Created the Real Analysis Theorems and Definitions packet
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\chapter{Continuous Functions}
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\subimport{./}{continuous-functions.tex}
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\subimport{./}{combinations-of-continuous-functions.tex}
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\subimport{./}{continuous-functions-on-intervals.tex}
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\subimport{./}{uniform-continuity.tex}
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\subimport{./}{continuity-and-gauges.tex}
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\subimport{./}{monotone-and-inverse-functions.tex}
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\section{Combinations of Continuous Functions}
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\begin{theorem}
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Let $A \subseteq \R$, let $f$ and $g$ be functions on $A$ to $\R$, and let $b \in \R$. Suppose that $c \in A$ and that $f$ and $g$ are continuous at $c$.
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\begin{enumerate}
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\item Then $f+g,\ f-g,\ fg$, and $bf$ are continuous at $c$.
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\item If $h:A \rightarrow \R$ is continuous at $c \in A$ and if $h(x) \neq 0$ for all $x \in A$, then the quotient $f/h$ is continuous at $c$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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Let $A \subseteq \R$, let $f$ and $g$ be continuous on $A$ to $\R$, and let $b \in \R$.
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\begin{enumerate}
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\item The functions $f+g,\ f-g,\ fg$, and $bf$ are continuous on $A$.
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\item If $h:A \rightarrow \R$ is continuous on $A$ and $h(x) \neq 0$ for $x \in A$, then the quotient $f/h$ is continuous on $A$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $|f|$ be defined by $|f|(x) := |f(x)|$ for $x \in A$
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\begin{enumerate}
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\item If $f$ is continuous at at point $c \in A$, then $|f|$ is continuous at $c$.
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\item If $f$ is continuous on $A$, then $|f|$ is continuous on $A$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $f(x) \geq 0$ for all $x \in A$. We let $\sqrt{f}$ be defined for $x \in A$ by $(\sqrt{f})(x) := \sqrt{f(x)}$.
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\begin{enumerate}
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\item If $f$ is continuous at at point $c \in A$, then $\sqrt{f}$ is continuous at $c$.
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\item If $f$ is continuous on $A$, then $\sqrt{f}$ is continuous on $A$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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Let $A,B \subseteq \R$ and let $f:A \rightarrow \R$ and $g:B \rightarrow \R$ be functions such that $f(A) \subseteq B$. If $f$ is continuous at a point $c \in A$ and g is continuous at $b= f(c) \in B$, then the composition $g \circ f:A \rightarrow \R$ is continuous $c$.
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\end{theorem}
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\begin{theorem}
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Let $A,B \subseteq \R$, let $f:A \rightarrow \R$, be continuous on $A$, and let $g:B \rightarrow \R$ be continuous on $B$. If $f(A) \subseteq B$, then the composite function $g \circ f:A \rightarrow \R$ is continuous on $A$.
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\end{theorem}
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\section{Continuity and Gauges}
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\begin{definition}
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A \textbf{partition} of an interval $I := [a,b]$ is a collection $\mathbb{P} = \{I_1, \dots, I_n\}$ of non-overlapping closed intervals whose union is $[a,b]$. We ordinarily denote the intervals by $I_i:=[x_{i-1}, x_i]$, where
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\[a=x_0 < \dots < x_{i-1} < x_i < \dots < x_n = b.\]
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The points $x_i\ (i-0, \dots, n)$ are called the \textbf{partition points} of $\mathbb{P}$. If a point $t_i$ has been chosen from each interval $I_i$ for $i=1, \dots, n$, then the points $t_i$ are called the \textbf{tags} and the set of ordered pairs
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\[\mathbb{P} = \{(I_1, t_1), \dots, (I_n,t_n)\}\]
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is called a \textbf{tagged partition} of $I$. (The dot signifies that the partition is tagged.)
