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\noindent {\bf Question}

\noindent For each subset $S$ of ${\bf R}$ given below, determine
whether $S$ is bounded above, bounded below, bounded, or neither.
If $S$ is bounded above, determine $\sup(S)$, and decide whether
or not $\sup(S)$ is an element of $S$. If $S$ is bounded below,
determine $\inf(S)$, and decide whether or not $\inf(S)$ is an
element of $S$.
\begin{enumerate}
\item $S = \left\{ \frac{1}{n}\: |\: n\in {\bf Z} -\{ 0\} \right\}$;
\item $S = \left\{ 2^x\: |\: x\in {\bf Z} \right\}$;
\item $S = [-1,1]\cup \{ 5\} =\left\{ x\in {\bf R}\: |\: -1\le x\le 1
\right\}\cup\left\{ 5\right\}$;
\item $S = \left\{ \frac{x}{2^y}\: |\: x,\; y\in {\bf N} \right\}$;
\item $S = \left\{ \frac{n+1}{n} \: |\: n\in {\bf N} \right\}$;
\item $S = \left\{ (-1)^n \left( 1+\frac{1}{n}\right)\: |\:
n\in {\bf N} \right\}$;
\item $S = \left\{ x\in {\bf Q}\: |\: x^2&lt;10 \right\}$;
\item $S = \left\{ x\in {\bf R}\: |\: |x|&gt;2 \right\}$;
\end{enumerate}

\medskip
\noindent {\bf Answer}
\begin{enumerate}
\item Bounded above by $1$ (since for $n\in {\bf Z} -\{ 0\}$, either
$n\ge 1$ in which case $\frac{1}{n}\le 1$, or $n\le -1$, in which
case $\frac{1}{n}\le 0$), and so has a supremum.  Since $1$ is an
upper bound for $S$ and since $1\in S$, $1 = \sup(S)$. In this
case, $\sup(S)\in S$.

\medskip
\noindent Bounded below by $-1$ (since for $n\in {\bf Z} -\{ 0\}$,
either $n\ge 1$, in which case $0 &lt;\frac{1}{n}$, or $n\le -1$, in
which case $\frac{1}{n}\ge \frac{1}{-1} = -1$), and so has an
infimum. Since $-1$ is a lower bound for $S$ and since $-1\in S$,
$-1 = \inf(S)$. In this case, $\inf(S)\in S$.

\medskip
\noindent Since $S$ is both bounded above and bounded below, it is
bounded.
\item Bounded below by $0$ (since $2^x &gt;0$ for all $x\in {\bf R}$, we
certainly have that $2^x &gt;0$ for all $x\in {\bf Z}$), and so has
an infimum.  Given any $\varepsilon &gt;0$, we can always find $x$ so
that $2^x &lt; \varepsilon$, namely take $\log_2$ of both sides, and
take $x$ to be any integer less than $\log_2(\varepsilon)$. Hence,
there is no positive lower bound, and so the greatest lower bound,
the infimum, is $\inf(S) =0$.  Since there are no solutions to
$2^x =0$, in this case $\inf(S)\not\in S$.

\medskip
\noindent Since $2^x &gt;x$ for positive integers $x$, given any $C
&gt;0$ we can find an $x$ so that $2^x &gt;C$, and so there is no upper
bound.  That is, $S$ is not bounded above.

\medskip
\noindent Since $S$ is not bounded above, it is not bounded.
\item Bounded below by $-1$ (since $[-1,1] =\{ x\in {\bf R}\: |\:
-1\le x\le 1\}$ and since $-1 &lt; 5$), and so has an infimum. Since
$-1$ is a lower bound for $S$ and since $-1\in S$, $-1 = \inf(S)$.
In this case, $\inf(S)\in S$.

\medskip
\noindent Bounded above by $5$, and so has a supremum. Since $5$
is an upper bound for $S$ and since $5\in S$, $5 = \sup(S)$. In
this case, $\sup(S)\in S$.

\medskip
\noindent Since $S$ is both bounded above and bounded below, it is
bounded.
\item Considering the subset of $S$ in which $y = 1$, we have that $S$
contains the natural numbers ${\bf N}$, and hence $S$ is not
bounded above.

\medskip
\noindent Since $x$ and $2^y$ are both positive for $x$, $y\in
{\bf N}$, we have that $\frac{x}{2^y} &gt;0$ for all $x$, $y\in {\bf
N}$.  Therefore, $S$ is bounded below by $0$, and so has an
infimum.  Considering the subset of $S$ in which $x = 1$, we have
that $S$ contains $\frac{1}{2^y}$ for all $y\in {\bf N}$.  In
particular, for each $\varepsilon &gt;0$, we can find $y\in {\bf N}$
so that $\frac{1}{2^y} &lt;\varepsilon$, namely take $\log_2$ of both
sides to get $-y &lt;\log_2(\varepsilon)$, or equivalently $y
&gt;\log_2(\varepsilon)$.  Hence, there is no positive lower bound,
and so $0 = \inf(S)$.  Since $\frac{x}{2^y}$ is never $0$ for $x$,
$y&gt;0$, in this case $\inf(S)\not\in S$.

