From 6423bfc8e832c6925074bee71d69b54b862ecf1b Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Tue, 15 Feb 2022 21:08:55 +0100
Subject: Poprawki banacha mazura

---
 lic_malinka.tex | 110 ++++++++++++++++++++++++++++++++++----------------------
 1 file changed, 68 insertions(+), 42 deletions(-)

(limited to 'lic_malinka.tex')

diff --git a/lic_malinka.tex b/lic_malinka.tex
index b1906c8..24e3008 100644
--- a/lic_malinka.tex
+++ b/lic_malinka.tex
@@ -124,7 +124,7 @@
     \end{definition}
     
     % \begin{example}
-    Every countable set is nowhere dense in any $T_1$ space, so, for example, $\bQ$ is meagre in $\bR$ (though being dense), which means that the set of irrationals is comeagre. Another example is...
+    Every countable set is meagre in any $T_1$ space, so, for example, $\bQ$ is meagre in $\bR$ (though being dense), which means that the set of irrationals is comeagre. Another example is...
     % \end{example}
 
     \begin{definition}
@@ -136,54 +136,75 @@
     \end{definition}
     
     \begin{definition}
-        Let $X$ be a nonempty topological space and let $A\subseteq X$. The \emph{Banach-Mazur game of $A$}, denoted as $G^{\star\star}(A)$ is defined as follows: Players $I$ and $II$ take turns in playing nonempty open sets $U_0, V_0, U_1, V_1,\ldots$ such that $U_0 \supseteq V_0 \supseteq U_1 \supseteq V_1 \supseteq\ldots$. We say that player $II$ wins the game if $\bigcap_{n}V_n \subseteq A$.
+        Let $X$ be a nonempty topological space and let $A\subseteq X$. The \emph{Banach-Mazur game of $A$}, denoted as $G^{\star\star}(A)$ is defined as follows: Players $I$ and $\textit{II}$ take turns in playing nonempty open sets $U_0, V_0, U_1, V_1,\ldots$ such that $U_0 \supseteq V_0 \supseteq U_1 \supseteq V_1 \supseteq\ldots$. We say that player $\textit{II}$ wins the game if $\bigcap_{n}V_n \subseteq A$.
     \end{definition}
 
     There is an important theorem on the Banach-Mazur game: $A$ is comeagre
-    iff $II$ can always choose sets $V_0, V_1, \ldots$ such that it wins. Before we prove it we need to define notions necessary to formalise and prove the theorem.
+    iff $\textit{II}$ can always choose sets $V_0, V_1, \ldots$ such that it wins. Before we prove it we need to define notions necessary to formalise and prove the theorem.
     
     \begin{definition}
         $T$ is \emph{the tree of all legal positions} in the Banach-Mazur game $G^{\star\star}(A)$ when $T$ consists of all finite sequences $(W_0, W_1,\ldots, W_n)$, where $W_i$ are nonempty open sets such that $W_0\supseteq W_1\supseteq\ldots\supseteq W_n$. In another words, $T$ is a pruned tree on $\{W\subseteq X\mid W \textrm{is open nonempty}\}$. 
     
-        By $[T]$ we denote the set of all "infinite branches" of $T$, i.e. infinite sequences $(U_0, V_0, \ldots)$ such that $(U_0, V_0, \ldots U_n, V_n)\in T$ for any $n\in \bN$.
+    \end{definition}
+    
+    \begin{definition}
+        We say that $\sigma$ is \emph{a pruned subtree} of the tree of all legal positions $T$ if $\sigma\subseteq T$ and for any $(W_0, W_1, \ldots, W_n)\in\sigma, n\ge 0$ there is a $W$ such that $(W_0, W_1,\ldots, W_n, W)\in\sigma$ (it simply means that there's no finite branch in~$\sigma$). 
+    \end{definition}
+
+    \begin{definition}
+        Let $\sigma$ be a pruned subtree of the tree of all legal positions $T$. By $[\sigma]$ we denote \emph{the set of all infinite branches of $\sigma$}, i.e. infinite sequences $(W_0, W_1, \ldots)$ such that $(W_0, W_1, \ldots W_n)\in \sigma$ for any $n\in \bN$.
     \end{definition}
 
