gist for lectures 14
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@ 58,7 +58,7 @@ However the converse of this does not hold.


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]



@ 76,7 +76,7 @@ However the converse of this does not hold.


\end{itemize}


\end{fact}


\begin{fact}


Compact\footnote{It is not clear whether compact means compact and Hausdorff in this lecture.} Hausdorff spaces are \vocab{normal} (T4)


Compact Hausdorff spaces are \vocab{normal} (T4)


i.e.~two disjoint closed subsets can be separated


by open sets.


\end{fact}



@ 114,7 +114,7 @@ However the converse of this does not hold.


\end{absolutelynopagebreak}




\subsection{Some facts about polish spaces}




\gist{%


\begin{fact}


Let $(X, \tau)$ be a topological space.


Let $d$ be a metric on $X$.



@ 130,9 +130,10 @@ 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 denote $\min(a,b)$ by $a \wedge b$.


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


\end{notation}




\begin{proposition}



@ 142,18 +143,20 @@ suffices to show that open balls in one metric are unions of open balls in the o




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.



@ 164,11 +167,11 @@ suffices to show that open balls in one metric are unions of open balls in the o


(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


Define $D$ on $A^\omega$ by


\[


D\left( (x_n), (y_n) \right) \coloneqq


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.





@ 194,7 +197,7 @@ suffices to show that open balls in one metric are unions of open balls in the o


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 $f: X \hookrightarrow [0,1]^{\omega}$


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}



@ 252,15 +255,15 @@ suffices to show that open balls in one metric are unions of open balls in the o


\begin{proposition}


Closed subspaces of Polish spaces are Polish.


\end{proposition}


\gist{}{


\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}


}


\end{proof}%


}{}




\begin{definition}


Let $X$ be a topological space.





@ 72,41 +72,51 @@


\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + y_1  y_2\]


metric is complete.




$f_U$ is an embedding of $U$ into $X \times \R$\gist{:


\begin{itemize}


\item It is injective because of the first coordinate.


\item It is continuous since $d(x, U^c)$ is continuous


and only takes strictly positive values. % TODO


\item The inverse is continuous because projections


are continuous.


\end{itemize}


}{.}


$f_U$ is an embedding of $U$ into $X \times \R$%


\gist{:


\begin{itemize}


\item It is injective because of the first coordinate.


\item It is continuous since $d(x, U^c)$ is continuous


and only takes strictly positive values. % TODO


\item The inverse is continuous because projections


are continuous.


\end{itemize}


}{.}




So we have shown that $U$ and


the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$


are homeomorphic.


The graph is closed \gist{in $U \times \R$,


because $\tilde{f_U}$ is continuous.


It is closed}{} in $X \times \R$ \gist{because


$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.


\todo{Make this precise}


\gist{%


So we have shown that $U$ and


the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$


are homeomorphic.


The graph is closed \gist{in $U \times \R$,


because $\tilde{f_U}$ is continuous.


It is closed}{} in $X \times \R$ \gist{because


$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.


\todo{Make this precise}




Therefore we identified $U$ with a closed subspace of


the Polish space $(X \times \R, d_1)$.


}{%


So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$


and the RHS is a close subspace of the Polish space


$(X \times \R, d_1)$.


}




Therefore we identified $U$ with a closed subspace of


the Polish space $(X \times \R, d_1)$.


\end{refproof}




Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.


Take


Consider


\begin{IEEEeqnarray*}{rCl}


f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\


x &\longmapsto &


\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)


\end{IEEEeqnarray*}




As for an open $U$, $f_Y$ is an embedding.


Since $X \times \R^{\N}$


is completely metrizable,


so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.


\gist{


As for an open $U$, $f_Y$ is an embedding.


Since $X \times \R^{\N}$


is completely metrizable,


so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.


}{}




\begin{claim}


\label{psubspacegdelta:c2}



@ 123,36 +133,35 @@


\item $\diam_d(U) \le \frac{1}{n}$,


\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.


