Josia Pietsch
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\lecture{09}{20231114}{}




\begin{theorem}


Let $X$ be an uncountable Polish space.


Then for all $\xi < \omega_1$,


we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$.


\end{theorem}


\begin{proof}


\gist{%


Fix $\xi < \omega_1$.


Towards a contradiction assume $\Sigma^0_\xi(X) = \Pi^0_\xi(X)$.


By \autoref{thm:cantoruniversal},


there is a $X$universal set $\cU$ for $\Sigma^0_\xi(X)$.




Take $A \coloneqq \{y \in X : (y,y) \not\in \cU\}$.


Then $A \in \Pi^0_\xi(X)$.\footnote{cf.~\yaref{s7e1} and use that $\{(x,x) \in X^2\} \cong X$.}


By assumption $A \in \Sigma^0_\xi(X)$,


i.e.~there exists some $z \in X$ such that $A = \cU_z$.


We have


\[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\]


But by the definition of $A$,


we have $z \in A \iff (z,z) \not\in \cU \lightning$.


}{%


Let $\cU$ be $X$universal for $\Sigma^0_\xi(X)$.


Consider $\{y \in X : (y,y) \not\in \cU\} \in \Pi^0_\xi(X) \setminus \Sigma^0_\xi(X)$.


}


\end{proof}






\begin{definition}


Let $X$ be a Polish space.


A set $A \subseteq X$


is called \vocab{analytic}


iff


\[


\exists Y \text{ Polish}.~\exists B \in \cB(Y).~


\exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~


f(B) = A.


\]


\end{definition}


\gist{%


Trivially, every Borel set is analytic.


We will see that not every analytic set is Borel.


}{}


\begin{remark}


In the definition we can replace the assertion that


$f$ is continuous


by the weaker assertion of $f$ being Borel.%


\footnote{use \yaref{thm:clopenize}, cf.~\yaref{s6e2}}


\end{remark}




\begin{theorem}


\label{thm:borel}


Let $X$ be Polish,


$\emptyset \neq A \subseteq X$.


Then the following are equivalent:


\begin{enumerate}[(i)]


\item $A$ is analytic.


\item There exists a Polish space $Y$


and $f\colon Y \to X$


continuous\footnote{or Borel}


such that $A = f(Y)$.


\item There exists $h\colon \cN \to X$


continuous with $h(\cN) = A$.


\item There is $F \overset{\text{closed}}{\subseteq} X \times \cN$


such that $A = \proj_X(F)$.


\item There is a Borel set $B \subseteq X \times Y$


for some Polish space $Y$,


such that $A = \proj_X(B)$.


\end{enumerate}


\end{theorem}


\begin{proof}


\gist{%


To show (i) $\implies$ (ii):


take $B \in \cB(Y')$


and $f\colon Y' \to X$


continuous with $f(B) = A$.


Take a finer Polish topology $\cT$ on $Y'$


adding no Borel sets,


such that $B$ is clopen with respect to the new topology.


Then let $g = f\defon{B}$


and $Y = (B, \cT\defon{B})$.


}{(i) $\implies$ (ii):


Clopenize the Borel set, then restrict.


}




(ii) $\implies$ (iii):


Any Polish space is the continuous image of $\cN$.


\gist{%


Let $g_1: \cN \to Y$


and $h \coloneqq g \circ g_1$.


}{}




(iii) $\implies$ (iv):


Let $h\colon \cN \to X$ with $h(\cN) = A$.


Let $G(h) \coloneqq \{(a,b) : h(a) = b\} \overset{\text{closed}}{\subseteq} \cN \times X$


be the \vocab{graph} of $h$.


Take $F \coloneqq G(h)^{1} \coloneqq \{(c,d)  (d,c) \in G(h)\}$




Clearly (iv) $\implies$ (v).




(v) $\implies$ (i):


Take $f \coloneqq \proj_X$.


\end{proof}






\begin{theorem}


Let $X,Y$ be Polish spaces.


Let $f\colon X \to Y$ be \vocab{Borel}


(i.e.~preimages of open sets are Borel).


\begin{enumerate}[(a)]


\item The image of an analytic set is analytic.


\item The preimage of an analytic set is analytic.


\item Analytic sets are closed under countable unions


and countable intersections.


\end{enumerate}


\end{theorem}


\begin{proof}


\begin{enumerate}[(a)]


\item Let $A \subseteq X$ analytic.


Then there exists $Z$ Polish and $g\colon Z \to X$


continuous


with $g(Z) = A$.


We have that $f(A) = (f \circ g)(Z)$


and $f \circ g$ is Borel.


\item Let $f\colon X \to Y$ be Borel


and $B \subseteq Y$ analytic.




Take $Z$ Polish


and $B_0 \subseteq Y \times Z$


such that $\proj_Y(B_0) = B$.


Take $f^+\colon X \times Z \to Y \times Z, f^+ = f \times \id$.


Then


\[f^{1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]


\item See \yaref{s7e2}.


\end{enumerate}


\end{proof}




\begin{notation}


Let $X$ be Polish.


Let $\Sigma^1_1(X)$ denote the set of all analytic


subsets of $X$.


$\Pi^1_1(X) \coloneqq \{B \subseteq X : X \setminus B \in \Sigma^1_1(X)\}$


is the set of \vocab{coanalytic} sets.


\end{notation}


We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.






\begin{theorem}


\label{thm:universals11}


Let $X,Y$ be uncountable Polish spaces.


There exists a $Y$universal $\Sigma^1_1(X)$ set.


\end{theorem}


\begin{proof}


Take $\cU \subseteq Y \times X \times \cN$


which is $Y$universal for $\Pi^0_1(X \times \cN)$.


Let $\cV \coloneqq \proj_{Y \times X}(\cU)$.


Then $\cV$ is $Y$universal for $\Sigma^1_1(X)$:


\begin{itemize}


\item $\cV \in \Sigma^1_1(Y \times X)$


since $\cV$ is a projection of a closed set.


\item All sections of $\cV$ are analytic.


Let $A \in \Sigma^1_1(X)$.


Let $C \subseteq X \times \cN$


be closed such that $\proj_X(C) = A$.


There is $y \in Y$


such that $\cU_y = C$,


hence $\cV_y = A$.


\end{itemize}


\end{proof}


\begin{remark}


In the same way that we proved


$\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$ for $\xi < \omega_1$,


we obtain that $\Sigma^1_1(X) \neq \Pi^1_1(X)$.




In fact if $\cU$ is universal for $\Sigma^1_1(X)$,


then $\{y : (y,y) \in \cU\} \in \Sigma^1_1(X) \setminus \Pi^1_1(X)$.


In particular, this set is not Borel.


\end{remark}






\begin{remark}+


Showing that there exist sets that don't have the Baire property


requires the axiom of choice.


An example of such a set is constructed in \yaref{s5e4}.


\end{remark}