     \begin{definition}
-        A \emph{strategy} for $II$ in $G^{\star\star}(A)$ is a subtree $\sigma\subseteq T$ such that
+        A \emph{strategy} for $\textit{II}$ in $G^{\star\star}(A)$ is a pruned subtree $\sigma\subseteq T$ such that
         \begin{enumerate}[label=(\roman*)]
             \item $\sigma$ is nonempty,
-            \item if $(U_0, V_0, \ldots, V_n)\in\sigma$, then for all open nonempty $U_{n+1}\subseteq V_n$, $(U_0, V_0, \ldots, V_n, U_{n+1})\in\sigma$,
-            \item if $(U_0, V_0, \ldots, U_{n})\in\sigma$, then for unique $V_n$, $(U_0, V_0, \ldots,  U_{n}, V_n)\in\sigma$. 
+            \item if $(U_0, V_0, \ldots, U_n, V_n)\in\sigma$, then for all open nonempty $U_{n+1}\subseteq V_n$, $(U_0, V_0, \ldots, U_n, V_n, U_{n+1})\in\sigma$,
+            \item if $(U_0, V_0, \ldots, U_{n})\in\sigma$, then for a unique $V_n$, $(U_0, V_0, \ldots,  U_{n}, V_n)\in\sigma$. 
         \end{enumerate}
     \end{definition}
 
-    Intuitively, a strategy $\sigma$ works as follows: $I$ starts playing $U_0$ as any open subset of $X$, then $II$ plays unique (by (iii)) $V_0$ such that $(U_0, V_0)\in\sigma$. Then $I$ responds by playing any $U_1\subseteq V_0$ and $II$ plays uniqe $V_1$ such that $(U_0, V_0, U_1, V_1)\in\sigma$, etc.
+    Intuitively, a strategy $\sigma$ works as follows: $I$ starts playing $U_0$ as any open subset of $X$, then $\textit{II}$ plays unique (by (iii)) $V_0$ such that $(U_0, V_0)\in\sigma$. Then $I$ responds by playing any $U_1\subseteq V_0$ and $\textit{II}$ plays uniqe $V_1$ such that $(U_0, V_0, U_1, V_1)\in\sigma$, etc.
 
     \begin{definition}
-        A strategy $\sigma$ is a \emph{winning strategy for $II$} if for any game $(U_0, V_0\ldots)\in [\sigma]$ (where $[\sigma]$ is defined analogically to $[T]$) player $II$ wins, i.e. $\bigcap_{n}V_n \subseteq A$.
+        A strategy $\sigma$ is a \emph{winning strategy for $\textit{II}$} if for any game $(U_0, V_0\ldots)\in [\sigma]$ player $\textit{II}$ wins, i.e. $\bigcap_{n}V_n \subseteq A$.
     \end{definition}
 
     Now we can state the key theorem.
 
-    \begin{theorem}
+    \begin{theorem}[Banach-Mazur, Oxtoby]
         \label{theorem:banach_mazur_thm}
-        Let $X$ be a nonempty topological space and let $A\subseteq X$. Then A is comeagre $\Leftrightarrow$  $II$ has a winning strategy in $G^{\star\star}(A)$.
+        Let $X$ be a nonempty topological space and let $A\subseteq X$. Then A is comeagre $\Leftrightarrow$  $\textit{II}$ has a winning strategy in $G^{\star\star}(A)$.
     \end{theorem}
 