\end{enumerate}


\gist{


We want to show that $Y = \bigcap_{n \in \N} V_n$.


For $x \in Y$, $n \in \N$ we have $x \in V_n$,


as we can choose two neighbourhoods


$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,


such that $\diam_{d_Y}(U) < \frac{1}{n}$


and $U_2 \cap Y = U_1$.


Additionally choose $x \in U_3$ open in $X$


with $\diam_{d}(U_3) < \frac{1}{n}$.


Then consider $U_2 \cap U_3 \subseteq V_n$.


Hence $Y \subseteq \bigcap_{n \in \N} V_n$.


\gist{%


We want to show that $Y = \bigcap_{n \in \N} V_n$.


For $x \in Y$, $n \in \N$ we have $x \in V_n$,


as we can choose two neighbourhoods


$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,


such that $\diam_{d_Y}(U) < \frac{1}{n}$


and $U_2 \cap Y = U_1$.


Additionally choose $x \in U_3$ open in $X$


with $\diam_{d}(U_3) < \frac{1}{n}$.


Then consider $U_2 \cap U_3 \subseteq V_n$.


Hence $Y \subseteq \bigcap_{n \in \N} V_n$.




Now let $x \in \bigcap_{n \in \N} V_n$.


For each $n$ pick $x \in U_n \subseteq X$ open


satisfying (i), (ii), (iii).


From (i) and (ii) it follows that $x \in \overline{Y}$,


since we can consider a sequence of points $y_n \in U_n \cap Y$


and get $y_n \xrightarrow{d} x$.


For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$


is an open set containing $x$,


hence $U_n' \cap Y \neq \emptyset$.


Thus we may assume that the $U_i$ form a decreasing sequence.


We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.


If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,


since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$


and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.


The sequence $y_n$ converges to the unique point in


$\bigcap_{n} \overline{U_n \cap Y}$.


Since the topologies agree, this point is $x$.


}{Then $Y = \bigcap_n U_n$.}


Now let $x \in \bigcap_{n \in \N} V_n$.


For each $n$ pick $x \in U_n \subseteq X$ open


satisfying (i), (ii), (iii).


From (i) and (ii) it follows that $x \in \overline{Y}$,


since we can consider a sequence of points $y_n \in U_n \cap Y$


and get $y_n \xrightarrow{d} x$.


For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$


is an open set containing $x$,


hence $U_n' \cap Y \neq \emptyset$.


Thus we may assume that the $U_i$ form a decreasing sequence.


We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.


If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,


since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$


and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.


The sequence $y_n$ converges to the unique point in


$\bigcap_{n} \overline{U_n \cap Y}$.


Since the topologies agree, this point is $x$.


}{Then $Y = \bigcap_n U_n$.}


\end{refproof}


\end{refproof}







@ 2,7 +2,7 @@




\subsection{Trees}






\gist{%


\begin{notation}


Let $A \neq \emptyset$, $n \in \N$.


Then



@ 34,7 +34,7 @@


$s\defon{m} \coloneqq (s_0,\ldots,s_{m1})$.




Let $s,t \in A^{<\N}$.


We say that $s$ is an \vocab{initial segment}


We say that $s$ is an \vocab{initial segment}


of $t$ (or $t$ is an \vocab{extension} of $s$)


if there exists an $n$ such that $s = t\defon{s}$.


We write this as $s \subseteq t$.



@ 44,8 +44,8 @@


Otherwise the are \vocab{incompatible},


we denote that as $s \perp t$.




The \vocab{concatenation}


of $s = (s_0,\ldots, s_{n1})$


The \vocab{concatenation}


of $s = (s_0,\ldots, s_{n1})$


and $t = (t_0,\ldots, t_{m1})$


is the sequence


$s\concat t \coloneqq (s_0,\ldots,s_{n1}, t_0,\ldots, t_{n1})$



@ 59,10 +59,11 @@


define extension, initial segments


and concatenation of a finite sequence with an infinite one.


