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\fancyhead[L]{Math F651: Homework 1 Solutions}
\fancyhead[R]{January 25, 2007}
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%% Make some helpful macros for commonly used stuff.  You can also put these commands in
%% a separate file (e.g. macros.tex) and include it in your homework file with
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\newcommand{\localhead}[1]{\medskip\par\textbf{#1}\\\nobreak\unskip}
\newcommand\solution{\localhead{Solution:}}
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\begin{document}

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\hproblem Prove that every ball $B_r(x)$ in a metric space $(X,d)$ is an open set.

\solution
Let $y\in B_r(x)$ and let $\rho = r-d(x,y)$.  Suppose $z\in B_\rho(y)$.  Then
$$
d(x,z) \le d(x,y)+d(y,z) < d(x,y)+\rho  = d(x,y)+(r-d(x,y)) = r.
$$
Hence $z\in B_r(x)$ and we conclude that $B_\rho(y)\subseteq B_r(x)$.
Since $y$ is arbitrary, we conclude that $B_r(x)$ is open.

\hproblem Let $V$ be a subset of a metric space $(X,d)$.  The set of limit
points of $V$ are those points $x$ that can be written as a the limit of a 
sequence of points in $V$.  Show that a set $V\subseteq X$ is closed if and only if
it contains its limit points.
\solution
First, we claim that a point $x$ is not a limit point of $V$ if and only if there
exists a ball of radius $r$ such that $B_r(x)\cap V=\emptyset$.  One direction is clear:
if such a ball exists, then no sequence in $V$ can converge to $x$ and hence
$x$ is not a limit point.  On the other hand, suppose that $x$ is a limit point.  Then
there is a sequence $\{x_n\}$ in $V$ converging $x$, and every ball about $x$ contains a 
tail of this sequence.  Hence every ball about $x$ intersects $V$.

Now suppose $V$ is a set that contains its limit points and let $x\in V^c$.  Then
$x$ is not a limit point, and hence there is a radius $r$ such that $B_r(x)\subseteq V^c$.
Hence $V^c$ is open and $V$ is closed.

Suppose conversely that $V$ is closed.  If $x\in V^c$, then since $V^c$ is open there exists
a ball about $x$ disjoint from $V$.  Hence $x$ is not a limit point of $V$, and $V$ contains its limit points

\hproblem Let $d_1$ and $d_2$ be two metrics on a set $X$.  Show that the following
conditions are equivalent.
\begin{enumerate}
\renewcommand{\labelenumii}{\arabic{enumii})}
\item For every sequence $\{ p_i\}_{i=1}^\infty$, if $p_i\converges{d_2} p$ 
then $p_i\converges{d_1} p$.
\item For every function $f:X\ra \Reals$, if $f$ is continuous with 
respect to $d_1$ then $f$ is continuous with respect to $d_2$.
\item For every set $V$, if $V$ is closed with respect to $d_1$ then
$V$ is closed with respect to $d_2$.
\item For every set $U$, if $U$ is open with respect to $d_1$ then
$U$ is open with respect to $d_2$.
\end{enumerate}

\solution
We start by proving $1\iff 2$. Suppose condition 1 holds, and 
that  $f:X\ra \Reals$ is continuous with respect to $d_1$. If 
$p_i\converges{d_2} p$, then by condition 1 $p_i\converges{d_1} p$. 
So by continuity of $f$ with respect to $d_1$ it follows that 
$f(p_i)\converges{} f(p)$.  Hence
$f$ is continuous with respect to $d_2$.

Conversely, suppose condition 2 holds.  Fix $p\in X$ and let $f(q)=d_1(p,q)$.
Then $f$ is continuous with respect to $d_1$ and therefore also with
respect to $d_2$.  So if $p_i\converges{d_2} p$ then $f(p_i)\ra f(p)=0$.
But $f(p)=d_1(p_i,p)$, so $p_i\converges{d_1} p$.  Since the choice of $p$
is arbitrary, condition 1 holds.

We next prove $1$ implies $3$.  Suppose condition 1 holds and that $V$ is
closed with respect to $d_1$. To show $V$ is closed with respect to $d_2$,
it is enough by problem {\emph 2} to show that it contains its ($d_2$) limit points.
Suppose $p$ is a limit point (with respect to $d_2$)
of $V$.  Then there is a sequence $\{p_i\}$ in $V$ that converges
with respect to $d_2$ to $p$.  By condition 1, the sequence converges with
respect to $d_1$. Since $V$ is closed with respect to $d_1$, $p\in V$.
Hence $V$ is closed with respect to $d_2$.

