Josia Pietsch
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\lecture{01}{20231010}{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 secondcountable,


\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 $G_{\delta}$


}{}
