gist for lectures 1-4
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@ -58,7 +58,7 @@ However the converse of this does not hold.
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Take $x_0 \in X$ and consider the topology given by
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Take $x_0 \in X$ and consider the topology given by
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\[
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\[
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\tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}.
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\tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}.
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\]
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\]
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Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
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Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
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\end{example}
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\end{example}
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\begin{example}[Sorgenfrey line]
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\begin{example}[Sorgenfrey line]
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@ -76,7 +76,7 @@ However the converse of this does not hold.
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\end{itemize}
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\end{itemize}
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\end{fact}
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\end{fact}
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\begin{fact}
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\begin{fact}
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Compact\footnote{It is not clear whether compact means compact and Hausdorff in this lecture.} Hausdorff spaces are \vocab{normal} (T4)
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Compact Hausdorff spaces are \vocab{normal} (T4)
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i.e.~two disjoint closed subsets can be separated
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i.e.~two disjoint closed subsets can be separated
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by open sets.
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by open sets.
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\end{fact}
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\end{fact}
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@ -114,7 +114,7 @@ However the converse of this does not hold.
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\end{absolutelynopagebreak}
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\end{absolutelynopagebreak}
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\subsection{Some facts about polish spaces}
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\subsection{Some facts about polish spaces}
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\gist{%
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\begin{fact}
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\begin{fact}
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Let $(X, \tau)$ be a topological space.
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Let $(X, \tau)$ be a topological space.
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Let $d$ be a metric on $X$.
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Let $d$ be a metric on $X$.
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@ -130,9 +130,10 @@ To show that $\tau_d = \tau_{d'}$
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for two metrics $d, d'$,
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for two metrics $d, d'$,
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suffices to show that open balls in one metric are unions of open balls in the other.
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suffices to show that open balls in one metric are unions of open balls in the other.
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\end{fact}
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\end{fact}
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}{}
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\begin{notation}
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\begin{notation}
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We sometimes denote $\min(a,b)$ by $a \wedge b$.
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We sometimes\footnote{only in this subsection?} denote $\min(a,b)$ by $a \wedge b$.
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\end{notation}
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\end{notation}
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\begin{proposition}
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\begin{proposition}
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@ -142,18 +143,20 @@ suffices to show that open balls in one metric are unions of open balls in the o
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Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
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Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
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\end{proposition}
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\end{proposition}
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\gist{%
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\begin{proof}
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\begin{proof}
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To check the triangle inequality:
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To check the triangle inequality:
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right) \wedge 1\\
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d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right) \wedge 1\\
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&\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right).
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&\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right).
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$
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For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$
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and for $\epsilon > 1$, $B'_\epsilon(x) = X$.
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and for $\epsilon > 1$, $B'_\epsilon(x) = X$.
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Since $d$ is complete, we have that $d'$ is complete.
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Since $d$ is complete, we have that $d'$ is complete.
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\end{proof}
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\end{proof}
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}{}
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\begin{proposition}
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\begin{proposition}
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Let $A$ be a Polish space.
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Let $A$ be a Polish space.
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Then $A^{\omega}$ Polish.
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Then $A^{\omega}$ Polish.
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@ -164,11 +167,11 @@ suffices to show that open balls in one metric are unions of open balls in the o
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(Consider the basic open sets of the product topology).
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(Consider the basic open sets of the product topology).
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Let $d \le 1$ be a complete metric on $A$.
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Let $d \le 1$ be a complete metric on $A$.
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Define $D$ on $A^\omega$ by
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Define $D$ on $A^\omega$ by
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\[
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\[
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D\left( (x_n), (y_n) \right) \coloneqq
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D\left( (x_n), (y_n) \right) \coloneqq
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\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
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\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
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\]
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\]
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Clearly $D \le 1$.
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Clearly $D \le 1$.
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It is also clear, that $D$ is a metric.
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It is also clear, that $D$ is a metric.
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@ -194,7 +197,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
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Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
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Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
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Then $X$ topologically embeds into the
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Then $X$ topologically embeds into the
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\vocab{Hilbert cube},
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\vocab{Hilbert cube},
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i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$
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i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$
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such that $f: X \to f(X)$ is a homeomorphism.
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such that $f: X \to f(X)$ is a homeomorphism.
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\end{proposition}
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\end{proposition}
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\begin{proof}
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\begin{proof}
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@ -252,15 +255,15 @@ suffices to show that open balls in one metric are unions of open balls in the o
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\begin{proposition}
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\begin{proposition}
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Closed subspaces of Polish spaces are Polish.
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Closed subspaces of Polish spaces are Polish.
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\end{proposition}
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\end{proposition}
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\gist{}{
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\gist{%
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\begin{proof}
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\begin{proof}
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Let $X$ be Polish and $V \subseteq X$ closed.
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Let $X$ be Polish and $V \subseteq X$ closed.
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Let $d$ be a complete metric on $X$.
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Let $d$ be a complete metric on $X$.
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Then $d\defon{V}$ is complete.
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Then $d\defon{V}$ is complete.
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Subspaces of second countable spaces
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Subspaces of second countable spaces
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are second countable.