     In order to prove it we add an auxilary definition and lemma.
     \begin{definition}
-        Let $S$ be a pruned subtree of a strategy $\sigma$ and let $p=(U_0, V_0,\ldots, V_n)\in S$. We say that S is \emph{comprehensive for p} if the family $\cV_p = \{V_{n+1}\mid (U_0, V_0,\ldots, V_n, U_{n+1}, V_{n+1})\in S\}$ (it may be that $n=-1$, which means $p=\emptyset$) is pairwise disjoint and $\bigcup\cV_p$ is dense in $V_n$ (where we think that $V_{-1} = X$).
+        Let $S\subseteq\sigma$ be a pruned subtree of tree of all legal positions $T$ and let $p=(U_0, V_0,\ldots, V_n)\in S$. We say that S is \emph{comprehensive for p} if the family $\cV_p = \{V_{n+1}\mid (U_0, V_0,\ldots, V_n, U_{n+1}, V_{n+1})\in S\}$ (it may be that $n=-1$, which means $p=\emptyset$) is pairwise disjoint and $\bigcup\cV_p$ is dense in $V_n$ (where we think that $V_{-1} = X$).
 
-        We say that $S$ is \emph{comprehensive} if it is comprehensive for any $p=(U_0, V_0,\ldots, V_n)\in S$.
+        We say that $S$ is \emph{comprehensive} if it is comprehensive for each $p=(U_0, V_0,\ldots, V_n)\in S$.
     \end{definition}
     
+    \begin{fact}
+        If $\sigma$ is a winnig strategy for $\mathit{II}$ then there exists a nonempty comprehensive $S\subseteq\sigma$.
+    \end{fact}
+
+    \begin{proof}
+        We construct $S$ recursively as follows:
+        \begin{enumerate}
+            \item $\emptyset\in S$,
+            \item if $(U_0, V_0, \ldots, U_n)\in S$, then $(U_0, V_0, \ldots, U_n, V_n)\in S$ for the unique $V_n$ given by the strategy $\sigma$,
+            \item let $p = (U_0, V_0, \ldots, V_n)\in S$. For a possible player $I$'s move $U_{n+1}\subseteq V_n$ let $U^\star_{n+1}$ be the unique set player $\mathit{II}$ would respond with by $\sigma$. Now, by Zorn's Lemma, let $\cU_p$ be a maximal collection of nonempty open subsets $U_{n+1}\subseteq V_n$ such that the set $\{U^\star_{n+1}\mid U_{n+1}\in\cU_p\}$ is pairwise disjoint. Then put in $S$ all $(U_0, V_0, \ldots, V_{n}, U_{n+1})$ such that $U_{n+1} \in \cU_p$. This way $S$ is comprehensive for $p$: the family $\cV_p = \{V_{n+1}\mid (U_0, V_0,\ldots, V_n, U_{n+1}, V_{n+1})\in S\}$ is exactly $\{U^\star_{n+1}\mid U_{n+1}\in\cU_p\}$, which is pairwise disjoint and $\bigcup\cV_p$ is obviously dense in $V_n$ by its maximality -- if there was any open set $\tilde{U}_{n+1}\subseteq V_n$ disjoint from $\bigcup\cU_p$, then the family $\cU_p\cup\{\tilde{U}_{n+1}\}$ would violate the maximality of $\cU_p$.
+            \qedhere
+        \end{enumerate}
+    \end{proof}
+
     \begin{lemma}
         \label{lemma:comprehensive_lemma}
-        Let $S$ be a comprehensive pruned subtree of a strategy $\sigma$. Then:
+        Let $S$ be a nonempty comprehensive pruned subtree of a strategy $\sigma$. Then:
         \begin{enumerate}[label=(\roman*)]
-            \item For any $V_n$ such that there is $p=(U_0, V_0, \ldots, V_n)\in S$, this $p$ is unique.
-            \item Let $W_n = \{V_n\mid (U_0, V_0, \ldots, V_n)\in S\}$, i.e. $W_n$ is a family of all possible choices player $II$ can make in its $n$-th move. Then $\bigcup W_n$ is open and dense in $X$. 
-            \item There exists such $S$.
+            \item For any open $V_n\subseteq X$ there is at most one $p=(U_0, V_0, \ldots, U_n, V_n)\in S$.
+            \item Let $S_n = \{V_n\mid (U_0, V_0, \ldots, V_n)\in S\}$ for $n\in\bN$ (i.e. $S_n$ is a family of all possible choices player $\textit{II}$ can make in its $n$-th move according to $S$). Then $\bigcup S_n$ is open and dense in $X$.
+            \item $S_n$ is a family of pairwise disjoint sets.
         \end{enumerate}
     \end{lemma}
 
