w23-logic-3/inputs/tutorial_04.tex

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2023-11-14 13:22:05 +01:00
\tutorial{04}{2023-11-14}{}
\subsection{Sheet 4}
% 14 / 20
\subsubsection{Exercise 1}
\begin{enumerate}[(a)]
\item $\langle X_\alpha : \alpha\rangle$
is a descending chain of closed sets (transfinite induction).
Since $X$ is second countable, there cannot exist
uncountable strictly decreasing chains of closed sets:
Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$
was such a sequence,
then $X \setminus X_{\alpha}$ is open for every $\alpha$,
Let $\{U_n : n < \omega\}$ be a countable basis.
Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$,
is a strictly ascending chain in $\omega$.
\item We need to show that $X_{\alpha_0}$ is perfect and closed.
It is closed since all $X_{\alpha}$ are,
and perfect, as a closed set $F$ is perfect
iff it coincides $F'$.
$X \setminus X_{\alpha_0}$
is countable:
$X_{\alpha} \setminus X_{\alpha + 1}$ is
countable as for every $x$ there exists a basic open set $U$,
such that $U \cap X_{\alpha} = \{x\}$,
and the space is second countable.
Hence $X \setminus X_{\alpha_0}$
is countable as a countable union of countable sets.
\end{enumerate}
\subsection{Exercise 3}
\begin{itemize}
\item Let $Y \subseteq \R$ be $G_\delta$
such that $Y$ and $\R \setminus Y$ are dense in $\R$.
Then $Y \cong \cN$.
$Y$ is Polish, since it is $G_\delta$.
$Y$ is 0-dimensional,
since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$
form a clopen basis.
Each compact subset of $Y$ has empty interior:
Let $K \subseteq Y$ be compact
and $U \subseteq K$ be open in $Y$.
Then we can find cover of $U$ that has no finite subcover $\lightning$.
\item Let $Y \subseteq \R$ be $G_\delta$ and dense
such that $\R \setminus Y$ is dense as well.
Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\} \subseteq \R^2$.
Then $Z$ is dense in $\R^2$
and $\R^2 \setminus \Z$ is dense in $\R^2$.
We have that for every $y \in Y$
$\partial B_y(0) \subseteq Z$.
Other example:
Consider $\R^2 \setminus \Q^2$.
\end{itemize}