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are second countable.
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\end{proof}
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\end{proof}%
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}
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}{}
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\begin{definition}
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\begin{definition}
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Let $X$ be a topological space.
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Let $X$ be a topological space.
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@ -72,41 +72,51 @@
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\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
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\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
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metric is complete.
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metric is complete.
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$f_U$ is an embedding of $U$ into $X \times \R$\gist{:
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$f_U$ is an embedding of $U$ into $X \times \R$%
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\begin{itemize}
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\gist{:
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\item It is injective because of the first coordinate.
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\begin{itemize}
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\item It is continuous since $d(x, U^c)$ is continuous
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\item It is injective because of the first coordinate.
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and only takes strictly positive values. % TODO
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\item It is continuous since $d(x, U^c)$ is continuous
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\item The inverse is continuous because projections
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and only takes strictly positive values. % TODO
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are continuous.
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\item The inverse is continuous because projections
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\end{itemize}
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are continuous.
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}{.}
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\end{itemize}
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}{.}
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So we have shown that $U$ and
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\gist{%
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the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
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So we have shown that $U$ and
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are homeomorphic.
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the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
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The graph is closed \gist{in $U \times \R$,
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are homeomorphic.
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because $\tilde{f_U}$ is continuous.
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The graph is closed \gist{in $U \times \R$,
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It is closed}{} in $X \times \R$ \gist{because
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because $\tilde{f_U}$ is continuous.
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$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
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It is closed}{} in $X \times \R$ \gist{because
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\todo{Make this precise}
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$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
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\todo{Make this precise}
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Therefore we identified $U$ with a closed subspace of
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the Polish space $(X \times \R, d_1)$.
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}{%
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So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$
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and the RHS is a close subspace of the Polish space
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$(X \times \R, d_1)$.
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}
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Therefore we identified $U$ with a closed subspace of
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the Polish space $(X \times \R, d_1)$.
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\end{refproof}
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\end{refproof}
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Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
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Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
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Take
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Consider
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
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f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
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x &\longmapsto &
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x &\longmapsto &
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\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
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\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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As for an open $U$, $f_Y$ is an embedding.
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\gist{
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Since $X \times \R^{\N}$
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As for an open $U$, $f_Y$ is an embedding.
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is completely metrizable,
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Since $X \times \R^{\N}$
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so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
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is completely metrizable,
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so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
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}{}
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\begin{claim}
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\begin{claim}
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\label{psubspacegdelta:c2}
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\label{psubspacegdelta:c2}
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@ -123,36 +133,35 @@
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\item $\diam_d(U) \le \frac{1}{n}$,
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\item $\diam_d(U) \le \frac{1}{n}$,
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\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
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\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
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\end{enumerate}
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\end{enumerate}
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\gist{
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\gist{%
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We want to show that $Y = \bigcap_{n \in \N} V_n$.
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We want to show that $Y = \bigcap_{n \in \N} V_n$.
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For $x \in Y$, $n \in \N$ we have $x \in V_n$,
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For $x \in Y$, $n \in \N$ we have $x \in V_n$,
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as we can choose two neighbourhoods
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as we can choose two neighbourhoods
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$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
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$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
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such that $\diam_{d_Y}(U) < \frac{1}{n}$
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such that $\diam_{d_Y}(U) < \frac{1}{n}$
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and $U_2 \cap Y = U_1$.
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and $U_2 \cap Y = U_1$.
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Additionally choose $x \in U_3$ open in $X$
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Additionally choose $x \in U_3$ open in $X$
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with $\diam_{d}(U_3) < \frac{1}{n}$.
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with $\diam_{d}(U_3) < \frac{1}{n}$.
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Then consider $U_2 \cap U_3 \subseteq V_n$.
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Then consider $U_2 \cap U_3 \subseteq V_n$.
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Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
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Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
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Now let $x \in \bigcap_{n \in \N} V_n$.
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Now let $x \in \bigcap_{n \in \N} V_n$.
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For each $n$ pick $x \in U_n \subseteq X$ open
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For each $n$ pick $x \in U_n \subseteq X$ open
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satisfying (i), (ii), (iii).
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satisfying (i), (ii), (iii).
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From (i) and (ii) it follows that $x \in \overline{Y}$,
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From (i) and (ii) it follows that $x \in \overline{Y}$,
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since we can consider a sequence of points $y_n \in U_n \cap Y$
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since we can consider a sequence of points $y_n \in U_n \cap Y$
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and get $y_n \xrightarrow{d} x$.
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and get $y_n \xrightarrow{d} x$.
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For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
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For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
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is an open set containing $x$,
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is an open set containing $x$,
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hence $U_n' \cap Y \neq \emptyset$.
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hence $U_n' \cap Y \neq \emptyset$.
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Thus we may assume that the $U_i$ form a decreasing sequence.
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Thus we may assume that the $U_i$ form a decreasing sequence.
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We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
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We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
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If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
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If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
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since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
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since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
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and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
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and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
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The sequence $y_n$ converges to the unique point in
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The sequence $y_n$ converges to the unique point in
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$\bigcap_{n} \overline{U_n \cap Y}$.
