ch 1 - the real numbers

there is no rational number whose square is 2 (proof by contradiction)
contrapositive: $-q \to -p$ - logically equivalent
triangle inequality: $|a+b| \leq |a| + |b|$ (often use |a-b| = |(a-c)+(c-b)|)
axiom of completeness - every nonempty set $A \subseteq \mathbb{R}$ that is bounded above has a least upper bound
- doesn’t work for $\mathbb{Q}$
supremum = supA = least upper bound (similarly, infimum)
1. supA is an upper bound of A
2. if $s \in \mathbb{R}$ is another u.b. then $s \geq supA$
  - can be restated as $\forall \epsilon > 0, \exists a \in A$ $s-\epsilon < a$
nested interval property - for each $n \in N$, assume we are given a closed interval $I_n = [a_n,b_n]={ x \in \mathbb{R} : a_n \leq x \leq b_n }$ Assume also that each $I_n$ contains $I_{n+1}$. Then, the resulting nested sequence of nonempty closed intervals $I_1 \supseteq I_2 \supseteq …$ has a nonempty intersection use AoC with x = sup{$a_n: n \in \mathbb{N}$} in the intersection of all sets
archimedean property
1. $\mathbb{N}$ is unbounded above (sup $\mathbb{N}=\infty$)
2. $\forall x \in \mathbb{R}, x>0, \exists n \in \mathbb{N}, 0 < \frac{1}{n} < x$
$\mathbb{Q}$ is dense in $\mathbb{R}$ - for every $a,b \in \mathbb{R}, a<b$, $\exists r \in \mathbb{Q}$ s.t. $a<r<b$
- pf: want $a < \frac{m}{n} < b$
  - by Archimedean property, want $\frac{1}{n} < b-a$
- corollary: the irrationals are dense in $\mathbb{R}$
there exists a real number $r \in \mathbb{R}$ satisfying $r^2 = 2$
- pf: let r = $sup { t \in \mathbb{R} : t^2 < 2 }$. disprove $r^2<2, r^2>2$ by considering $r+\frac{1}{n},r-\frac{1}{n}$
A ~ B if there exists f:A->B that is 1-1 and onto
A is finite - there exists n $\in \mathbb{N}$ s.t. $\mathbb{N}_n$~A
countable = $\mathbb{N}$~A.
- uncountable - inifinite set that isn’t countable
- Q is countable
  - pf: Let $A_n = { \pm \frac{p}{q}:$ where p,q $\in \mathbb{N}$ are in lowest terms with p+q=n}
- R is uncountable
  - pf: Assume we can enumerate $\mathbb{R}$ Use NIP to exclude one point from $\mathbb{R}$ each time. The intersection is still nonempty, so we didn’t succesfully enumerate $\mathbb{R}$
- $\frac{x}{x^2-1}$ maps (0,1) $\to \mathbb{R}$
- countable union of countable sets is countable
if $A \subseteq B$ and B countable, then A is either countable or finite
if $A_n$ is a countable set for each $n \in \mathbb{N}$, then their union is countable
the open interval (0,1) = ${ x \in \mathbb{R} : 0 < x < 1 }$ is uncountable
- pf: diagonalization - assume there exists a function from (0,1) to $\mathbb{R}$. List the decimal expansions of these as rows of a matrix. Complement of diagonal does not exist.
cantor’s thm - Given any set A, there does not exist a function f:$A \to P(A)$ that is onto
- P(A) is the set of all subsets of A