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\end{definition}
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\begin{definition}
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A \textbf{gauge} on $I$ is a strictly positive function defined on $I$. If $\delta$ is a gauge on $I$, then a (tagged) partition $\mathbb{P}$ is said to be $\delta$-\textbf{fine} if
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\[t_i \in I_i \subseteq [t_i - \delta(t_i), t_i + \delta(t_i)],\ \ \ \ \ \text{for}\ \ \ \ \ i=1, \dots, n.\]
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We note that the notion of $\delta$-fineness requires that the partition be tagged, so we do not need to say ``tagged partition" in this case.
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\end{definition}
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\begin{lemma}
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If a partition $\mathbb{P}$ of $I:=[a,b]$ is $\delta$-fine and $x \in I$, then there exists a tag $t_i$ in $\mathbb{P}$ such that $|x-t_i| \leq \delta (t_i)$.
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\end{lemma}
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\begin{theorem}
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If $\delta$ is a gauge defined on the interval $[a,b]$, then there exists a $\delta$-fine partition of $[a,b]$.
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\end{theorem}
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\section{Continuous Functions on Intervals}
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\begin{definition}
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A function $f:A \rightarrow \R$ is said to be \textbf{bounded on} $A$ if there exists a constant $M > 0$ such that $|f(x)| \leq M$ for all $x \in A$.
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\end{definition}
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\begin{theorem}[\textbf{Boundedness Theorem}]
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Let $I:=[a,b]$ be a closed bounded interval and let $f: I \rightarrow \R$ be continuous on $I$. Then $f$ is bounded on $I$.
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\end{theorem}
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\begin{definition}
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Let $A \subseteq \R$ and let $f:A \rightarrow \R$. We say that $f$ \textbf{has an absolute maximum} on $A$ if there is a point $x^* \in A$ such that
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\[f(x^*) \geq f(x)\ \ \ \ \text{for all}\ \ \ \ x \in A.\]
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We say that $f$ \textbf{has an absolute minimum} on $A$ if there is a point $x_* \in A$ such that
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\[f(x_*) \leq f(x)\ \ \ \ \text{for all}\ \ \ \ x \in A.\]
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We say that $x^*$ is an \textbf{absolute maximum point} for $f$ on $A$, and that $x_*$ is an \textbf{absolute minimum point} for $f$ on $A$, if they exist.
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\end{definition}
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\begin{theorem}[\textbf{Maximum-Minimum Theorem}]
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Let $I := [a,b]$ be a closed bounded interval and let $f: I \rightarrow \R$ be continuous on $I$. Then $f$ has an absolute maximum and an absolute minimum on $I$.
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\end{theorem}
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\begin{theorem}[\textbf{Location of Roots Theorem}]
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Let $I=[a,b]$ and let $f:I \rightarrow \R$ be continuous on $I$. If $f(a) < 0 < f(b)$, or if $f(a) > 0 > f(b)$, then there exists a number $c \in (a,b)$ such that $f(c)=0$.
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\end{theorem}
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\begin{theorem}[\textbf{Bolzano's Intermediate Value Theorem}]
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Let $I$ be an interval and let $f:I \rightarrow \R$ be continuous on $I$. If $a,b \in I$ and if $k \in \R$ satisfies $f(a) < k<f(b)$, then there exists a point $c \in I$ between $a$ and $b$ such that $f(c) = k$.
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\end{theorem}
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\begin{corollary}
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Let $I=[a,b]$ be a closed, bounded interval and let $f:I \rightarrow \R$ be on $I$. If $k \in \R$ is any number satisfying
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\[\inf f(I) \leq k \leq \sup f(I),\]
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then there exists a number $c \in I$ such that $f(c) = k$.
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\end{corollary}
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\begin{theorem}
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Let $I$ be a closed bounded interval and let $f:I \rightarrow \R$ be continuous on $I$. Then the set $f(I):= \{f(x): x \in I\}$ is a closed bounded interval.
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\end{theorem}
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\begin{theorem}[\textbf{Preservation of Intervals Theorem}]
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Let $I$ be an interval and let $f: I \rightarrow \R$ be continuous on $I$. Then the set $f(I)$ is an interval.
