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\documentclass{amsart}
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\usepackage{sagetex}
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\usepackage{amsmath,amssymb,nopageno,ulem}
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\usepackage{mathtools}
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\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
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%\usepackage[all]{xy}
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\usepackage[shortlabels]{enumitem}
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\renewcommand{\dot}{\bullet}
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\newtheorem{define}{Definition}
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\newtheorem{lemma}{Lemma}
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\newtheorem{ex}{Exercise}
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\newcommand{\aut}[1]{\mathbf{Aut}{#1}}
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\newcommand{\conj}[2][g]{{#1}^{-1}{#2}{#1}}
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\newcommand{\Aut}{\operatorname{Aut}}
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\newcommand{\Inn}{\operatorname{Inn}}
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\newcommand{\Out}{\operatorname{Out}}
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\newcommand{\gens}[1]{\left\langle{#1}\right\rangle}
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\newcommand{\epi}{\twoheadrightarrow}
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\newcommand{\epic}{\twoheadrightarrow}
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\newcommand{\mono}{\hookrightarrow}
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\newcommand{\Z}{\mathbf{Z}}
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\newcommand{\C}{\mathbf{C}}
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\newcommand{\R}{\mathbf{R}}
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\newcommand{\F}{\mathbf{F}}
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\newcommand{\N}{\mathbf{N}}
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\newcommand{\Q}{\mathbf{Q}}
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\newcommand{\Int}{\overset{\circ}}
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\def\GL{\text{GL}}
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\def\SL{\text{SL}}
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\def\nin{\not\in}
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\def\F{\mathbf{F}}
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\def\ker{\text{ker}}
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\def\Re{\mbox{Re}}
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\def\inf{\mbox{inf}}
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\def\sup{\mbox{sup}}
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\def\ne{\not =}
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\newcommand{\inv}[1]{{#1}^{-1}}
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\newcommand{\ndef}{\overset{\text{def}}=}
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\def\smat{[\begin{smallmatrix} b&c\\ a&b \end{smallmatrix}]}
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\newcommand*\colvec[3][]{
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\begin{pmatrix}\ifx\relax#1\relax\else#1\\\fi#2\\#3\end{pmatrix}
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}
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\renewcommand{\theenumi}{\alph{enumi}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{document}
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\section*{Ch2 Exercises}
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\ex{Prove that the empty set is a subset of every set}
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\begin{proof}
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For $B \subseteq S \in \mathbf{Set}, B^c \cap B = \emptyset \subseteq
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S$
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\end{proof}
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\ex{Prove the set of all algebraic numbers are countable. \\
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Hint: For $\sum^n_0{a_k z^k} = 0, N \in \N$, there are only finitely many equations with \[n + \sum|a_k|=N\]
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}
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\begin{proof}
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Write $\bold{A}$ for the algebraic numbers. Since each algebraic number is determined by its (finite) integral coefficients, then for some $n$ we have inclusions
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\[A \mono \Z[X] \mono \Z^n, a \mapsto a_{n-1}x^{n-1}+...+a_0 \mapsto (a_0,...,a_{n-1})\]
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Since $\Q \subset \bold{A}$, the algebraic numbers are countably infinite.
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\end{proof}
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\ex{Prove that there exist real numbers which are not algebraic.}
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\begin{lemma}[Gelfond-Schneider Thm] $\alpha \ne 0,1, \beta \nin \Q, \alpha^\beta$ is transcendental.
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\end{lemma}
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\ex{Is the set of all irrational real numbers countable?}
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\begin{proof}
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Note there is an inclusion $\R/\Q \overset{\iota}\mono \R \setminus
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\Q$ of irrational numbers with rational multiple exactly 1. If $\R/\Q$ is countable, $\R$ is a countable union of countable sets $\{xq\}_{q \in \Q}$ and therefore countable, which we know to be untrue. Therefore $\R/\Q$ is uncountable. Since $\iota(\R/\Q) \subset \R \setminus \Q$, the irrational numbers must also be uncountable.
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\end{proof}
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\ex{Construct a bounded set of real numbers with exactly three limit points.}
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\begin{proof}
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For $n \in \N$,
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\[
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A_0 \ndef \{1/n\}, A_1 \ndef \left\{\frac{n-1}{n}\right\}, A_2 \ndef
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\left\{\frac{-n+1}{n}\right\}
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\]
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Let $A$ be the union of $A_0, A_1, A_2$, then $A \subset (-1,1)$ and
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its limit points are exactly $\{\pm 1, 0\}$.
