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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)}}
+\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 \#4, Part II}}
+
+\question{Give an example of each of the following, or argue that such a
+request is impossible.}
+
+\item{a.} A sequence that has a subsequence that is bounded but contains
+no subsequence that converges.
+
+\intro{Disproof}
+This is impossible.
+Let $(a_n)$ be a sequence bounded by $M.$
+
+We will show by contradiction that for any $\epsilon > 0,$ there must be
+a susbequence $(a_{\phi(n)})$ such that $|a_n-l|<\epsilon$ for some $l.$
+
+First, we break the interval $[-M,M] \supseteq \{a_n\}$ into the
+partition $$[-M,-M+\epsilon)\cup
+[-M+\epsilon,-M+2\epsilon)\cup\cdots\cup [M-\epsilon,M] =
+S_1\cup S_2\cup\cdots\cup S_n.$$
+For contradiction, we assume that there is no subsequence meeting the
+Cauchy criterion for $\epsilon.$
+
+This means $(a_n)\cap S_i$ is finite for all $S_i$ because otherwise
+these would be subsequences satisfying the Cauchy Criterion for
+$\epsilon.$
+$$S_1\cup\ldots\cup S_n = [-M,M],$$
+so $[-M,M]\cap (a_n) = (a_n)$ has a length less than or equal to the sum
+of lengths across $S_i\cap (a_n),$ which makes it finite, but the
+sequence is infinite, giving a contradiction.
+
+\endproof
+
+\vfil\eject
+\item{b.} A sequence that does not contain $0$ or $1$ as a term but
+contains subsequences converging to each of these values.
+
+\proof
+Let $$a_n = 1/2 + (-1)^n{n\over 2(n+1)}.$$
+\endproof
+
+\vfil\eject
+\item{c.} A sequence that contains subsequences converging to every
+point in the infinite set $\{1, 1/2, 1/3, 1/4, 1/5,\ldots\},$ and no
+subsequences converging to points outside of this set.
+
+\intro{Disproof}
+Such a sequence would necessarily contain a subsequence converging to
+$0.$
+
+Let $\epsilon > 0.$
+We can easily choose some element in the set $0<s = 1/n < \epsilon/2.$
+Then, because we have a subsequence converging to this point, for some
+$N,$ we get $$|a_N - s| < \epsilon/2 \Longrightarrow s-\epsilon/2 < a_N
+< s+\epsilon/2.$$
+We already know $0 < s < \epsilon/2,$ so $|s-\epsilon/2| < \epsilon/2 <
+|s+\epsilon/2| < \epsilon,$ and $|a_N| <
+\max\{|s-\epsilon/2|,|s+\epsilon/2|\} < \epsilon,$ showing that we must
+have a subsequence converging to $0.$
+\endproof
+
+\vfil\eject
+\item{d.} A sequence that contains subsequences converging to every
+point in the infinite set below, and no subsequences converging to
+points outside of this set.
+
+$$\{2^{-m} + 2^{-n} : 1\leq m < n\} = \bigcup_{m=1}^\infty \{2^{-m} +
+2^{-m-1}, 2^{-m}+2^{-m-2},\ldots\}.$$
+
+\intro{Disproof}
+For a very similar reason as (c), we cannot have such a sequence
+containing such subsequences without containing their limit point $0$
+(and possibly others).
+
+We will dispense with the long $\epsilon$ proof and simply note that we
+must have some $m$ such that $2^{-m+1}<\epsilon/4,$ and
+$2^{-m}+2^{-{m+1}} = 3(2^{-m+1})<3\epsilon/4,$ and choosing a point
+$a_N$ from
+our sequence within $\epsilon/4$ because of the subsequence already
+established, we again get an $a_N<\epsilon,$ showing it converges to
+$0,$ which is outside the set.
+
+\bye