\end{notation}


}{}




\begin{definition}


A \vocab{tree}


on a set $A$ is a subset $T \subseteq A^{<\N}$


A \vocab{tree}


on a set $A$ is a subset $T \subseteq A^{<\N}$


closed under initial segments,


i.e.~if $t \in T, s \subseteq t \implies s \in T$.


Elements of trees are called \vocab{nodes}.



@ 70,27 +71,27 @@


A \vocab{leave} is an element of $T$,


that has no extension in $t$.




An \vocab{infinite branch} of a tree $T$


is $x \in A^{\N}$


An \vocab{infinite branch} of a tree $T$


is $x \in A^{\N}$


such that $\forall n.~x\defon{n} \in T$.




The \vocab{body} of $T$ is the set of all


infinite branches:


\[


[T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.


\]


\]




We say that $T$ is \vocab{pruned},


iff


\[


\forall t\in T.\exists s \supsetneq t.~s \in T.


\]


\]


\end{definition}




\begin{definition}


A \vocab{Cantor scheme}


on a set $X$ is a family


$(A_s)_{s \in 2^{< \N}}$


on a set $X$ is a family


$(A_s)_{s \in 2^{< \N}}$


of subsets of $X$ such that


\begin{enumerate}[i)]


\item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.



@ 99,7 +100,7 @@


\end{definition}




\begin{definition}


A topological space


A topological space


is \vocab{perfect}


if it has no isolated points,


i.e.~for any $U \neq \emptyset$ open,



@ 108,15 +109,15 @@




\begin{theorem}


\label{thm:cantortopolish}


Let $X \neq \emptyset$


Let $X \neq \emptyset$


be a perfect Polish space.


Then there is an embedding


of the Cantor space $2^{\N}$


of the Cantor space $2^{\N}$


into $X$.


\end{theorem}


\begin{proof}


We will define a Cantor scheme


$(U_s)_{s \in 2^{<\N}}$


$(U_s)_{s \in 2^{<\N}}$


such that $\forall s \in 2^{< \N}$.


\begin{enumerate}[(i)]


\item $U_s \neq \emptyset$ and open,



@ 127,39 +128,46 @@




We define $U_s$ inductively on the length of $s$.




For $U_{\emptyset}$ take any nonempty open set


with small enough diameter.


\gist{%


For $U_{\emptyset}$ take any nonempty open set


with small enough diameter.




Given $U_s$, pick $x \neq y \in U_s$


and let $U_{s \concat 0} \ni x$,


$U_{s \concat 1} \ni y$


be disjoint, open,


of diameter $\le \frac{1}{2^{s +1}}$


and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.


Given $U_s$, pick $x \neq y \in U_s$


and let $U_{s \concat 0} \ni x$,


$U_{s \concat 1} \ni y$


be disjoint, open,


of diameter $\le \frac{1}{2^{s +1}}$


and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.


}{}




\gist{%


Let $x \in 2^{\N}$.


Then let $f(x)$ be the unique point in $X$


Then let $f(x)$ be the unique point in $X$


such that


\[


\{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}.


\]


\]


(This is nonempty as $X$ is a completely metrizable space.)


It is clear that $f$ is injective and continuous.


% TODO: more details


$2^{\N}$ is compact, hence $f^{1}$ is also continuous.


}{Consider $f\colon 2^{\N} \hookrightarrow X, x \mapsto y$, where $\{y\} = \bigcap_n U_{x\defon n}$.


By compactness of $2^{\N}$, we get that $f^{1}$ is continuous.}


\end{proof}




\begin{corollary}


\label{cor:perfectpolishcard}


Every nonempty perfect Polish


space $X$ has cardinality $\fc = 2^{\aleph_0}$


% TODO: eulerscript C ?


space $X$ has cardinality $\fc = 2^{\aleph_0}$


\end{corollary}


\begin{proof}


Since the cantor space embeds into $X$,


we get the lower bound.


