `alpha - epsilon < alpha` is not an upper bound for `G`  so there is a number `c in G` with `alpha-epsilon < c<alpha`.
  
  
  
  
  Since `c in G`, the interval `[a,c]` is covered by a finite subset of the members of `O`.
  But then add `U_{alpha}` to this family of members of `O` and we see that `alpha = b` is also in `G`.   Thus `[a,b]` is (topologically) compact.
  EOP
  

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  See also : Closed_Bounded_Subset_of_Real_Numbers_is_Compact
  

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• Theorem: A closed subset of a compact set is compact.
  Proof: Supposed ${\displaystyle C}$ is a closed set and ${\displaystyle K}$ is a compact set with ${\displaystyle C\subset K}$. 
  Now suppose `O` is a family of open sets where ${\displaystyle `uuu` O\supset \mathrm{C`}}$.
  Since ${\displaystyle \mathrm{`C`}}$ is closed, let `U = R-C` and ${\displaystyle \hat{O}=O\cup \left\{U\right\}}$.
  Then ${\displaystyle \hat{O}}$ covers ${\displaystyle K}$ and by compactness a finite subcover of ${\displaystyle \hat{O}}$ covers ${\displaystyle K\supset C}$.
  But then if needed that finite subcover less ${\displaystyle U}$ will be a finite subcover of ${\displaystyle O}$ that covers ${\displaystyle C}$.
  So ${\displaystyle C}$ is compact. EOP.
  
  
• Corollary: If K is a closed and bounded subset of R, then K is compact.
  Proof: K is bounded- so K ${\displaystyle \subset [m,M]}$ for some ${\displaystyle m}$ and ${\displaystyle M}$. But by HB, ${\displaystyle [a,b]}$ is compact, so K is a closed subset of a compact set and thus K is compact. EOP
  
  4-15
  
• Theorem: If ${\displaystyle K}$ is a (topologically) compact subset of ${\displaystyle R}$, then ${\displaystyle K}$ is closed and bounded. [HB in R]
  Proof:
  (i) ${\displaystyle K}$ is bounded: Consider the family of open sets, ${\displaystyle O=\{N(0,n)\text{for}n=1,2,3....\}}$. Certainly this family covers any subset of ${\displaystyle R}$, so it covers ${\displaystyle K}$. But a finite subcover will be bounded by ${\displaystyle n+1}$ for the largest ${\displaystyle n}$ of the sets in that finite subcover. Thus ${\displaystyle K}$ is bounded.
  
  (ii) ${\displaystyle K}$ is closed.  Let ${\displaystyle U}$ = `R -K`, the complement of  `K`. We will show that ${\displaystyle U}$ is open. 
  Suppose ${\displaystyle p\in U}$  and ${\displaystyle x\in K}$. 
  Then let ${\displaystyle r\left(x\right)=\frac{|x-p|}{3}}$ so ${\displaystyle r\left(x\right)>0}$ and ${\displaystyle N(p,r\left(x\right))\mathrm{`nn`\; N}(x,r\left(x\right))=\varnothing }$.
  Let ${\displaystyle O}$ be the family of open sets ${\displaystyle O=\{N(x,r\left(x\right)):x\in K\}}$. 
  Then ${\displaystyle O}$ covers ${\displaystyle K}$.
  Because ${\displaystyle K}$ is compact there a finite number of points ${\displaystyle x_1,...,x_m}$ where the family of open sets ${\displaystyle \{N(x_k,r\left(x_k\right)):k=1,2,...,m\}}$ covers ${\displaystyle K}$.
  Then let ${\displaystyle r=min\{r\left(x_k\right):k=1,2,...,m\}}$ and we have that ${\displaystyle N(p,r)`nn`N(x_k,r\left(x_k\right))=\varnothing }$ for all k.
  Hence ${\displaystyle N(p,r)`nn`K=\varnothing }$
  Thus ${\displaystyle N(p,r)\subset U}$ and ${\displaystyle U}$ is an open set.  EOP.

• The Ratio and Root Tests and other stuff.
  4-26
  
• Alternating Series Test.
  If `a_k> a_{k+1} > 0` for all `k` and `lim_{k->oo}a_k = 0` then `sum_{k=0}^{oo} (-1)^k a_k` converges.
  
• Proof: plan: Let `S_n = sum_{k=0}^n (-1)^k a_k`. Show `{S_{2n}}` is decreasing and bounded below by `S_1`. So `{S_{2n}}` will converge. But then `{S_{2n+1}}` will converge to the same limit.
  Conditional Convergence
• Definition. `sum_{k=0}^{oo} a_k` converges conditionally if `sum_{k=0}^{oo} a_k` converges but `sum_{k=0}^{oo} |a_k|` diverges.
  
• Rearrangements of series and conditional convergence