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\end{theorem}
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\section{Continuous Functions}
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\begin{definition}
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Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $c \in A$. We say that $f$ is \textbf{continuous at} $c$ if, given any number $\varepsilon > 0$, there exists $\delta > 0$ such that if $x$ is any point of $A$ satisfying $|x-c|<\delta$, then $|f(x)-f(c)|<\varepsilon$.
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\\If $f$ fails to be continuous at $c$, then we say that $f$ is \textbf{discontinuous at} $c$.
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\end{definition}
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\begin{theorem}
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A function $f:A \rightarrow \R$ is continuous at a point $c \in A$ if and only if given any $\varepsilon$-neighborhood $V_\varepsilon (f(c))$ of $f(c)$ there exists a $\delta$-neighborhood $V_\varepsilon(c)$ of $c$ such that if $x$ is any point of $A \cap V_\delta(c)$, then $f(x)$ belongs to $V_\varepsilon (f(c))$, that is
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\[f(A \cap V_\varepsilon (c)) \subseteq V_\varepsilon (f(c))\]
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\end{theorem}
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\begin{theorem}[\textbf{Sequential Criterion for Continuity}]
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A function $f:A \rightarrow \R$ is continuous at the point $c \in A$ if and only if for every sequence $(x_n)$ in $A$ that converges to $c$, the sequence $(f(x_n))$ converges to $f(c)$.
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\end{theorem}
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\begin{theorem}[\textbf{Discontinuity Criterion}]
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Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $c \in A$. Then $f$ is discontinuous at $c$ if and only if there exists a sequence $(x_n)$ in $A$ such that $(x_n)$ converges to $c$, but the sequence $(f(x_n))$ does not converge to $f(c)$.
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\end{theorem}
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\begin{definition}
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Let $A \subseteq \R$ and let $f: A \rightarrow \R$. If $B$ is a subset of $A$, we say that $f$ is \textbf{continuous on the set} $B$ if $f$ is continuous at every point of $B$.
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\end{definition}
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\section{Monotone and Inverse Functions}
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\begin{definition}
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Recall that if $A \subseteq \R$, then a function $f: A \to \R$ is said to be \textbf{increasing on} $A$ if whenever $x_1,x_2 \in A$ and $x_1 \leq x_2$, then $f(x_1) \leq f(x_2)$. The function $f$ is said to be \textbf{strictly increasing on} $A$ if whenever $x_1,x_2 \in A$ and $x_1<x_2$, then $f(x_1) < f(x_2)$. Similarly, $g:A \to \R$ is said to be \textbf{decreasing on} $A$ if whenever $x_1,x_2 \in A$ and $x_1 \leq x_2$ then $g(x_1) \geq g(x_2)$. The function $g$ is said to be \textbf{strictly decreasing on} $A$ if whenever $x_1, x_2 \in A$ and $x_1 < x_2$ then $g(x_1) > g(x_2)$.
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\qquad If a function is either increasing or decreasing on $A$, we say that it is \textbf{monotone} on $A$. If $f$ is either strictly increasing or strictly decreasing on $A$, we say that $f$ is \textbf{strictly monotone} on $A$.
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\end{definition}
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\begin{theorem}
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Let $I \subseteq \R$ be an interval and let $f:I \rightarrow \R$ be increasing on $I$. Suppose that $c \in I$ is not an endpoint of $I$. Then
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\begin{enumerate}
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\item $\lim\limits_{x\to c^-} f = \sup \{f(x): x \in I,\ x < c\}$,
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\item $\lim\limits_{x\to c^+} f = \inf \{f(x): x \in I,\ x > c\}$.
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\end{enumerate}
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\end{theorem}
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\begin{corollary}
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Let $I \subseteq \R$ be an interval and let $f:I \rightarrow \R$ be increasing on $I$. Suppose that $c \in I$ is not an endpoint of $I$. Then the following statements are equivalent.
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\begin{enumerate}
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\item $f$ is continuous at $c$.
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\item $\lim\limits_{x\to c^-} f = f(c) = \lim\limits_{x\to c^+}$.