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$\bigcap_{n} \overline{U_n \cap Y}$.
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Since the topologies agree, this point is $x$.
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Since the topologies agree, this point is $x$.
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}{Then $Y = \bigcap_n U_n$.}
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}{Then $Y = \bigcap_n U_n$.}
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\end{refproof}
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\end{refproof}
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\end{refproof}
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\end{refproof}
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@ -2,7 +2,7 @@
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\subsection{Trees}
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\subsection{Trees}
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\gist{%
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\begin{notation}
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\begin{notation}
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Let $A \neq \emptyset$, $n \in \N$.
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Let $A \neq \emptyset$, $n \in \N$.
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Then
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Then
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@ -34,7 +34,7 @@
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$s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$.
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$s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$.
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Let $s,t \in A^{<\N}$.
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Let $s,t \in A^{<\N}$.
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We say that $s$ is an \vocab{initial segment}
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We say that $s$ is an \vocab{initial segment}
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of $t$ (or $t$ is an \vocab{extension} of $s$)
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of $t$ (or $t$ is an \vocab{extension} of $s$)
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if there exists an $n$ such that $s = t\defon{|s|}$.
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if there exists an $n$ such that $s = t\defon{|s|}$.
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We write this as $s \subseteq t$.
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We write this as $s \subseteq t$.
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@ -44,8 +44,8 @@
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Otherwise the are \vocab{incompatible},
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Otherwise the are \vocab{incompatible},
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we denote that as $s \perp t$.
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we denote that as $s \perp t$.
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The \vocab{concatenation}
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The \vocab{concatenation}
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of $s = (s_0,\ldots, s_{n-1})$
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of $s = (s_0,\ldots, s_{n-1})$
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and $t = (t_0,\ldots, t_{m-1})$
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and $t = (t_0,\ldots, t_{m-1})$
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is the sequence
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is the sequence
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$s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$
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$s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$
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@ -59,10 +59,11 @@
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define extension, initial segments
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define extension, initial segments
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and concatenation of a finite sequence with an infinite one.
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and concatenation of a finite sequence with an infinite one.
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\end{notation}
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\end{notation}
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}{}
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\begin{definition}
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\begin{definition}
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A \vocab{tree}
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A \vocab{tree}
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on a set $A$ is a subset $T \subseteq A^{<\N}$
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on a set $A$ is a subset $T \subseteq A^{<\N}$
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closed under initial segments,
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closed under initial segments,
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i.e.~if $t \in T, s \subseteq t \implies s \in T$.
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i.e.~if $t \in T, s \subseteq t \implies s \in T$.
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Elements of trees are called \vocab{nodes}.
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Elements of trees are called \vocab{nodes}.
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@ -70,27 +71,27 @@
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A \vocab{leave} is an element of $T$,
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A \vocab{leave} is an element of $T$,
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that has no extension in $t$.
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that has no extension in $t$.
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An \vocab{infinite branch} of a tree $T$
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An \vocab{infinite branch} of a tree $T$
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is $x \in A^{\N}$
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is $x \in A^{\N}$
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such that $\forall n.~x\defon{n} \in T$.
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such that $\forall n.~x\defon{n} \in T$.
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The \vocab{body} of $T$ is the set of all
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The \vocab{body} of $T$ is the set of all
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infinite branches:
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infinite branches:
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\[
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\[
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[T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.
|
[T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.
|
||||||
\]
|
\]
|
||||||
|
|
||||||
We say that $T$ is \vocab{pruned},
|
We say that $T$ is \vocab{pruned},
|
||||||
iff
|
iff
|
||||||
\[
|
\[
|
||||||
\forall t\in T.\exists s \supsetneq t.~s \in T.
|
\forall t\in T.\exists s \supsetneq t.~s \in T.
|
||||||
\]
|
\]
|
||||||
\end{definition}
|
\end{definition}
|
||||||
|
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A \vocab{Cantor scheme}
|
A \vocab{Cantor scheme}
|
||||||
on a set $X$ is a family
|
on a set $X$ is a family
|
||||||
$(A_s)_{s \in 2^{< \N}}$
|
$(A_s)_{s \in 2^{< \N}}$
|
||||||
of subsets of $X$ such that
|
of subsets of $X$ such that
|
||||||
\begin{enumerate}[i)]
|
\begin{enumerate}[i)]
|
||||||
\item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
|
\item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
|
||||||
@ -99,7 +100,7 @@
|
|||||||
\end{definition}
|
\end{definition}
|
||||||
|
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A topological space
|
A topological space
|
||||||
is \vocab{perfect}
|
is \vocab{perfect}
|
||||||
if it has no isolated points,
|
if it has no isolated points,
|
||||||
i.e.~for any $U \neq \emptyset$ open,
|
i.e.~for any $U \neq \emptyset$ open,
|
||||||
@ -108,15 +109,15 @@
|
|||||||
|
|
||||||
\begin{theorem}
|
\begin{theorem}
|
||||||
\label{thm:cantortopolish}
|
\label{thm:cantortopolish}
|
||||||
Let $X \neq \emptyset$
|
Let $X \neq \emptyset$
|
||||||
be a perfect Polish space.
