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\begin{notes}{Mods Analysis I - Sequences \& Series}{}{}

\tableofcontents

\part{Definitions}

\section{Axioms of $\mathbb{R}$}

See Richard Earl's clear listing of axioms.

\section{Completeness of $\mathbb{R}$}

\begin{defn}
	Let $B \subseteq \mathbb{R}$. Then
	\begin{enumerate}
		\item[(i)] $b_1$ is the \textbf{least element} or \textbf{minimum} $B$ if $b_1 \in B$ and $b_1 \leq b\ \forall b \in B$.  We write $b_1 = \min B$.
		\item[(ii)] $b_2$ is the \textbf{greatest element} or \textbf{maximum} $B$ if $b_2 \in B$ and $b_2 \geq b\ \forall b \in B$.  We write $b_2 = \max B$.
	\end{enumerate}
\end{defn}

\begin{defn}
	Let $B \subseteq \mathbb{R}$. Then
	\begin{enumerate}
		\item[(i)] $h_1$ is a \textbf{lower bound} $B$ if $h_1 \leq b\ \forall b \in B$.
		\item[(ii)] $h_2$ is an \textbf{upper bound} $B$ if $h_2 \geq b\ \forall b \in B$.
	\end{enumerate}
\end{defn}

\begin{defn}
	Let $B \subseteq \mathbb{R}$.
	\begin{enumerate}
		\item[(i)] $B$ is \textbf{bounded above} if it has an upper bound.
		\item[(ii)] $B$ is \textbf{bounded below} if it has a lower bound.
		\item[(iii)] $B$ is \textbf{bounded} (or \textbf{bdd.}) if it has upper and lower bounds.
	\end{enumerate}
\end{defn}

\subsection{Suprema}

\noindent \textbf{Axiom C}$\ $ Let $E \subr$ be a non-empty, bounded above set.  Then the set of upper bounds $U$ of $E$ has a least element.

\begin{defn}
	The \textbf{least upper bound} or \textbf{supremum} of $E$, $\sup E$, is $\min U$.
\end{defn}

\subsection{Infima}

\begin{defn}
	Let $F$ be a non-empty set which is bounded below.  Then the
	greatest element of the set of lower bounds of $F$ is called the
	\textbf{greatest lower bound} or \textbf{infimum}, and we write
	$\inf F$.

	It is trivial to show that infima are unique, and that if $F$ has a
	minimum then $\min F = \inf F$.
\end{defn}

%\begin{align*}
%	a + b &= b + a \tag{+ commutes} \\
%\end{align*}

\section{Countability}

Georg Cantor (1845 - 1918).

\begin{defn}
	Let $A,\ B$ be sets.  If there exists a bijection $\theta: A \to B$ then we say that $A$ and $B$ are \textbf{equinumerous} and we write $A \approx B$.
\end{defn}

\begin{rmk}
	$\approx$ is an equivalence relation, so it is reflexive, symmetric and transitive.
\end{rmk}

\begin{defn}
	A set $A$ is \textbf{finite} if $A = \emptyset$ or $A \approx \{1,\
	2,...\, k\}$ for $k \in \mathbb{N}_1$. In the former case we say
	that $A$ has \textbf{size} 0, written as $|A|$.  In the latter case,
	$|A| = k$. A set which is not finite is called \textbf{infinite}.
\end{defn}

\begin{rmk}
	$A$ is an infinite set $\iff$ $B \subsetneq A$ and $A \approx B$.
\end{rmk}

\begin{defn}
	A set $A$ is \textbf{countably infinite} or \textbf{denumerable} if $A \approx \mathbb{N}$.
\end{defn}

\begin{defn}
	A set $A$ is \textbf{countable} if $A$ is finite or $A$ is countably infinite.  A set which is not countable is \textbf{uncountable}.
\end{defn}

\begin{defn}
	Two sets are \textbf{disjoint} if they have no elements in common; i.e. if their intersection is the empty set.
\end{defn}

\begin{notation}
	``Aleph-null'', $\aleph_0$, denotes the cardinality of $\mathbb{N}$ or of any other countably infinite set.
\end{notation}

\begin{notation}
	$c$, which stands for \emph{continuum}, denotes $|\mathbb{R}|$.
\end{notation}

\section{Sequences in $\mathbb{R}$ and $\mathbb{C}$}

Our task is to make precise the idea that successive approximations to
irrational numbers approach the real numbers they represent.