ch 2 - sequences and series

a sequence $(a_n)$ converges to a real number if $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $\forall n\geq N, |a_n-a| < \epsilon$
- otherwise it diverges
if a limit exists, it is unique
a sequence $(x_n)$ is bounded if there exists a number M > 0 such that $|x_n|\leq M \forall n \in \mathbb{N}$
- every convergent sequence is bounded
algebraic limit thm - let lim $a_n = a$ and lim $b_n$ = b. Then
1. lim($ca_n$) = ca
2. lim($a_n+b_n$) = a+b
3. lim($a_n b_n$) = ab
4. lim($a_n/b_n$) = a/b, provided b $\neq$ 0
  - pf 3: use triangle inequality, $|a_nb_n-ab|=|a_nb_n-ab_n+ab_n-ab|=…=|b_n||a_n-a|+|a||b_n-b|$
  - pf 4: show $(b_n) \to b$ implies $(\frac{1}{b_n}) \to \frac{1}{b}$
order limit thm - Assume lim $a_n = a$ and lim $b_n$ = b.
1. If $a_n \geq 0$ $\forall n \in \mathbb{N}$, then $a \geq 0$
2. If $a_n \leq b_n$ $\forall n \in \mathbb{N}$, then $a \leq b$
3. If $\exists c \in \mathbb{R}$ for which $c \leq b_n$ $\forall n \in \mathbb{N}$, then $c \leq b$
  - pf 1: by contradiction
monotone - increasing or decreasing (not strictly)
monotone convergence thm - if a sequence is monotone and bounded, then it converges
convergence of a series
- define $s_m=a_1+a_2+…+a_m$
- $\sum_{n=1}^\infty a_n$ converges to A $\iff (s_m)$ converges to A
cauchy condensation test - suppose $a_n$ is decreasing and satisfies $a_n \geq 0$ for all $n \in \mathbb{N}$. Then, the series $\sum_{n=1}^\infty a_n$ converges iff the series $\sum_{n=1}^\infty 2^na_{2^n}$ converges
- p-series $\sum_{n=1}^\infty 1/n^p$ converges iff p > 1

2.5

let $(a_n)$ be a sequence and $n_1<n_2<…$ be an increasing sequence of natural numbers. Then $(a_{n_1},a_{n_2},…)$ is a subsequence of $(a_n)$
subsequences of a convergent sequence converge to the same limit as the original sequence
- can be used as divergence criterion
bolzano-weierstrass thm - every bounded sequence contains a convergent subsequence
- pf: use NIP, keep splitting interval into two

2.6

$(a_n)$ is a cauchy sequence if $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $\forall m,n\geq N, |a_n-a_m| < \epsilon$
cauchy criterion - a sequence converges $\iff$ it is a cauchy sequence
- cauchy sequences are bounded
overview: AoC $\iff$ NIP $\iff$ MCT $\iff$ BW $\iff$ CC

2.7

algebraic limit thm - let $\sum_{n=1}^\infty a_n$ = A, $\sum_{n=1}^\infty b_n$ = B
1. $\sum_{n=1}^\infty ca_n$ = cA
2. $\sum_{n=1}^\infty a_n+b_n$ = A+B
  1. cauchy criterion for series - series converges $\iff$ $(s_m)$ is a cauchy sequence
if the series $\sum_{n=1}^\infty a_n$ converges then lim $a_n=0$
1. comparison test
2. geometric series - $\sum_{n=0}^\infty a r^n = \frac{a}{1-r}$
- $s_m = a+ar+…+ar^{m-1} = \frac{a(1-r^m)}{1-r}$
  1. absolute convergence test
  2. alternating series test 1. decreasing 2. lim $a_n$ = 0
- then, $\sum_{n=1}^\infty (-1)^{n+1} a_n$ converges
rearrangements: there exists one-to-one correspondence
if a series converges absolutely, any rearrangement converges to same limit

ch 3 - basic topology of R

3.1 cantor set

C has small length, but its cardinality is uncountable
discussion of dimensions, doubling sizes leads to 2^dimension sizes
- Cantor set is about dimension .631

