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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 \#6}}
+
+\question{(3 points each) Answer True or False. No justification
+needed.}
+
+\item{a)} A set can be both open and closed.
+\tf
+\true
+\endtf
+
+\item{b)} For a set $A\subset\bb R,$ the set of limit points of $A$ is closed.
+\tf
+\true
+\endtf
+
+\item{c)} The Cantor set is uncountable.
+\tf
+\true
+\endtf
+
+\item{d)} Make a `thin' Cantor set by starting with $[0,1],$ remove the middle
+$2/3.$ From the two intervals remaining, remove the middle $3/4.$ From
+the four remaining intervals the middle $4/5,$ and so on. The remaining
+set is uncountable.
+\tf
+\true
+\endtf
+
+\item{e)} If $\sum_n a_n$ converges absolutely, then $\sum_n a_n^2$ converges
+absolutely.
+\tf
+\true
+\endtf
+
+\item{f)} A countable subset of $\bb R$ must have a limit point.
+\tf
+\false
+\endtf
+
+\item{g)} An uncountable subset of $\bb R$ must have a limit point.
+\tf
+\true
+\endtf
+
+\question{Decide whether the following statements are true or false.
+Provide counterexamples for those that are false, and supply proofs for
+those that are true.}
+
+\item{a)} An open set that contains every rational number must
+necessarily be all of $\bb R.$
+
+\intro{Disproof}
+False. Let us have the bijection $\phi:\bb N\to\bb Q.$
+$$\bigcup_{n=0}^\infty (\phi(n)-2^{-n},\phi(n)+2^n)$$
+is a union of arbitrary (countably many) open intervals, and it must
+contain every rational number, but their length is clearly less than the
+length of the reals (the sum of their lengths, which is greater than the
+length of their union, is 4).
+\endproof
+
+\item{b)} The Nested Interval Property remains true if the term ``closed
+interval'' is replaced by ``closed set.''
+
+\proof{Disproof}
+False. The integers are a closed set, but the decreasing subsets
+$A_n = \bb N\setminus\{1,\ldots, n\}$ will not limit to one number.
+\endproof
+
+\item{c)} Every nonempty open set contains a rational number.
+
+\proof
+True. Let $x\in A$ where $A$ is an open set. For some $\epsilon>0,$
+$V_\epsilon(x)\subseteq A,$ and there must be $y\in\bb Q\cap
+V_\epsilon(x)$ by density of the rationals in the reals.
+$y\in A,$ so every nonempty open set contains a rational number.
+\endproof
+
+\item{d)} Every bounded infinite closed set contains a rational number.
+
+\intro{Disproof}
+$\{(1-2^{-n})\sqrt2 : n\in\bb N\}\cup \{\sqrt2\}$ contains its limit
+point ($\sqrt2$), and it is infinite and bounded by $\sqrt 2,$ but it
+does not contain a rational number since a rational times an irrational
+is irrational.
+\endproof
+
+\question{Given $A\subset\bb R,$ let $L$ be the set of all limit points
+of $A.$}
+
+\item{a)} Show that the set $L$ is closed.
+
+\proof
+Let $l$ be a limit point of $L.$
+We will show $l$ is a limit point of $A,$ and thus $l\in L.$
+
+Let $\epsilon > 0.$
+
+By definition of limit point, $V_{\epsilon/2}(l)\cap L$ isn't empty.
+We choose $x$ from that set.
+Again, by definition of limit point, $V_{\epsilon/2}(x)\cap A$ isn't
+empty.
+We have thus shown, by the triangle inequality, that $V_\epsilon(l)\cap
+A$ isn't empty, so $l\in L,$ meaning $L$ contains all of its limit
+points and is therefore closed.
+\endproof
+
+\item{b)} Argue that if $x$ is a limit point of $A\cup L,$ then $x$ is a
+limit point of $A.$
+
+\proof
+Let $l$ be a limit point of $A\cup L.$
+
+Then, for all $\epsilon > 0,$ there must be a distinct $x\in
+V_\epsilon(l)\cap(A\cup L).$
+If $x\in A,$ then we have shown that $x$ is a limit point of $A.$
+If $x\in L,$ then we can reuse the theorem in the last answer, so $x$ is
+a limit point of $A.$
+\endproof
+
+\question{A set $A$ is called an $F_\sigma$ set if it can be written as
+the countable union of closed sets. A set $B$ is called a $G_\delta$ set
+if it can be written as the countable intersection of open sets.}
+
+\item{(1)} Show that a closed interval $[a,b]$ is a $G_\delta$ set.
+
+\proof
+Let $A_n = (a-1/n,b+1/n).$
+All of these intervals contain $[a,b],$ but they do not contain any
+element greater than $b$ or less than $a,$ so their intersection is
+$[a,b],$ proving that it is a $G_\delta$ set.
+\endproof
+
+\item{(2)} Show that the half-open interval $(a,b]$ is both a $G_\delta$
+and an $F_\sigma$ set.
+
+\proof
+Let $A_n = (a,b+1/n).$
+The intersection of these sets is $(a,b],$ so it is a $G_\delta$ set.
+
+Let $B_n = [a+1/n,b].$
+The union of these sets is $(a,b],$ so it is a $F_\sigma$ set.
+\endproof
+
+\item{(3)} Show that $\bb Q$ is an $F_\sigma$ set, and the set of
+irrationals $\bb I$ forms a $G_\delta$ set. (We will see in section 3.5
+that $\bb Q$ is not a $G_\delta$ set, nor is $\bb I$ an $F_\sigma$
+set.)
+
+\proof
+We know $\bb Q$ is countable, so we take the union of $\{q\}$ over every
+$q\in\bb Q,$ generating a countable union, so $\bb Q$ is an $F_\sigma$
+set.
+
+The complement of a closed set is open, so $\bb R\setminus\{q\}$ is an
+open set, and the intersection of these must exclude every rational
+number, constructing the set of irrationals as a $G_\delta$ set.
+\endproof
+
+\bye