Since $X$ is second countable and Hausdorff,


we get the upper bound.


\gist{%


Since the cantor space embeds into $X$,


we get the lower bound.


Since $X$ is second countable and Hausdorff,


we get the upper bound.%


}{Lower bound: $2^{\N} \hookrightarrow X$,


upper bound: \nth{2} countable and Hausdorff.


\end{proof}




\begin{theorem}



@ 173,13 +181,13 @@




\begin{definition}


A \vocab{Lusin scheme} on a set $X$


is a family $(A_s)_{s \in \N^{<\N}}$


is a family $(A_s)_{s \in \N^{<\N}}$


of subsets of $X $


such that


\begin{enumerate}[(i)]


\item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$


for all $j \neq i \in \N$, $s \in \N^{<\N}$.


\item $A_{s \concat i} \subseteq A_s$


\item $A_{s \concat i} \subseteq A_s$


for all $i \in \N, s \in \N^{<\N}$.


\end{enumerate}


\end{definition}



@ 191,11 +199,11 @@


\[


D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN


% TODO correct N for the Baire space?


\]


\]


and a continuous bijection from


$D$ onto $X$ (the inverse does not need to be continuous).




Moreover there is a continuous surjection $g: \cN \to X$


Moreover there is a continuous surjection $g: \cN \to X$


extending $f$.


\end{theorem}


\begin{definition}



@ 203,12 +211,14 @@


countable union of closed sets,


i.e.~the complement of a $G_\delta$ set.


\end{definition}


\gist{%


\begin{observe}


\begin{itemize}


\item Any open set is $F {\sigma}$.


\item In metric spaces the intersection of an open and closed set is $F_\sigma$.


\end{itemize}


\end{observe}


}{}


\begin{refproof}{thm:bairetopolish}


Let $d$ be a complete metric on $X$.


W.l.o.g.~$\diam(X) \le 1$.



@ 220,7 +230,7 @@


\item $F_\emptyset = X$,


\item $F_s$ is $F_\sigma$ for all $s$.


\item The $F_{s \concat i}$ partition $F_s$,


i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$. % TODO change notation?


i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.




Furthermore we want that


$\overline{F_{s \concat i}} \subseteq F_s$



@ 228,6 +238,7 @@


\item $\diam(F_s) \le 2^{s}$.


\end{enumerate}




\gist{%


Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.


We need to construct a partition $(F_i)_{i \in \N}$


of $F$ with $\overline{F_i} \subseteq F$



@ 252,6 +263,7 @@


The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i1})$


are $F_\sigma$, disjoint


and $F_i^0 = \bigcup_{j} D_j$.


}{Induction.}











@ 5,6 +5,7 @@


\end{remark}




\begin{refproof}{thm:bairetopolish}


\gist{%


Take


\[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]





@ 12,16 +13,19 @@


we have


\[


\bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.


\]


\]


}{}




$f\colon D \to X$ is determined by


\[


\{f(x)\} = \bigcap_{n} F_{x\defon{n}}


\]


\]




$f$ is injective and continuous.


The proof of this is exactly the same as in


\yaref{thm:cantortopolish}.


\gist{%


$f$ is injective and continuous.


The proof of this is exactly the same as in


\yaref{thm:cantortopolish}.


}{}




\begin{claim}


\label{thm:bairetopolish:c1}



@ 47,7 +51,7 @@


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



@ 55,20 +59,20 @@


i.e.~$y \in D$ and $y = f(x)$.


\end{refproof}




We extend $f$ to $g\colon\cN \to X$


We extend $f$ to $g\colon\cN \to X$


in the following way:




Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.


Clearly $S$ is a pruned tree.


Moreover, since $D$ is closed, we have that\todo{Proof this (homework?)}


Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1})


\[


D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.