@@ -194,40 +215,45 @@
             \item $U_k\neq U'_k$: by the comprehensiveness of $S$ we know that for $q =(U_0, V_0, \ldots, U_{k-1}, V_{k-1})$ the set $\cV_q$ is pairwise disjoint. Thus $V_k\cap V'_k=\emptyset$, because $V_k, V'_k\in \cV_q$. But this leads to a contradiction -- $V_n$ cannot be a nonempty subset of both $V_k, V'_k$.
         \end{itemize}
 
-        (ii): The lemma is proved by induction on $n$. For $n=0$ it follows trivially from the definition of comprehensiveness. Now suppose the lemma is true for $n$. Then the set $\bigcup_{V_n\in W_n}\bigcup\cV_{p_{V_n}}$ ($p_{V_n}$ is given uniquely from (i)) is dense and open in $X$ by the induction hypothesis. But $\bigcup W_{n+1}$ is its superset, thus $\bigcup W_{n+1}$ is dense and open in $X$.
+        (ii): The lemma is proved by induction on $n$. For $n=0$ it follows trivially from the definition of comprehensiveness. Now suppose the lemma is true for $n$. Then the set $\bigcup_{V_n\in S_n}\bigcup\cV_{p_{V_n}}$ (where $p_{V_n}$ is given uniquely from (i)) is dense and open in $X$ by the induction hypothesis. But $\bigcup S_{n+1}$ is exactly this set, thus it is dense and open in $X$.
 
-        (iii): We construct $S$ recursively as follows:
-        \begin{enumerate}
-            \item $\emptyset\in S$,
-            \item if $(U_0, V_0, \ldots, U_n)\in S$, then $(U_0, V_0, \ldots, U_n, V_n)\in S$ for the unique $V_n$ given by the strategy $\sigma$,
-            \item let $p = (U_0, V_0, \ldots, V_n)\in S$, let $U^\star_{n+1}$ be the unique set player $II$ would play by $\sigma$ given that player $I$ played $U_{n+1}\subseteq V_n$. Now, by Zorn's Lemma, let $\cU_p$ be a maximal collection of nonempty open subsets $U_{n+1}\subseteq V_n$ such that the set $\{U^\star_{n+1}\mid U_{n+1}\in\cU_p\}$ is pairwise disjoint. Then put in $S$ all $(U_0, V_0, \ldots, V_{n}, U_{n+1})$ such that $U_{n+1} \in \cU_p$. This way $S$ is comprehensive for $p$: the family $\cV_p = \{V_{n+1}\mid (U_0, V_0,\ldots, V_n, U_{n+1}, V_{n+1})\in S\}$ is exactly $\{U^\star_{n+1}\mid U_{n+1}\in\cU_p\}$, which is pairwise disjoint and $\bigcup\cV_p$ is obviously dense in $V_n$ by it's maximality -- if there was any open set $\tilde{U}_{n+1}\subseteq V_n$ disjoint from $\bigcup\cU_p$, then the family $\cU_p\cup\{\tilde{U}_{n+1}\}$ violates the maximality of $\cU_p$.
-        \end{enumerate}
+        (iii): We will prove it by induction on $n$. Once again, the case $n = 0$ follows from the comprehensiveness of $S$. Now suppose that the sets in $S_n$ are pairwise disjoint. Take some $x \in V_{n+1}\in S_{n+1}$. Of course $\bigcup S_n \supseteq \bigcup S_{n+1}$, thus by the inductive hypothesis $x\in V_{n}$ for the unique $V_n\in S_n$. It must be that $V_{n+1}\in \cV_{p_{V_n}}$, because $V_n$ is the only superset of $V_{n+1}$ in $S_n$. But $\cV_{p_{V_n}}$ is disjoint, so there is no other $V'_{n+1}\in \cV_{p_{V_n}}$ such that $x\in V'_{n+1}$. Moreover, there is no such set in $S_{n+1}\setminus\cV_{p_{V_n}}$, because those sets are disjoint from $V_{n}$. Hence there is no $V'_{n+1}\in S_{n+1}$ other than $V_n$ such that $x\in V'_{n+1}$. We chosed $x$ and $V_{n+1}$ arbitrarily, so $S_{n+1}$ is pairwise disjoint.
     \end{proof}
 
