From 59b1eb92fb9c2d09029d7fe9fcc4b5a55cf5cdfc Mon Sep 17 00:00:00 2001 From: "Jiri (George) Lebl" Date: Sun, 23 Sep 2018 00:14:35 -0500 Subject: [PATCH] A little bit of cleanup, fix erratum --- ch-contfunc.tex | 2 +- ch-der.tex | 2 +- ch-metric.tex | 43 ++++++++++++++++++++++--------------------- ch-seq-ser.tex | 2 +- 4 files changed, 25 insertions(+), 24 deletions(-) diff --git a/ch-contfunc.tex b/ch-contfunc.tex index e352c0b..fa17f98 100644 --- a/ch-contfunc.tex +++ b/ch-contfunc.tex @@ -1803,7 +1803,7 @@ \subsection{Exercises} \begin{enumerate}[a)] \item Prove that if there is a $c$ such that $f(c)f(-c) < 0$, -then there is a $d$ such that $f(d) = 0$. +then there is a $d \in \R$ such that $f(d) = 0$. \item Find a continuous function $f$ such that $f(\R) = \R$, but $f(x)f(-x) \geq 0$ for all $x \in \R$. diff --git a/ch-der.tex b/ch-der.tex index 17279a9..9b61ff5 100644 --- a/ch-der.tex +++ b/ch-der.tex @@ -1473,7 +1473,7 @@ \subsection{Exercises} \begin{exercise} Prove the more general version of the second derivative test. Suppose $f \colon (a,b) \to \R$ is differentiable and $x_0 \in (a,b)$ -is such that $f''(x_0)$ exists and $f''(x_0) > 0$. +is such that, $f'(x_0) = 0$, $f''(x_0)$ exists, and $f''(x_0) > 0$. Prove that $f$ has a strict relative minimum at $x_0$. Hint: Consider the limit definition of $f''(x_0)$. \end{exercise} diff --git a/ch-metric.tex b/ch-metric.tex index 42dfadf..ebc026c 100644 --- a/ch-metric.tex +++ b/ch-metric.tex @@ -2143,7 +2143,7 @@ \subsection{Compactness} \begin{thm} \label{thm:mscompactisseqcpt} -Let $(X,d)$ be a metric space. Then $K \subset X$ is a compact set if +Let $(X,d)$ be a metric space. Then $K \subset X$ is compact if and only if every sequence in $K$ has a subsequence converging to a point in $K$. \end{thm} @@ -2226,11 +2226,11 @@ \subsection{Compactness} of radius $\delta$ are drawn.\label{fig:seqcompactiscompact}} \end{myfigureht} -Either at some point we obtain a finite subcover of $K$ +Either at some point we obtain a finite subcover of $K$, or we obtain an infinite sequence $\{ x_n \}$ as above. -For contradiction suppose that +For contradiction, suppose that there is no finite subcover and we have the sequence $\{ x_n \}$. For all $n$ and $k$, $n \not= k$, we have $d(x_n,x_k) \geq \delta$, @@ -2281,7 +2281,7 @@ \subsection{Compactness} a condition that is much easier to check. Let us reiterate that the Heine--Borel theorem only holds for $\R^n$ and not for metric spaces in general. In general, compact implies closed and -bounded, but not vice versa. +bounded, but not vice-versa. \begin{proof} For $\R = \R^1$ if $K \subset \R$ is closed and bounded, then @@ -2297,7 +2297,7 @@ \subsection{Compactness} that $B$ is compact. Then $K$, being a closed subset of a compact $B$, is also compact. -Let $\{ (x_k,y_k) \}_{k=1}^\infty$ be a sequence in $B$. That is, +Let $\bigl\{ (x_k,y_k) \bigr\}_{k=1}^\infty$ be a sequence in $B$. That is, $a \leq x_k \leq b$ and $c \leq y_k \leq d$ for all $k$. A bounded sequence of real numbers has a convergent @@ -2508,7 +2508,7 @@ \subsection{Exercises} \begin{exercise} \label{exercise:relativelycompactseq} Let $(X,d)$ be a complete metric space. -We say a set $S$ is \emph{\myindex{relatively compact}} +We say a set $S \subset X$ is \emph{\myindex{relatively compact}} if the closure $\widebar{S}$ is compact. Prove that $S \subset X$ is relatively compact if and only if given any sequence $\{ x_n \}$ in $S$, there exists a subsequence @@ -2680,11 +2680,11 @@ \subsection{Compactness and continuity} \end{thm} \begin{proof} -As $X$ is compact and $f$ is continuous, we have -that $f(X) \subset \R$ is compact. Hence $f(X)$ is closed +As $X$ is compact and $f$ is continuous, then +$f(X) \subset \R$ is compact. Hence $f(X)$ is closed and bounded. In particular, $\sup f(X) \in f(X)$ and -$\inf f(X) \in f(X)$, because both the sup and inf +$\inf f(X) \in f(X)$, because both the