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\def\classtitle{Real Analysis}
\def\classnumber{Math 70100}
\def\examtitle{Homework 2: Extra problems}

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\begin{document}
\begin{center}
{\large \bf \examtitle}\\
{\large Prof. Hooper \qquad \today}\\
\end{center}

\ifprintanswers 
\noindent
{\bf Solutions:}
\else
\noindent
%{\bf Instructions:}
%Think about how to write clear and concise answers. Write the solution in your own words.
\fi

\begin{questions}
\question Let $X$ be a set and let $\{\sT_i:i\in \Lambda\}$ be an arbitrary collection of topologies on $X$. Prove that
$\bigcap_{i \in \Lambda} \sT_i$ is a topology on $X$. (This is the finest topology which is coarser than each $\sT_i$, or the strongest topology which is weaker than each $\sT_i$.)

\begin{solution}
\end{solution}

\question If $(X, \| \cdot \|)$ is a normed vector space, then the {\em $1$-ball} about the origin is 
$$B_1 = \{\x \in X:~\|x\|<1\}.$$
Show that $B_1$ is centrally symmetric, convex, and has the property that for all $\by \in X$ with $\by \neq 0$, there is an $s>0$ so that $s \by \in B_1$. Show that conversely if a set $B$ has these properties, it is a $1$-ball for some norm on $X$ uniquely determined by $B$. (When I say it is a $1$-ball, I mean that it contains the open $1$-ball about the origin and is contained in the closed $1$-ball about the origin.)

\begin{solution}
\end{solution}

\question{\em (Folland \S 4.1 \#2)} If $X$ is an infinite set, the cofinite topology on $X$ is $T_1$ but not $T_2$, and is first
countable iff $X$ is countable.
\begin{solution}
\end{solution}

%\question{\em (Folland \S 4.1 \#3)} Every metric space is normal. (If $A$, $B$ are closed sets in the metric space $(X, \rho)$,
%consider the sets of points $x$ where $\rho(x, A) < \rho(x, B)$ or $\rho(x, A) > \rho(x, B)$.)
%\begin{solution}
%\end{solution}

\question{\em (Folland \S 4.1 \#5)} Every separable metric space is second countable.
\begin{solution}
\end{solution}


\question{\em (Folland \S 4.1 \#8)} If $X$ is an infinite set with the cofinite topology and ${x_j}$ is a sequence of distinct
points in $X$, then $x_j \to x$ for every $x \in X$.
\begin{solution}
\end{solution}

\question{\em (Folland \S 4.1 \#9)} If $X$ is a linearly ordered set, the topology $\sT$ generated by the sets $\{x :~ x < a\}$
and $\{x :~ x > a\}$ for every $a \in X$ is called the {\em order topology}.
\begin{parts}
\part If $a, b \in X$ and $a < b$, there exist $U, V \in \sT$ with $a \in U$, $b \in V$, and $x < y$
for all $x \in U$ and $y \in V$. The order topology is the weakest topology with this property.
\begin{solution}
\end{solution}
\part If $Y \in X$, the order topology on $Y$ is never stronger than, but may be weaker
than, the relative topology on $Y$ induced by the order topology on $X$.
\begin{solution}
\end{solution}
\part The order topology on $\mathbb R$ is the usual topology.
\begin{solution}
\end{solution}
\end{parts}

\question{\em (Folland \S 4.1 \#11)} If $E_1, \ldots, E_n$ are subsets of a topological space, the closure of $\bigcup_{i=1}^k E_j$ is $\bigcup \overline{E}_j$.

\begin{solution}
\end{solution}

\question{\em (Folland \S 4.1 \#13)} If $X$ is a topological space, $U$ is open in $X$, and $A$ is dense in $X$, then ${\overline U} = \overline{U \cap A}$.
\begin{solution}
\end{solution}


\end{questions}
\end{document}