     Now we can move to the proof of the Banach-Mazur theorem.
 
-    \begin{proof}
-        $\Rightarrow$: Let $(A_n)$ be a sequence of dense open sets with $\bigcap_n A_n\subseteq A$. The simply $II$ plays $V_n = U_n\cap A_n$, which is nonempty by the denseness of $A_n$.
+    \begin{proof}[Proof of theorem \ref{theorem:banach_mazur_thm}]
+        $\Rightarrow$: Let $(A_n)$ be a sequence of dense open sets with $\bigcap_n A_n\subseteq A$. The simply $\textit{II}$ plays $V_n = U_n\cap A_n$, which is nonempty by the denseness of $A_n$.
 
-        $\Leftarrow$: Suppose $II$ has a winning strategy $\sigma$. We will show that $A$ is comeagre. Take a comprehensive $S\subseteq \sigma$. We claim that $\mathcal{W} = \bigcap_n\bigcup W_n \subseteq A$. By \ref{lemma:comprehensive_lemma}, (ii) sets $\bigcup W_n$ are open and dense, thus $A$ must be comeagre. Now we prove the claim. 
+        $\Leftarrow$: Suppose $\textit{II}$ has a winning strategy $\sigma$. We will show that $A$ is comeagre. Take a comprehensive $S\subseteq \sigma$. We claim that $\mathcal{S} = \bigcap_n\bigcup S_n \subseteq A$. By the lemma~\ref{lemma:comprehensive_lemma}, (ii) sets $\bigcup S_n$ are open and dense, thus $A$ must be comeagre. Now we prove the claim towards contradiction. 
         
-        (A.a.) Suppose there is $x\in \mathcal{W}$ that is not in $A$. We will prove by induction that for any $n$ there is exactly one $V_n\in W_n$ such that $x\in V_n$. For $n = 0$ this follows trivially by the comprehensiveness of $S$. Now suppose that there is exactly one $V_n\in W_n$ such that $x\in V_n$. By our assumption there is a $V'_{n+1}\in W_{n+1}$ such that $x\in V'_{n+1}$. By \ref{lemma:comprehensive_lemma} we have unique $p_{V'_{n+1}}=(U'_0, V'_0, \ldots, V'_{n+1})\in S$. It must be that $x\in V'_n$, so by the induction hypothesis $V'_n = V_n$, thus $V'_{n+1}\in \cV_{p_{V_{n}}}$. But the family $\cV_{p_{V_{n}}}$ is disjoint, hence $V_{n+1} = V'_{n+1}$ is unique.
-
-        Now the game $(U_0, V_0, U_1, V_1,\ldots) = \bigcup_n p_{V_n}\in [S]\subseteq [\sigma]$ where $x\in V_0, V_1,\ldots$ is not winning for player $II$, which contradicts the assumption that $\sigma$ is a winning strategy.
+        Suppose there is $x\in \mathcal{S}\setminus A$. By the lemma \ref{lemma:comprehensive_lemma}, (iii) for any $n$ there is unique $x\in V_n\in S_n$. It follows that $p_{V_0}\subset p_{V_1}\subset\ldots$. Now the game $(U_0, V_0, U_1, V_1,\ldots) = \bigcup_n p_{V_n}\in [S]\subseteq [\sigma]$ is not winning for player $\textit{II}$, which contradicts the assumption that $\sigma$ is a winning strategy.
     \end{proof}
 
