\subsection{Sheet 2}

\tutorial{03}{2023-10-31}{}


% 15 / 16

\begin{remark}
    $F_\sigma$ stands for \vocab{ferm\'e sum denumerable}.
\end{remark}


\nr 1

Let $(U_i)_{i < \omega}$ be a countable base of $X$.
Define
\begin{IEEEeqnarray*}{rCl}
    f\colon X &\longrightarrow & 2^{\omega} \\
    x &\longmapsto & (x_i)_{i < \omega}
\end{IEEEeqnarray*}
where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
is open.
We have that $f$ is injective since $X$ is T1.

Let $f\colon X \hookrightarrow 2^\omega$ be such that
$f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\})$.

Let $V \subseteq 2^{\omega}$ be closed.
Then $2^{\omega} \setminus V$ is open, i.e.~has the form
$\bigcup_{i \in I} ((\prod_{j<n_j} X_{i,j}) \times 2^{\omega})$
for some $X_{i,j} \subseteq 2$.
As $2^{\omega}$ is second countable, we may assume $I$ to be countable.

Then $V = \bigcap_{i \in  I} \left(2^{\omega} \setminus  ((\prod_{i <n_j} X_{i,j}) \times 2^{\omega})\right)$.
Since  $f$ is injective, we have $f^{-1}(\bigcap_{a \in A} a) = \bigcap_{a \in A} f^{-1}(a)$.
Thus it suffices to show that $f^{-1}(2^{\omega} \setminus  ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
is $G_{\delta}$, as a countable intersection of $G_{\delta}$-sets
is $G_{\delta}$.

We have that $U_k \coloneqq f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 1\})$
is open. Since $f$ is injective
$f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 0\}) = X \setminus U_k$
is closed, in particular it is $G_\delta$.
Let $x = (x_1,\ldots, x_n) \in 2^{n} \setminus (\prod_{i < n} X_i)$.
Then $f^{-1}(\{x\} \times 2^\omega) = \bigcap_{i < n}\bigcap U'_n$
is $G_{\delta}$, were $U'_i = U_i$ if $x_k = i$
and $U'_i = X \setminus U_i$ otherwise.

Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$
is finite, we get that
$f^{-1}(2^{\omega} \setminus  ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
is $G_\delta$ as a finite union of $G_{\delta}$ sets.



\nr 2
\todo{handwritten solution}
(b)
Let $f(x^{(i)})$ be a sequence in $f(X)$.
Suppose that $f(x^{(i)}) \to y$.
We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.


\nr 3
\begin{example}
    Consider
    \begin{IEEEeqnarray*}{rCl}
        f\colon \R &\longrightarrow & [0,1] \\
        \frac{p}{q} &\longmapsto & \frac{1}{q}\\
        \R \setminus \Q \ni x &\longmapsto & 0
    \end{IEEEeqnarray*}
    Then $\osc_f(\frac{p}{q}) = \frac{1}{q}$
    and $\osc_f(x) = 0$ for $x \not\in \Q$.
\end{example}

\begin{definition}
    We say that $f\colon X \to Y$ is continuous
    at $a \in X$,
    if for $N$ a neighbourhood of $f(a)$ (i.e.~there exists
    $f(a) \in U \overset{\text{open}}{\subseteq}  N$,
    then $f^{-1}(N)$ is a neighbourhood of $a$.
\end{definition}

\begin{theorem}[Kuratowski]
    Let $X$ be metrizable, $Y$ completely metrizable,
    $S \subseteq X$ and $f\colon S \to Y$ continuous.
    Then $f$ can be extended to a continuous function $\tilde{f}$
    on a $G_\delta$ set  $G$ with $S \subseteq  G \subseteq \overline{S}$.
\end{theorem}
\begin{proof}
    Let $G \coloneqq  \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$.
    Clearly $S \subseteq G$ as $f$ is continouos on $f$.
    \begin{claim}
        $G$ is $G_\delta$
    \end{claim}
    \begin{subproof}
        $\overline{S}$ is closed
        and
        \[
        \bigcap_{n \ge 1} \{x : \osc_f(x) <\frac{1}{n}\}
        \]
        is an intersection of open sets.
    \end{subproof}
    is an intersection of open sets.

