Created the Real Analysis Theorems and Definitions packet
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\chapter{Preliminaries}
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\subimport{./}{sets-and-functions.tex}
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\subimport{./}{mathematical-induction.tex}
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\subimport{./}{finite-and-infinite-sets.tex}
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\section{Finite and Infinite Sets}
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\begin{definition}
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\begin{enumerate}
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\item[]
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\item The empty set $\emptyset$ is said to have $0$ \textbf{elements}.
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\item If $n \in \N$, a set $S$ is said to have $n$ \textbf{elements} if there exists a bijection from the set $\N_n := \{1, 2, \dots, n\}$ onto $S$.
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\item A set $S$ is said to be \textbf{finite} if it is either empty or it has $n$ elements for some $n \in \N$.
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\item A set $S$ is said to be \textbf{infinite} if it is not finite.
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\end{enumerate}
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\end{definition}
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\begin{theorem}[\textbf{Uniqueness Theorem}]
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If $S$ is a finite set, then the number of elements is $S$ is a unique number in $\N$.
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\end{theorem}
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\begin{theorem}
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The set $\N$ of natural numbers is an infinite set.
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\end{theorem}
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\begin{theorem}
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\begin{enumerate}
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\item[]
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\item If $A$ is a set with $m$ elements and $B$ is a set with $n$ elements and if $A \cap B = \emptyset$, then $A \cup B$ has $m +n$ elements.
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\item If $A$ is a set with $m \in \N$ elements and $C \subseteq A$ is a set with $1$ element, then $A \setminus C$ is a set with $m-1$ elements.
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\item If $C$ is an infinite set and $B$ is a finite set, then $C \setminus B$ is an infinite set.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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Suppose that $S$ and $T$ are sets and that $T \subseteq S$.
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\begin{enumerate}
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\item If $S$ is a finite set, then $T$ is a finite set.
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\item If $T$ is an infinite set, then $S$ is an infinite set.
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\end{enumerate}
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\end{theorem}
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\begin{definition}
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\begin{enumerate}
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\item[]
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\item A set $S$ is said to be \textbf{denumerable} (or \textbf{countably infinite}) if there exists a bijection of $\N$ onto $S$.
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\item A set $S$ is said to be \textbf{countable} if it is either finite or denumerable.
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\item A set $S$ is said to be \textbf{uncountable} if it is not countable.
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\end{enumerate}
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\end{definition}
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\begin{theorem}
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The set $\N \times \N$ is denumerable.
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\end{theorem}
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\begin{theorem}
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Suppose that $S$ and $T$ are sets and that $T \subseteq S$.
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\begin{enumerate}
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\item If $S$ is a countable set, then $T$ is a countable set.
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\item If $T$ is an uncountable set, then $S$ is an uncountable set.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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The following statements are equivalent:
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\begin{enumerate}
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\item $S$ is a countable set.
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\item There exists a surjection of $\N$ onto $S$.
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\item There exists an injection of $S$ into $\N$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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The set $\Q$ of all rational numbers is denumerable.
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\end{theorem}
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\begin{theorem}
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If $A_m$ is a countable set for each $m \in \N$, then the union $A:= \bigcup\limits_{m=1}^{\infty} A_m$ is countable.
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\end{theorem}
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\begin{theorem}[\textbf{Cantor's Theorem}]
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If $A$ is any set, then there is no surjection of $A$ onto the set $\mathcal{P}(A)$ of all subsets of $A$.
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\end{theorem}
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\section{Mathematical Induction}
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\begin{theorem}[\textbf{Well-Ordering Property of $\N$}]
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Every nonempty subset of $\N$ has a least element.
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\end{theorem}
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A more detailed statement of this property is as follows: If $S$ is a subset of $\N$ and if $S \neq \emptyset$, then there exists $m \in S$ such that $m \leq k$ for all $k \in S$.
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\begin{theorem}[\textbf{Principle of Mathematical Induction}]
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Let $S$ be a subset of $\N$ that possesses the two properties:
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\begin{enumerate}
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\item The number $1 \in S$.
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\item For every $k \in \N$, if $k \in S$, then $k + 1 \in S$.
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\end{enumerate}
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Then we have $S = \N$.
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\end{theorem}
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\begin{theorem}[\textbf{Principle of Mathematical Induction (second version)}]
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Let $n_0 \in \N$ and let $P(n)$ be a statement for each natural number $n \geq n_0$. Suppose that:
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\begin{enumerate}
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\item The statement $P(n_0)$ is true.
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\item For all $k \geq n_0$, the truth of $P(k)$ implies the truth of $P(k+1)$.
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\end{enumerate}
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Then $P(n)$ is true for all $n \geq n_0$.
