Merge branch 'release/release-1.0' into develop

abstract.tex

@@ -6,23 +6,23 @@
 \begin{abstract}
 
 	These solutions were done as a student who is fresh out of university.
-	Now I'm studying the book when serving my mandatory military service.
+	Now I'm studying the book when doing my mandatory military service.
 
 	\vspace{2ex}
 
-	The solutions might be brief, however I state them as clearly as possible I can.
-	For the computation parts, I might omit details about them because they are tedious but simple.
+	The solutions might be brief. However, I state them as clearly as possible I can.
+	For the computation part, I might omit details because it is tedious but simple.
 
 	\vspace{2ex}
 
-	This material now only contains some selected solutions. 
+	This manual now only contains some selected solutions which I can afford.
 	If there is something vague or incredible, it is possible that it doesn't make sense since it is wrong.
 	If you have any suggestions or corrections to solutions, please direct to email "chengmao.lee@gmail.com".
 	Your intelligence will be highly appreciated.
 
 	\vspace{2ex}
 
-	Hope this material could share some constructive ideas or help you that given important hints to solve problems by yourselves, or just verify whether the answers are consistent or not.
+	Hope this manual could share some constructive ideas or help you that given important hints to solve problems by yourselves, or just verify whether the answers are consistent or not.
 
 	\vspace{2ex}
 

chapter2/exercises/ex_2.10.tex

@@ -1,23 +1,23 @@
 % === Exercise 2.10 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item $\mathbf{Prove\ that\ this\ is\ a\ metric.}$
+		\item \textbf{Prove that this is a metric.}
 		\begin{proof}
 			It suffices to prove three properties as following one by one.
 			\begin{itemize}
-				\item $\mathbf{Property\ 1:\ Positive\ Definite}$
+				\item \textbf{Property $1$ : Positive Definite}
 
 				For $p=q$, $d(p,q) = d(p,p) = 0$. For $p\neq q$, $d(p,q) = 1 > 0$.
 
 				Hence $d(p,q)\geq 0$ with equality if and only if $p=q$.
 
-				\item $\mathbf{Property\ 2:\ Symmetric}$
+				\item \textbf{Property $2$ : Symmetric}
 
 				For $p=q$, $d(p,q) = 0 = d(q,p)$. For $p\neq q$, $d(p,q) = 1 = d(q,p)$.
 
 				Hence $d(p,q) = d(q,p)$.
 
-				\item $\mathbf{Property\ 3:\ Triangle\ Inequality}$
+				\item \textbf{Property $3$ : Triangle Inequality}
 
 				Let $r\in X$, then we consider four cases as following.
 

chapter2/exercises/ex_2.11.tex

@@ -8,7 +8,7 @@
 		In order to determine whether these functions are metric or not, the verification is similar to Exercise 2.10. Here we only show something nontrivial. Notice that we suppose $z\in\mathbb{R}^1$.
 
 		\begin{itemize}
-			\item $\mathbf{d_1\ violates\ the\ Triangle\ Inequality \ property}$
+			\item \textbf{$d_1$ violates the Triangle Inequality property.}
 
 			Pick $x=4$, $y=0$, $z=1$. Then
 			\begin{alignat*}{7}
@@ -18,8 +18,8 @@
 			\end{alignat*}
 			This is a contradiction.
 
-			\item $\mathbf{d_2\ obeys\ the\ Triangle\ Inequality\ property}$
-			
+			\item \textbf{$d_2$ obeys the Triangle Inequality property.}
+
 			\begin{alignat*}{7}
 			\quad&& |x-y| &\leq |x-z| + |z-y| \\
 			\implies&& |x-y| &\leq |x-z| + |z-y| + 2\sqrt{|x-z| |z-y|} \\
@@ -29,18 +29,18 @@
 			\end{alignat*}
 			This implies what we need to show.
 
