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\noindent {\bf Question}
\noindent For each of the following, either give an example of a
subset $S$ of ${\bf R}$ satisfying the stated property, or prove
that no such set exists.
\begin{enumerate}
\item $S$ has a rational lower bound and $\inf(S)$ is irrational;
\item $S$ has a rational lower bound and $\inf(S)$ is rational;
\item $S$ has an irrational lower bound and $\inf(S)$ is rational;
\item $S$ has an irrational lower bound and $\inf(S)$ is irrational;
\end{enumerate}
\medskip
\noindent {\bf Answer}
\noindent [Note that each of these exercises has many, many
possible solutions. And yes, it is a very silly question.]
\begin{enumerate}
\item Take $S =\{ x\in {\bf R}\: |\: x > \sqrt{2} \}$, so that
$\inf(S) =\sqrt{2}$, which is irrational, and $S$ is also bounded
below by $0$, which is rational. (In fact, any set of real
numbers that is bounded below has both infinitely many rational
lower bounds and infinitely many irrational lower bounds.)
\item Take $S = (0, \infty)$, so that $\inf(S) = 0$, which is
rational, and $S$ is bounded below by $-1$, which is also
rational.
\item Take $S =(2, 4)$, so that $\inf(S) =2$, which is rational, and
$S$ is also bounded below by $-\pi$, which is irrational.
\item Take $S =(\sqrt{3}, \infty)$, so that $\inf(S) =\sqrt{3}$, which
is irrational, and $S$ is also bounded below by $\sqrt{2}$, which
is also irrational.
\end{enumerate}
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