\lecture{01}{2023-10-10}{Introduction}
\section{Introduction}

\begin{definition}
    Let $X$ be a nonempty topological space.
    We say that $X$ is a \vocab{Polish space}
    if $X$ is
    \begin{itemize}
        \item \vocab{separable},
            i.e.~there exists a countable dense subset, and
        \item \vocab{completely metrisable},
            i.e.~there exists a complete metric on $X$
            which induces the topology.
    \end{itemize}
\end{definition}
Note that Polishness is preserved under homeomorphisms,
i.e.~it is really a topological property.
\gist{%
\begin{example}
    \begin{itemize}
        \item $\R$ is a Polish space,
        \item $\R^n$ for finite $n$ is Polish,
        \item $[0,1]$,
        \item any countable discrete topological space,
        \item the completion of any separable metric space
            considered as a topological space.
    \end{itemize}
\end{example}
}{}
\gist{%
Polish spaces behave very nicely.
We will see that uncountable polish spaces have size $2^{\aleph_0}$. % TODO: mathfrak c for continuum
There are good notions of big (comeager)
and small (meager).
}{}

\subsection{Topology background}
Recall the following notions:
\begin{definition}[\vocab{product topology}]
        Let $(X_i)_{i \in I}$ be a family of topological spaces.
        Consider the set  $\prod_{i \in I} X_i$ 
        and the topology induced by basic open sets
        $\prod_{i \in I} U_i$ with $U_i \subseteq X_i$ open
        and $U_i \subsetneq X_i$ for only finitely many $i$.
\end{definition}
\begin{fact}
    Countable products of separable spaces are separable.
\end{fact}
\begin{definition}
    A topological space $X$ is \vocab{second countable},
    if it has a countable base.
\end{definition}
Let $X$ be a topological space.
If $X$ is second countable, it is also separable.
However the converse of this does not hold.

\begin{example}
    Let $X$ be an uncountable set.
    Take $x_0 \in X$ and consider the topology given by
    \[
    \tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}.
    \]
    Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
\end{example}
\begin{example}[Sorgenfrey line]
    Consider $\R$ with the topology given by the basis $\{[a,b) : a,b \in \R\}$.
    This is T3, but not second countable
    and not metrizable.
\end{example}

\begin{fact}
    For metric spaces, the following are equivalent:
    \begin{itemize}
        \item separable,
        \item second-countable,
        \item \vocab{Lindelöf} (every open cover has a countable subcover).
    \end{itemize}
\end{fact}
\gist{%
\begin{fact}
    Compact Hausdorff spaces are \vocab{normal} (T4)
    i.e.~two disjoint closed subsets can be separated
    by open sets.
\end{fact}
}{}
\begin{fact}
    For a metric space, the following are equivalent:
    \begin{itemize}
        \item compact,
        \item \vocab{sequentially compact}
            (every sequence has a convergent subsequence),
        \item complete and \vocab{totally bounded}
            (for all $\epsilon > 0$ we can cover the space with finitely many $\epsilon$-balls).
    \end{itemize}
\end{fact}
\begin{theorem}[Urysohn's metrisation theorem]
    Let $X$ be a topological space.
    If $X$ is
    \begin{itemize}
        \item second countable,
        \item Hausdorff and
        \item regular (T3)
    \end{itemize}
    then $X$ is metrisable.
\end{theorem}

\begin{absolutelynopagebreak}
\begin{fact}
    If $X$ is a compact Hausdorff space,
    the following are equivalent:
    \begin{itemize}
        \item $X$ is Polish,
        \item $X$ is metrisable,
        \item $X$ is second countable.
    \end{itemize}
\end{fact}
\end{absolutelynopagebreak}

\subsection{Some facts about polish spaces}
\gist{%
\begin{fact}
Let $(X, \tau)$ be a topological space.
Let $d$ be a metric on $X$.
We will denote the topology induces by this metric as $\tau_d$.
To show that $\tau = \tau_d$,
it is equivalent to show that
\begin{itemize}
    \item every open $d$-ball is in $\tau$ ($\implies \tau_d \subseteq d$ )
        and
    \item every open set in $\tau$ is a union of open $d$-balls.
\end{itemize}
To show that $\tau_d = \tau_{d'}$
for two metrics  $d, d'$,
suffices to show that open balls in one metric are unions of open balls in the other.
\end{fact}
}{}

\begin{notation}
    We sometimes\footnote{only in this subsection?} denote $\min(a,b)$ by $a \wedge b$.
\end{notation}

\begin{proposition}
    \label{prop:boundedmetric}
        Let $(X, \tau)$ be a topological space,
        $d$ a metric on $ X$ compatible with $\tau$ (i.e.~it induces $\tau$).

        Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
\end{proposition}
\gist{%
\begin{proof}
    To check the triangle inequality:
    \begin{IEEEeqnarray*}{rCl}
        d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right)  \wedge 1\\
                        &\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right).
    \end{IEEEeqnarray*}

    For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$
    and for $\epsilon > 1$, $B'_\epsilon(x) = X$.

    Since $d$ is complete, we have that $d'$ is complete.
\end{proof}
}{}
\begin{proposition}
    Let $A$ be a Polish space.
    Then $A^{\omega}$ Polish.
\end{proposition}
\begin{proof}
    Let $A$ be separable.
    Then $A^{\omega}$ is separable.
    (Consider the basic open sets of the product topology).

    Let $d \le 1$ be a complete metric on $A$.
    Define $D$ on $A^\omega$ by
    \[
        D\left( (x_n), (y_n) \right) \coloneqq
        \sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
    \]
\gist{%
        Clearly $D \le 1$.
    It is also clear, that $D$ is a metric.

    We need to check that $D$ is complete:
    Let $(x_n)^{(k)}$ be a Cauchy sequence in $A^{\omega}$.
    Consider the pointwise limit $(a_n)$.
    This exists since $x_n^{(k)}$ is Cauchy for every fixed $n$.
    Then $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
}{Clearly $D$ is a complete metric.}
\end{proof}

\begin{definition}[Our favourite Polish spaces]
    \leavevmode
    \begin{itemize}
        \item $2^{\N}$ is called the \vocab{Cantor set}.
            (Consider $2$ with the discrete topology)
        \item $\cN \coloneqq \N^{\N}$ is called the \vocab{Baire space}.
            ($\N$ with descrete topology)
        \item $\mathbb{H} \coloneqq [0,1]^{\N}$ is called the \vocab{Hilbert cube}.
            ($[0,1] \subseteq \R$ with the usual topology)
    \end{itemize}
\end{definition}
\begin{proposition}
    Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
    Then $X$ topologically embeds into the
    \vocab{Hilbert cube},
    i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$
    such that $f: X \to f(X)$ is a homeomorphism.
\end{proposition}
\begin{proof}
    \gist{$X$ is separable, so it has some countable dense subset,
    which we order as a sequence $(x_n)_{n \in \omega}$.}%
    {Let $(x_n)_{n \in \omega}$ be a countable dense subset.}

    \gist{Let $d$ be a metric on $X$ which is compatible with the topology.
    W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).}%
    {Let $d \le 1$ be a metric of $X$.}
    Define
    \begin{IEEEeqnarray*}{rCl}
        f\colon X &\longrightarrow & [0,1]^{\omega} \\
         x&\longmapsto & (d(x,x_n))_{n < \omega}
    \end{IEEEeqnarray*}
\gist{
    \begin{claim}
        $f$ is injective.
    \end{claim}
    \begin{subproof}
        Suppose that $f(x) = f(y)$.
        Then $d(x,x_n) = d(y,y_n)$ for all $n$.
        Hence $d(x,y) \le  d(x,x_n) + d(y,x_n) = 2 d(x,x_n)$.
        Since $(x_n)$ is dense, we get $d(x,y) = 0$.
    \end{subproof}
    \begin{claim}
        $f$ is continuous.
    \end{claim}
    \begin{subproof}
        Consider a basic open set in $[0,1]^{\omega}$,
        i.e. specify open sets $U_1, \ldots, U_n$ on finitely many coordinates.
        $f^{-1}(U_1 \times  \ldots \times  U_n \times  \ldots)$
        is a finite intersection of open sets,
        hence it is open.
    \end{subproof}
    \begin{claim}
        $f^{-1}$ is continuous.
    \end{claim}
    \begin{subproof}
        Consider $B_{\epsilon}(x_n) \subseteq X$ for some $n \in \N$, $\epsilon > 0$.
        Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$
        is open\footnote{as a subset of $f(X)$!}.
    \end{subproof}
}{}
\end{proof}
\begin{proposition}
    Countable disjoint unions of Polish spaces are Polish.
\end{proposition}
\gist{
\begin{proof}
    Define a metric in the obvious way.
\end{proof}
}{}

\begin{proposition}
    Closed subspaces of Polish spaces are Polish.
\end{proposition}
\gist{%
\begin{proof}
    Let $X$ be Polish and $V \subseteq  X$ closed.
    Let $d$ be a complete metric on $X$.
    Then $d\defon{V}$ is complete.
    Subspaces of second countable spaces
    are second countable.
\end{proof}%
}{}

\begin{definition}
    Let $X$ be a topological space.
    A subspace $A \subseteq X$ is called
    $G_\delta$\footnote{Gebietdurchschnitt}, if it is a countable intersection of open sets.
\end{definition}
\gist{
Next time: Closed sets are $G_\delta$.
A subspace of a Polish space is Polish iff it is $G_{\delta}$
}{}