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\end{theorem}
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\begin{theorem}[\textbf{Principle of Strong Induction}]
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Let $S$ be a subset of $\N$ such that
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\begin{enumerate}
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\item $1 \in S$.
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\item For every $k \in \N$, if $\{1, 2, \dots \} \subseteq S$, then $k+1 \in S$.
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\end{enumerate}
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Then $S = \N$.
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\end{theorem}
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\section{Sets and Functions}
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\begin{definition}
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Two sets $A$ and $B$ are said to be \textbf{equal,} and we write $A=B$ if they contain the same elements.
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\\\\Thus, to prove that the sets $A$ and $B$ are equal, we must show that
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\[A \subseteq B \text{ and } B \subseteq A\]
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\end{definition}
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\begin{definition}
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\begin{enumerate}
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\item[]
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\item The \textbf{union} of sets $A$ and $B$ is the set
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\[A \cup B := \{x:x \in A \text{ or } x \in B\}.\]
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\item The \textbf{intersection} of the sets $A$ and $B$ is the set
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\[A \cap B := \{x:x \in A \text{ and } x \in B \}.\]
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\item The \textbf{complement of $B$ relative to $A$} is the set
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\[A \setminus B := \{x:x \in A \text{ and } x \notin B\}\]
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\end{enumerate}
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\end{definition}
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\begin{theorem}
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If $A$, $B$, $C$ are sets, then
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\begin{enumerate}
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\item $A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C),$
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\item $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C).$
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\end{enumerate}
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\end{theorem}
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\begin{definition}
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If $A$ and $B$ are nonempty sets, then the \textbf{Cartesian product} $A \times B$ of $A$ and $B$ is the set of all ordered pairs $(a,b)$ with $a \in A$ and $b \in B$. That is,
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\[A \times B := \{(a,b): a \in A,\ b \in B\}.\]
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\end{definition}
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\begin{definition}
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Let $A$ and $B$ be sets. Then a \textbf{function} from $A$ to $B$ is a set $f$ of ordered pairs in $A \times B$ such that for each $a \in A$ there exists a unique $B \in B$ with $(a,b) \in f$. (In other words, if $(a,b) \in f$ and $(a, b') \in f$, then $b = b'$.)
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\end{definition}
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\begin{definition}
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If $E$ is a subset of $A$, then the \textbf{direct image} of $E$ under $f$ is the subset $f(E)$ of $B$ given by
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\[f(E):=\{f(x):x \in E\}\]
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If $H$ is a subset of $B$, then the \textbf{inverse image} of $H$ under $f$ is the subset $f^{-1}(H)$ of $A$ given by
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\[f^{-1}(H):=\{x \in A:f(x) \in H\}\]
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\end{definition}
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\begin{definition}
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Let $f:A \rightarrow B$ be a function from $A$ to $B$.
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\begin{enumerate}
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\item The function $f$ is said to be \textbf{injective} (or to be \textbf{one-one}) if whenever $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. If $f$ is an injective function, we also say that $f$ is an \textbf{injection}.
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\item The function $f$ is said to be \textbf{surjective} (or to map $A$ \textbf{onto} $B$) if $f(A)=B$; that is, if the range $R(f)=B$. If $f$ is a surjective function, we also say that $f$ is a \textbf{surjection}.
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\item If $f$ is both injective and surjective, then $f$ is said to be \textbf{bijective}. If $f$ is bijective, we also say that $f$ is a \textbf{bijection}.
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\end{enumerate}
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\end{definition}
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\begin{definition}
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If $f: A \rightarrow B$ is a bijection of $A$ onto $B$, then
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\[g := \{(b,a) \in B \times A: (a,b) \in f\}\]
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is a function on $B$ into $A$. This function is called the \textbf{inverse function} of $f$, and is denoted by $f^{-1}$. The function $f^{-1}$ is also called the \textbf{inverse} of $f$.
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\\We can also express the connection between $f$ and its inverse $f^{-1}$ by noting that $D(f)=R(f^{-1})$ and $R(f)=D(f^{-1})$ and that
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\[b=f(a) \text{ if and only if } a=f^{-1}(b)\]
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\end{definition}
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\begin{definition}
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If $f: A \rightarrow B$ and $g:B \rightarrow C$, and if $R(f) \subseteq D(g) = B$, then the \textbf{composite function} $g \circ f$ (note the order!) is the function from $A$ into $C$ defined by
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\[(g \circ f)(x) := g(f(x)) \text{ for all } x \in A\]
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\end{definition}
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\begin{theorem}
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Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be functions and let $H$ be a subset of $C$. Then we have
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\[(g \circ f)^{-1}(H) = f^{-1}(g^{-1}(H)).\]
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Note the reversal in the order of the functions.
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\end{theorem}
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