|
be a perfect Polish space.
|
||||||
Then there is an embedding
|
Then there is an embedding
|
||||||
of the Cantor space $2^{\N}$
|
of the Cantor space $2^{\N}$
|
||||||
into $X$.
|
into $X$.
|
||||||
\end{theorem}
|
\end{theorem}
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
We will define a Cantor scheme
|
We will define a Cantor scheme
|
||||||
$(U_s)_{s \in 2^{<\N}}$
|
$(U_s)_{s \in 2^{<\N}}$
|
||||||
such that $\forall s \in 2^{< \N}$.
|
such that $\forall s \in 2^{< \N}$.
|
||||||
\begin{enumerate}[(i)]
|
\begin{enumerate}[(i)]
|
||||||
\item $U_s \neq \emptyset$ and open,
|
\item $U_s \neq \emptyset$ and open,
|
||||||
@ -127,39 +128,46 @@
|
|||||||
|
|
||||||
We define $U_s$ inductively on the length of $s$.
|
We define $U_s$ inductively on the length of $s$.
|
||||||
|
|
||||||
For $U_{\emptyset}$ take any non-empty open set
|
\gist{%
|
||||||
with small enough diameter.
|
For $U_{\emptyset}$ take any non-empty open set
|
||||||
|
with small enough diameter.
|
||||||
|
|
||||||
Given $U_s$, pick $x \neq y \in U_s$
|
Given $U_s$, pick $x \neq y \in U_s$
|
||||||
and let $U_{s \concat 0} \ni x$,
|
and let $U_{s \concat 0} \ni x$,
|
||||||
$U_{s \concat 1} \ni y$
|
$U_{s \concat 1} \ni y$
|
||||||
be disjoint, open,
|
be disjoint, open,
|
||||||
of diameter $\le \frac{1}{2^{|s| +1}}$
|
of diameter $\le \frac{1}{2^{|s| +1}}$
|
||||||
and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
|
and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
|
||||||
|
}{}
|
||||||
|
|
||||||
|
\gist{%
|
||||||
Let $x \in 2^{\N}$.
|
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
|
such that
|
||||||
\[
|
\[
|
||||||
\{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}.
|
\{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}.
|
||||||
\]
|
\]
|
||||||
(This is nonempty as $X$ is a completely metrizable space.)
|
(This is nonempty as $X$ is a completely metrizable space.)
|
||||||
It is clear that $f$ is injective and continuous.
|
It is clear that $f$ is injective and continuous.
|
||||||
% TODO: more details
|
% TODO: more details
|
||||||
$2^{\N}$ is compact, hence $f^{-1}$ is also continuous.
|
$2^{\N}$ 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}
|
\end{proof}
|
||||||
|
|
||||||
\begin{corollary}
|
\begin{corollary}
|
||||||
\label{cor:perfectpolishcard}
|
\label{cor:perfectpolishcard}
|
||||||
Every nonempty perfect Polish
|
Every nonempty perfect Polish
|
||||||
space $X$ has cardinality $\fc = 2^{\aleph_0}$
|
space $X$ has cardinality $\fc = 2^{\aleph_0}$
|
||||||
% TODO: eulerscript C ?
|
|
||||||
\end{corollary}
|
\end{corollary}
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
Since the cantor space embeds into $X$,
|
\gist{%
|
||||||
we get the lower bound.
|
Since the cantor space embeds into $X$,
|
||||||
Since $X$ is second countable and Hausdorff,
|
we get the lower bound.
|
||||||
we get the upper 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}
|
\end{proof}
|
||||||
|
|
||||||
\begin{theorem}
|
\begin{theorem}
|
||||||
@ -173,13 +181,13 @@
|
|||||||
|
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A \vocab{Lusin scheme} on a set $X$
|
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 $
|
of subsets of $X $
|
||||||
such that
|
such that
|
||||||
\begin{enumerate}[(i)]
|
\begin{enumerate}[(i)]
|
||||||
\item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$
|
\item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$
|
||||||
for all $j \neq i \in \N$, $s \in \N^{<\N}$.
|
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 $i \in \N, s \in \N^{<\N}$.
|
for all $i \in \N, s \in \N^{<\N}$.
|
||||||
\end{enumerate}
|
\end{enumerate}
|
||||||
\end{definition}
|
\end{definition}
|
||||||
@ -191,11 +199,11 @@
|
|||||||
\[
|
\[
|
||||||
D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN
|
D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN
|
||||||
% TODO correct N for the Baire space?
|
% TODO correct N for the Baire space?
|
||||||
\]
|
\]
|
||||||
and a continuous bijection from
|
and a continuous bijection from
|
||||||
$D$ onto $X$ (the inverse does not need to be continuous).
|
$D$ 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 $f$.
|
extending $f$.
|
||||||
\end{theorem}
|
\end{theorem}
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
@ -203,12 +211,14 @@
|
|||||||
countable union of closed sets,
|
countable union of closed sets,
|
||||||
i.e.~the complement of a $G_\delta$ set.
|
i.e.~the complement of a $G_\delta$ set.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
|
\gist{%
|
||||||
\begin{observe}
|
\begin{observe}
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item Any open set is $F {\sigma}$.
|
\item Any open set is $F {\sigma}$.
|
||||||
\item In metric spaces the intersection of an open and closed set is $F_\sigma$.
|
\item In metric spaces the intersection of an open and closed set is $F_\sigma$.
