\documentclass[11pt]{article}
\usepackage{fullpage}

\usepackage{times}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{microtype}

\newtheorem{theorem}{Theorem}
\newtheorem{definition}{Definition}
\newtheorem{problem}{Problem}

\newcommand{\TF}{\item\textnormal{\textbf{T} or \textbf{F}:\ }}

%%%%%%%%%%%%%%%%
\begin{document}


\section{Uniform}
\begin{definition}[Uniform Continuity]
A function $f\!:\!D\to\mathbb{R}$ is \emph{uniformly continuous on $D$} iff
for any $\varepsilon>0$ there is a $\delta>0$ s.t.~for all $\vec{x}_1,\vec{x}_2\in D$,
if $\|\vec{x}_1-\vec{x}_2\|<\delta$, then $|f(\vec{x}_1)-f(\vec{x}_2)|<\varepsilon$.
\end{definition}

\begin{theorem}
If $f$ is continuous on $D$, and $D$ is closed \& bounded (compact), then
\begin{enumerate}
\item $f$ is bounded,
\item $f$ attains extreme values (max and min),
\item $f$ is uniformly continuous on $D$.
\end{enumerate}
\end{theorem}

\begin{proof}[Proof (Homework).] (Keys) \\
\begin{enumerate}
\item Hint: Assume not, then look at $f^{-1}(a_n)$ where $a_n\to\infty$. 
\item \emph{Bolzano-Weierstrass in action.} 
\item Hint: Assume not. Create sequences $\vec{x}_n$, $\vec{y}_n$ that converge to $\vec{a}$, but have $|f(\vec{x}_n)-f(\vec{y}_n)|>\varepsilon$. Cont gives a contradiction.
\end{enumerate}
\end{proof}



\section{Fun with Functions}
\begin{problem}[Functions]
Let $f\!:\!\mathbb{R}^n\to\mathbb{R}$ be a function. Let $A$ and $B$ be subsets of the domain and range of $f,$ respectively. Then
\begin{align*}
	f(A) &= \left\{y\in\mathbb{R}\,|\, f(a)=y \mbox{ for some $a\in A$} \right\} \subseteq \mathrm{range}(f) \\
	f^{-1}(B) &= \left\{x\in\mathbb{R}^n\,|\, f(x)=b \mbox{ for some $b\in B$} \right\} \subseteq \mathrm{dom}(f)
\end{align*}
Give an example justifying your answer.
%\begin{multicols}{2}
\begin{enumerate}	\setlength{\itemsep}{10pt}
\TF $A \subseteq f^{-1}\!\left(\rule{0pt}{9pt}f(A)\right)$ 
\TF $A = f^{-1}\!\left(\rule{0pt}{9pt}f(A)\right)$ 
\TF $A \supseteq f^{-1}\!\left(\rule{0pt}{9pt}f(A)\right)$ or $f^{-1}\!\left(\rule{0pt}{9pt}f(A)\right) \subseteq A$
\TF $B \subseteq f\!\left(\rule{0pt}{9pt}f^{-1}(B)\right)$ 
\TF $B = f\!\left(\rule{0pt}{9pt}f^{-1}(B)\right)$ 
\TF $B \supseteq f\!\left(\rule{0pt}{9pt}f^{-1}(B)\right)$ or $f\!\left(\rule{0pt}{9pt}f^{-1}(B)\right) \subseteq B$
\end{enumerate}
%\end{multicols}
\end{problem}




\end{document}
\end