\]


We construct a \vocab{retraction} $r\colon\cN \to D$


\]


We construct a \vocab{retraction} $r\colon\cN \to D$


(i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).


Then $g \coloneqq f \circ r$.




To construct $r$, we will define




To construct $r$, we will define


$\phi\colon \N^{<\N} \to S$ by induction on the length


such that


\begin{itemize}



@ 76,21 +80,27 @@


\item $s = \phi(s)$,


\item if $s \in S$, then $\phi(s) = s$.


\end{itemize}


Let $\phi(\emptyset) = \emptyset$.


Suppose that $\phi(t)$ is defined.


If $t\concat a \in S$, then set


$\phi(t\concat a) \coloneqq t\concat a$.


Otherwise take some $b$ such that


$t\concat b \in S$ and define


$\phi(t\concat a) \coloneqq \phi(t)\concat b$.


\gist{%


Let $\phi(\emptyset) = \emptyset$.


Suppose that $\phi(t)$ is defined.


If $t\concat a \in S$, then set


$\phi(t\concat a) \coloneqq t\concat a$.


Otherwise take some $b$ such that


$t\concat b \in S$ and define


$\phi(t\concat a) \coloneqq \phi(t)\concat b$.%


}{}%


This is possible since $S$ is pruned.




Let $r\colon \cN = [\N^{<\N}] \to [S] = D$


be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.


\gist{%


Let $r\colon \cN = [\N^{<\N}] \to [S] = D$


be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.


}{}




$r$ is continuous, since


$d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz


It is immediate that $r$ is a retraction.


\gist{%


It is immediate that $r$ is a retraction.


}{}


\end{refproof}




\section{Meager and Comeager Sets}



@ 110,42 +120,48 @@


then $A \cap U$ is not dense in $U$).


\end{itemize}




A set $B \subseteq X$ is \vocab{meager}


A set $B \subseteq X$ is \vocab{meager}


(or \vocab{first category}),


iff it is a countable union of nwd sets.




The complement of a meager set is called


\vocab{comeager}.


\end{definition}


\gist{%


\begin{example}


$\Q \subseteq \R$ is meager.


\end{example}


}{}


\begin{notation}


Let $A, B \subseteq X$.


We write $A =^\ast B$


We write $A =^\ast B$


iff the \vocab{symmetric difference},


$A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$,


is meager.


\end{notation}


\gist{%


\begin{remark}


$=^\ast$ is an equivalence relation.


\end{remark}


}{}


\begin{definition}


A set $A \subseteq X$


has the \vocab{Baire property} (\vocab{BP})


if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.


\end{definition}


\gist{%


Note that open sets and meager sets have the Baire property.


}{}






\gist{%


\begin{example}


\begin{itemize}


\item $\Q \subseteq \R$ is $F_\sigma$.


\item $\R \setminus \Q \subseteq \R$ is $G_\delta$.


\item $\Q \subseteq \R$ is not $G_{\delta}$.


(It is dense and meager,


\item $\Q \subseteq \R$ is not $G_{\delta}$:


It is dense and meager,


hence it can not be $G_\delta$


by the Baire category theorem).


by the \yaref{thm:bct}.


\end{itemize}




\end{example}


}





@ 6,10 +6,9 @@


\item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense.


\item Any meager set $B$ is contained in a meager $F_{\sigma}$set.


\end{itemize}






\end{fact}


\begin{proof} % remove?


\gist{%


\begin{proof}


\begin{itemize}


\item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.


\item Trivial.



@ 17,7 +16,9 @@


Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.


\end{itemize}


\end{proof}


}{}




\gist{%


\begin{definition}


A \vocab{$\sigma$algebra} on a set $X$


is a collection of subsets of $X$



@ 32,14 +33,15 @@


Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$


we have that $\sigma$algebras are closed under countable intersections.


\end{fact}


}{}




\begin{theorem}


\label{thm:bairesigma}


Let $X$ be a topological space.