-			\item $\mathbf{d_3\ violates\ the\ Positive\ Definite\ property}$
+			\item \textbf{$d_3$ violates the Positive Definite property.}
 
 			Pick $x=1$, $y=-1$, then $d_3(x,y) = 0$ with $x\neq y$.
 
-			\item $\mathbf{d_4\ violates\ the\ Symmetry\ property}$
+			\item \textbf{$d_4$ violates the Symmetry property.}
 
 			Pick $x=0$, $y=1$, then
 			$$
 			d_4(x,y) = |x-2y| = 2 \neq 1 = |y-2x| = d_4(y,x).
 			$$
 
-			\item $\mathbf{d_5\ obeys\ the\ Triangle\ Inequality\ property}$
+			\item \textbf{$d_5$ obeys the Triangle Inequality property.}
 
 			Consider
 			$$

chapter2/exercises/ex_2.15.tex

@@ -1,7 +1,7 @@
 % === Exercise 2.15 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item $\mathbf{The\ term\ is\ replaced\ by\ "closed"}$.
+		\item \textbf{The term is replaced by "\textit{closed}".}
 	\end{itemize}
 	\begin{proof}
 		Let $C_n := \left\{m\geq n:m\in\mathbb{N}\right\}$ which is closed for each $n\in\mathbb{N}$ since $C_n$ has no limit points; therefore every limit point of $C_n$ is in $C_n$ vacuously. Consider any finite subcollection $\{C_i\}_{i\in I}$ of $\{C_n\}$ where $I\subset \mathbb{N}$ and $|I|$ is finite.
@@ -23,7 +23,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item $\mathbf{The\ term\ is\ replaced\ by\ "bounded"}$.
+		\item \textbf{The term is replaced by "\textit{bounded}".}
 	\end{itemize}
 	\begin{proof}
 		Let $B_n := \left(0,\frac{1}{n}\right)$ which is bounded for each $n\in\mathbb{N}$. Consider any finite subcollection $\{B_i\}_{i\in I}$ of $\{B_n\}$ where $I\subset \mathbb{N}$ and $|I|$ is finite.

chapter2/exercises/ex_2.17.tex

@@ -1,7 +1,7 @@
 % === Exercise 2.17 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item $\mathbf{Is\ E\ countable?}$
+		\item \textbf{Is $E$ countable?}
 	\end{itemize}
 	\begin{answer}
 		No.
@@ -15,7 +15,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item $\mathbf{Is\ E\ dense\ in\ [0,1]?}$
+		\item \textbf{Is $E$ dense in $[0,1]$?}
 	\end{itemize}
 	\begin{answer}
 		No.
@@ -27,7 +27,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item $\mathbf{Is\ E\ compact?}$
+		\item \textbf{Is $E$ compact?}
 	\end{itemize}
 	\begin{answer}
 		Yes.
@@ -64,7 +64,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item $\mathbf{Is\ E\ perfect?}$
+		\item \textbf{Is $E$ perfect?}
 	\end{itemize}
 	\begin{answer}
 		Yes.

chapter2/exercises/ex_2.20.tex

@@ -1,7 +1,7 @@
 % === Exercise 2.20 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item $\mathbf{Are\ closures\ of\ connected\ sets\ always\ connected?}$
+		\item \textbf{Are closures of connected sets always connected?}
 	\end{itemize}
 	\begin{answer}
 		Yes.
@@ -15,7 +15,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item $\mathbf{Are\ interiors\ of\ connected\ sets\ always\ connected?}$
+		\item \textbf{Are interiors of connected sets always connected?}
 	\end{itemize}
 	\begin{answer}
 		No.

chapter2/exercises/ex_2.5.tex

@@ -3,7 +3,7 @@
 	\begin{proof}
 		Let $E := \left\{ 1+\frac{1}{n} : n\in \{2,3,\cdots\} \right\}$.
 		\begin{itemize}
-			\item $\mathbf{Claim\ only\ 1\ is\ a\ limit\ point}$
+			\item \textbf{Claim only $1$ is a limit point.}
 
