Josia Pietsch
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276 lines
8.9 KiB
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276 lines
8.9 KiB
TeX
\lecture{01}{2023-10-10}{Introduction}
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\section{Introduction}
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\begin{definition}
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Let $X$ be a nonempty topological space.
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We say that $X$ is a \vocab{Polish space}
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if $X$ is
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\begin{itemize}
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\item \vocab{separable},
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i.e.~there exists a countable dense subset, and
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\item \vocab{completely metrisable},
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i.e.~there exists a complete metric on $X$
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which induces the topology.
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\end{itemize}
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\end{definition}
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Note that Polishness is preserved under homeomorphisms,
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i.e.~it is really a topological property.
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\begin{example}
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\begin{itemize}
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\item $\R$ is a Polish space,
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\item $\R^n$ for finite $n$ is Polish,
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\item $[0,1]$,
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\item any countable discrete topological space,
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\item the completion of any separable metric space
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considered as a topological space.
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\end{itemize}
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\end{example}
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\gist{%
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Polish spaces behave very nicely.
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We will see that uncountable polish spaces have size $2^{\aleph_0}$. % TODO: mathfrak c for continuum
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There are good notions of big (comeager)
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and small (meager).
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}{}
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\subsection{Topology background}
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Recall the following notions:
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\begin{definition}[\vocab{product topology}]
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Let $(X_i)_{i \in I}$ be a family of topological spaces.
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Consider the set $\prod_{i \in I} X_i$
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and the topology induced by basic open sets
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$\prod_{i \in I} U_i$ with $U_i \subseteq X_i$ open
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and $U_i \subsetneq X_i$ for only finitely many $i$.
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\end{definition}
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\begin{fact}
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Countable products of separable spaces are separable.
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\end{fact}
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\begin{definition}
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A topological space $X$ is \vocab{second countable},
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if it has a countable base.
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\end{definition}
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Let $X$ be a topological space.
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If $X$ is second countable, it is also separable.
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However the converse of this does not hold.
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\begin{example}
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Let $X$ be an uncountable set.
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Take $x_0 \in X$ and consider the topology given by
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\[
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\tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}.
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\]
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Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
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\end{example}
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\begin{example}[Sorgenfrey line]
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Consider $\R$ with the topology given by the basis $\{[a,b) : a,b \in \R\}$.
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This is T3, but not second countable
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and not metrizable.
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\end{example}
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\begin{fact}
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For metric spaces, the following are equivalent:
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\begin{itemize}
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\item separable,
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\item second-countable,
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\item \vocab{Lindelöf} (every open cover has a countable subcover).
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\end{itemize}
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\end{fact}
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\begin{fact}
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Compact Hausdorff spaces are \vocab{normal} (T4)
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i.e.~two disjoint closed subsets can be separated
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by open sets.
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\end{fact}
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\begin{fact}
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For a metric space, the following are equivalent:
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\begin{itemize}
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\item compact,
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\item \vocab{sequentially compact}
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(every sequence has a convergent subsequence),
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\item complete and \vocab{totally bounded}
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(for all $\epsilon > 0$ we can cover the space with finitely many $\epsilon$-balls).
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\end{itemize}
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\end{fact}
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\begin{theorem}[Urysohn's metrisation theorem]
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Let $X$ be a topological space.
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If $X$ is
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\begin{itemize}
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\item second countable,
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\item Hausdorff and
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\item regular (T3)
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\end{itemize}
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then $X$ is metrisable.
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\end{theorem}
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\begin{absolutelynopagebreak}
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\begin{fact}
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If $X$ is a compact Hausdorff space,
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the following are equivalent:
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\begin{itemize}
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\item $X$ is Polish,
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\item $X$ is metrisable,
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\item $X$ is second countable.
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\end{itemize}
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\end{fact}
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\end{absolutelynopagebreak}
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\subsection{Some facts about polish spaces}
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\gist{%
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\begin{fact}
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Let $(X, \tau)$ be a topological space.
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Let $d$ be a metric on $X$.
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We will denote the topology induces by this metric as $\tau_d$.
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To show that $\tau = \tau_d$,
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it is equivalent to show that
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\begin{itemize}
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\item every open $d$-ball is in $\tau$ ($\implies \tau_d \subseteq d$ )
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and
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\item every open set in $\tau$ is a union of open $d$-balls.
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\end{itemize}
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To show that $\tau_d = \tau_{d'}$
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for two metrics $d, d'$,
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suffices to show that open balls in one metric are unions of open balls in the other.
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\end{fact}
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}{}
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\begin{notation}
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We sometimes\footnote{only in this subsection?} denote $\min(a,b)$ by $a \wedge b$.
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\end{notation}
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\begin{proposition}
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\label{prop:boundedmetric}
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Let $(X, \tau)$ be a topological space,
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$d$ a metric on $ X$ compatible with $\tau$ (i.e.~it induces $\tau$).