Then the collection of sets with the Baire property


is a $\sigma$algebra on $X$.


is \gist{a $\sigma$algebra on $X$.




It is the smallest $\sigma$algebra


It is}{} the smallest $\sigma$algebra


containing all meager and open sets.


\end{theorem}


\begin{refproof}{thm:bairesigma}



@ 274,9 +276,11 @@ but for meager sets:


% \end{refproof}


% TODO fix claim numbers




\gist{%


\begin{remark}


Suppose that $A$ has the BP.


Then there is an open $U$ such that


$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.


Then $A = U \symdif M$.


\end{remark}


}{}





@ 98,14 +98,13 @@ Let $X$ be a topological space.


Then define


\[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\]


\[


\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq


\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq


\{X \setminus A  A \in \Sigma^0_\alpha(X)\},


\]


% \todo{Define $\lnot$ (elementwise complement)}


\]


and for $\alpha > 1$


\[


\Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.


\]


\]




Note that $\Pi_1^0$ is the set of closed sets,


$\Sigma^0_2 = F_\sigma$,





@ 16,17 +16,17 @@


We have that


\[


\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.


\]


\]


Hence $\Sigma^0_\xi(X) \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.




We have


\[


\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).


\]


\]


Hence


\[


\cB(X) \le \omega_1 \cdot \fc = \fc.


\]


\]


\end{proof}




\begin{proposition}[Closure properties]



@ 37,54 +37,58 @@


\item \begin{itemize}


\item $\Sigma^0_\xi(X)$ is closed under countable unions.


\item $\Pi^0_\xi(X)$ is closed under countable intersections.


\item $\Delta^0_\xi(X)$ is closed under complements,


countable unions and


countable intersections.


\item $\Delta^0_\xi(X)$ is closed under complements.


\end{itemize}


\item \begin{itemize}


\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.


\item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.


\item $\Delta^0_\xi(X)$ is closed under finite unions and


finite intersections.


\end{itemize}




\end{enumerate}


\end{proposition}


\gist{%


\begin{proof}


\begin{enumerate}[(a)]


\item This follows directly from the definition.


\item This follows directly from the definition.


Note that a countable intersection can be written


as a complement of the countable union of complements:


\[


\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.


\]


\]


\item If suffices to check this for $\Sigma^0_{\xi}(X)$.


Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$


and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.


Then


\[


A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)


\]


\]


and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.




\end{enumerate}


\end{proof}


}{}


\begin{example}


Consider the cantor space $2^{\omega}$.


We have that $\Delta^0_1(2^{\omega})$


is not closed under countable unions


(countable unions yield all open sets, but there are open


sets that are not clopen).


is not closed under countable unions%


\gist{ (countable unions yield all open sets, but there are open


sets that are not clopen)}{}.


\end{example}




\subsection{Turning Borels Sets into Clopens}




\begin{theorem}%


\gist{%


\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''


unfortunately seems to be nonstandard vocabulary.


Our tutor repeatedly advised against using it in the final exam.


Contrary to popular belief


the very same tutor was \textit{not} the one first to introduce it,


as it would certainly be spelled ``to clopenise'' if that were the case.


}


}%


}{}%


\label{thm:clopenize}


Let $(X, \cT)$ be a Polish space.


For any Borel set $A \subseteq X$,



@ 163,7 +167,7 @@


such that $\cT_n \supseteq \cT$


and $\cB(\cT_n) = \cB(\cT)$.


Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$


is still Polish


is Polish


and $\cB(\cT_\infty) = \cB(T)$.


\end{lemma}


\begin{proof}



@ 183,7 +187,51 @@


definition of $\cF$ belong to


a countable basis of the respective $\cT_n$).




\todo{This proof will be finished in the next lecture}


% Proof was finished in lecture 8


Let $Y = \prod_{n \in \N} (X, \cT_n)$.


Then $Y$ is Polish.