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\item $\sup \{f(x):x \in I,\ x < c\} = f(c) = \inf \{f(x) : x \in I,\ x > c\}$.
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\end{enumerate}
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\end{corollary}
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\begin{definition}
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If $f:I \to \R$ is increasing on $I$ and if $c$ is not an endpoint of $I$, we define the \textbf{jump of $f$ at $c$} to be $j_f(c):=\lim\limits_{x \to c^+} f-\lim\limits_{x \to c^-} f$. It follows from \textit{Theorem 5.6.1} that
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\[j_f(c)=\inf \{f(x)\ :\ x \in I,\ x > c\}- \sup \{f(x)\ :\ x \in I,\ x < c\}\]
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for an increasing function. If the left endpoint $a$ of $I$ belongs to $I$, we define the \textbf{jump of $f$ at $a$} to be $j_f(a):=\lim\limits_{x \to a^+} f-f(a)$. If the right endpoint $b$ belongs to $I$, we define the \textbf{jump of $f$ at $b$} to be $j_f(b):=f(b)-\lim\limits_{x \to b^-} f$.
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\end{definition}
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\begin{theorem}
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Let $I \subseteq \R$ be an interval and let $f:I \rightarrow \R$ be increasing on $I$. If $c \in I$, then $f$ is continuous at $c$ if and only if $j_f(c)=0$.
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\end{theorem}
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\begin{theorem}
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Let $I \subseteq \R$ be an interval and let $f: I \rightarrow \R$ be monotone on $I$. Then the set of points $D \subseteq I$ at which $f$ is discontinuous is a countable set.
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\end{theorem}
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\begin{theorem}[\textbf{Continuous Inverse Theorem}]
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Let $I \subseteq \R$ be an interval and let $f:I \rightarrow \R$ be strictly monotone and continuous on $I$. Then the function $g$ inverse to $f$ is strictly monotone and continuous on $J:=f(I)$.
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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 If $m,n \in \N$ and $x \geq 0$, we define $x^{m/n} := (x^{1/n})^m$.
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\item If $m,n \in N$ and $x > 0$, we define $x^{-m/n} := (x^{1/n})^{-m}$.
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\end{enumerate}
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\end{definition}
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\begin{theorem}
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If $m \in \Z,\ n \in \N$, and $x > 0$, then $x^{m/n}=(x^m)^{1/n}$.
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\end{theorem}
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\section{Uniform Continuity}
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\begin{definition}
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Let $A \subseteq \R$ and let $f:A \rightarrow \R$. We say that $f$ is \textbf{uniformly continuous} on $A$ if for each $\varepsilon > 0$ there is a $\delta (\varepsilon) > 0$ such that if $x,u \in A$ are any numbers satisfying $|x-u|<\delta (\varepsilon)$, then $|f(x)-f(u)| < \varepsilon$.
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\end{definition}
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\begin{theorem}[\textbf{Nonuniform Continuity Criteria}]
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Let $A \subseteq \R$ and let $f:A \rightarrow \R$. Then the following statements are equivalent:
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\begin{enumerate}
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\item $f$ is not uniformly continuous on $A$.
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\item There exists an $\varepsilon_0 > 0$ such that for every $\delta > 0$ there are points $x_\delta, u_\delta$ in $A$ such that $|x_\delta - u_\delta|<\delta$ and $|f(x_\delta) - f(u_\delta)| \geq \varepsilon_0$.
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\item There exists an $\varepsilon_0 > 0$ and two sequences $(x_n)$ and $(u_n)$ in $A$ such that $\lim (x_n - u_n)=0$ and $|f(x_n)-f(u_n)|\geq \varepsilon_0=1$ for all $n \in \N$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[\textbf{Uniform Continuity Theorem}]
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Let $I$ be a closed bounded interval and let $f:I \rightarrow \R$ be continuous on $I$. Then $f$ is uniformly continuous on $I$.