 			By Archimedean Property, we can pick radius $r$ such that $\frac{1}{n} < r$. 
 			Then $p_n = 1+\frac{1}{n}$ is a member of $E$ and $d(1, p_n) = \frac{1}{n} < r$. 

chapter2/exercises/ex_2.6.tex

@@ -1,7 +1,7 @@
 % === Exercise 2.6 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item $\mathbf{Prove\ E'\ is\ closed.}$
+		\item \textbf{Prove $E'$ is closed.}
 		\begin{proof}
 			By definition, it suffice to prove $(E')' \subseteq E'$. 
 
@@ -22,7 +22,7 @@
 			Hence for arbitrary radius $r$, there is a point $z\in E$ in the neighborhood $N_r(x)$, then we see that $x\in E'$. Moreover, $x$ is arbitrary, we know $(E')' \subseteq E'$ ; hence we conclude $E'$ is closed.
 		\end{proof}
 
-		\item $\mathbf{Prove\ E\ and\ E'\ have\ the\ same\ limit\ points.}$
+		\item \textbf{Prove $E$ and $E'$ have the same limit points.}
 
 		\begin{proof}
 			We will claim $(\overline{E})' \subseteq E'$ and $E' \subseteq (\overline{E})'$ to obtain $(\overline{E})' = E'$ which is asked to prove.
@@ -36,7 +36,7 @@
 			Finally, we conclude $(\overline{E})' = E'$ which means $E$ and $E'$ have the same limit points.
 		\end{proof}
 
-		\item $\mathbf{Do\ E\ and\ E'\ always\ have\ the\ same\ limit\ points?}$
+		\item \textbf{Do $E$ and $E'$ always have the same limit points?}
 		\begin{answer}
 			No.
 		\end{answer}

chapter2/exercises/ex_2.7.tex

@@ -75,7 +75,7 @@
 			$$
 
 			\begin{itemize}
-				\item $\mathbf{Claim:this\ inclusion\ can\ be\ proper.}$
+				\item \textbf{Claim this inclusion can be proper.}
 			\end{itemize}
 
 			For example, from Exercise 2.5, we let $A_i = \left\{ 1+\frac{1}{i+1} \right\}$ for all $i\in\mathbb{N}$ and $B=\cup_{i=1}^{\infty} A_i$. It follows that $\cup_{i=1}^{\infty}\overline{A_i} = \emptyset$ and $\overline{B} = \{1\}$; hence $\overline{B} \neq \bigcup_{i=1}^{\infty}\overline{A_i}$.

chapter2/exercises/ex_2.8.tex

@@ -1,7 +1,7 @@
 % === Exercise 2.8 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item $\mathbf{Is\ every\ point\ of\ every\ open\ set\ E \subset \mathbb{R}^2\ a\ limit\ point\ of\ E?}$
+		\item \textbf{Is every point of every open set $E \subset \mathbb{R}^2$ a limit point of $E$?}
 		\begin{answer}
 			Yes.
 		\end{answer}
@@ -18,7 +18,7 @@
 			Because $x$ was arbitrary, we conclude $E \subseteq E'$ for any open set $E\subset \mathbb{R}^2$.
 		\end{proof}
 
-		\item $\mathbf{Is\ every\ point\ of\ every\ closed\ set\ E \subset \mathbb{R}^2\ a\ limit\ point\ of\ E?}$
+		\item \textbf{Is every point of every closed set $E \subset \mathbb{R}^2$ a limit point of E?}
 		\begin{answer}
 			No.
 		\end{answer}

chapter2/exercises/ex_2.9.tex

@@ -41,7 +41,7 @@
 			We need to prove $(E^{\circ})^c = \overline{E^c}$ which means $(E^{\circ})^c \subseteq \overline{E^c}$ and $(E^{\circ})^c \supseteq \overline{E^c}$.
 
 			\begin{itemize}
-				\item $\mathbf{Claim:\ (E^{\circ})^c \subseteq \overline{E^c}}$
+				\item \textbf{Claim $(E^{\circ})^c \subseteq \overline{E^c}$.}
 
 				Let $x\in(E^{\circ})^c$ arbitrarily, then $x\notin E^{\circ}$.
 				This implies every neighborhood of $x$ contains no elements in $E$; hence every neighborhood of $x$ contains some elements $y\in E^c$.
@@ -54,7 +54,7 @@
 
 				By the arbitrary choice of $x$, we conclude $(E^{\circ})^c \subseteq \overline{E^c}$.
 
