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\noindent {\bf Question}
\noindent In this question, $A$ is a subset of ${\bf R}$. Define
$A^- =\{ -a \: |\: a\in A\}$. Show that each of the following
holds.
\begin{enumerate}
\item if $\sup(A)$ exists, then $\inf(A^-)$ exists and $\inf(A^-) =
-\sup(A)$;
\item if $\inf(A)$ exists, then $\sup(A^-)$ exists and $\sup(A^-) =
-\inf(A)$.
\end{enumerate}
\medskip
\noindent {\bf Answer}
\begin{enumerate}
\item Since $\sup(A)$ exists, the set $A$ is bounded above. Let $u$
be any upper bound for $A$, so that $a\le u$ for all $a\in A$.
Multiplying through by $-1$, this becomes $-a\ge -u$ for all $a\in
A$. Since $-a$ ranges over all of $A^-$ as $a$ ranges over $A$,
this yields that $-u$ is a lower bound for $A^-$, and so
$\inf(A^-)$ exists. In particular, taking $u =\sup(A)$, we have
that $-\sup(A)$ is a lower bound for $A^-$.
\medskip
\noindent To see that there is no lower bound for $A^-$ that is
greater than $-\sup(A)$, note that $t$ is a lower bound for $A^-$
if and only if $-t$ is an upper bound for $A$. Therefore, a lower
bound for $A^-$ greater than $-\sup(A)$ exists if and only if an
upper bound for $A$ less than $\sup(A)$ exists, but by the
definition of supremum no such upper bound can exist. Hence,
$-\sup(A)$ is the greatest lower bound for $A^-$, or in other
words, $-\sup(A) =\inf(A^-)$, as desired.
\item Since $\inf(A)$ exists, the set $A$ is bounded below. Let $t$
be any lower bound for $A$, so that $a\ge t$ for all $a\in A$.
Multiplying through by $-1$, this becomes $-a\le -t$ for all $a\in
A$. Since $-a$ ranges over all of $A^-$ as $a$ ranges over $A$,
this yields that $-t$ is an upper bound for $A^-$, and so
$\sup(A^-)$ exists. In particular, taking $t =\inf(A)$, we have
that $-\inf(A)$ is an upper bound for $A^-$.
\medskip
\noindent To see that there is no upper bound for $A^-$ that is
less than $-\inf(A)$, note that $u$ is an upper bound for $A^-$ if
and only if $-u$ is a lower bound for $A$. Therefore, an upper
bound for $A^-$ less than $-\inf(A)$ exists if and only if a lower
bound for $A$ greater than $\inf(A)$ exists, but by the definition
of infimum no such lower bound can exist. Hence, $-\inf(A)$ is
the least upper bound for $A^-$, or in other words, $-\inf(A)
=\sup(A^-)$, as desired.
\end{enumerate}
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