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\end{proof}
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\ex{Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed. Prove that $E,\bar{E}$ have the same limit points. Do $E,E'$ always have the same limit points?}
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For any limit point $\ell \in E'$, we have $\ell: N(\ell) \cap E'\subseteq E'; \ne \emptyset$
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\ex{$A_1,A_2,A_3,\dots$ subsets of a metric space. \\
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\begin{enumerate}
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\item For $B_n \ndef \bigcup^n A_i$, show $\bar{B_n} = \bigcup^n \bar{A_i}$
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\item For $B \ndef \bigcup^\Z A_i$, show $\bar{B} \subset \bigcup \bar{A_i}$
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\end{enumerate} Show this inclusion can be proper.}
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\begin{enumerate}
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\item Consider $ n=2: \bar{A}_1 \cup \bar{A}_2$
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\end{enumerate}
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\ex{Are points of each open set $E \subseteq \R^2$ limit points? For closed sets?}
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For $x \in E, \exists U_x \subset E$, its limit points would be $\ell:N(\ell) \cap E \setminus \{\ell\} \ne \emptyset$.
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\ex{Let $\Int E$ denote the interior of $E$.
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\begin{enumerate}
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\item Prove $\Int E$ open
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\item Prove $E$ open iff $\Int E = E$
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\item For $G$ open, $G \subseteq E$, prove $G \subseteq \Int E$
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\item Prove $(\Int E)^c = \bar{E^c}$
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\item Do $E,\bar E $ always have the same interiors?
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\item Do $E, \Int E$ always have the same closures?
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\end{enumerate}
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}
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\begin{enumerate}
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\item By definition, $A$ is open if for each point $p, \exists N(p)\subseteq A$, and the interior of $A$ is exactly such a set.
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\item
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\end{enumerate}
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\ex{Let $X$ be infinite. For $p,q \in X$, define
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\[
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d(p,q) \ndef
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\begin{cases}
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1 & p \ne q \\
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0 & p=q
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\end{cases}
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\]
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Prove this is a metric. What are its open, closed and compact subsets?}
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\begin{proof}
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$x=y \iff d(x,y) = 0$ so positive definite; if $x \ne y, y \ne x \implies d(x,y) = d(y,x)$ so it's symmetric. To show the triangle inequality holds, break the right-hand side into cases:
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\[
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d(p,r)+d(r,q) =
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\begin{cases}
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2 & p \ne r \text{ and } r \ne q \\
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1 & p=r \ne q \text{ or } p \ne r = q \\
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0 & p=r=q
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\end{cases}
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\]
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Then for $q,r,p$ pairwise distinct the sum (RHS) is always greater. Otherwise both sides equal $1$ (or $0$ if all terms equal).
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Neighborhoods can't have zero radius, therefore $N(p) = \{ q:d(p,q) = 0\} = \{p\}, N(p) \cap A \subseteq \{p\}$ so no limit points, i.e., $A = \bar{A}$. Since $p \in A \implies N(p) \subseteq A, A = \overset{\circ}A$, and all subsets are clopen.
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Let $\mathcal{U}$ be a cover of $A$. If $A$ is a finite subset with $n$ elements, we can finitely cover $A$ with
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$\{U_k \in \mathcal{U}:x_k \in U_k\}_{j:1\leq j \leq n}$
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, i.e., (finite) open sets each containing a point of $A$. Otherwise, for uncountable $A$, consider the cover $\mathcal{U}'\ndef \{U_j = x_j\}$. Any finite subcovering contains only finite points of $A$.
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Thus, finite sets are compact and infinite sets noncompact.
