tutorial 10

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@ -61,7 +61,9 @@ However the converse of this does not hold.
Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
\end{example}
\begin{example}[Sorgenfrey line]
\todo{Counterexamples in Topology}
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}
@ -98,6 +100,7 @@ However the converse of this does not hold.
then $X$ is metrisable.
\end{theorem}
\begin{absolutelynopagebreak}
\begin{fact}
If $X$ is a compact Hausdorff space,
the following are equivalent:
@ -107,6 +110,7 @@ However the converse of this does not hold.
\item $X$ is second countable.
\end{itemize}
\end{fact}
\end{absolutelynopagebreak}
\subsection{Some facts about polish spaces}
@ -175,6 +179,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
\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)
@ -185,7 +190,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
\end{itemize}
\end{definition}
\begin{proposition}
Let $X$ be a separable, metrisable topological space.
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 $f: X \hookrightarrow [0,1]^{\omega}$
@ -227,7 +232,9 @@ suffices to show that open balls in one metric are unions of open balls in the o
$f^{-1}$ is continuous.
\end{claim}
\begin{subproof}
\todo{Exercise!}
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}

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@ -13,7 +13,6 @@
\begin{proof}
Let $C \subseteq X$ be closed.
Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
\todo{Exercise}
Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
Let $x \in \bigcap U_{\frac{1}{n}}$.
Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
@ -54,7 +53,7 @@
then there exists a complete metric on $Y$.
\end{claim}
\begin{refproof}{psubspacegdelta:c1}
Let $Y = U$be open in $X$.
Let $Y = U$ be open in $X$.
Consider the map
\begin{IEEEeqnarray*}{rCl}
f_U\colon U &\longrightarrow &

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@ -1,6 +1,6 @@
\lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}
% ? \subsection{Trees} TODO
\subsection{Trees}
\begin{notation}
@ -255,9 +255,7 @@
[To be continued]
\phantom\qedhere
\end{refproof}

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@ -1,6 +1,7 @@
\lecture{04}{2023-10-20}{}
\begin{remark}
Some of $F_s$ might be empty.
Some of the $F_s$ might be empty.
\end{remark}
\begin{refproof}{thm:bairetopolish}
@ -29,7 +30,6 @@
\begin{refproof}{thm:bairetopolish:c1}
Let $x_n$ be a series in $D$
converging to $x$ in $\cN$.
Then $x \in \cN$.
\begin{claim}
$(f(x_n))$ is Cauchy.
\end{claim}
@ -40,21 +40,19 @@
$x_m\defon{N} = x\defon{N}$.
Then for all $m, n \ge M$,
we have that $f(x_m), f(x_n) \in F_{x\defon{N}}$.
So $d(f(x_m), f(x_n)) < \epsilon$
we have that $(f(x_n))$ is Cauchy.
Since $(X,d)$ is complete,
there exists $y = \lim_n f(x_n)$.
So $d(f(x_m), f(x_n)) < \epsilon$, i.e.~$(f(x_n))$ is Cauchy.
Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
we get that $y \in \overline{F_{x\defon{N}}}$.
Note that for $N' > N$ by the same argument
we get $y \in \overline{F_{x\defon{N'}}}$.
Hence
\[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
i.e.~$y \in D$ and $y = f(x)$.
\end{subproof}
Since $(X,d)$ is complete,
there exists $y = \lim_n f(x_n)$.
Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
we get that $y \in \overline{F_{x\defon{N}}}$.
Note that for $N' > N$ by the same argument
we get $y \in \overline{F_{x\defon{N'}}}$.
Hence
\[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
i.e.~$y \in D$ and $y = f(x)$.
\end{refproof}
We extend $f$ to $g\colon\cN \to X$
@ -70,7 +68,7 @@
(i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).
Then $g \coloneqq f \circ r$.
To construct $r$, we will define by induction
To construct $r$, we will define
$\phi\colon \N^{<\N} \to S$ by induction on the length
such that
\begin{itemize}

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@ -13,9 +13,9 @@
\begin{fact}
\begin{itemize}
\item Let $X$ be a topological space.
Then $X$ 2nd countable $\implies$ X separable.
Then $X$ \nth{2} countable $\implies$ X separable.
\item If $X$ is a metric space and separable,
then $X$ is 2nd countable.
then $X$ is \nth{2} countable.
\end{itemize}
\end{fact}
\begin{proof}
@ -32,18 +32,18 @@
\begin{fact}
Let $X$ be a metric space.
If $X$ is Lindelöf,
then it is 2nd countable.
then it is \nth{2} countable.
\end{fact}
\begin{proof}
For all $q \in \Q$
Consider the cover $B_q(x), x \in X$
consider the cover $B_q(x), x \in X$
and choose a countable subcover.
The union of these subcovers is
a countable base.
\end{proof}
\begin{fact}
Let $X$ be a topological space.
If $X$ is 2nd countable,
If $X$ is \nth{2} countable,
then it is Lindelöff.
\end{fact}
\begin{proof}
@ -60,7 +60,7 @@
\end{proof}
\begin{remark}
For metric spaces the notions
of being 2nd countable, separable
of being \nth{2} countable, separable
and Lindelöf coincide.
In arbitrary topological spaces,

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@ -50,3 +50,13 @@
\RequirePackage{mkessler-mathfixes} % Load this last since it renews behaviour
\DeclareMathOperator{\inter}{int} % interior
\newcommand{\defon}[1]{|_{#1}} % TODO
\RequirePackage[super]{nth}
% TODO MOVE
% https://tex.stackexchange.com/a/94702
\newenvironment{absolutelynopagebreak}
{\par\nobreak\vfil\penalty0\vfilneg
\vtop\bgroup}
{\par\xdef\tpd{\the\prevdepth}\egroup
\prevdepth=\tpd}

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@ -61,6 +61,7 @@
\input{inputs/tutorial_07}
\input{inputs/tutorial_08}
\input{inputs/tutorial_09}
\input{inputs/tutorial_10}
\section{Facts}
\input{inputs/facts}