3.2 open/closed sets

A set O $\subseteq \mathbb{R}$ is open if for all points a $\in$ O there exists an $\epsilon$-neighborhood $V_{\epsilon}(a) \subseteq O$
- $V_{\epsilon}(a)={ x \in R : |x-a| < \epsilon$}
  1. the union of an arbitrary collection of open sets is open
  2. the intersection of a finite collection of open sets is open
a point x is a limit point of a set A if every $\epsilon$-neighborhood $V_{\epsilon}(x)$ of x intersects the set A at some point other than x
a point x is a limit point of a set A if and only if x = lim $a_n$ for some sequence ($a_n$) contained in A satisfying $a_n \neq x$ for all n $\in$ N
isolated point - not a limit point
set $F \subseteq \mathbb{R}$ closed - contains all limit points
- closed iff every Cauchy sequence contained in F has a limit that is also an element of F
density of Q in R - for every $y \in \mathbb{R}$, there exists a sequence of rational numbers that converges to y
closure - set with its limit points
- closure $\bar{A}$ is smallest closed set containing A
iff set open, complement is closed
- R and $\emptyset$ are both open and closed
  1. the union of a finite collection of closed sets is closed
  2. the intersection of an arbitrary collection of closed sets is closed

3.3

a set K $\subseteq \mathbb{R}$ is compact if every sequence in K has a subsequence that converges to a limit that is also in K
Nested Compact Set Property - intersection of nested sequence of nonempty compact sets is not empty
let A $\subseteq \mathbb{R}$. open cover for A is a (possibly infinite) collection of open sets whose union contains the set A.
given an open cover for A, a finite subcover is a finite sub-collection of open sets from the original open cover whose union still manages to completely contain A
Heine-Borel thm - let K $\subseteq \mathbb{R}$. All of the following are equivalent
1. K is compact
2. K is closed and bounded
3. every open cover for K has a finite subcover

ch 4 - functional limits and continuity

4.1

dirichlet function: 1 if r $\in \mathbb{Q}$ 0 otherwise

4.2 functional limits

def 1. Let f:$A \to R$, and let c be a limit point of the domain A. We say that $lim_{x \to c} f(x) = L$ provided that for all $\epsilon$ > 0, there exists a $\delta$ > 0 s.t. whenever 0 < |x-c| < $\delta$ (and x $\in$ A) it follows that |f(x)-L|< $\epsilon$
def 2. Let f:$A \to R$, and let c be a limit point of the domain A. We say that $lim_{x \to c} f(x) = L$ provided that for every $\epsilon$-neighborhood $V_{\epsilon}(L)$ of L, there exists a $\delta$-neighborhood $V_{\delta}($c) around c with the property that for all x $\in V_{\delta}($c) different from c (with x $\in$ A) it follows that f(x) $\in V_{\epsilon}(L)$.
sequential criterion for functional limits - Given function f:$A \to R$ and a limit point c of A, the following 2 statements are equivalent:
1. $lim_{x \to c} f(x) = L$
2. for all sequences $(x_n) \subseteq$ A satisfying $x_n \neq$ c and $(x_n) \to c$, it follows that $f(x_n) \to L$.
algebraic limit thm for functional limits
divergence criterion for functional limits

4.3 continuous functions

a function f:$A \to R$ is continuous at a point c $\in$ A if, for all $\epsilon$>0, there exists a $\delta$>0 such that whenever |x-c|<$\delta$ (and x$\in$ A) it follows that $|f(x)-f( c)|<\epsilon$. F is continous if it is continuous at every point in the domain A
characterizations of continuouty
criterion for discontinuity
algebraic continuity theorem
if f is continuous at c and g is continous at f( c) then g $\circ$ f is continuous at c

4.4 continuous functions on compact sets

preservation of compact sets - if f continuous and K compact, then f(K) is compact as well
extreme value theorem - if f if continuous on a compact set K, then f attains a maximum and minimum value. In other words, there exist $x_0,x_1 \in K$ such that $f(x_0) \leq f(x) \leq f(x_1)$ for all x $\in$ K
f is uniformly continuous on A if for every $\epsilon$>0, there exists a $\delta$>0 such that for all x,y $\in$ A, $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$
- a function f fails to be uniformly continuous on A iff there exists a particular $\epsilon_o$ > 0 and two sequences $(x_n),(y_n)$ in A sastisfying $|x_n - y_n| \to 0$ but $|f(x_n)-f(y_n)| \geq \epsilon_o$
a function that is continuous on a compact set K is uniformly continuous on K