%\begin{defn}
%	A \textbf{real sequence} is a function $\alpha:\mathbb{N} \to
%	\mathbb{R}$; a \textbf{complex sequence} is a function
%	$\beta:\mathbb{N} \to \mathbb{C}$.
%\end{defn}
%
%\begin{notation}
%	The \textbf{\textit{n}\textsuperscript{th} term} of a sequence
%	$\alpha$ is $\alpha(n)$ which is written $\alpha_n$.  We can refer
%	to the whole sequence as $(\alpha_n)$.
%\end{notation}

The sequences $(\alpha_n + \beta_n)$, $(c\alpha_n)$ for some $c \in
\mathbb{C}$ \&c. are defined as one would expect.

\subsection{Convergence}

\boxthm{\begin{defn}
	Let $(a_n)$ be a real sequence and let $L \isreal$. Then \textbf{$(a_n)$ converges to \textit{L}} if
	\[ \forall \eps > 0\quad\exists N_\eps \in \mathbb{N}\quad\forall n \geq N_\eps\quad|a_n - L| < \eps. \]
	A sequence that is not convergent is said to be \textbf{divergent}.
\end{defn}
\begin{notation}
	If a real sequence $(a_n)$ converges to $L$ we say that \textbf{$(a_n)$ tends to \textit{L}} and we can write
	\[ (a_n) \to L,\quad a_n \to L\text{ as }n \to \infty \quad\text{or}\quad a_n \to L.\]
\end{notation}}

\begin{defn}
	If $(a_n) \to L$ then $L$ is the \textbf{limit} of $(a_n)$ and we write
	\[ L = \lim_{n\to\infty}a_n\quad\text{or simply}\quad L = \lim a_n. \]
\end{defn}

\begin{defn}
	We have that
	\begin{multline*}
		(a_n) \text{ converges} \iff \\ \exists L \isreal\quad\forall \eps > 0\quad\exists N_\eps \in \mathbb{N}\quad\forall n \geq N_\eps\quad|a_n - L| < \eps,
	\end{multline*}
	and
	\begin{multline*}
		(a_n) \text{ diverges} \iff \neg((a_n) \text{ converges}) \iff \\ \forall L \isreal\quad\exists\eps > 0\quad\forall N_\eps \in \mathbb{N}\quad\exists n \geq N_\eps\quad |a_n - L| \geq \eps,
	\end{multline*}
	by DeMorgan's laws.
\end{defn}

\part{Theorems \& Propositions}
\setcounter{section}{0}

\section{Axioms of $\mathbb{R}$}

Easy properties (let $x,y,a,b\isreal$):
\begin{gather*}
	a + x = a \implies x = 0 \text{ (additive identity is unique)} \\
	a + x = a + y \implies x = y \text{ (additive inverses are unique)} \\
	-(-a) = a \\
	-(a + b) = (-a) + (-b) \\
	-0 = 0 \\
	a \cdot x = a \implies x = 1 \\
	a \ne 0 \text{ and } a \cdot x = a \cdot y \implies x = y \\
	a \ne 0 \implies 1/(1/a) = a \\
	a \ne 0 \ne b \implies 1/(ab) = (1/a) \cdot (1/b) \\
	(a + b) \cdot c = a \cdot c + b \cdot c \\
	a \cdot 0 = 0 \\
	a \cdot b = 0 \implies a = 0 \text{ or } b = 0 \\
	a \cdot (-b) = -(a \cdot b) \\
	(-1) \cdot (-1) = 1 \\
	a > b \iff -a < -b \\
\end{gather*}

There are some miscellaneous propositions relating to inequalities (esp.
shifting) worth looking at.

\subsection{The Triangle Law}

\boxthm{\begin{thm}
	Let $a,b\isreal$.  Then
	\begin{equation*}
		|a + b| \leq |a| + |b| \tag{$\Delta$} \\
	\end{equation*}
	with equality iff ($a \geq 0$ and $b \geq 0$) or ($a < 0$ and $b < 0$).
	
	\begin{proof}
		Consider eight possible cases by trichotomy.
	\end{proof}
\end{thm}}

\subsection{Bernoulli's Inequality}

\begin{thm}
	Let $x > -1$ and $n \iszp$.  Then
	\[
		(1 + x)^n \geq 1 + nx.
	\]

	\begin{proof}
		Use induction on $n$, with base case $n = 1$.
	\end{proof}
\end{thm}

