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Josia Pietsch 2024-01-17 01:15:13 +01:00
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@ -58,7 +58,7 @@
\end{proof} \end{proof}
\subsection{Parametrizations} \subsection{Parametrizations}
\todo{choose better title} %\todo{choose better title}
Let $\Gamma$ denote a collection of sets in some space. Let $\Gamma$ denote a collection of sets in some space.
@ -109,9 +109,9 @@ where $X$ is a metrizable, usually second countable space.
put $(y,x) \in \cU$ iff put $(y,x) \in \cU$ iff
$x \in \bigcup \{V_n : y_n = 1\}$. $x \in \bigcup \{V_n : y_n = 1\}$.
$\cU$ is open. $\cU$ is open.
Let $V = \bigcup \{V_n : V_n \subseteq V\}$. For any $V \overset{\text{open}}{\subseteq} X$,
Pick $y \in 2^\omega$ define $y \in 2^\omega$
and let $y_n = 1$ iff $V_n \subseteq V$. by $y_n = 1$ iff $V_n \subseteq V$.
Then $\cU_y = V$. Then $\cU_y = V$.
@ -125,7 +125,7 @@ where $X$ is a metrizable, usually second countable space.
Recall that $\eta_1 \le \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$. Recall that $\eta_1 \le \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$.
Note that if $A = \bigcup_n A_n$, with $A_n \in \Pi^0_{\eta_n}(X)$ Note that if $A = \bigcup_n A_n$, with $A_n \in \Pi^0_{\eta_n}(X)$
some $\eta_n < \xi$, for some $\eta_n < \xi$,
we also have we also have
$A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$. $A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$.
@ -146,7 +146,7 @@ where $X$ is a metrizable, usually second countable space.
Since $2^{\omega}$ embeds Since $2^{\omega}$ embeds
into any uncountable polish space $Y$ into any uncountable polish space $Y$
such that the image is closed, such that the image is closed,
we can $2^{\omega}$ by $Y$ we can replace $2^{\omega}$ by $Y$
in the statement of the theorem.% in the statement of the theorem.%
\footnote{By definition of the subspace topology \footnote{By definition of the subspace topology
and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.} and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}