Apply fixes for errata from Thomas Mahoney

ch-contfunc.tex

@@ -2397,7 +2397,7 @@ \subsection{Limits at infinity}
 \begin{equation*}
 \abs{f(x) - L} < \epsilon 
 \end{equation*}
-whenever $x \geq M$, then we say $f(x)$
+whenever $x \in S$ and $x \geq M$, then we say $f(x)$
 \emph{converges}\index{converges!function} to $L$
 as $x$ goes to $\infty$.  We call $L$ the \emph{limit}
 and write
@@ -2414,7 +2414,7 @@ \subsection{Limits at infinity}
 \begin{equation*}
 \abs{f(x) - L} < \epsilon 
 \end{equation*}
-whenever $x \leq M$, then we say $f(x)$ \emph{converges} to $L$
+whenever $x \in S$ and $x \leq M$, then we say $f(x)$ \emph{converges} to $L$
 as $x$ goes to $-\infty$.  We call $L$ the \emph{limit} and write
 \begin{equation*}
 \lim_{x \to -\infty} f(x) := L .
@@ -2481,7 +2481,7 @@ \subsection{Limits at infinity}
 \begin{equation*}
 \lim_{n\to\infty} f(x_n) = L
 \end{equation*}
-for all sequences $\{ x_n \}$ such that $\lim\limits_{n\to\infty} x_n = \infty$.
+for all sequences $\{ x_n \}$ in $S$ such that $\lim\limits_{n\to\infty} x_n = \infty$.
 \end{lemma}
 
 The lemma holds for the limit as $x \to -\infty$.
@@ -2500,8 +2500,8 @@ \subsection{Limits at infinity}
 We prove the converse by contrapositive.  Suppose $f(x)$ does
 not go to $L$ as $x \to \infty$.
 This means that there exists an $\epsilon > 0$,
-such that for every $M \in \N$, there exists an $x \in S$, $x \geq M$, let
-us call it $x_M$, such that $\abs{f(x_M)-L} \geq \epsilon$.
+such that for every $n \in \N$, there exists an $x \in S$, $x \geq n$, let
+us call it $x_n$, such that $\abs{f(x_n)-L} \geq \epsilon$.
 Consider the sequence $\{ x_n \}$.  Clearly 
 $\{ f(x_n) \}$ does not converge to $L$.  It remains to note
 that $\lim\, x_n = \infty$, because $x_n \geq n$ for all $n$.
@@ -2805,7 +2805,7 @@ \subsection{Continuity of monotone functions}
 we have $\abs{f(x)-a} < \epsilon$.
 \end{proof}
 
-Suppose $f \colon S \to \R$, $c \in S$, and
+Suppose $f \colon S \to \R$ is increasing, $c \in S$, and
 that both one-sided limits exist.
 Since $f(x) \leq f(c) \leq f(y)$
 whenever $x < c < y$, taking the limits we obtain
@@ -3084,11 +3084,12 @@ \subsection{Exercises}
 {\ }
 \begin{enumerate}[a)]
 \item
-Let $S \subset \R$ be any subset.  If $f \colon S \to \R$ is increasing,
+Let $S \subset \R$ be any subset.  If $f \colon S \to \R$ is increasing and
+bounded,
 then show that there exists an increasing $F \colon \R \to \R$
 such that $f(x) = F(x)$ for all $x \in S$.
 \item
-Find an example of a strictly increasing $f \colon S \to \R$ such that
+Find an example of a strictly increasing bounded $f \colon S \to \R$ such that
 an increasing $F$ as above is never strictly increasing.
 \end{enumerate}
 \end{exercise}

ch-der.tex

@@ -917,9 +917,9 @@ \subsection{Continuity of derivatives and the intermediate value theorem}
 $\frac{g(x)-g(b)}{x-b} < 0$ or that $g(x) > g(b)$, thus
 $g$ cannot possibly have a maximum at $b$.
 Therefore, $c \in (a,b)$,
-and \hyperref[thm:rolle]{Rolle's theorem} applies: As $g$ attains a maximum
+and \lemmaref{relminmax:lemma} applies: As $g$ attains a maximum
 at $c$ we find $g'(c) = 0$
-and $f'(c) = y$.
+and so $f'(c) = y$.
 
 Similarly, if $f'(a) > y > f'(b)$, consider $g(x) := f(x)- yx$.
 \end{proof}

ch-riemann.tex

@@ -202,7 +202,7 @@ \subsection{Partitions and lower and upper integrals}
 $k_0 < k_1 < \cdots < k_n$ such that $x_j = \widetilde{x}_{k_j}$ for
 $j=0,1,2,\ldots,n$.
 