4.5 intermediate value theorem

intermediate value theorem - Let f:[a,b]$ \to R$ be continuous. If L is a real number satisfying f(a) < L < f(b) or f(a) > L > f(b), then there exists a point c $\in (a,b)$ where f( c) = L
a function f has the intermediate value property on an inverval [a,b] if for all x < y in [a,n] and all L between f(x) and f(y), it is always possible to find a point c $\in (x,y)$ where f( c)=L.

ch 5 - the derivative

5.2 derivatives and the intermediate value property

let g: A -> R be a function defined on an interval A. Given c $\in$ A, the derivative of g at c is defined by g’( c) = $\lim_{x \to c} \frac{g(x) - g( c)}{x-c}$, provided this limit exists. Then g is differentiable at c. If g’ exists for all points in A, we say g is differentiable on A
identity: $x^n-c^n = (x-c)(x^{n-1}+cx^{n-2}+c^2x^{n-3}+…+c^{n-1}$)
differentiable $\implies$ continuous
algebraic differentiability theorem
1. adding
2. scalar multiplying
3. product rule
4. quotient rule
chain rule: let f:A-> R and g:B->R satisfy f(A)$\subseteq$ B so that the composition g $\circ$ f is defined. If f is differentiable at c in A and g differentiable at f( c) in B, then g $\circ$ f is differnetiable at c with (g$\circ$f)’( c)=g’(f( c))*f’( c)
interior extremum thm - let f be differentiable on an open interval (a,b). If f attains a maximum or minimum value at some point c $\in$ (a,b), then f’( c) = 0.
Darboux’s thm - if f is differentiable on an interval [a,b], and a satisfies f’(a) < $\alpha$ < f’(b) (or f’(a) > $\alpha$ > f’(b)), then there exists a point c $\in (a,b)$ where f’( c) = $\alpha$
- derivative satisfies intermediate value property

5.3 mean value theorems

mean value theorem - if f:[a,b] -> R is continuous on [a,b] and differentiable on (a,b), then there exists a point c $\in$ (a,b) where $f’( c) = \frac{f(b)-f(a)}{b-a}$
- Rolle’s thm - f(a)=f(b) -> f’( c)=0
- if f’(x) = 0 for all x in A, then f(x) = k for some constant k
if f and g are differentiable functions on an interval A and satisfy f’(x) = g’(x) for all x $\in$ A, then f(x) = g(x) + k for some constant k
generalized mean value theorem - if f and g are continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c $\in (a,b)$ where |f(b)-f(a)|g’( c) = |g(b)-g(a)|f’( c). If g’ is never 0 on (a,b), then can be restated $\frac{f’( c)}{g’( c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$
given g: A -> R and a limit point c of A, we say that $lim_{x \to c} g(x) = \infty$ if, for every M > 0, there exists a $\delta$> 0 such that whenever 0 < |x-c| < $\delta$ it follows that g(x) ≥ M
L’Hospital’s Rule: 0/0 - let f and g be continuous on an interval containing a, and assume f and g are differentiable on this interval with the possible exception of the point a. If f(a) = g(a) = 0 and g’(x) ≠ 0 for all x ≠ a, then $lim_{x \to a} \frac{f’(x)}{g’(x)} = L \implies lim_{x \to a} \frac{f’(x)}{g’(x)} = L$
L’Hospital’s Rule: $\infty / \infty$ - assume f and g are differentiable on (a,b) and g’(x) ≠ 0 for all x in (a,b). If $lim_{x \to a} g(x) = \infty $, then $lim_{x \to a} \frac{f’(x)}{g’(x)} = L \implies lim_{x \to a} \frac{f’(x)}{g’(x)} = L$

