From 87b6137f59c74e6e52fa4bf761b5ec1ad8fcab7a Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Tue, 15 Feb 2022 21:50:25 +0100
Subject: Poprawa usterki faktu

---
 lic_malinka.tex | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

(limited to 'lic_malinka.tex')

diff --git a/lic_malinka.tex b/lic_malinka.tex
index 24e3008..0b03eb7 100644
--- a/lic_malinka.tex
+++ b/lic_malinka.tex
@@ -193,7 +193,7 @@
         \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 $p = (U_0, V_0, \ldots, V_n)\in S$. 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 $\cU_p$.
+            \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 the maximality of $\cU_p$ -- if there was any open set $\tilde{U}_{n+1}\subseteq V_n$ disjoint from $\bigcup\cV_p$, then $\tilde{U}^{\star}_{n+1}\subseteq \tilde{U}_{n+1}$ would be also disjoint from $\bigcup\cV_p$, so the family $\cU_p\cup\{\tilde{U}_{n+1}\}$ would violate the maximality of $\cU_p$ .
             \qedhere
         \end{enumerate}
     \end{proof}
@@ -217,7 +217,7 @@
 
         (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 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.
+        (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 $V'_{n+1}\in \cV_{p_{V_n}}$ suc h 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.
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