Josia Pietsch 6c2a76d838
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improved lecture 17
2024-02-05 20:30:13 +01:00

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Let $X$ be an uncountable Polish space.
Then for all $\xi < \omega_1$,
we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$.
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 $\{y \in X : (y,y) \not\in \cU\} \in \Pi^0_\xi(X) \setminus \Sigma^0_\xi(X)$.
Let $X$ be a Polish space.
A set $A \subseteq X$
is called \vocab{analytic}
\exists Y \text{ Polish}.~\exists B \in \cB(Y).~
\exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~
f(B) = A.
Trivially, every Borel set is analytic.
We will see that not every analytic set is Borel.
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}}
Let $X$ be Polish,
$\emptyset \neq A \subseteq X$.
Then the following are equivalent:
\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 $A = \proj_X(B)$.
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 $\cN$.
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 $f \coloneqq \proj_X$.
Let $X,Y$ be Polish spaces.
Let $f\colon X \to Y$ be \vocab{Borel}
(i.e.~preimages of open sets are Borel).
\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.
\item Let $A \subseteq X$ analytic.
Then there exists $Z$ Polish and $g\colon Z \to X$
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 $f^+\colon X \times Z \to Y \times Z, f^+ = f \times \id$.
\[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
\item See \yaref{s7e2}.
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.
We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
Let $X,Y$ be uncountable Polish spaces.
There exists a $Y$-universal $\Sigma^1_1(X)$ set.
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 $\Sigma^1_1(X)$:
\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 $\cV_y = A$.
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.
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}.