ch 6 - sequences and series of function

6.2 uniform convergence of a sequence of functions

for each n $\in \mathbb{N}$ let $f_n$ be a function defined on a set A$\subseteq R$. The sequence ($f_n$) of functions converges pointwise on A to a function f if, for all x in A, the sequence of real numbers $f_n(x)$ converges to f(x)
let ($f_n$) be a sequence of functions defined on a set A$\subseteq$R. Then ($f_n$) converges unformly on A to a limit function f defined on A if, for every $\epsilon$>0, there exists an N in $\mathbb{N}$ such that $\forall n ≥N, x \in A , |f_n(x)-f(x)|<\epsilon$
- Cauchy Criterion for uniform convergence - a sequence of functions $(f_n)$ defined on a set A $\subseteq$ R converges uniformly on A iff $\forall \epsilon > 0 \exists N \in \mathbb{N}$ s.t. whenever m,n ≥N and x in A, $|f_n(x)-f_m(x)|<\epsilon$
continuous limit thm - Let ($f_n$) be a sequence of functions defined on A that converges uniformly on A to a function f. If each $f_n$ is continuous at c in A, then f is continuous at c

6.3 uniform convergence and differentiation

differentiable limit theorem - let $f_n \to f$ pointwise on the closed interval [a,b], and assume that each $f_n$ is differentiable. If $(f’_n)$ converges uniformly on [a,b] to a function g, then the function f is differentiable and f’=g
let ($f_n$) be a sequence of differentiable functions defined on the closed interval [a,b], and assume $(f’_n)$ converges uniformly to a function g on [a,b]. If there exists a point $x_0 \in [a,b]$ for which $f_n(x_0)$ is convergent, then ($f_n$) converges uniformly. Moreover, the limit function f = lim $f_n$ is differentiable and satisfies f’ = g

6.4 series of functions

term-by-term continuity thm - let $f_n$ be continuous functions defined on a set A $\subseteq$ R and assume $\sum f_n$ converges uniformly on A to a function f. Then, f is continuous on A.
term-by-term differentiability thm - let $f_n$ be differentiable functions defined on an interval A, and assume $\sum f’_n(x)$ converges uniformly to a limit g(x) on A. If there exists a point $x_0 \in [a,b]$ where $\sum f_n(x_0)$ converges, then the series $\sum f_n(x)$ converges uniformly to a differentiable function f(x) satisfying f’(x) = g(x) on A. In other words, $f(x) = \sum f_n(x)$ and $f’(x) = \sum f’_n(x)$
Cauchy Criterion for uniform convergence of series - A series $\sum f_n$ converges uniformly on A iff $\forall \epsilon > 0 \exists N \in N$ s.t. whenever n>m≥N, x in A $|f_{m+1}(x) + f_{m+2}(x) + f_{m+3}(x) + …+f_n(x)| < \epsilon$
- Wierstrass M-Test - For each n in N, let $f_n$ be a function defined on a set A $\subseteq$ R, and let $M_n > 0$ be a real number satisfying $|f_n(x)| ≤ M_n$ for all x in A. If $\sum M_n$ converges, then $\sum f_n$ converges uniformly on A

6.5 power series

power series f(x) = $\sum_{n=0}^\infty a_n x^n = a_0 + a_1 x_1 + a_2 x^2 + a_3 x^3 + …$
if a power series converges at some point $x_0 \in \mathbb{R}$, then it converges absolutely for any x satisfying |x|<|$x_0$|
if a power series converges pointwise on the set A, then it converges uniformly on any compact set K $\subseteq$ A
- if a power series converges absolutely at a point $x_0$, then it converges uniformly on the closed interval [-c,c], where c = |$x_0$|
- Abel’s thm - if a power series converges at the point x = R > 0, the the series converges uniformly on the interval [0,R]. A similar result holds if the series converges at x = -R
if $\sum_{n=0}^\infty a_n x^n$ converges for all x in (-R,R), then the differentiated series $\sum_{n=0}^\infty n a_n x^{n-1}$ converges at each x in (-R,R) as well. Consequently the convergence is uniform on compact sets contained in (-R,R).
- can take infinite derivatives