sup and the inf can be achieved by sequences in $f(X)$ and $f(X)$ is closed. Therefore, there is some $x \in X$ such that $f(x) = \sup f(X)$ and some $y \in X$ such that $f(y) = \inf f(X)$. @@ -2787,7 +2787,7 @@ \subsection{Uniform continuity} \begin{thm} \label{thm:Xcompactfunifcont} Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. -Suppose $f \colon X \to Y$ is continuous and $X$ compact. Then +Suppose $f \colon X \to Y$ is continuous and $X$ is compact. Then $f$ is uniformly continuous. \end{thm} @@ -2874,7 +2874,7 @@ \subsection{Uniform continuity} In applications, if we are interested in continuity at $y_0$, we just need to apply the proposition in $[a,b] \times [y_0-\epsilon,y_0+\epsilon]$ -for some small $\epsilon > 0$. So for example if $f$ is continuous in +for some small $\epsilon > 0$. For example, if $f$ is continuous in $[a,b] \times \R$, then $g$ is continuous on $\R$. \begin{example} @@ -2918,14 +2918,15 @@ \subsection{Cluster points and limits of functions} \setminus \{ p \}$ is not empty. \end{defn} -It is not just that $p$ is in the closure of $S$, but an equivalent -definition is that $p$ is in the closure of $S \setminus \{ p \}$ (exercise). -Therefore, $p$ is a cluster point if and only if there exists a sequence in $S \setminus +It is not enough that $p$ is in the closure of $S$, +it must be in the closure of +$S \setminus \{ p \}$ (exercise). +So, $p$ is a cluster point if and only if there exists a sequence in $S \setminus \{ p \}$ that converges to $p$. \begin{defn} \index{limit!of a function in a metric space}% -Let $(X,d_X)$, $(Y,d_Y)$ be a metric space, $S \subset X$, $p$ a cluster point of $S$, +Let $(X,d_X)$, $(Y,d_Y)$ be metric spaces, $S \subset X$, $p \in X$ a cluster point of $S$, and $f \colon S \to Y$ a function. Suppose there exists an $L \in Y$ and for every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $x \in S \setminus \{ p \}$ @@ -2933,9 +2934,9 @@ \subsection{Cluster points and limits of functions} \begin{equation*} d_Y\bigl(f(x),L\bigr) < \epsilon . \end{equation*} -In this case we say $f(x)$ +Then we say $f(x)$ \emph{converges}\index{converges!function in a metric space} to $L$ as $x$ goes -to $p$. We say $L$ is the \emph{limit} of $f(x)$ as $x$ +to $p$, and $L$ is the \emph{limit} of $f(x)$ as $x$ goes to $p$. We write \glsadd{not:limfunc} \begin{equation*} @@ -2945,7 +2946,7 @@ \subsection{Cluster points and limits of functions} \emph{diverges}\index{diverges!function in a metric space} at $p$. \end{defn} -And as usual, we have used the definite article without showing that the +As usual, we used the definite article without showing that the limit is unique. The proof is a direct translation of the proof from \chapterref{lim:chapter}, so we leave it as an exercise. @@ -2956,10 +2957,10 @@ \subsection{Cluster points and limits of functions} the limit of $f(x)$ as $x$ goes to $p$ is unique. \end{prop} -In any metric space, just like in $\R$ continuous limits may be +In any metric space, just like in $\R$, continuous limits may be replaced by sequential limits. The proof is again a direct translation -of the proof from \chapterref{lim:chapter}, and we leave that as an -exercise. The upshot is that if we really only need to prove things for +of the proof from \chapterref{lim:chapter}, and we leave it as an +exercise. The upshot is that we really only need to prove things for sequential limits. \begin{lemma}\label{ms:seqflimit:lemma} diff --git a/ch-seq-ser.tex b/ch-seq-ser.tex index bf279a2..19d4c25 100644 --- a/ch-seq-ser.tex +++ b/ch-seq-ser.tex @@ -2768,7 +2768,7 @@ \subsection{Exercises} \end{exercise} \begin{exercise} -True/False prove or find a counterexample: If $\{ x_n \}$ is a Cauchy sequence then there exists an $M$ +True or false, prove or find a counterexample: If $\{ x_n \}$ is a Cauchy sequence then there exists an $M$ such that for all $n \geq M$ we have $\abs{x_{n+1}-x_n} \leq