real analysis view markdown
Some notes on real analysis, following the textbook Understanding analysis
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 ab = (ac)+(cb))
 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)
 supA is an upper bound of A
 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
 $\mathbb{N}$ is unbounded above (sup $\mathbb{N}=\infty$)
 $\forall x \in \mathbb{R}, x>0, \exists n \in \mathbb{N}, 0 < \frac{1}{n} < x$
 $\mathbb{Q}$ 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 $\frac{1}{n} < ba$
 corollary: the irrationals are dense in $\mathbb{R}$
 pf: want $a < \frac{m}{n} < b$
 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 11 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 $\mathbb{R}$
 $\frac{x}{x^21}$ 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 $\forall n\geq N, a_na < \epsilon$
 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
 lim($ca_n$) = ca
 lim($a_n+b_n$) = a+b
 lim($a_n b_n$) = ab
 lim($a_n/b_n$) = a/b, provided b $\neq$ 0
 pf 3: use triangle inequality, $a_nb_nab=a_nb_nab_n+ab_nab=…=b_na_na+ab_nb$
 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.
 If $a_n \geq 0$ $\forall n \in \mathbb{N}$, then $a \geq 0$
 If $a_n \leq b_n$ $\forall n \in \mathbb{N}$, then $a \leq b$
 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
 pseries $\sum_{n=1}^\infty 1/n^p$ 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
 bolzanoweierstrass 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 $\forall m,n\geq N, a_na_m < \epsilon$
 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
 $\sum_{n=1}^\infty ca_n$ = cA
 $\sum_{n=1}^\infty a_n+b_n$ = A+B
 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 $a_n=0$
 comparison test
 geometric series  $\sum_{n=0}^\infty a r^n = \frac{a}{1r}$
 $s_m = a+ar+…+ar^{m1} = \frac{a(1r^m)}{1r}$
 absolute convergence test
 alternating series test 1. decreasing 2. lim $a_n$ = 0
 then, $\sum_{n=1}^\infty (1)^{n+1} a_n$ converges
 rearrangements: there exists onetoone 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 $V_{\epsilon}(a) \subseteq O$
 $V_{\epsilon}(a)={ x \in R : xa < \epsilon$}
 the union of an arbitrary collection of open sets is open
 the intersection of a finite collection of open sets is open
 $V_{\epsilon}(a)={ x \in R : xa < \epsilon$}
 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
 the union of a finite collection of closed sets is closed
 the intersection of an arbitrary collection of closed sets is closed
 R and $\emptyset$ are both open and 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 subcollection of open sets from the original open cover whose union still manages to completely contain A
 HeineBorel thm  let K $\subseteq \mathbb{R}$. All of the following are equivalent
 K is compact
 K is closed and bounded
 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 < xc < $\delta$ (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:
 $lim_{x \to c} f(x) = L$
 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 xc<$\delta$ (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, $xy < \delta \implies f(x)f(y) < \epsilon$
 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) = $\lim_{x \to c} \frac{g(x)  g( c)}{xc}$, 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: $x^nc^n = (xc)(x^{n1}+cx^{n2}+c^2x^{n3}+…+c^{n1}$)
 differentiable $\implies$ continuous
 algebraic differentiability theorem
 adding
 scalar multiplying
 product rule
 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 $f’( c) = \frac{f(b)f(a)}{ba}$
 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 < xc < $\delta$ 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
 termbyterm 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.
 termbyterm 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 MTest  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 $\sum_{n=0}^\infty n a_n x^{n1}$ 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 Riemannintegrable 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
 sequences
 series
 functional limits
 normal, uniform
 sequence of funcs
 pointwise, uniform
 series of funcs
 pointwise, uniform
 integrability
 sequential criterion  usually good for proving discontinuous
 limit points
 functional limits
 continuity
 absence of uniform continuity
 algebraic limit theorem ~ scalar multiplication, addition, multiplication, division
 limit thm
 sequences
 series  can’t multiply / divide these
 functional limits
 continuity
 differentiability
 ~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