\section{Completeness of $\mathbb{R}$}

\begin{propn}
	A maximum (if it exists) is unique.  Similarly a minimum is unique.

	\begin{proof}
		Use fact that
		\[ a \geq b,\ b \geq a \implies a = b. \]
	\end{proof}
\end{propn}

\subsection{Suprema}

\begin{propn}
	If $E \subr$ has a maximum then $\max E = \sup E$.

	\begin{proof}
		Now $\max E \geq x\ \forall x \in E$ by definition of $\max E$
		being a maximum.  Further if $l$ is an upper bound for $E$ then
		$l \geq \max E$ because $\max E \in E$.  Hence $\max E$ is the
		least upper bound.
	\end{proof}
\end{propn}

\boxthm{\begin{propn}
	\textnormal{(The Approximation Property)}
	Let $E \subr$ have a supremum, and let $\varepsilon > 0$.  Then there exists $e \in E$ such that
	\[ \sup E - \varepsilon < e \leq \sup E.\]

	\begin{proof}
		If not, then $\sup E - \varepsilon$ is an upper bound of $E$
		less than the least upper bound, which is a contradiction.
	\end{proof}
\end{propn}}

\subsection{Infima}

\begin{thm}
	Let $F \subr$ be a non-empty, bounded below set.  Then the set of lower bounds of $F$ has a greatest element.

	\begin{proof}
		Show that $E := \set{-x}{x \in F}$ has a supremum using $x \leq y \iff -y \leq -x$ to show bounded above.
		Then show $-\sup E$ is a lower bound of $F$ and that any other lower bounds of $F$, $l$, are such that $l \leq -\sup E$.
	\end{proof}
\end{thm}

\begin{corr}
	\textnormal{(The Approximation Property for infima)}
	Let $F \subr$ have an infimum, and let $\varepsilon > 0$.  Then there exists $f \in E$ such that
	\[ \inf F \leq f < \inf F + \varepsilon. \]
\end{corr}

\subsection{$\sqrt{2}$ exists}

\begin{thm}
	There exists a unique positive number $\alpha$ such that $\alpha^2 = 2$.
	\begin{proof}
		Epicness.
	\end{proof}
\end{thm}

\begin{notation}
	We write $\sqrt{2}$ for $\alpha$.
\end{notation}

\begin{thm}
	Let $a$ be any positive real number.  Then there exists a unique real number - denoted by $\sqrt{a}$ - whose square is $a$.
	
	\begin{proof}
		A refinement of the above suffices.
	\end{proof}
\end{thm}

\subsection{Archimedean Property}

\begin{thm}
	\textnormal{(Archimedean Property of the natural numbers)}
	Let $x \isreal$.  Then there exists $n \in \mathbb{N}$ such that $x < n$; i.e. $\mathbb{N}$ is not bounded above.
	\begin{proof}
		Assume otherwise, and let $\xi = \sup \mathbb{N}$.  Then use the
		approximation property with $\varepsilon = 1$ to get $\xi < \xi$ which
		contradicts trichotomy.
	\end{proof}
\end{thm}

\begin{corr}
	Let $\eps > 0$.  Then $0 < \nicefrac{1}{n} < \eps$ for some $n \in \mathbb{N}$.
	
	\begin{proof}
		Apply the Archimedean Property to $\nicefrac{1}{\eps}$.
	\end{proof}
\end{corr}

\section{Countability}

\begin{rmk}
	\begin{itemize}
		\item If $A \subseteq B$ then $|A| \leq |B|$.
		\item If $\varphi:A \to B$ is injective then $|A| \leq |B|$.
		\item If $\psi: A \to B$ is surjective then $|A| \geq |B|$.
	\end{itemize}
\end{rmk}

\begin{propn}
	A set $A$ is countable iff there exists an injection $\theta:A \to
	\mathbb{N}$.  `We can find a natural number for every member of $A$,
	even if we don't have to use them all.'
\end{propn}

\begin{corr}
	If $B \subseteq A$ and $A$ is countable then $B$ is also countable.  The contrapositive is also true.
\end{corr}

\begin{propn}
	Suppose $A$ and $B$ are countably infinite, and $A \cap B = \emptyset$.  Then $A \cup B$ is countably infinite.