-				\item $\mathbf{Claim:\ (E^{\circ})^c \supseteq \overline{E^c}}$
+				\item \textbf{Claim $(E^{\circ})^c \supseteq \overline{E^c}$.}
 
 				Let $x\in\overline{E^c}$ arbitrarily, then $x\in E^c$ or $x\in (E^c)'$.
 

chapter3/exercises/ex_3.3.tex

@@ -4,7 +4,7 @@
 		It suffices to prove $\{s_n\}$ is increasing and bounded above by $2$. By Theorem 3.14, it follows that $\{s_n\}$ is convergent.
 
 		\begin{itemize}
-			\item $\mathbf{\{s_n\}\ is\ increasing.}$ 
+			\item \textbf{$\{s_n\}$ is increasing.}
 		\end{itemize}
 		It is clear that $0<s_1=\sqrt{2}<2$ holds trivially.
 		Consider
@@ -18,7 +18,7 @@
 		also holds. By induction, $s_{n+1}>s_n$ holds for all $n\in\mathbb{N}$; hence $\{s_n\}$ is increasing.
 
 		\begin{itemize}
-			\item $\mathbf{\{s_n\}\ is\ bounded\ above\ by\ 2.}$ 
+			\item \textbf{$\{s_n\}$ is bounded above by $2$.}
 		\end{itemize}
 		It is clear that $s_1 = \sqrt{2} < 2$. Then suppose for $n=k$, $s_k < 2$ holds. For $n=k+1$, we consider
 		\begin{alignat*}{7}

chapter3/exercises/ex_3.9.tex

@@ -50,7 +50,7 @@
 
 		\item
 		\begin{answer}
-			$\frac{1}{2}$.
+			$\dfrac{1}{2}$.
 		\end{answer}
 		\begin{solution}
 			Let $a_n := \frac{2^n}{n^2}$. 

chapter4/exercises/ex_4.2.tex

@@ -15,7 +15,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item $f(\overline{E})$ can be a proper subset of $\overline{f(E)}$.
+		\item \textbf{$f(\overline{E})$ can be a proper subset of $\overline{f(E)}$.}
 	\end{itemize}
 	\begin{solution}
 		We define $f:\mathbb{Z}\to\mathbb{R}$ by $f(x) = \frac{1}{x}$. Pick $\delta$ such that $0 < \delta < 1$. Given $\epsilon>0$, for all $x,y\in \mathbb{Z}$, we have

chapter4/exercises/ex_4.21.tex

@@ -17,7 +17,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item The conclusion may fail for two disjoint closed sets if neither is compact.
+		\item \textbf{The conclusion may fail for two disjoint closed sets if neither is compact.}
 	\end{itemize}
 	\begin{solution}
 		Let

chapter4/exercises/ex_4.4.tex

@@ -1,7 +1,7 @@
 % === Exercise 4.4 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item Prove that $f(E)$ is dense in $f(X)$.
+		\item \textbf{Prove that $f(E)$ is dense in $f(X)$.}
 	\end{itemize}
 	\begin{proof}
 		Let $y := f(x)$ arbitrarily, then we know $x\in E$ or $x\in E'$ by hypothesis that $E$ is dense. It suffices to prove $y\in f(E)$ or $y\in f(E)'$ so that $f(E)$ is dense in $f(X)$.
@@ -12,7 +12,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item Prove that $g(p) = f(p)$ for all $p\in X$.
+		\item \textbf{Prove that $g(p) = f(p)$ for all $p\in X$.}
 	\end{itemize}
 	\begin{proof}
 		By hypothesis, we suppose $g(q) = f(q)$ for all $q\in E$. Given $\epsilon>0$. Let $p\in X$ arbitrarily. Since $f$ is continuous on $X$ and $E\subseteq X$, then $f$ is continuous at $q\in E$. There exists $\delta_1>0$ such that