-    Pytania:
-    \begin{itemize}
-        \item Czy da się coś zrobić, żeby $\mathcal{V}$ nie było takie brzydkie?
-        \item Jak to napisać, że się zrzyna z książki?
-        \item Dodatkowy przykład pod def 2.2
-        \item $G^{\star\star}(A)$ czy $G^{**}(A)$? Czy może $G^{**}(X, A)?$ Jakiś skrót na to?
-        \item w \ref{lemma:comprehensive_lemma} (i), jak to ładniej sformułować?
-        \item w \ref{lemma:comprehensive_lemma} (iii), może to wyodrębnić? Może to dać jako pierwsze, a pierwsze dwa później?
-        \item dodać tytuł do \ref{theorem:banach_mazur_thm}
-        \item czy w dowodzie twierdzenia napisać jeszcze raz co to jest $W_n$?
-        \item ostatni akapit dowodu twierdzenia, czy taka suma tych $p_{V_n}$ to jest sensowny napis? Jak to inaczej napisać?
-    \end{itemize}
+    % Pytania:
+    % \begin{itemize}
+    %     \item Czy da się coś zrobić, żeby $\mathcal{V}$ nie było takie brzydkie?
+    %     \item Jak to napisać, że się zrzyna z książki?
+    %     \item Dodatkowy przykład pod def 2.2
+    %     \item DONE W/E $G^{\star\star}(A)$ czy $G^{**}(A)$? Czy może $G^{**}(X, A)?$ Jakiś skrót na to?
+    %     \item w \ref{lemma:comprehensive_lemma} (i), jak to ładniej sformułować?
+    %     \item w \ref{lemma:comprehensive_lemma} (iii), może to wyodrębnić? Może to dać jako pierwsze, a pierwsze dwa później?
+    %     \item DONE dodać tytuł do \ref{theorem:banach_mazur_thm}
+    %     \item czy w dowodzie twierdzenia napisać jeszcze raz co to jest $W_n$?
+    %     \item DONE TAK ostatni akapit dowodu twierdzenia, czy taka suma tych $p_{V_n}$ to jest sensowny napis? Jak to inaczej napisać?
+    % \end{itemize}
+
+    % Odpowiedzi:
+    % \begin{itemize}
+    %     % \item dodać osobną definicję na $[\sigma]$ (tzn dla dowolnego poddrzewa).
+    %     \item $II$ -> $\mathit{II}$
+    %     \item w ogóle dodać def poddrzewa (co to znaczy pruned?)
+    %     \item w \ref{lemma:comprehensive_lemma} (i) zmienić na "dla każdego otwartego podzbioru X jest co najwyżej jedno takie p"
+    %     \item powiedzieć że S musi być niepuste
+    %     \item w \ref{lemma:comprehensive_lemma} zamienić kolejność w pierwszym zdaniu gwiazdki z nie-gwiazdką
+    %     \item zmienić $W_n -> S_n$
+    %     \item rozwinąć \ref{lemma:comprehensive_lemma} (ii), poza tym $W_{n+1}$ jest dokładnie równy, a nie tylko superset, dodać kwantyfikator na n, a może wplecić jeszcze to że te rodziny $W_n$ też są rozłączne?
+    % \end{itemize}
 
     \subsection{Fraïssé classes}
     \begin{fact}[Fraïssé theorem]
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