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\parbox{2.25in}{\bf Math 3220}
\hfill {\Large \bf Homework \#2} \hfill
\parbox{2.25in}{\bf \hfill Due: Mon., Feb. $4^\mathrm{th}$, 2019}

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%\noindent {\bf\large Name:} \underline{\qquad\sc\large This is my fabulous name! \qquad}
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\begin{description}
\item[\#1 A Selection of Tasty Sups] For each of the following sets determine if its supremum and/or infimum exists. If either fails to exist, explain why. If either does exist, find it and determine if it belongs to the set or not. 

{\it Example:} If $S=(-\infty,0)$, then $\mathrm{inf}(S)$ does not exist since $S$ is not bounded below. On the other hand, $\mathrm{sup}(S)=0$ but $\mathrm{sup}(S) \not\in S$ since the interval is open.

\begin{enumerate}[(a)]

\item $\displaystyle A = [0,2) \cup [4,6) \cup [8,10) \cup \cdots = \bigcup_{k=0}^\infty [4k,4k+2)$

\item $\displaystyle B = \left\{ 5-\dfrac{2}{n} \;\Big|\; n \in \mathbb{N} \right\}$ \qquad {\it Note:} $\mathbb{N} = \mathbb{Z}_{>0} = \{ 1,2,3,\dots \}$

\item $\displaystyle C = \left\{ -2n^2+\dfrac{1}{(n^2+1)^2} \;\Big|\; n \in \mathbb{Z} \right\}$

{\it Note:} Sometimes doing some numerical experiments is very important when looking for a solution. Here's some Maple code which computes decimal approximations (notice the ``evalf'') of the members set $C$ for integers $n$ where $-3 \leq n \leq 5$. You could also just do this on any calculator.

{\color{red}
\verb|[> evalf(seq(-2*n^2+1/(n^2+1)^2,n=-3..5));|
}

\end{enumerate}

\item[\#2 Don't Let Your Sup Get Away] (Zorn 1.8 \#20) Let $S \subseteq \mathbb{R}$ be non-empty and bounded above with $\beta = \mathrm{sup}(S)$. Show that $S$ is {\it not} bounded away from $\beta$. 

\item[\#3 A Supremely Infimum Problem] (Zorn 1.9 \#8) Let $S \subseteq \mathbb{R}$ be non-empty and bounded below. Let $-S = \{ -x \;|\; x \in S\}$. Show that $\mathrm{sup}(-S)$ exists. Then show that $-\mathrm{inf}(S) = \mathrm{sup}(-S)$.

This problem shows that the completeness axiom guaranteeing the existence of supremums implies a similar statement about the existence of infimums. Write down an ``infimum'' version of the completeness axiom.

\item[\#4 Supreme Addition] Let $S$ and $T$ be non-empty subsets of $\mathbb{R}$ and assume both $S$ and $T$ are bounded above. Let $S+T = \{ s+t \;|\; s \in S \mbox{ and } t \in T\}$. Explain why $\mathrm{sup}(S)$, $\mathrm{sup}(T)$, and $\mathrm{sup}(S+T)$ exist. Then show that $\mathrm{sup}(S+T) \leq \mathrm{sup}(S)+\mathrm{sup}(T)$.


\item[\#5 Closing In] (Zorn 1.9 \#10) Show that every closed interval $[a,b]$ is the intersection of a nested family: $I_1 \supseteq I_2 \supseteq \cdots$ of {\it open} intervals.

{\it Note:} This should feel more like a calculation than a proof.

\item[RESUBMIT] Type up Homework \#1 Problem \#4 and its solution in \LaTeX. 

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