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\end{theorem}
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\begin{definition}
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Let $A \subseteq \R$ and let $f:A \rightarrow \R$. If there exists a constant $K > 0$ such that
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\[(4)\ \ \ \ \ \ \ \ |f(x)-f(u)| \leq K|x-u|\]
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for all $x,u \in A$, then $f$ is said to be a \textbf{Lipschitz function} (or to satisfy a \textbf{Lipschitz condition}) on $A$.\\
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The condition $(4)$ that a function $f: I \to \R$ on an interval $I$ is a Lipschitz function can be interpreted geometrically as follows. If we write the condition as
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\[\abs{\frac{f(x)-f(u)}{x-u}}\leq K,\ x,u \in I,\ x \neq u,\]
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then the quantity inside the absolute values is the slope of a line segment joining the points $(x,f(x))$ and $(u,f(u))$. Thus a function $f$ satisfies a Lipschitz condition if and only if the slopes of all line segments joining two points on the graph of $y=f(x)$ over $I$ are bounded by some number $K$.
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\end{definition}
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\begin{theorem}
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If $f:A \rightarrow \R$ is a Lipschitz function, then $f$ is uniformly continuous on $A$.
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\end{theorem}
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\begin{theorem}
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If $f:A \rightarrow \R$ is uniformly continuous on a subset $A$ of $\R$ and if $(x_n)$ is a Cauchy sequence in $A$, then $(f(x_n))$ is a Cauchy sequence in $\R$.
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\end{theorem}
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\begin{theorem}[\textbf{Continuous Extension Theorem}]
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A function $f$ is uniformly continuous on the interval $(a,b)$ if and only if it can be defined at the endpoints $a$ and $b$ such that the extended function is continuous on $[a,b]$.
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\end{theorem}
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\begin{definition}
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A function $s:[a,b] \rightarrow \R$ is called a \textbf{step function} if $[a,b]$ is the union of a finite number of nonoverlapping intervals $I_1, I_2, \dots, I_n$ such that $s$ is constant on each interval, that is, $s(x)=c_k$ for all $x \in I_k,\ k=1,2, \dots, n$.
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\end{definition}
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\begin{theorem}
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Let $I$ be a closed bounded interval and let $f:I \rightarrow \R$ be continuous on $I$. If $\varepsilon > 0$, then there exists a step function $s_\varepsilon : I \rightarrow \R$ such that $|f(x) - s_\varepsilon(x)| < \varepsilon$ for all $x \in I$.
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\end{theorem}
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\begin{corollary}
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Let $I:=[a,b]$ be a closed bounded interval and let $f: I \rightarrow \R$ be continuous on $I$. If $\varepsilon > 0$, there exists a natural number $m$ such that if we divide $I$ into $m$ disjoint intervals $I_k$ having length $h:= (b-a /m)$, then the step function $s_\varepsilon$ defined in equation (5) satisfies $|f(x) - s_\varepsilon (x)| < \varepsilon$ for all $x \in I$.
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\end{corollary}
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\begin{definition}
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Let $I:= [a,b]$ be an interval. Then a function $g:I \rightarrow \R$ is said to be \textbf{piecewise linear} on $I$ if $I$ is the union of a finite number of disjoint intervals $I_1, \dots, I_m$, such that the restriction of $g$ to each interval $I_k$ is a linear function.
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\end{definition}
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\begin{theorem}
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Let $I$ be a closed bounded interval and let $:I \rightarrow \R$ be continuous on $I$. If $\varepsilon > 0$, then there exists a continuous piecewise linear function $g_\varepsilon : I \rightarrow \R$ such that $|f(x) - g_\varepsilon (x)| < \varepsilon$ for all $ x \in I$.
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\end{theorem}
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\begin{theorem}[\textbf{Wierstrass Approximation Theorem}]
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Let $I=[a,b]$ and let $f: I \rightarrow \R$ be a continuous function. If $\varepsilon > 0$ is given, then there exists a polynomial function $p_\varepsilon$ such that $|f(x) - p_\varepsilon (x)| < \varepsilon$ for all $ x \in I$.
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\end{theorem}
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