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\begin{document}


\begin{center}
MATH 770 HOMEWORK 3 \\
{\it Semester I, 2006-2007} \\
{\it Due October 19}
\end{center} 


\bigskip


\textbf{11.} Prove the two statements on ``Exponentiation'' 
on page 40; that is,
$\alpha^{\beta + \gamma} = \alpha^\beta \cdot \alpha^\gamma $ and 
$ a^{\beta \cdot \gamma} = (\alpha^\beta)^\gamma $.
  You may use the other facts stated there in your proof.

\bigskip


\textbf{12.} If $X$ is a topological space, let $X'$ be the set of limit
(= non-isolated) points of $X$.
Define the \textit{Cantor-Bendixson sequence} $X^{(\alpha)}$ for
$\alpha \in \ON$ by:
\begin{itemizz}
\renewcommand\labelitemi{\ding{43}}
\item $X^{(0)} = X$
\item $X^{(\alpha + 1)} = (X^{(\alpha)})'$
\item $X^{(\gamma) } = \bigcap_{\alpha < \gamma}X^{(\alpha)}$
for limit $\gamma$.
\end{itemizz}
So, $X = X^{(0)} \supseteq X^{(1)} \supseteq X^{(2)} \cdots$.
Note (by induction) that each $X^{(\alpha)}$ is closed in $X$.
Now, assume that $X$ is a non-empty compact subset of $\RRR$.  Prove
that $X^{(\alpha)} = X^{(\alpha+1)}$ for some countable $\alpha$.
\textit{Remark.}  Note that this $X^{(\alpha)}$ is either empty
or perfect (is non-empty and has no isolated points).
If $X^{(\alpha)}$ is empty, then $X$ is countable.
If $X^{(\alpha)}$ is perfect, then $|X| = \cccc = 2^{\aleph_0}$.

\bigskip

\textbf{13.}
A \textit{Bernstein set} is a set $B \subseteq \RRR$
such that both $B$ and $\RRR\backslash B$ meet every perfect
subset of $\RRR$.  Prove that you can partition $\RRR$
into $\cccc = 2^{\aleph_0}$ disjoint Bernstein sets;
$\RRR = \dot{\bigcup}_{\alpha< \cccc} B_\alpha$.
You may use fact that there are
only $\cccc = 2^{\aleph_0}$ perfect sets and each one
has size $\cccc$.
\textit{Hint.}  List the perfect sets as $\{P_\xi : \xi < \cccc\}$
Note that it's enough that each $B_\alpha \cap P_\xi \ne \emptyset$,
so there's $|\cccc \times \cccc| = \cccc$ tasks.
\textit{Remark.}
A Bernstein set cannot be Lebesgue measurable,
since both $B$ and $\RRR\backslash B$ have inner measure $0$.

\bigskip

\textbf{14.}
A  {\it two point set\/} is a  $T \subset \RRR \times \RRR$
such that each line meets $T$ in exactly two points.
Prove that there is a two point set (Mazurkiewicz, 1914).
\textit{Hint.}  There's only $\cccc$ lines and each line has size $\cccc$.
\textit{Famous open problem.}  Is there a Borel two point set?
Can one prove in $\ZF$ that there is a two point set?

\bigskip


\textbf{15.}
Let $\Sigma_1$ be the family of open subsets of $\RRR$, and
let $\Pi_1$ be the family of closed subsets of $\RRR$.
For $\beta \ge 2$,
let $\Sigma_{\beta}$ be the family of all countable unions
of sets from $\bigcup_{1 \le \alpha < \beta} \Pi_\alpha$,
and let $\Pi_{\beta}$ be the family of all countable intersections
of sets from $\bigcup_{1 \le \alpha < \beta} \Sigma_\alpha$.
So the sets in $\Sigma_2$ are the $F_\sigma$ sets and
the sets in $\Pi_2$ are the $G_\delta$ sets.
Note that in $\RRR$, all closed sets are $G_\delta$ sets and
all open sets are $F_\sigma$ sets.
Prove that $\Sigma_{\omega_1} = \Pi_{\omega_1}$ is the family
of all Borel sets.

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