chapter4/exercises/ex_4.5.tex

@@ -20,7 +20,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item Show that the result becomes false if the word "closed" is omitted.
+		\item \textbf{Show that the result becomes false if the word "closed" is omitted.}
 	\end{itemize}
 	\begin{proof}
 		Consider $f(x) = \frac{1}{x}$ on $(0,1)$.
@@ -29,7 +29,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item Extend the result to vector-valued functions.
+		\item \textbf{Extend the result to vector-valued functions.}
 	\end{itemize}
 	\begin{proof}
 		Let $f:E\to\mathbb{R}^k$ be a continuous vector-valued function defined by $f = \big( f_1, f_2, \cdots, f_k)$.

chapter4/exercises/ex_4.7.tex

@@ -1,7 +1,7 @@
 % === Exercise 4.7 ===
 \begin{Exercise}
 	\begin{itemize}
-		\item Prove that $f$ is bounded on $\mathbb{R}^2$.
+		\item \textbf{Prove that $f$ is bounded on $\mathbb{R}^2$.}
 	\end{itemize}
 	\begin{proof}
 		For $(x,y) \neq (0,0)$, we consider $(x-y^2)^2 \geq 0$. Then
@@ -17,7 +17,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item Prove that $g$ is unbounded in every neighborhood of $(0,0)$.
+		\item \textbf{Prove that $g$ is unbounded in every neighborhood of $(0,0)$.}
 	\end{itemize}
 	\begin{proof}
 		Let $x := t^3$ and $y := t$, we observe $\lim_{t\to 0} x = 0$ and $\lim_{t\to 0} y = 0$. Then for $t\neq 0$, we know $(x,y) \neq (0,0)$. Consider
@@ -28,7 +28,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item Prove that $f$ is not continuous at $(0,0)$.
+		\item \textbf{Prove that $f$ is not continuous at $(0,0)$.}
 	\end{itemize}
 	\begin{proof}
 		Given $\epsilon = \frac{1}{2}$, for $(x,y)\neq (0,0)$, there exists $\delta > 0$ such that
@@ -43,7 +43,7 @@
 	\end{proof}
 
 	\begin{itemize}
-		\item The restrictions of both $f$ and $g$ to every straight line in $\mathbb{R}^2$ are continuous. 
+		\item \textbf{The restrictions of both $f$ and $g$ to every straight line in $\mathbb{R}^2$ are continuous.} 
 	\end{itemize}
 	\begin{proof}
 		Since a straight line which doesn't pass through $(0,0)$ is always continuous trivially, then here we only prove the lines passing through $(0,0)$ are continuous.

chapter4/exercises/ex_4.8.tex

@@ -16,7 +16,7 @@
 		Hence $f$ is bounded on $E$.
 	\end{proof}
 	\begin{itemize}
-		\item Show that the conclusion is false if boundedness of $E$ is omitted.
+		\item \textbf{Show that the conclusion is false if boundedness of $E$ is omitted.}
 	\end{itemize}
 	\begin{proof}
 		Consider $f(x) = x$ and $E = \mathbb{R}$.

chapter5/exercises/ex_5.11.tex

@@ -21,7 +21,7 @@
 		which is a desired result.
 	\end{proof}
 	\begin{itemize}
-		\item An example that the limit may exist even if $f''(x)$ does not.
+		\item \textbf{An example that the limit may exist even if $f''(x)$ does not.}
 	\end{itemize}
 	\begin{answer}
 		$f(x) = \begin{cases}