It is obvious that $3$ implies $4$.  To complete the proof, we show that $4$ implies $1$.
Suppose condition 4 holds and that 
$p_i\converges{d_2} p$.  To show that $p_i\converges{d_1} p$, fix $\epsilon>0$
and consider $U=B_{d_1,\epsilon}(p)$.  Since $U$ is open with respect to 
$d_1$ it is open with respect to $d_2$ and therefore contains a ball
$B_{d_2,\epsilon'}(p)$ for some $\epsilon'>0$.  Since the sequence converges
to $p$ with respect to $d_1$, there is an $N$ such that if $i>N$, 
$p_i\in B_{d_2,\epsilon'}(p)\subset B_{d_1,\epsilon}(p)$.  Since $\epsilon$
is arbitrary, we have shown that $p_i\converges{d_1} p$.

\hproblem Munkres 13.1 (To be written up nicely and typeset)
\solution
Let $A\subseteq X$ be set such that for each $x\in A$ there is an open set $U_x$ such that 
$x\in U_x$ and $U_x\subseteq A$.  Let $\hat A = \cup_x U_x$.  Clearly $\hat A$ 
is open as it is the union of open sets.  It contains $A$
since $\hat A$  contains each point of $A$.  Moreover, $\hat A$ 
is contained by $A$ since it is the union of sets contained in $A$.  
Hence $\hat A = A$ and hence $A$ is open.

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\hproblem Munkres 13.4 

\localhead{\bf Part a.}
If $\{\calT_\alpha\}$ is a family of topologies, let 
$\calT= \cap_\alpha \calT_\alpha$.  Then both $X$ and $\emptyset$ 
are elements of $\calT$ since they are elements of each $\calT_\alpha$.
If $\{U_\beta\}$ is a family of sets in $\calT$, then $U_\beta\in\calT_\alpha$
for every $\alpha$ and $\beta$.  Hence $\cup_\beta U_\beta\in \calT_\alpha$
for each $\alpha$, and hence $\cup_\beta\in\calT$.  Finally, if $U_1,\cdots,U_n$
are in $\calT$, then $U_i\in \calT_\alpha$ for every $i$ and $\alpha$.
Hence $\cap_{i=1}^n U_i \in \calT_\alpha$ for every $\alpha$.  Hence 
$\cap_{i=1}^n U_i\in \calT$.  This proves $\calT$ is a topology.

On the other hand, the union of topologies need not be a topology.
Consider $X=\{a,b,c\}$, $\calT_1=\{\emptyset, \{a\}, X\}$, and
$\calT_2=\{\emptyset, \{b\}, X\}$.  It is easy to check that
these are both topologies.  But $\calT_1\cup \calT_2$ is not
a toplology ($\{a\}$ and $\{b\}$ are both in the union, but
$\{a,b\}$ is not.

\localhead{\bf Part b.}
Let $\{\calT_\alpha\}$ be a family of topologies on $X$.
Let $\calT_1$ be the intersection of all topologies that contain each 
$\calT_\alpha$.  We know that $\calT_1$ is non-empty since the
discrete topology contains each $\calT_\alpha$. From part 4a we know
that $\calT_1$ is a topology and it is clear that it contains $\calT_\alpha$.
Moreover, if $\calT'$
is some other topology such that $\calT'\supset \calT_\alpha$ for each 
$\alpha$, then $\calT'\supset \calT_1$ by construction.  So
$\calT_1$ must be the unique smallest topology containing each
$\calT_\alpha$.  A similar argument shows that the unique largest
topology contained in each $\calT_\alpha$ is just $\cap_\alpha \calT_\alpha$.

\localhead{\bf Part c.}
The smallest topology containing $\calT_1$ and 
$\calT_2$ is 
$$
\{ \emptyset, X, \{a\}, \{b\}, \{a,b\}, \{b,c\}\}.
$$
The largest topology contained in $\calT_1$ and $\calT_2$ is 
$\{ \emptyset, X, \{a\} \}$.

\hproblem Munkres 13.5 
\solution
We will prove the problem for a subbasis, since every
basis {\it is} a subbasis for the topology it generates. 
Let $\calT'$ be the intersection of all
topologies containing $\calA$, and let $\calT$ be the topology
generated by $\calA$.  Clearly $\calT\supset\calT'$.  So we need only
show that $\calT'\supset\calT$.  Suppose $U\in\calT$.  Then
we can write $U=\cup_\alpha U_\alpha$ where each $U_\alpha$ has the form
$ U_\alpha = \cap_{i=1}^{n_\alpha} A_i$, and where each $A_i\in \calA$.  
Now each $A_i\in \calT'$, and topologies are closed under finite intersections,
so each $U_\alpha\in \calT'$. But topologies are also closed under arbitrary unions,
so $U\in\calT'$.  Hence $\calT'\supset\calT$.

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