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\noindent
\parbox{2.25in}{\bf Math 3510}
\hfill {\Large \bf Homework \#5} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Oct. $31^\mathrm{st}$, 2018}

\vspace{0.2in}

%\noindent {\bf\large Name:} \underline{\qquad\sc\large This is my fabulous name! \qquad}
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\begin{description}
\item[\#1 Connected and Compact] For each of the following sets, decide whether they are connected and/or compact. Briefly explain your answers.
\begin{enumerate}[(a)]
\item $A=[-2,\infty)$
\item $B=[-5,2)$
\item $C=\left\{ x \;\Big|\; 1 \leq |x| \leq 2 \right\}$
\item $D=\left\{ (x,y) \;\Big|\; 4 \leq (x-1)^2+(y+2)^2 \leq 9 \right\}$
\item $E=\left\{ (x,y) \;\Big|\; |x-y| \leq 1 \right\}$
\end{enumerate}

\item[\#2 Continuity] Let $A \subset \mathbb{R}^n$ and $f:A \to \mathbb{R}^m$. Recall that $f$ is continuous at $\B a \in A$ if and only if for every $\epsilon > 0$ there is some $\delta > 0$ such that every $\B a \in A$ (the domain) and $\| \B x - \B a \|<\delta$ we have $\| f(\B x) - f(\B a) \| < \epsilon$.

Let $A$ and $B$ be topological spaces and $g:A \to B$. We say that $g$ is continuous at $a \in A$ if and only if for every open set $V \subseteq B$ with $f(a) \in V$ (i.e., $V$ is an open neighborhood of $f(a)$), there exists some open set $U \subseteq A$ such that $a \in U$ (i.e., $U$ is an open neighborhood of $a$) and $f(U) \subseteq V$.

\begin{enumerate}[(a)]
\item Briefly explain why $g$ is continuous (in the topology sense) if and only if $g$ is continuous at every $a \in A$ (in the topological sense).
\item Show that $f$ is continuous at $\B a \in A$ (in the analysis sense) if and only if $f$ is continuous at $\B a \in A$ (in the topological sense).
\end{enumerate}

\item[\#3 Inverse Function Theorem] The inverse function theorem says that a smooth function $f$ with an invertible Jacobian matrix $Df(\B a)$ is a diffeomorphism when restricted to a small enough neighborhood of $\B a$. Moreover, the Jacobian of the inverse of this restriction of $f$ is $(Df(\B a))^{-1}$ (the inverse of the Jacobian matrix $Df(\B a)$). In other words, when $f$'s derivative is invertible, $f$ is locally invertible and the derivative of the inverse is the inverse of the derivative.

Prove the following converse: Let $f:U \to V$ be a diffeomorphism (where $U$ and $V$ are open subsets of $\mathbb{R}^n$) and $\B a \in U$. Show that $Df(\B a)$ is invertible.

{\it Hint:} This is really easy. Note that the chain rule says for differentiable functions $D(f \circ g) = Df \circ Dg$.

\item[\#4 Linear Stuff] Let $f:\mathbb{R}^n \to \mathbb{R}^m$. Recall that $f$ is linear if and only if $f(\B x + \B y) = f(\B x) + f(\B y)$ and $f(c\B x) = cf(\B x)$. It follows from {\sc Linear Algebra}! that $f(\B x) = A\B x$ for some $m \times n$ matrix $A$. In coordinate notation: $f^j(x^1,\dots,x^n) = A^j_ix^i$ (implied summation).

Show that $Df = A$. This means that $Df(\B x) = A\B x$. So what is $D(Df)$? 

{\it Hint:} This is also very easy. If you're working hard, you're overthinking this one.

\end{description}

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