w23-logic-3/inputs/lecture_01.tex
Josia Pietsch c986475c77
Some checks failed
Build latex and deploy / checkout (push) Failing after 1m4s
gist for lectures 1-4
2024-01-23 21:52:45 +01:00

276 lines
8.9 KiB
TeX

\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.
\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}
\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
\[
D\left( (x_n), (y_n) \right) \coloneqq
\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
\]
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 $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
\end{proof}
\begin{definition}[Our favourite Polish spaces]
\leavevmode
\begin{itemize}
\item $2^{\omega}$ is called the \vocab{Cantor set}.
(Consider $2$ with the discrete topology)
\item $\omega^{\omega}$ is called the \vocab{Baire space}.
($\omega$ with descrete topology)
\item $[0,1]^{\omega}$ 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 $G_{\delta}$
}{}