    For $x \in G$, as $x \in  \overline{S}$,
    there exists $(x_n)_{x_n < \omega}$, $x_n \in S$
    such that $x_n \to x$.
    We have that $(f(x_n))_n$ is Cauchy,
    as $\osc_f(x) = 0$.

    \todo{Something is missing here}
\end{proof}

\begin{corollary}
    Let $X$ be Polish and $Y \subseteq X$ Polish.
    Then $Y$ is $G_{\delta}$.
\end{corollary}
\begin{proof}
    \todo{TODO}
    % Consider the identity $f\colon Y \hookrightarrow X$.
    % Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
    % with $Y \subseteq G \subseteq  \overline{Y}$.
    % $\tilde{f}$ and $\id_G$ agree on $Y$.
    % $Y$ is dense in $G$ and the codomain of $f$ is Hausdorff.\todo{????}
    % So $f = \id_G$, i.e.~$G = Y$.
\end{proof}

\nr 4
\begin{enumerate}[(1)]
    \item $f$ is a topological embedding:
        Consider a basic open set
        $B = \prod_{i < n} X_i \times \omega^{\omega}$
        for some $X_i \subseteq \omega$.

        Then $f(B) = \left(\bigcup_{x \in \prod_{i<n} X_i} B_x \right) \cap f(\omega^{\omega})$
        is open in $f(\omega^{\omega})$,
        where $B_x \coloneqq \{0^{x_0}10^{x_1}1\ldots 10^{x_n-1}1\} \times 2^{\omega}$.

        On the other hand let $C = \{x_0x_1x_2x_3x_4 \ldots x_{n-1}\}\times 2^{\omega}$
        be some basic open set of $2^{\omega}$.
        W.l.o.g.~$x_0x_1\ldots x_{n-1}$
        has the form $0^{a_0}10^{a_1}1\ldots 10^{a_k}x_{n-1}$.
        If $x_{n-1} = 1$,
        we get
        \[
            f^{-1}(C \cap f(\omega^{\omega})) = \{(a_0,a_1,\ldots,a_k)\} \times \omega^{\omega}.
        \]
        In the case of $x_{n-1} = 0$,
        it is
        \[
        f^{-1}(C \cap f(\omega^\omega)) = \bigcup_{b > a_k} \{(a_0,a_1,\ldots,a_{k-1}, b)\} \times \omega^{\omega}.
        \]
        In both cases the preimage is open.


    \item  $C \coloneqq 2^{\omega} \setminus f(\omega^\omega)$
        is countable and dense in $2^{\omega}$.

        We have $C = \{x \in 2^{\omega} | x_i = 0 \text{ for all but finitely many $i$}\} = \bigcup_{i < \omega} (2^{i} \times 1^{\omega})$.
        Clearly this is countable.

        For denseness take some $x \in 2^\omega$.
        Let $x^{(n)}$ be defined by $x^{(n)}_i = x_i$ for $i < n$
        and $x^{(n)}_i = 0$ for $i \ge n$.
        Then $x^{(n)} \in C$ for all $n$,
        and $x^{(n)}$ converges to $x$.

    \item $f(\omega^\omega)$ is $G_\delta$:

        We have
        \begin{IEEEeqnarray*}{rCl}
            f(\omega^\omega) &=& 2^\omega \setminus \left(\bigcup_{i < \omega} (2^{i} \times 1^{\omega})\right)\\
                             &=& \bigcap_{i < \omega} \left(2^{\omega} \setminus(2^{i} \times 1^{\omega})\right).
        \end{IEEEeqnarray*}
    \item $C$ as in (2) is homeomorphic to $\Q$.

        Go to the right in the even digits, go to the left for the odd digits,
        i.e.~let $C = (1,-1,1,-1, \ldots)$
        and set $x < y$ iff $C \cdot  x <_{\text{lex}} C \cdot y$,
        where $<_{\text{lex}}$ denotes the lexicographical ordering.
        Note that the order topology of $<$ on $C$ 
        agrees with the subspace topology from $2^\omega$.

        By Cantor's theorem for countable, unbounded, dense linear
        linear orders,
        we get an order isomorphism $C \leftrightarrow \Q$.
        This is also a homeomorphism, as the topologies
        on $C$ and $\Q$ are the respective order topologies.
\end{enumerate}