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
\end{observe}
|
\end{observe}
|
||||||
|
}{}
|
||||||
\begin{refproof}{thm:bairetopolish}
|
\begin{refproof}{thm:bairetopolish}
|
||||||
Let $d$ be a complete metric on $X$.
|
Let $d$ be a complete metric on $X$.
|
||||||
W.l.o.g.~$\diam(X) \le 1$.
|
W.l.o.g.~$\diam(X) \le 1$.
|
||||||
@ -220,7 +230,7 @@
|
|||||||
\item $F_\emptyset = X$,
|
\item $F_\emptyset = X$,
|
||||||
\item $F_s$ is $F_\sigma$ for all $s$.
|
\item $F_s$ is $F_\sigma$ for all $s$.
|
||||||
\item The $F_{s \concat i}$ partition $F_s$,
|
\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
|
Furthermore we want that
|
||||||
$\overline{F_{s \concat i}} \subseteq F_s$
|
$\overline{F_{s \concat i}} \subseteq F_s$
|
||||||
@ -228,6 +238,7 @@
|
|||||||
\item $\diam(F_s) \le 2^{-|s|}$.
|
\item $\diam(F_s) \le 2^{-|s|}$.
|
||||||
\end{enumerate}
|
\end{enumerate}
|
||||||
|
|
||||||
|
\gist{%
|
||||||
Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
|
Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
|
||||||
We need to construct a partition $(F_i)_{i \in \N}$
|
We need to construct a partition $(F_i)_{i \in \N}$
|
||||||
of $F$ with $\overline{F_i} \subseteq F$
|
of $F$ with $\overline{F_i} \subseteq F$
|
||||||
@ -252,6 +263,7 @@
|
|||||||
The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
|
The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
|
||||||
are $F_\sigma$, disjoint
|
are $F_\sigma$, disjoint
|
||||||
and $F_i^0 = \bigcup_{j} D_j$.
|
and $F_i^0 = \bigcup_{j} D_j$.
|
||||||
|
}{Induction.}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
@ -5,6 +5,7 @@
|
|||||||
\end{remark}
|
\end{remark}
|
||||||
|
|
||||||
\begin{refproof}{thm:bairetopolish}
|
\begin{refproof}{thm:bairetopolish}
|
||||||
|
\gist{%
|
||||||
Take
|
Take
|
||||||
\[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
|
\[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
|
||||||
|
|
||||||
@ -12,16 +13,19 @@
|
|||||||
we have
|
we have
|
||||||
\[
|
\[
|
||||||
\bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
|
\bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
|
||||||
\]
|
\]
|
||||||
|
}{}
|
||||||
|
|
||||||
$f\colon D \to X$ is determined by
|
$f\colon D \to X$ is determined by
|
||||||
\[
|
\[
|
||||||
\{f(x)\} = \bigcap_{n} F_{x\defon{n}}
|
\{f(x)\} = \bigcap_{n} F_{x\defon{n}}
|
||||||
\]
|
\]
|
||||||
|
|
||||||
$f$ is injective and continuous.
|
\gist{%
|
||||||
The proof of this is exactly the same as in
|
$f$ is injective and continuous.
|
||||||
\yaref{thm:cantortopolish}.
|
The proof of this is exactly the same as in
|
||||||
|
\yaref{thm:cantortopolish}.
|
||||||
|
}{}
|
||||||
|
|
||||||
\begin{claim}
|
\begin{claim}
|
||||||
\label{thm:bairetopolish:c1}
|
\label{thm:bairetopolish:c1}
|
||||||
@ -47,7 +51,7 @@
|
|||||||
there exists $y = \lim_n f(x_n)$.
|
there exists $y = \lim_n f(x_n)$.
|
||||||
Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
|
Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
|
||||||
we get that $y \in \overline{F_{x\defon{N}}}$.
|
we get that $y \in \overline{F_{x\defon{N}}}$.
|
||||||
|
|
||||||
Note that for $N' > N$ by the same argument
|
Note that for $N' > N$ by the same argument
|
||||||
we get $y \in \overline{F_{x\defon{N'}}}$.
|
we get $y \in \overline{F_{x\defon{N'}}}$.
|
||||||
Hence
|
Hence
|
||||||
@ -55,20 +59,20 @@
|
|||||||
i.e.~$y \in D$ and $y = f(x)$.
|
i.e.~$y \in D$ and $y = f(x)$.
|
||||||
\end{refproof}
|
\end{refproof}
|
||||||
|
|
||||||
We extend $f$ to $g\colon\cN \to X$
|
We extend $f$ to $g\colon\cN \to X$
|
||||||
in the following way:
|
in the following way:
|
||||||
|
|
||||||
Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
|
Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
|
||||||
Clearly $S$ is a pruned tree.
|
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})
|
||||||
\[
|
\[
|
||||||
D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
|
D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
|
||||||
\]
|
\]
|
||||||
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).
|
(i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).
|
||||||
Then $g \coloneqq f \circ r$.
|
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
|
$\phi\colon \N^{<\N} \to S$ by induction on the length
|
||||||
such that
|
such that
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
@ -76,21 +80,27 @@
|
|||||||
\item $|s| = \phi(|s|)$,
|
\item $|s| = \phi(|s|)$,
|
||||||
\item if $s \in S$, then $\phi(s) = s$.
|
\item if $s \in S$, then $\phi(s) = s$.