6.6 taylor series

Taylor’s Formula $\sum_{n=0}^\infty a_n x^n = a_0 + a_1 x_1 + a_2 x^2 + a_3 x^3 + …$
- centered at 0: $a_n = \frac{f^{(n)}(0)}{n!}$
Lagrange’s Remainder thm - Let f be differentiable N+1 times on (-R,R), define $a_n = \frac{f^{(n)}(0)}{n!}…..$
not every infinitely differentiable function can be represented by its Taylor series (radius of convergence zero)

ch 7 - the Riemann Integral

7.2 def of Riemann integral

partition of [a,b] is a finite set of points from [a,b] that includes both a and b
lower sum - sum all the possible smallest rectangles
a partition Q is a refinement of a partition P if $P \subseteq Q$
if $P \subseteq Q$, then L(f,P)≤L(f,Q) and U(f,P)≥U(f,Q)
a bounded function f on the interval [a,b] is Riemann-integrable if U(f) = L(f) = $\int_a^b f$
- iff $\forall \epsilon >0$, there exists a partition P of [a,b] such that $U(f,P)-L(f,P)<\epsilon$
- U(f) = inf{U(f,P)} for all possible partitions P
if f is continuous on [a,b] then it is integrable

7.3 integrating functions with discontinuities

if f:[a,b]->R is bounded and f is integrable on [c,b] for all c in (a,b), then f is integrable on [a,b]

7.4 properties of Integral

assume f: [a,b]->R is bounded and let c in (a,b). Then, f is integrable on [a,b] iff f is integrable on [a,c] and [c,b]. In this case we have $\int_a^b f = \int_a^c f + \int_c^b f.$F
integrable limit thm - Assume that $f_n \to f$ uniformly on [a,b] and that each $f_n$ is integarble. Then, f is integrable and $lim_{n \to \infty} \int_a^b f_n = \int_a^b f$.

7.5 fundamental theorem of calculus

If f:[a,b] -> R is integrable, and F:[a,b]->R satisfies F’(x) = f(x) for all x $\in$ [a,b], then $\int_a^b f = F(b) - F(a)$
Let f: [a,b]-> R be integrable and for x $\in$ [a,b] define G(x) = $\int_a^x g$. Then G is continuous on [a,b]. If g is continuous at some point $c \in [a,b]$ then G is differentiable at c and G’(c) = g(c).

overview

convergence
1. sequences
2. series
3. functional limits
  - normal, uniform
4. sequence of funcs
  - pointwise, uniform
5. series of funcs
  - pointwise, uniform
6. integrability
sequential criterion - usually good for proving discontinuous
1. limit points
2. functional limits
3. continuity
4. absence of uniform continuity
algebraic limit theorem ~ scalar multiplication, addition, multiplication, division
1. limit thm
2. sequences
3. series - can’t multiply / divide these
4. functional limits
5. continuity
6. differentiability
7. ~integrability~
limit thms
- continuous limit thm - Let ($f_n$) be a sequence of functions defined on A that converges uniformly on A to a function f. If each $f_n$ is continuous at c in A, then f is continuous at c
- differentiable limit theorem - let $f_n \to f$ pointwise on the closed interval [a,b], and assume that each $f_n$ is differentiable. If $(f’_n)$ converges uniformly on [a,b] to a function g, then the function f is differentiable and f’=g
  - convergent derivatives almost proves that $f_n \to f$
  - let ($f_n$) be a sequence of differentiable functions defined on the closed interval [a,b], and assume $(f’_n)$ converges uniformly to a function g on [a,b]. If there exists a point $x_0 \in [a,b]$ for which $f_n(x_0) \to f(x_0)$ is convergent, then ($f_n$) converges uniformly
- integrable limit thm - Assume that $f_n \to f$ uniformly on [a,b] and that each $f_n$ is integarble. Then, f is integrable and $lim_{n \to \infty} \int_a^b f_n = \int_a^b f$.
functions are continuous at isolated points, but limits don’t exist there
uniform continuity: minimize $|f(x)-f(y)|$
derivative doesn’t have to be continuous
integrable if finite amount of discontinuities and bounded