From 24f8c733a6082ad815009d53f418d44b382a2dfc Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Sat, 12 Nov 2022 14:43:58 -0500
Subject: 8 homeworks from analysis

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+\font\bigbf=cmbx12 at 24pt
+\newfam\bbold
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\def\bb#1{{\fam\bbold #1}}
+\input color
+
+\newcount\pno
+\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}\par}
+\def\endtf{\egroup\medskip}
+\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}}
+\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}}
+
+\newcount\qno
+\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}}
+
+\def\intro#1{\medskip\noindent{\it #1.}\quad}
+\def\proof{\intro{Proof}}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+
+\let\bu\bullet
+
+{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad
+HOMEWORK \#7}}
+
+\question{(3 points each) Answer True or False. No justification
+needed.}
+
+\item{a)} The arbitrary intersection of compact sets is compact.
+\tf
+\true
+\endtf
+
+\item{b)} The arbitrary union of compact sets is compact.
+\tf
+\false
+\endtf
+
+\item{c)} Let $A$ be arbitrary, and let $K$ be compact. Then, the
+intersection $A\cap K$ is compact.
+\tf
+\false % open set \subset closed bounded set
+\endtf
+
+\item{d)} If $F_1\supset F_2\supset F_3\supset F_4\supset\cdots$ is a
+nested sequence of nonempty closed sets, then the intersection $\cap_n
+F_n$ is not empty.
+\tf
+\false
+\endtf
+
+\item{e)} The intersection of a perfect set and a compact set is
+perfect.
+\tf
+\false
+\endtf
+
+\item{f)} The rationals contain a perfect set.
+\tf
+\true % empty set
+\endtf
+
+\question{(3.3.10) Let $K = [a,b]$ be a closed interval. Let ${\cal O} =
+\{O_\lambda | \lambda\in \Lambda\}$ be an open cover for $[a,b].$ Show
+that there is a finite subcover using this strategy: define $S$ to be
+the set of all $x\in [a,b]$ such that $[a,x]$ has a finite subcover from
+${\cal O}.$}
+
+\item{a)} Argue that $S$ is nonempty and bounded, and thus $s = \sup S$
+exists.
+
+\proof
+Let $x = a.$
+We know that $[a,a] = \{a\}\subseteq [a,b],$ so there must exist, in
+${\cal O},$ an $O_\lambda$ such that $a\in O_\lambda.$
+Taking $\{O_\lambda\}$ as the subcover, we have shown that $S$ is
+nonempty.
+
+From the definition, $S\subseteq [a,b],$ so $|S| \leq \max\{|a|,|b|\},$
+making $S$ bounded.
+Therefore, $s = \sup S$ exists.
+\endproof
+
+\item{b)} Show that $s = b.$ Explain why this implies ${\cal O}$ admits
+a finite subcover. That is, every closed interval is compact.
+
+\proof
+From the definition $S\subseteq [a,b],$ and for all $y\in[a,b],$ we know
+$y \leq b,$ so $s\leq b.$
+
+For the sake of contradiction, assume $s < b.$
+From the open cover, we know there must be an open interval
+$V_\epsilon(s)$ containing $s,$ with some size $\epsilon,$ and from the
+definition of supremum, there must be $x\in [s-\epsilon,s)\cap S,$ so
+by taking the finite cover of $x$ and appending $V_\epsilon(s),$ we
+construct a finite cover which includes $y\in (s,s+\epsilon)\cap[a,b],$
+and this is nonempty since $s < b,$ and $y > s,$ giving us a
+contradiction of our assumption about the supremum.
+
+This conclusively tells us that $s = b,$ so an open cover of $[a,b]$
+must admit a finite subcover.
+This is the definition of compactness.
+\endproof
+
+\item{c)} Use the fact that closed intervals are compact to conclude
+that a set $F$ that is closed and bounded is compact.
+
+\proof
+Let ${\cal O}$ be an open cover for $F.$
+We are given that $F$ is bounded by some $M,$ so we construct
+$${\cal O}\cup\left\{\left(\bigcup_{\lambda\in\Lambda}
+O_\lambda\right)^c\right\},$$
+The appended set is an open set because the complement of a closed set
+is an open set.
+We thus have an open cover for $[-M,M]\supseteq F,$
+and we have shown that $[-M, M]$ admits a finite subcover, so taking the
+obtained finite subcover sans the appended set, we have a finite
+subcover for $F.$
+
+\endproof
+
+\question{We proved: the countable intersection of compact nonempty
+$K_1\supset K_2\supset \cdots$ is not empty. We did so using a
+diagonalization and the limit definition of compact sets. Prove this
+theorem using the open cover definition of a compact set. (Assume
+otherwise, and construct an open cover of $K_1$ without finite
+subcover.)}
+
+\proof
+Assume for the sake of contradiction that the intersection of these sets
+is empty.
+Since $K_i$ is closed (assuming we are working in a Hausdorff space),
+$K_i^c$ is open, and $C = \{K_2^c,K_3^c,\ldots\}$ is an open cover,
+since $\bigcup_{2\leq i} K_i^c = \left(\bigcap_{2\leq i} K_i\right)^c =
+X,$ where $X$ is the space.
+
+This set is therefore an open cover of $K_1,$ but any finite subcover
+${\cal O}$ must omit, for some $N,$ all $K_n$ where $n>N.$
+Let $x\in K_n.$
+For $m\leq n,$ $K_n\subseteq K_m,$ so $x\in K_m,$ and $x\not\in K_m^c,$
+so $x\not\in \bigcup_{o\in\cal O} o.$
+But $x\in K_1,$ contradicting the existence of a finite subcover from
+this open cover.
+
+Therefore, $K_1$ is not compact, but that is a contradiction.
+\endproof
+
+\bye
-- 
cgit