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
Let $\phi(\emptyset) = \emptyset$.
|
\gist{%
|
||||||
Suppose that $\phi(t)$ is defined.
|
Let $\phi(\emptyset) = \emptyset$.
|
||||||
If $t\concat a \in S$, then set
|
Suppose that $\phi(t)$ is defined.
|
||||||
$\phi(t\concat a) \coloneqq t\concat a$.
|
If $t\concat a \in S$, then set
|
||||||
Otherwise take some $b$ such that
|
$\phi(t\concat a) \coloneqq t\concat a$.
|
||||||
$t\concat b \in S$ and define
|
Otherwise take some $b$ such that
|
||||||
$\phi(t\concat a) \coloneqq \phi(t)\concat b$.
|
$t\concat b \in S$ and define
|
||||||
|
$\phi(t\concat a) \coloneqq \phi(t)\concat b$.%
|
||||||
|
}{}%
|
||||||
This is possible since $S$ is pruned.
|
This is possible since $S$ is pruned.
|
||||||
|
|
||||||
Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
|
\gist{%
|
||||||
be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
|
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
|
$r$ is continuous, since
|
||||||
$d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz
|
$d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz
|
||||||
It is immediate that $r$ is a retraction.
|
\gist{%
|
||||||
|
It is immediate that $r$ is a retraction.
|
||||||
|
}{}
|
||||||
\end{refproof}
|
\end{refproof}
|
||||||
|
|
||||||
\section{Meager and Comeager Sets}
|
\section{Meager and Comeager Sets}
|
||||||
@ -110,42 +120,48 @@
|
|||||||
then $A \cap U$ is not dense in $U$).
|
then $A \cap U$ is not dense in $U$).
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
|
|
||||||
A set $B \subseteq X$ is \vocab{meager}
|
A set $B \subseteq X$ is \vocab{meager}
|
||||||
(or \vocab{first category}),
|
(or \vocab{first category}),
|
||||||
iff it is a countable union of nwd sets.
|
iff it is a countable union of nwd sets.
|
||||||
|
|
||||||
The complement of a meager set is called
|
The complement of a meager set is called
|
||||||
\vocab{comeager}.
|
\vocab{comeager}.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
|
\gist{%
|
||||||
\begin{example}
|
\begin{example}
|
||||||
$\Q \subseteq \R$ is meager.
|
$\Q \subseteq \R$ is meager.
|
||||||
\end{example}
|
\end{example}
|
||||||
|
}{}
|
||||||
\begin{notation}
|
\begin{notation}
|
||||||
Let $A, B \subseteq X$.
|
Let $A, B \subseteq X$.
|
||||||
We write $A =^\ast B$
|
We write $A =^\ast B$
|
||||||
iff the \vocab{symmetric difference},
|
iff the \vocab{symmetric difference},
|
||||||
$A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$,
|
$A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$,
|
||||||
is meager.
|
is meager.
|
||||||
\end{notation}
|
\end{notation}
|
||||||
|
\gist{%
|
||||||
\begin{remark}
|
\begin{remark}
|
||||||
$=^\ast$ is an equivalence relation.
|
$=^\ast$ is an equivalence relation.
|
||||||
\end{remark}
|
\end{remark}
|
||||||
|
}{}
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A set $A \subseteq X$
|
A set $A \subseteq X$
|
||||||
has the \vocab{Baire property} (\vocab{BP})
|
has the \vocab{Baire property} (\vocab{BP})
|
||||||
if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
|
if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
|
\gist{%
|
||||||
Note that open sets and meager sets have the Baire property.
|
Note that open sets and meager sets have the Baire property.
|
||||||
|
}{}
|
||||||
|
|
||||||
|
\gist{%
|
||||||
\begin{example}
|
\begin{example}
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item $\Q \subseteq \R$ is $F_\sigma$.
|
\item $\Q \subseteq \R$ is $F_\sigma$.
|
||||||
\item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
|
\item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
|
||||||
\item $\Q \subseteq \R$ is not $G_{\delta}$.
|
\item $\Q \subseteq \R$ is not $G_{\delta}$:
|
||||||
(It is dense and meager,
|
It is dense and meager,
|
||||||
hence it can not be $G_\delta$
|
hence it can not be $G_\delta$
|
||||||
by the Baire category theorem).
|
by the \yaref{thm:bct}.
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
|
|
||||||
\end{example}
|
\end{example}
|
||||||
|
}
|
||||||
|
@ -6,10 +6,9 @@
|
|||||||
\item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense.
|
\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.
|
\item Any meager set $B$ is contained in a meager $F_{\sigma}$-set.