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Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
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\end{proposition}
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\gist{%
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\begin{proof}
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To check the triangle inequality:
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\begin{IEEEeqnarray*}{rCl}
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d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right) \wedge 1\\
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&\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right).
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\end{IEEEeqnarray*}
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For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$
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and for $\epsilon > 1$, $B'_\epsilon(x) = X$.
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Since $d$ is complete, we have that $d'$ is complete.
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\end{proof}
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}{}
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\begin{proposition}
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Let $A$ be a Polish space.
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Then $A^{\omega}$ Polish.
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\end{proposition}
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\begin{proof}
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Let $A$ be separable.
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Then $A^{\omega}$ is separable.
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(Consider the basic open sets of the product topology).
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Let $d \le 1$ be a complete metric on $A$.
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Define $D$ on $A^\omega$ by
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\[
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D\left( (x_n), (y_n) \right) \coloneqq
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\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
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\]
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Clearly $D \le 1$.
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It is also clear, that $D$ is a metric.
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We need to check that $D$ is complete:
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Let $(x_n)^{(k)}$ be a Cauchy sequence in $A^{\omega}$.
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Consider the pointwise limit $(a_n)$.
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This exists since $x_n^{(k)}$ is Cauchy for every fixed $n$.
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Then $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
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\end{proof}
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\begin{definition}[Our favourite Polish spaces]
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\leavevmode
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\begin{itemize}
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\item $2^{\omega}$ is called the \vocab{Cantor set}.
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(Consider $2$ with the discrete topology)
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\item $\omega^{\omega}$ is called the \vocab{Baire space}.
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($\omega$ with descrete topology)
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\item $[0,1]^{\omega}$ is called the \vocab{Hilbert cube}.
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($[0,1] \subseteq \R$ with the usual topology)
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\end{itemize}
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\end{definition}
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\begin{proposition}
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Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
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Then $X$ topologically embeds into the
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\vocab{Hilbert cube},
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i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$
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such that $f: X \to f(X)$ is a homeomorphism.
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\end{proposition}
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\begin{proof}
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\gist{$X$ is separable, so it has some countable dense subset,
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which we order as a sequence $(x_n)_{n \in \omega}$.}%
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{Let $(x_n)_{n \in \omega}$ be a countable dense subset.}
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\gist{Let $d$ be a metric on $X$ which is compatible with the topology.
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W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).}%
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{Let $d \le 1$ be a metric of $X$.}
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Define
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\begin{IEEEeqnarray*}{rCl}
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f\colon X &\longrightarrow & [0,1]^{\omega} \\
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x&\longmapsto & (d(x,x_n))_{n < \omega}
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\end{IEEEeqnarray*}
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\gist{
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\begin{claim}
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$f$ is injective.
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\end{claim}
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\begin{subproof}
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Suppose that $f(x) = f(y)$.
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Then $d(x,x_n) = d(y,y_n)$ for all $n$.
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Hence $d(x,y) \le d(x,x_n) + d(y,x_n) = 2 d(x,x_n)$.
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Since $(x_n)$ is dense, we get $d(x,y) = 0$.
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\end{subproof}
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\begin{claim}
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$f$ is continuous.
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\end{claim}
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\begin{subproof}
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Consider a basic open set in $[0,1]^{\omega}$,
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i.e. specify open sets $U_1, \ldots, U_n$ on finitely many coordinates.
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$f^{-1}(U_1 \times \ldots \times U_n \times \ldots)$
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is a finite intersection of open sets,
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hence it is open.
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\end{subproof}
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\begin{claim}
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$f^{-1}$ is continuous.
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\end{claim}
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\begin{subproof}
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Consider $B_{\epsilon}(x_n) \subseteq X$ for some $n \in \N$, $\epsilon > 0$.
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Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$
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is open\footnote{as a subset of $f(X)$!}.
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\end{subproof}
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}{}
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\end{proof}
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\begin{proposition}
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Countable disjoint unions of Polish spaces are Polish.
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\end{proposition}
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\gist{
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\begin{proof}
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Define a metric in the obvious way.
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\end{proof}
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}{}
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\begin{proposition}
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Closed subspaces of Polish spaces are Polish.
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\end{proposition}
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\gist{%
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\begin{proof}
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Let $X$ be Polish and $V \subseteq X$ closed.
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Let $d$ be a complete metric on $X$.
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Then $d\defon{V}$ is complete.
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Subspaces of second countable spaces
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are second countable.
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\end{proof}%
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}{}
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\begin{definition}
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Let $X$ be a topological space.
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A subspace $A \subseteq X$ is called
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$G_\delta$\footnote{Gebietdurchschnitt}, if it is a countable intersection of open sets.
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\end{definition}
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\gist{
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Next time: Closed sets are $G_\delta$.
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A subspace of a Polish space is Polish iff it is $G_{\delta}$
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}{}
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