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\begin{center}
\textbf{Mathematics 468} \\
\textbf{Homework 2 solutions}
\end{center}
\vspace{-0.05in}

1. Prove that in $\mathbf{R}^{n}$, the only sets which are both open
and closed are the empty set and all of $\mathbf{R}^{n}$.  (If you
can't figure this out in general, try to do it when $n=1$.)

\textsc{Answer:} I'll start with the $n=1$ case, so suppose that $U$
is a nonempty open subset of $\mathbf{R}^{1}$, and assume that its
complement is nonempty; I will show that $U$ cannot be closed.  Since
$U$ is nonempty, I can pick a point $x$ in $U$.  Let $A$ denote the
complement of $U$, and pick a point $z$ in $A$.  I'll assume that
$x<z$ (if $x>z$, then the same argument works, with slightly different
notation).  Since $U$ is open, there is a $y$, $x < y < z$, so that
the interval $(x,y)$ is completely contained in $U$.  In fact, there
are many such $y$'s.  Let $y_{0}$ be the ``largest such $y$'': let
\[
y_{0} = \sup \{y \in (x,z) \, | \, (x,y) \subseteq U\}.
\]
Then every neighborhood of $y_{0}$ intersects $A$, and every
neighborhood of $y_{0}$ intersects $U$.  If $y_{0}$ is in $U$, then
$U$ is not open; therefore, $y_{0}$ must be in $A$.  But then $A$ is
not open, and so $U$ is not closed.  This is what I wanted to show.
(This argument shows, in fact, that any interval---$(a,b)$, $(a,b]$,
$[a,b]$, or $[a,b)$---is connected.)

Now I'll do the general case.  The setup is the same: let $U$ be a
nonempty open set, and assume that its complement $A=U^{c}$ is also
nonempty.  Pick points $x \in U$ and $z \in A$, and consider the
straight line segment $L$ connecting them.  This line segment is
connected, by the $n=1$ case.  Note that neither $U \cap L$ nor $A
\cap L$ is empty.  If $U$ is both open and closed, then $U \cap L$ is
both open and closed as a subset of $L$.  Since $L$ is connected, this
can't happen; hence $\mathbf{R}^{n}$ has no proper subsets which are
both open and closed.

2. If $U_{1}$, $U_{2}$, \dots, $U_{n}$ are all open, is their
intersection?  Prove or give a counterexample.

\textsc{Answer:} Yes, the intersection is open.  This follows by
induction on $n$.  When $n=1$, the intersection of one open set
$U_{1}$ with itself is just $U_{1}$, which is open by assumption.  (If
you want a slightly less trivial case, then when $n=2$, we proved in
class that the intersection of two open sets is open.)  Now, assume
that the intersection of any $n-1$ open sets is open, and consider 
\[
U_{1} \cap U_{2} \cap \dots \cap U_{n-1} \cap U_{n}.
\]
I can rewrite this as $(U_{1} \cap \dots \cap U_{n-1}) \cap U_{n}$.
By the inductive hypothesis, the set $U_{1} \cap \dots \cap U_{n-1}$
is open, so by the $n=2$ case (which we did in class), the
intersection of this with $U_{n}$ is still open.

3. If $U_{1}$, $U_{2}$, \dots, $U_{n}$, \dots are all open, is their
intersection?  (The difference between this problem and the previous
one is that in this case there are infinitely many sets $U_{i}$; in
problem 2 there were only finitely many.)

\textsc{Answer:} No, the intersection doesn't have to be open.  For
example, if I let $U_{n}$ be the open interval $(0,1+\frac{1}{n})$, then 
\[
\bigcap_{n=1}^{\infty} U_{n} = \bigcap_{n=1}^{\infty}
(0,1+\frac{1}{n}) = (0,1]. 
\]
The set $(0,1]$ is certainly not open.

4. Suppose $f: \mathbf{R}^{n} \rightarrow \mathbf{R}^{k}$ is a
function.  What does it mean for $f$ to be continuous?  Try to explain
it in words; also try to give a precise mathematical definition.

\textsc{Answer:} Both parts of this have several correct answers.  For
the precise mathematical definition, the usual one is that $f$ is
continuous if for every $x \in \mathbf{R}^{n}$ and every $\varepsilon
> 0$, there is a $\delta > 0$ so that whenever $|y-x| < \delta$, then
$|f(y)-f(x)| < \varepsilon$.  (A less common, but equally useful,
definition is that $f$ is continuous if the preimage of every open set
in $\mathbf{R}^{k}$ is open in $\mathbf{R}^{n}$.)

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