Let $\delta\colon (X, \cT_\infty) \to Y$


defined by $\delta(x) = (x,x,x,\ldots)$.


\begin{claim}


$\delta$ is a homeomorphism.


\end{claim}


\begin{subproof}


Clearly $\delta$ is a bijection.


We need to show that it is continuous and open.




Let $U \in \cT_i$.


Then


\[


\delta^{1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,


\]


hence $\delta$ is continuous.


Let $U \in \cT_\infty$.


Then $U$ is the union of sets of the form


\[


V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}


\]


for some $n_1 < n_2 < \ldots < n_u$


and $U_{n_i} \in \cT_i$.




Thus is suffices to consider sets of this form.


We have that


\[


\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.


\]


\end{subproof}




This will finish the proof since


\[


D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y


\]


Why? Let $(x_n) \in Y \setminus D$.


Then there are $i < j$ such that $x_i \neq x_j$.


Take disjoint open $x_i \in U$, $x_j \in V$.


Then


\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]


is open in $Y\setminus D$.


Hence $Y \setminus D$ is open, thus $D$ is closed.


It follows that $D$ is Polish.


\end{proof}




We need to show that $A$ is closed under countable unions.





@ 1,61 +1,8 @@


\lecture{08}{20231110}{}




\todo{put this lemma in the right place}


\begin{lemma}[Lemma 2]


Let $(X, \cT)$ be a Polish space.


Let $\cT_n \supseteq \cT$ be Polish


with $\cB(X, \cT_n) = \cB(X, \cT)$.


Let $\cT_\infty$ be the topology generated


by $\bigcup_n \cT_n$.


Then $(X, \cT_\infty)$ is Polish


and $\cB(X, \cT_\infty) = \cB(X, \cT)$.


\end{lemma}


\begin{proof}


Let $Y = \prod_{n \in \N} (X, \cT_n)$.


Then $Y$ is Polish.


Let $\delta\colon (X, \cT_\infty) \to Y$


defined by $\delta(x) = (x,x,x,\ldots)$.


\begin{claim}


$\delta$ is a homeomorphism.


\end{claim}


\begin{subproof}


Clearly $\delta$ is a bijection.


We need to show that it is continuous and open.




Let $U \in \cT_i$.


Then


\[


\delta^{1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,


\]


hence $\delta$ is continuous.


Let $U \in \cT_\infty$.


Then $U$ is the union of sets of the form


\[


V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}


\]


for some $n_1 < n_2 < \ldots < n_u$


and $U_{n_i} \in \cT_i$.




Thus is suffices to consider sets of this form.


We have that


\[


\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.


\]


\end{subproof}




This will finish the proof since


\[


D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y


\]


Why? Let $(x_n) \in Y \setminus D$.


Then there are $i < j$ such that $x_i \neq x_j$.


Take disjoint open $x_i \in U$, $x_j \in V$.


Then


\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]


is open in $Y\setminus D$.


Hence $Y \setminus D$ is open, thus $D$ is closed.


It follows that $D$ is Polish.


\end{proof}


\lecture{08}{20231110}{}\footnote{%


In the beginning of the lecture, we finished


the proof of \yaref{thm:clopenize:l2}.


This has been moved to the notes on lecture 7.%


}




\subsection{Parametrizations}


%\todo{choose better title}





@ 27,8 +27,10 @@




\begin{definition}


A topological space is \vocab{Lindelöf}


if every open cover has a countable subcover.


iff every open cover has a countable subcover.


\end{definition}






\begin{fact}


Let $X$ be a metric space.


If $X$ is Lindelöf,



@ 64,5 +66,12 @@


and Lindelöf coincide.




In arbitrary topological spaces,


Lindelöf is the strongest of these notions.


Lindelöf is the weakest of these notions.


\end{remark}




\begin{definition}+


A metric space $X$ is \vocab{totally bounded}


iff for every $\epsilon > 0$ there exists


a finite set of points $x_1,\ldots,x_n$


such that $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$.


\end{definition}




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