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\documentclass{amsart}
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\newcommand{\R}{\mathbf{R}}
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\newcommand{\N}{\mathbf{N}}
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\newcommand{\Z}{\mathbf{Z}}
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\newcommand{\Q}{\mathbf{Q}}
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\newcommand{\I}{\mathbf{I}}
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\newcommand{\ndef}{\overset{\mathrm{def}}{=\joinrel=}}
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\usepackage{amssymb}
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\newtheorem{define}{Definition}
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\let\oldemptyset\emptyset
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\let\emptyset\varnothing
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\begin{document}
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\section*{$\R$-analysis (Brown)}
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\emph{$f \in C^0\{a\}$ short for $f$ continuous at $a$} \\
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\emph{intervals $(x - r, x+ r)$ sometimes notated $(x \pm r)$ or $B^r_x$} \\
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\emph{subsets proper unless $\subseteq$}
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\subsection*{1.1}
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\subsubsection*{1.1.1 Show equivalance of the following:} \begin{itemize}
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\item (a) $N \in Op(x)$ (a neighborhood of $x$)
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\item (b) $ [x \pm \delta] \subset N$
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\item (c) $[x \pm n^{-1}] \subset N$
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\end{itemize}
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\begin{proof}$\exists \delta': (x - \delta',x+ \delta') \subset N$. \\ Defining $\delta = \frac{1}{3}\delta', [x - \delta, x + \delta] \subset x - \delta', x + \delta') \subset N$ \\
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$(a) \implies (b)$ \\
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Fixing $\delta < 1, \exists n: 1 < n\delta \implies n^{-1} < \delta$ and $[x \pm n^{-1}] \subset [x \pm \delta] \subset N$ \\
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$(b) \implies (c)$ \\
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$(c) \implies (a)$ \\ \end{proof}
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\subsubsection*{1.1.2 Prove $Int(\bold{FinSet}) = \emptyset$}
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\begin{proof} let $x,y$ be distinct points and fix $\epsilon = \frac{1}{3}\inf_{x,y \in S}d(x,y)>0$. \\
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$d(x,x + \epsilon) < \inf_S d(x,y) \implies x \pm \epsilon \notin S$.
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\end{proof}
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\subsubsection*{1.1.3 Prove $Int(A \cap B ) = Int A \cap Int B$ }
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\begin{proof}Holds trivially for empty sets. elements satifying $x+\epsilon_2 \leq x+\epsilon' \in Int A, x+\epsilon'' \in Int B$ are exactly $Int A \cap Int B$. The interior of the intersection are elements $x+\epsilon_1:\epsilon_1 = \inf\{\epsilon',\epsilon''\}$ ie simultaneously interior $A\cap B$ \\
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Clearly $x + \epsilon_2 \leq x + \epsilon_1 \implies Int A \cap Int B \subset Int(A \cap B )$ and for nonempty $A,B, Int(A \cap B ) \subset Int A \cap Int B$. Therefore, $Int(A \cap B ) = Int A \cap Int B$ \end{proof}
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\subsubsection*{1.1.4 Does the converse, $Int(A \cup B ) = Int A \cup Int B$ hold? }
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No. \\ Consider $A= [0,1),B=[1,2]: Int(A \cup B ) = (0,2) \not= Int A \cup Int B = (0,2) - \{1\}$
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\subsubsection*{1.1.5 Does $Int\bigcap A_i = \bigcap Int A_i$?}
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\begin{proof}
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Use induction. from 1.1.3, $Int(A \cap B ) = Int A \cap Int B$.\\
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$Int \left(A_i \cap (\bigcap A_{i-1})\right) = Int(\bigcap A_{i})$. Conversely, $Int A_i \cap Int \bigcap A_{i-1} = \bigcap Int A_i$.