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\end{proof}
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\ex{For $x,y \in \R$, define
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\begin{itemize}
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\item $d_1 \ndef (x-y)^2$
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\item $d_2 \ndef \sqrt{|x-y|}$
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\item $d_3 \ndef |x^2 - y^2|$
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\item $d_4 \ndef |x-2y| $
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\item $d_5 \ndef \frac{|x-y|}{1+|x-y|}$
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\end{itemize}
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Which are metrics?}
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%\begin{proof}
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Immediately, $d_4$ isn't symmetric [$d(1,4) = 7 \ne 2 = d(4,1)$] and
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$d_3$ isn't positive definite [let $y = -x$]. Then of the remaining three we check whether or not we have the triangle inequality $d(x,z)+d(z,y)-d(x,y) \geq 0$:
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\begin{itemize}
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\item $(x-z)^2+(z-y)^2 -(x-y)^2= x^2 + y^2 +2z^2 -2z(x+y)-x^2-y^2-2xy = z^2 + (z-(x+y))^2 + (x+y)-2xy$
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\item $$\sqrt{|x-z|} + \sqrt{|z-y|} = \frac{|x-z|-|z-y|}{\sqrt{|x-z|} - \sqrt{|z-y|}}$$
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\item $$\frac{|x-z|}{1+|x-z|} + \frac{|z-y|}{1+|z-y|}$$
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\end{itemize}
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%\end{proof}
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\ex[Exercise 12]{Prove $K\ndef \{0\} \cup \{1/n\}_{n \in \N}$ is compact directly from definition.}
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\begin{proof}
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Let $U$ be a open cover of $K$. For each $x_n\in K, \exists U_n \in U:x_n \in U_n$. Define $U_0:0 \in U_0$ to be the open set containing $0$. For each $\delta >0, \exists m \in \N: \{\frac{1}{m},\frac{1}{m+1},...\} \subset U_0 \cap N_\delta$. Thus, there are finitely many $U_n:n<m$ which, in addition to $U_0$, cover $K$.
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\end{proof}
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\ex{Construct a compact set of reals whose limit points are countable.}
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%\begin{proof}
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Something like Cantor set?
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\[f(x) \ndef
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\begin{cases}
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1/r & x \in \Q \\
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0 & x \nin \Q
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\end{cases}
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\]
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%\end{proof}
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\ex[14]{Give an example of an open cover of $(0,1)$ without finite subcover.}
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\begin{proof}
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Let $U \ndef \{(\frac{1}{n},1)\}_{n \in \N}$. Then for any $x \in (0,1), \exists M \in \N: x<1 \implies \frac{1}{M} < x$. So $U$ is a cover of $(0,1)$. However for any finite subset $U' \ndef \{(\frac{1}{n},1)\}_{n:1 \leq n \leq k } $ of $U$ with $k < M, \frac{1}{M} \nin U'$. Therefore $(0,1)$ is not compact.
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\end{proof}
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\ex[15]{Show that Thm 2.36 and its Corollary become false if compact is replaced with either closed or bounded.}
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\ex[16]{$E \ndef \{p \in \Q:p^2 \in (2,3)\}$. Show $E$ is closed and bounded in $\Q$, but not compact. Is $E$ open?}
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\begin{proof}
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$1^2 < p^2 < 2^2 \implies E \subset (1,2)$, so $E$ bounded.
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Assume $E$ is not closed, and note $\Q \setminus E = ub(E) \cup
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lb(E)$. Say $E$ has a limit point $\ell \in
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ub(E):(\ell-\delta,\ell+\delta) \cap E \ne \emptyset$. Then for some
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$\ell' \in ub(E): \ell > \ell'$, set $\delta = \ell-\ell'>0$. Then we have
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$(\ell-\delta,\ell+\delta) = (\ell',2\ell-\ell')$ and $
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(\ell',2\ell-\ell') \cap E \ne \emptyset$, which is impossible since
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$\ell'>p \in E$. A similar argument follows for $\ell \in lb(E)$,
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thus $E$ is closed.
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The cover $U \ndef \{(0,r- \frac{1}{n}):n \in \N, r^2 \in (2,3)\}$
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has no finite subcover, so $E$ is not compact.
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\end{proof}
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\ex{Decimal expansions on the unit interval with only 4's and 7's
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\[
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E \ndef \left\{\sum a_k 10^{-k}:a_k \in \{4,7\}\right\}
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\]
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Is $E$ countable?dense?compact?perfect?}
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\begin{proof}
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There are really two cases here: series including eventually zero terms like $0.44=0.44\bar{0}$, and nonterminating series without zeros after the decimal point.
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To determine cardinality, we map into sets of known cardinality. In the case including terminating series, $E$ is in surjection with binary decimals covering the unit interval $\sum a_k 10^{-k} \mapsto \sum b_k 2^{-k}$, otherwise there is a bijection in the latter case. So $E$ is uncountable either way.
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$E$ is not dense in the unit interval because there are no terms of $E$ in the interval $(0.5,0.6)$.
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We first show $E$ is perfect. Assume contrary, then there
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exists $\ell \nin E:N_\epsilon(\ell) \cap E \ne \emptyset$. Then
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$\ell$ has a decimal expansion with digits which aren't 4,7, or in
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certain cases, 0 [for example, $0.404$]. If the first such digit
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occurs at $k=n$, then for any $s \in E, |\ell -
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s|>10^{-m-1}$. Therefore $\ell$ cannot be a limit point, and $E$ is
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perfect.