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
|
|
||||||
|
|
||||||
\end{fact}
|
\end{fact}
|
||||||
\begin{proof} % remove?
|
\gist{%
|
||||||
|
\begin{proof}
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.
|
\item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.
|
||||||
\item Trivial.
|
\item Trivial.
|
||||||
@ -17,7 +16,9 @@
|
|||||||
Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.
|
Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
}{}
|
||||||
|
|
||||||
|
\gist{%
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A \vocab{$\sigma$-algebra} on a set $X$
|
A \vocab{$\sigma$-algebra} on a set $X$
|
||||||
is a collection of subsets of $X$
|
is a collection of subsets of $X$
|
||||||
@ -32,14 +33,15 @@
|
|||||||
Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$
|
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.
|
we have that $\sigma$-algebras are closed under countable intersections.
|
||||||
\end{fact}
|
\end{fact}
|
||||||
|
}{}
|
||||||
|
|
||||||
\begin{theorem}
|
\begin{theorem}
|
||||||
\label{thm:bairesigma}
|
\label{thm:bairesigma}
|
||||||
Let $X$ be a topological space.
|
Let $X$ be a topological space.
|
||||||
Then the collection of sets with the Baire property
|
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.
|
containing all meager and open sets.
|
||||||
\end{theorem}
|
\end{theorem}
|
||||||
\begin{refproof}{thm:bairesigma}
|
\begin{refproof}{thm:bairesigma}
|
||||||
@ -274,9 +276,11 @@ but for meager sets:
|
|||||||
% \end{refproof}
|
% \end{refproof}
|
||||||
% TODO fix claim numbers
|
% TODO fix claim numbers
|
||||||
|
|
||||||
|
\gist{%
|
||||||
\begin{remark}
|
\begin{remark}
|
||||||
Suppose that $A$ has the BP.
|
Suppose that $A$ has the BP.
|
||||||
Then there is an open $U$ such that
|
Then there is an open $U$ such that
|
||||||
$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
|
$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
|
||||||
Then $A = U \symdif M$.
|
Then $A = U \symdif M$.
|
||||||
\end{remark}
|
\end{remark}
|
||||||
|
}{}
|
||||||
|
@ -98,14 +98,13 @@ Let $X$ be a topological space.
|
|||||||
Then define
|
Then define
|
||||||
\[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\]
|
\[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\]
|
||||||
\[
|
\[
|
||||||
\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq
|
\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq
|
||||||
\{X \setminus A | A \in \Sigma^0_\alpha(X)\},
|
\{X \setminus A | A \in \Sigma^0_\alpha(X)\},
|
||||||
\]
|
\]
|
||||||
% \todo{Define $\lnot$ (element-wise complement)}
|
|
||||||
and for $\alpha > 1$
|
and for $\alpha > 1$
|
||||||
\[
|
\[
|
||||||
\Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.
|
\Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.
|
||||||
\]
|
\]
|
||||||
|
|
||||||
Note that $\Pi_1^0$ is the set of closed sets,
|
Note that $\Pi_1^0$ is the set of closed sets,
|
||||||
$\Sigma^0_2 = F_\sigma$,
|
$\Sigma^0_2 = F_\sigma$,
|
||||||
|
@ -16,17 +16,17 @@
|
|||||||
We have that
|
We have that
|
||||||
\[
|
\[
|
||||||
\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
|
\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
|
||||||
\]
|
\]
|
||||||
Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.
|
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
|
We have
|
||||||
\[
|
\[
|
||||||
\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).
|
\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).
|
||||||
\]
|
\]
|
||||||
Hence
|
Hence
|
||||||
\[
|
\[
|
||||||
|\cB(X)| \le \omega_1 \cdot \fc = \fc.
|
|\cB(X)| \le \omega_1 \cdot \fc = \fc.
|
||||||
\]
|
\]
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
|
||||||
\begin{proposition}[Closure properties]
|
\begin{proposition}[Closure properties]
|
||||||
@ -37,54 +37,58 @@
|
|||||||
\item \begin{itemize}
|
\item \begin{itemize}
|
||||||
\item $\Sigma^0_\xi(X)$ is closed under countable unions.
|
\item $\Sigma^0_\xi(X)$ is closed under countable unions.
|
||||||
\item $\Pi^0_\xi(X)$ is closed under countable intersections.
|
\item $\Pi^0_\xi(X)$ is closed under countable intersections.
|
||||||
\item $\Delta^0_\xi(X)$ is closed under complements,
|
\item $\Delta^0_\xi(X)$ is closed under complements.
|
||||||
countable unions and
|
|
||||||
countable intersections.
|
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
\item \begin{itemize}
|
\item \begin{itemize}
|
||||||
\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
|
\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
|
||||||
\item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
|
\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{itemize}
|
||||||
|
|
||||||
\end{enumerate}
|
\end{enumerate}
|
||||||
\end{proposition}
|
\end{proposition}
|
||||||
|
\gist{%
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
\begin{enumerate}[(a)]
|
\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
|
Note that a countable intersection can be written
|
||||||
as a complement of the countable union of complements:
|
as a complement of the countable union of complements:
|
||||||
\[
|
\[
|
||||||
\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
|
\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
|
||||||
\]
|
\]
|
||||||
\item If suffices to check this for $\Sigma^0_{\xi}(X)$.