\medskip
\noindent Since $S$ is not bounded above, it is not bounded.
\item Write $\frac{n+1}{n} =1+\frac{1}{n}$.  Bounded below by $1$,
since $\frac{1}{n} &gt;0$ for all $n\in {\bf N}$, and hence
$1+\frac{1}{n} &gt;1$ for all $n\in {\bf N}$. Moreover, since for
each $\varepsilon &gt;1$ we can find $n$ so that $1+\frac{1}{n}
&lt;\varepsilon$, there is no lower bound greater than $1$, and so
$\inf(S) =1$.  In this case, $\inf(S)\not\in S$, since
$1+\frac{1}{n}\ne 1$ for all $n\in {\bf N}$.

\medskip
\noindent Bounded above by $2$, since $\frac{1}{n}\le 1$ for all
$n\in {\bf N}$ and hence $1+\frac{1}{n}\le 2$.  In this case, $2
=1+\frac{1}{1}$ and so $2\in S$.  Since $2$ is an upper bound for
$S$ that is contained in $S$, we have that $2 =\sup(S)$ and so
$\sup(S) \in S$.

\medskip
\noindent Since $S$ is both bounded above and bounded below, it is
bounded.
\item Break $S$ up into two subsets, one of the positive terms (when
$n$ is even) and the negative terms (when $n$ is odd).  So, $S =\{
-2, -\frac{4}{3}, -\frac{6}{5},\ldots \}\cup\{ \frac{3}{2},
\frac{5}{4}, \frac{7}{6},\ldots\}$.

\medskip
\noindent The positive terms are all of the form $1+\frac{1}{n}$
where $n$ is even. Since $\frac{1}{n}$ decreases as $n$ increases,
the largest positive term is $1+\frac{1}{2} =\frac{3}{2}$, and so
$S$ is bounded above and hence has a supremum. Since $S$ is
bounded above by $\frac{3}{2}$ and since $\frac{3}{2}\in S$,
$\sup(S) =\frac{3}{2}$, and in this case $\sup(S)\in S$.

\medskip
\noindent The negative terms are all of the form $1+\frac{1}{n}$
where $n$ is odd. Since $\frac{1}{n}$ decreases as $n$ increases,
$-\frac{1}{n}$ increases as $n$ increases, and so the smallest
negative term is $\frac{-1-1}{1} =-2$, and so $S$ is bounded below
and hence has an infimum.  Since $S$ is bounded below by $-2$ and
since $-2\in S$, $\inf(S) = -2$, and in this case $\inf(S)\in S$.

\medskip
\noindent Since $S$ is both bounded above and bounded below, it is
bounded.
\item We can rewrite $S$ as $S =(-\sqrt{10}, \sqrt{10})\cap {\bf Q}$.
By the definition of $(-\sqrt{10}, \sqrt{10})$, $S$ is bounded
below by $-\sqrt{10}$, and hence has an infimum.  Since there are
rational numbers greater than $-\sqrt{10}$ but arbitrarily close
to $-\sqrt{10}$ (as can be seen by taking the decimal expansion of
$-\sqrt{10}$ and truncating it after some number of places to get
a rational number near $-\sqrt{10}$), there is no lower bound
greater than $-\sqrt{10}$, and so $\inf(S) =-\sqrt{10}$. In this
case, $\inf(S)\not\in S$.

\medskip
\noindent $S$ is bounded above by $\sqrt{10}$, and hence has a
supremum.  Since there are rational numbers less than $\sqrt{10}$
but arbitrarily close to $\sqrt{10}$ (as can be seen by taking the
decimal expansion of $\sqrt{10}$ and truncating it after some
number of places to get a rational number near $\sqrt{10}$), there
is no upper bound less than $\sqrt{10}$, and so $\sup(S)
=\sqrt{10}$.  In this case, $\sup(S)\not\in S$.

\medskip
\noindent Since $S$ is both bounded above and bounded below, it is
bounded.
\item Rewrite $S$ as $S =(-\infty, -2)\cup (2, \infty)$.  This set is
neither bounded above (since for each real number $r$, there is
$s\in S$ with $s &gt; r$, namely the larger of $3$ and $r+1$) nor
bounded below (since for each real number $r$, there is $s\in S$
with $s &lt; r$, namely the smaller of $-3$ and $r-1$).

\medskip
\noindent Since $S$ is not bounded below, it has no infimum. Since
$S$ is not bounded above, it has no supremum.

\medskip
\noindent Since $S$ is neither bounded above not bounded below, it
is not bounded.
\end{enumerate}
\end{document}
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