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Since $E$ is perfect, it contains its limit points and is therefore closed. Either $(0.4,0.7)$ or $(0.\bar{4},0.\bar{7}) = (\frac{4}{9}, \frac{7}{9})$ is a bound for $E$, and since $E$ is closed and bounded, $E$ is compact.
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\end{proof}
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\ex[18]{Is there a nonempty perfect set in $\R$ which contains no rational number?}
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Assume there is such a set $P$, then for $q \in \Q, \epsilon > 0$, the intersection
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\[
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N_\epsilon(q) \cap (P \setminus \{q\})
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\]
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must be empty.
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Since $P$ is nonempty, we have $p \in P:p \in (p - \epsilon, p+ \epsilon)$. We show that any open interval $(a,b)$ contains rational points. [WTS open segment has rationals]
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\ex[19]{Prove:
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\begin{enumerate}
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\item Disjoint closed sets $A,B$ in a metric space $X$ are separated
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\item Prove the same for disjoint open sets.
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\item For some $p\in X, A \ndef \{q:d(p,q)<\delta\}, B \ndef \{q:d(p,q)>\delta\}$. $A,B$ are separated.
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\item Every connected metric space with at least two points is uncountable.
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\end{enumerate}
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}
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\begin{enumerate}
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\item For disjoint closed sets, $A \cap B = \emptyset, A = \bar{A}, B = \bar{B}$. Then $\bar{A} \cap \bar{B} = \emptyset$
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\item For disjoint open subsets $A,B$, $A \cup B$ also open
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\end{enumerate}
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\ex[20]{Are closures and interiors of connected sets connected?}
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\ex[21]{Let $A,B \subset \R^k$ be separated subsets. Then for $a \in A, b \in B, t \in \R$, define
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\[
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p(t) = (1-t)a+tb, A_o=\inv{p}(A), B_0=\inv{p}(B)
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\]
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Prove
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\begin{enumerate}
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\item $A_0,B_0$ are separated subsets.
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\item $\exists t_0 \in (0,1):p(t_o)\nin A \cup B$
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\item Every convex subset of $\R^k$ is connected.
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\end{enumerate}
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}
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Rewrite as $p(t) = a+ t(b-a)$. On the closed unit interval, $p(t)$ deforms $a$ into $b$ and vice-versa.
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\ex[22]{A metric space is \emph{separable} if it contains a countable dense subset. Prove $\R^k$ is separable.}
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%\begin{proof}
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TBD:Show cart product of intervals interior $N_\delta(x) \subset \R^k$
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For $x = (x_1,...,x_k) \in \R^k, E \ndef \{(q_i,q_j)_{i,j:1 \leq i,j \leq k} \subseteq \Q^k:q_i,q_j \in \Q\}$, the cartesian product of open intervals, we will show $N_\delta(x) \cap E$ is an open subset of $\Q^k$ and $\R^k$ is thus separable. \\
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For each \( \delta >0, \exists n \in \N:1/n < \delta \). Then
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$B_{1/n}(x) \subset B_\delta (x)$.
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%\end{proof}
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\ex[23]{Prove that every separable metric space has a countable base.}
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\ex[24]{Let $X$ be a metric space in which every infinite subset has a limit point. Prove $X$ is separable.}
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\ex[25]{Prove every compact metric space $K$ has countable base, and that $K$ is therefore separable.}
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\ex[26]{Let $X$ be a metric space in which every infinite subset has a limit point. Prove $X$ is compact.}
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\ex[27]{For $E\subseteq \R^k$ uncountable and $P$ its condensation points, prove $P$ is perfect and $P^c \cap E$ at most countable.}
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\ex[28]{Prove that every closed set in a separable metric space is the union of a (perhaps empty) perfect set and an at most countable set. (corollary: every countable closed subset of $\R^k$ has isolated points) hint: ex27}
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\ex[29]{Prove that open sets in $\R$ are the union of at most countable intervals.}
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Define a relation on $U \subseteq \R$ as follows: $\forall x,y \in U, x \sim y \iff $ there is a closed interval of $U$ with endpoints $x,y$. This relation is reflexive and symmetric, and for $x \sim z, z \sim y$, the union of closed segments is again in $U$.