-Let $\Delta \widetilde{x}_p = \widetilde{x}_{p-1} - \widetilde{x}_p$.
+Let $\Delta \widetilde{x}_p = \widetilde{x}_p - \widetilde{x}_{p-1}$.
 See \figureref{fig:refinement}.
 We get 
 \begin{equation*}
@@ -2448,9 +2448,9 @@ \section{Improper integrals}
 requires the optional \sectionref{sec:limitatinf})}
 
 Often it is necessary to integrate over the
-entire real line, or a infinite interval of the form $[a,\infty)$ or
-$(\infty,b]$.  Also, we may wish to integrate unbounded functions
-defined on a finite interval $(a,b)$.
+entire real line, or a unbounded interval of the form $[a,\infty)$ or
+$(-\infty,b]$.  Also, we may wish to integrate unbounded functions
+defined on a open bounded interval $(a,b)$.
 Such functions are not Riemann integrable, but we may want to write down
 the integral anyway in the spirit of \lemmaref{lemma:boundedimpriemann}.
 These integrals are called \emph{\myindex{improper integrals}},
@@ -2521,8 +2521,7 @@ \section{Improper integrals}
 -
 \frac{1^{-p+1}}{-p+1}
 =
--
-\frac{1}{(p-1)b^{p-1}}
+\frac{-1}{(p-1)b^{p-1}}
 +
 \frac{1}{p-1} .
 \end{equation*}
@@ -2799,7 +2798,7 @@ \section{Improper integrals}
 if the limits exist.
 
 Suppose $f \colon \R \to \R$ is a function such that
-$f$ is Riemann integrable on all finite intervals $[a,b]$.  Then
+$f$ is Riemann integrable on all bounded intervals $[a,b]$.  Then
 we define
 \begin{equation*}
 \int_{-\infty}^\infty f := \lim_{c \to -\infty} \, \lim_{d \to \infty} \, \int_c^d f ,
@@ -2835,7 +2834,8 @@ \section{Improper integrals}
 exist.  Let us prove this fact only for the infinite limits.
 
 \begin{prop}
-If $f \colon \R \to \R$ is a function integrable on every interval.
+If $f \colon \R \to \R$ is a function integrable on every bounded interval
+$[a,b]$.
 Then 
 \begin{equation*}
 \lim_{a \to -\infty} \, \lim_{b \to \infty} \, \int_a^b f
@@ -2986,7 +2986,7 @@ \section{Improper integrals}
 \operatorname{sinc}(x) =
 \begin{cases}
 \frac{\sin(x)}{x} & \text{if $x \not= 0$} , \\
-0 & \text{if $x = 0$} .
+1 & \text{if $x = 0$} .
 \end{cases}
 \end{equation*}
 \begin{myfigureht}

ch-seq-ser.tex

@@ -2637,7 +2637,7 @@ \section{Cauchy sequences}
 
 Define $a := \limsup \, x_n$ and
 $b := \liminf \, x_n$.
-By \propref{seqconvsubseqconv:prop}, there exist subsequences
+By \thmref{subseqlimsupinf:thm}, there exist subsequences
 $\{ x_{n_i} \}$ and
 $\{ x_{m_i} \}$, such that
 \begin{equation*}
@@ -3214,7 +3214,7 @@ \subsection{Absolute convergence}
 \end{proof}
 
 If $\sum x_n$ converges absolutely, the limits of
-$\sum x_n$ and $\sum \abs{x_n}$ are different.  Computing one
+$\sum x_n$ and $\sum \abs{x_n}$ may be different.  Computing one
 does not help us compute the other.  However the computation above leads to
 a useful inequality for absolutely convergent series,
 a series version of the triangle inequality,
@@ -3241,7 +3241,7 @@ \subsection{Absolute convergence}
 \sum_{n=1}^\infty \frac{1}{n}
 \end{equation*}
 diverges.  Therefore,
-$\sum \frac{{(-1)}^n}{n}$ is a conditionally convergent subsequence.
+$\sum \frac{{(-1)}^n}{n}$ is a conditionally convergent series.
 
 \subsection{Comparison test and the \texorpdfstring{$p$}{p}-series}
 
@@ -3464,7 +3464,8 @@ \subsection{Ratio test}
 \end{prop}
 
 \begin{proof}
-From \lemmaref{seq:ratiotest} we note that if $L > 1$, then $x_n$
+If $L > 1$, then
+\lemmaref{seq:ratiotest} says that the sequence $\{ x_n \}$
 diverges.  Since it is a necessary condition for the convergence of series
 that the terms go to zero, we know that $\sum x_n$ must diverge.
 

figures/invcontfigA.xp

@@ -17,8 +17,8 @@ int main()
   line(P(0,1),P(1,2));
 
   dot_size(3);
-  dot(P(0,0));
-  circ(P(0,1));
+  circ(P(0,0));
+  dot(P(0,1));
 
   tikz_format();
   end();

figures/invcontfigB.xp

@@ -17,8 +17,8 @@ int main()
   line(P(1,0),P(2,1));
 
   dot_size(3);
-  dot(P(0,0));
-  circ(P(1,0));
+  circ(P(0,0));
+  dot(P(1,0));
 
   tikz_format();
   end();