|
\item If suffices to check this for $\Sigma^0_{\xi}(X)$.
|
||||||
Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
|
Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
|
||||||
and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
|
and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
|
||||||
Then
|
Then
|
||||||
\[
|
\[
|
||||||
A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)
|
A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)
|
||||||
\]
|
\]
|
||||||
and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
|
and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
|
||||||
|
|
||||||
\end{enumerate}
|
\end{enumerate}
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
}{}
|
||||||
\begin{example}
|
\begin{example}
|
||||||
Consider the cantor space $2^{\omega}$.
|
Consider the cantor space $2^{\omega}$.
|
||||||
We have that $\Delta^0_1(2^{\omega})$
|
We have that $\Delta^0_1(2^{\omega})$
|
||||||
is not closed under countable unions
|
is not closed under countable unions%
|
||||||
(countable unions yield all open sets, but there are open
|
\gist{ (countable unions yield all open sets, but there are open
|
||||||
sets that are not clopen).
|
sets that are not clopen)}{}.
|
||||||
\end{example}
|
\end{example}
|
||||||
|
|
||||||
\subsection{Turning Borels Sets into Clopens}
|
\subsection{Turning Borels Sets into Clopens}
|
||||||
|
|
||||||
\begin{theorem}%
|
\begin{theorem}%
|
||||||
|
\gist{%
|
||||||
\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
|
\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
|
||||||
unfortunately seems to be non-standard vocabulary.
|
unfortunately seems to be non-standard vocabulary.
|
||||||
Our tutor repeatedly advised against using it in the final exam.
|
Our tutor repeatedly advised against using it in the final exam.
|
||||||
Contrary to popular belief
|
Contrary to popular belief
|
||||||
the very same tutor was \textit{not} the one first to introduce it,
|
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.
|
as it would certainly be spelled ``to clopenise'' if that were the case.
|
||||||
}
|
}%
|
||||||
|
}{}%
|
||||||
\label{thm:clopenize}
|
\label{thm:clopenize}
|
||||||
Let $(X, \cT)$ be a Polish space.
|
Let $(X, \cT)$ be a Polish space.
|
||||||
For any Borel set $A \subseteq X$,
|
For any Borel set $A \subseteq X$,
|
||||||
@ -163,7 +167,7 @@
|
|||||||
such that $\cT_n \supseteq \cT$
|
such that $\cT_n \supseteq \cT$
|
||||||
and $\cB(\cT_n) = \cB(\cT)$.
|
and $\cB(\cT_n) = \cB(\cT)$.
|
||||||
Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
|
Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
|
||||||
is still Polish
|
is Polish
|
||||||
and $\cB(\cT_\infty) = \cB(T)$.
|
and $\cB(\cT_\infty) = \cB(T)$.
|
||||||
\end{lemma}
|
\end{lemma}
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
@ -183,7 +187,51 @@
|
|||||||
definition of $\cF$ belong to
|
definition of $\cF$ belong to
|
||||||
a countable basis of the respective $\cT_n$).
|
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}
|
\end{proof}
|
||||||
|
|
||||||
We need to show that $A$ is closed under countable unions.
|
We need to show that $A$ is closed under countable unions.
|
||||||
|
@ -1,61 +1,8 @@
|
|||||||
\lecture{08}{2023-11-10}{}
|
\lecture{08}{2023-11-10}{}\footnote{%
|
||||||
|
In the beginning of the lecture, we finished
|
||||||
\todo{put this lemma in the right place}
|
the proof of \yaref{thm:clopenize:l2}.
|
||||||
\begin{lemma}[Lemma 2]
|
This has been moved to the notes on lecture 7.%
|
||||||
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}
|
|
||||||
|
|
||||||
\subsection{Parametrizations}
|
\subsection{Parametrizations}
|
||||||
%\todo{choose better title}
|
%\todo{choose better title}
|
||||||
|
@ -27,8 +27,10 @@
|
|||||||
|
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A topological space is \vocab{Lindelöf}
|
A topological space is \vocab{Lindelöf}
|
||||||
if every open cover has a countable subcover.
|
iff every open cover has a countable subcover.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
|
|
||||||
|
|
||||||
\begin{fact}
|
\begin{fact}
|
||||||
Let $X$ be a metric space.
|
Let $X$ be a metric space.
|
||||||
If $X$ is Lindelöf,
|
If $X$ is Lindelöf,
|
||||||
@ -64,5 +66,12 @@
|
|||||||
and Lindelöf coincide.
|
and Lindelöf coincide.
|
||||||
|
|
||||||
In arbitrary topological spaces,
|
In arbitrary topological spaces,
|
||||||
Lindelöf is the strongest of these notions.
|
Lindelöf is the weakest of these notions.
|
||||||
\end{remark}
|
\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}
|
||||||
|
Loading…
Reference in New Issue
Block a user