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\textbf{Math 165: Honors Calculus I}

\textbf{Assignment 9}\quad \emph{Sept. 16, 1998}

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\begin{enumerate}
\item Problem 13 on p. 45.

\problemspace
\item Let $A$ and $B$ be non-empty sets of real numbers bounded above
and below, respectively.

  \begin{enumerate}[a)]
  \item Show that for any $k < \sup A$ there exists a number $a\in A$ such that
  $k< a$.

  \item Show that for any $h > \inf B$ there exists a number $b\in B$ such that
  $b < h$.
  \end{enumerate}

\problemspace
\item Let $A$ and $B$ be sets of real numbers with $A \ne \emptyset$ and
$A\subset B$.  Prove:
  \begin{enumerate}[a)]
  \item If $A$ and $B$ are bounded above, then $\sup A \le \sup B$.

  \item If $A$ and $B$ are bounded below, then $\inf B \le \inf A$.
  \end{enumerate}


\problemspace
\item For $0 \le x \le 1$ define
  \[
    f(x) = \left\{\begin{array}{ll}
           1 & \textrm{if $x$ is rational}\\
           0 & \textrm{if $x$ is irrational}
           \end{array}\right.
  \]
Prove:
  \begin{enumerate}[a)]
  \item If $s$ and $t$ are step functions satisfying
  $s(x) \le f(x) \le t(x)$ for all $0 \le x \le 1$, then $s(x) \le 0$
  and $t(x) \ge 1$, except possibly at partition points.

  \item $\underline{I}(f) = 0$, $\overline{I}(f) = 1$, and $f$ is
  not integrable.
  \end{enumerate}
\end{enumerate}

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