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\ex[30]{Imitate the proof of Thm 2.43 to obtain the following result: \\
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If $\R^k = \bigcup_1^\infty F_n$, where each $F_n$ is a closed subset then at least one $F_n$ has nonempty interior. \\
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Equivalent statement: If $G_n$ is a dense open subset of $\R^n$, then
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$\bigcap_1^\infty G_n \ne \emptyset$ (in fact, dense in $\R^n$).}
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\newpage
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\section*{definitions and results}
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\begin{itemize}
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\item [limit point] We say $\ell$ is a \emph{limit point} of $E$ if there exists $\{q:q \ne \ell\} \subset E \cap N_r(\ell)$
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\item [closure] The closure $\bar{A}$ of $A \subseteq X$ is $A$ plus its limit points.
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\item [compact] A subset $S \subseteq X$ is \emph{compact} if each cover $\{U_i\}_I$ admits a finite subindexing $\{U_i\}_{J\subseteq I}$, i.e., a finite subcover.
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\item [perfect] A subset is \emph{perfect} if it is closed and each point is a limit point.
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\item [dense] $A$ is \emph{dense} in $X$ if for any other subset $B, B \cap \bar{A} \ne \emptyset$.
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\item [condensation point] A point $p$ such that $|N_r(p) \cap E| = \mathfrak{c}$
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\item [separate] $A,B$ are \emph{separated} if $A \cap \bar{B} = \emptyset \text{ and } \bar{A} \cap B = \emptyset$
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\item [connected] $S$ is \emph{connected} if it is not the disjoint union of separable sets.
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\item [base] A collection of open subsets $\{V_a\}$ such that every open set is some union $\bigcup V_a$.
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\end{itemize}
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We want to show there exists a nontrivial open subset of $\Q^k$ interior to every open subset of $\R^k$. \\
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Assuming Euclidean metric, set $\delta = \frac{\sqrt{n}}{n^2}\epsilon$ and define $\Delta \ndef (\delta,...,\delta) \in \R^k$.
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\[
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d(x, x+\Delta) = \sqrt{\sum{\left(x_i-(x_i+\frac{\sqrt{n}}{n^2}\epsilon)\right)^2}} = \frac{\epsilon}{n} \in N_\epsilon(x)
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\]
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By the archimidean property of $\R$, there exists $M \in \N, M>1$:
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\[
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\frac{\sqrt{n}}{n^2}\epsilon<1 \implies \inv{M} < \frac{\sqrt{n}}{n^2}\epsilon
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\]
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Therefore, we have the open subset $E \subset \Q^k$:
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\[
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E \ndef \Q^k \cap \prod_{i=1}^n(x_i-\delta,x_i+\delta) \subset \prod_{i=1}^n(x_i-\delta,x_i+\delta) \subset N_\epsilon(x)
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\]
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Moreover for $(\inv{M},...,\inv{M}) \in N_\epsilon(0)$, so $E$ is nonempty. [show $N_\epsilon(x) \subset U(x)$] \\
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WTS (countable) rationals in any real open set. Any real open set is the disjoint product of at most countable open intervals. Since $\R \cong (0,1)$, it suffices to prove this on the unit interval. Let $(a,b) \subset (0,1)$. Then $0 < \frac{1}{M} < b-a<1, M \in \N$.
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From this we can show $b-\inv{M}>a, a+\inv{M}<b \implies a+\inv{M} -b+\inv{M} < b-a, -(b-a) + 2 $
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For any $x \in (a,b), x-a,b-x>0, 2x+b-a>0$
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WTS $\exists \delta>0:N_\delta(x) \cap U_x \ne emptyset$ \\
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\[
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\sage{latex(continued_fraction(pi).str(nterms=2))}
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\]
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\section*{things which are actually wrong}
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Let $U,L$ be the open upper and open lower halves of $\R^k$, i.e., given $x = (x_1,...,x_n),(0,...,0) \in \R^k,
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\begin{cases}
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x_1 > 0 \implies & x \in U \\
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x_1 < 0 \implies & x \in L \\
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x_1 = 0 \implies & x \nin U \cup L
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\end{cases}$. Then for $\delta >0, (\pm \delta,0,...,0) \in N_r(0) \cap (A \subset U) N_r(0) \cap (B \subset L), 0 \in \bar{U}\cap\bar{L}$ therefore by definition $(U \cap \bar{L}) \cap (\bar{U} \cap L) = \emptyset$.
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We will show these definitions are equivalent.
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\end{document}
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