	\begin{proof}
		We have two bijections for $A$ and for $B$ with
		$\mathbb{N}$.  Construct a bijection to the union from these using
		even/odd natural numbers.
	\end{proof}
\end{propn}

\begin{rmk}
	The above proposition still holds even if $A$ and $B$ aren't
	disjoint.  Essentially the same bijection $\varphi: \mathbb{N} \to A \cup B$ can
	be used to list them both, but we skip over any repetitions as they
	occur.  This can be used to show that $\mathbb{Z}$ is countably
	infinite.
\end{rmk}

\begin{propn}
	Suppose $A$ and $B$ are countably infinite.  Then $A \times B$ is countably infinite.
	
	\begin{proof}
		We can list the two sets as $a_n$ and $b_m$.  Then we can place
		the ordered pairs in a matrix-like grid with each row having
		constant $m$ and each column having constant $n$.  We can find a
		path through all of these that allows us to form a list of all
		the elements.
	\end{proof}
\end{propn}

\begin{propn}
	If $A_0,\ A_1,\ A_2,\ ...$ are countable sets then so is their union, \[\bigcup^\infty_{i = 0} A_i.\]

	\begin{proof}
		We can list each $A_i$ as $a_{i0},\ a_{i1},\ a_{i2},\ ...$ so we can
		get a square grid as before and we can count in a similar
		fashion, omitting any repetitions of elements that may arise.
	\end{proof}
\end{propn}

\begin{eg}
	$\mathbb{Q}$ is countable.

	\begin{solution}
		The set $\mathbb{Q}^+$ is countable as the map taking a rational
		$\nicefrac{m}{n}$ in its lowest form to $(m,n)$ is an injection from
		$\mathbb{Q}^+$ to $\mathbb{N}^2$, and we have $\mathbb{N}^2 \approx
		\mathbb{N}$.  So $\mathbb{Q} = \{0\} \cup \mathbb{Q}^+ \cup \set{-q}{q
		\in \mathbb{Q}^+}$ is also countable.
	\end{solution}
\end{eg}

\begin{thm}
	$\mathbb{R}$ is uncountable.

	\begin{proof}
		If $\mathbb{R}$ were countable, then [0,1] would also be
		countable.  Clearly [0,1] is not finite since $\nicefrac{1}{k}
		\in [0,1]\;\forall k \in \mathbb{N}$.  A proof by contradiction
		is then used to show that [0,1] is not countably infinite.

		Then we suppose that $\theta: \mathbb{N} \to [0,1]$ is a
		bijection and write $x_k = \theta(k)$.  We now find $a_i, b_j
		\isreal$ s.t. $0 \leq a_1 < a_2 < a_3 < ... < b_3 < b_2 < b_1
		\leq 1,\ x_i \notin [a_i,b_i].$

		Using the fact that $a_m \leq b_n\;\forall n,m \in \mathbb{N}$,
		show that $a_n \leq \sup\set{a_j}{j\in\mathbb{N}} \leq
		\inf\set{b_j}{j \in \mathbb{N}} \leq b_n\;\forall n \in
		\mathbb{N}$.  Then we have $\sup\set{a_j}{j\in\mathbb{N}} \ne
		x_n\;\forall n \in \mathbb{N}$ so $\theta$ is not a bijection,
		giving the required contradiction.
	\end{proof}
\end{thm}

\begin{corr}
	$\mathbb{C}$ is uncountable.
\end{corr}

\begin{rmk}
	It is harder to prove, but in fact, $\mathbb{C} \approx \mathbb{R}$.
\end{rmk}

\begin{thm}
	\textnormal{(Cantor's Theorem)}
	Let $A$ be a set.  Then there is an injection from $A$ to $\mathcal{P}(A)$
	but there is no bijection from $A$ to $\mathcal{P}(A)$, i.e.  \[ |A| <
	|\mathcal{P}(A)|. \]

	\begin{proof}
		$a \mapsto \{a\}$ is an injection.  Show that a general map $\pi:A \to
		\mathcal{P}(A)$ cannot be a surjection and so is not a bijection, by
		considering $X := \set{a \in A}{a \notin \pi(a)}$ and showing that $X
		\notin \pi(A)$ by a reduction to absurdity.
	\end{proof}
\end{thm}

\begin{eg}
	$\mathcal{P}(\mathbb{N}) \approx \mathbb{R}.$
\end{eg}

\section{Sequences in $\mathbb{R}$ and $\mathbb{C}$}

----
\begin{note}
	$A \subseteq B \iff A \cap B = A$
\end{note}

\end{notes}

\end{document}