Poprawki do dowodu

2 files changed, 6 insertions, 5 deletions

diff --git a/lic_malinka.pdf b/lic_malinka.pdf
index e25f4b6..a83d3d3 100644
--- a/lic_malinka.pdf
+++ b/lic_malinka.pdf
diff --git a/lic_malinka.tex b/lic_malinka.tex
index 0f916db..b1906c8 100644
--- a/lic_malinka.tex
+++ b/lic_malinka.tex
@@ -172,7 +172,7 @@
 

     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 $n=-1$) is pairwise disjoint and $\bigcup\cV_p$ is dense in $V_n$.

+        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 $V_{-1} = X$).

 

         We say that $S$ is \emph{comprehensive} if it is comprehensive for any $p=(U_0, V_0,\ldots, V_n)\in S$.

     \end{definition}

@@ -183,14 +183,14 @@
         \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 comprehensive $S\subseteq \sigma$.

+            \item There exists such $S$.

         \end{enumerate}

     \end{lemma}

 

     \begin{proof}

         (i): Suppose that there are some $p = (U_0, V_0,\ldots, U_n, V_n)$, $p'=(U'_0, V'_0, \ldots, U'_n, V'_n)$ such that $V_n = V'_n$ and $p \neq p'$. Let $k$ be the smallest index such that those sequences differ. We have two possibilities:

         \begin{itemize}

-            \item $U_k = U'_k$ and $V_k\neq V'_k$ -- this cannot be true simply by the fact that $S$ is a subset of a strategy.     

+            \item $U_k = U'_k$ and $V_k\neq V'_k$ -- this cannot be true simply by the fact that $S$ is a subset of a strategy (so $V_k$ is unique for $U_k$).

             \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}

 

@@ -200,7 +200,7 @@
         \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 $II$ whould 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 $(U_0, V_0, \ldots, V_{n}, U_{n+1}, U^\star_{n+1})$ such that $U_{n+1} \in \cU_p$. This way $S$ is comprehensive for $p$: the family $\cV_p = \{V_{n+1}\mid (U_0, V_0,\ldots, V_n, U_{n+1}, V_{n+1})\ S\}$ 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 $\cU_p$.

+            \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}

     \end{proof}

 

@@ -209,7 +209,7 @@
     \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 $A_n$.

 

-        $\Leftarrow$: Suppose $II$ has a winning strategy $\sigma$. We will show that $A$ is comeagre. Suppose we have 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 $II$ 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. 

         

         (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.

 

@@ -226,6 +226,7 @@
         \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}

 

     \subsection{Fraïssé classes}