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\end{proof}
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\subsubsection*{1.1.6-8. Show neighborhoods preserve addition, multiplication and inverse operations}
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\begin{proof}
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\emph{1.1.6 (addition)} Setting $$\epsilon_1 = \epsilon_2 = \frac{\epsilon}{3}$$
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$(a+\epsilon_1)+(b+\epsilon_2) = (a+b) + \frac{2}{3}\epsilon < (a+b) + \epsilon \qed$ \\
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\emph{1.1.7 (multiplication)} [works for $\epsilon \leq 1$] Setting $$\epsilon_1 = \epsilon_2 = \inf \left\{????, \frac{\epsilon}{2} \inf \left\{1,\frac{1}{a+b}\right\}\right\}$$
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$(a+\epsilon_1)(b+\epsilon_2) = \begin{cases} ab + \frac{1}{2}\epsilon + \frac{1}{4}\epsilon^2 < ab+\frac{3}{4}\epsilon & \frac{1}{a+b} \leq 1\\ ab + \frac{\epsilon}{2}(a+b) + \frac{1}{4}\epsilon^2 < ab + \frac{\epsilon}{2} + \frac{1}{4}\epsilon^2 &\frac{1}{a+b} > 1 \end{cases} \qed$ \\
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\emph{1.1.8 (multiplicative inverses)}
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\end{proof}
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\subsubsection*{1.1.9 Prove $Int(Int A) = Int A$}
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\begin{proof}
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asdf
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\end{proof}
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\subsubsection*{1.1.10 Show there are exactly 14 subsets generated by the complementation and Int operators from base sets $A_1,A_2,A_3$}
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\subsection*{1.2} $f$ \emph{continuous} at $a$: $\forall V_{f(a)} \exists U_a:f[U_a] \subset V \iff \lim_{x \to a}f(x) = f(a)$
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\subsubsection*{1.2.1 For $f,g \in C^0\{a\}, \exists? h,h'|a \in C^0: h=f+g, h'=f*g \in C^0, \frac{f}{g}$}
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\begin{proof}
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Fix $\delta := \min\{\delta',\delta''\}, h(x):=f(x)+g(x)$.
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Then $h(x + \delta) = f(x+\delta)+g(x + \delta) \leq f(x)+g(x) + \epsilon' + \epsilon''$
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\end{proof}
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\subsubsection*{1.2.2 Prove squeeze, demonstrate with $x\sin\frac{1}{x}$}
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\begin{proof}
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asdf
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\end{proof}
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\subsubsection*{1.2.3 Prove $f|_{A \cap N} \in C^0\{a\} \implies f \in C^0\{a\}$}
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\begin{proof}
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asdf
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\end{proof}
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\subsubsection*{1.2.4 Show $f(x):= x, x \in [0,1);f(x):=x-1, x \in [2,3]$ continuous and injective but $f^{-1} \not\in C^0\{1\}$}
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\begin{proof}
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asdf
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\end{proof}
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\subsubsection*{1.2.5 Prove monotone bij function $f: [a,b] \to [c,d]$ continuous}
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\begin{proof}
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asdf
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\end{proof}
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\subsubsection*{1.2.6 $f \in End(\R)$, show $f \in C^0\R \iff f^{-1}IntA \subset Int f^{-1}[A]$}
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\begin{proof}
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asdf
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\end{proof}
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\subsubsection*{1.2.7 Show equivalance of the following:} \begin{itemize}
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\item (a) $f \in C^0\{0\}$
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\item (b) $\forall \epsilon >0, \exists \delta: f(a - \delta, a + \delta) \subset (f(a) - \epsilon, f(a) + \epsilon)$
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\item (c) $\forall n \in \N, \exists m: f(a - m^{-1}, a + m^{-1}) \subset (f(a) - n^{-1}, f(a) + n^{-1})$
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\end{itemize}
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\begin{proof}
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asdf
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\end{proof}
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\subsubsection*{1.2.8 (Gluing theorem) Suppose there exists a function}
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$$f: A \to \R: f|_{A_1,A_2} \in C^0\{0\}, A = \bigcup A_i, a \in \bigcap A_i$$
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\begin{proof}
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poof
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\end{proof}
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\subsubsection*{1.2.9 Prove $\big| {\hom_{C^0}[\I,\R]} \big| \not\in \aleph_0$}
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\begin{proof}
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Try iso from R to I to R then compound map is an R-automorphism, the constant functions should be uncountable and continuous.
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\end{proof}
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{ }
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\subsection*{1.3 More Int, Ext and Fr}
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\end{document}
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