From 2b830cb2d9c2237fcb7809bab7c64966098ea6fb Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Sun, 17 Jul 2022 14:16:41 +0200
Subject: Oby ostatnie poprawki

---
 sections/conj_classes.tex | 72 +++++++++++------------------------------------
 1 file changed, 16 insertions(+), 56 deletions(-)

(limited to 'sections/conj_classes.tex')

diff --git a/sections/conj_classes.tex b/sections/conj_classes.tex
index c0120d3..9ec4b0c 100644
--- a/sections/conj_classes.tex
+++ b/sections/conj_classes.tex
@@ -46,7 +46,7 @@
 	$\Aut(M)$: take any finite $\gamma\colon M\to M$ such that $[\gamma]_{\Aut(M)}$
 	is nonempty. Then also 
 	$\bigcup_{\beta\in B_{n,k}} [\beta]_{\Aut(M)}\cap[\gamma]_{\Aut(M)}\neq\emptyset$,
-	one can easily construct a permutation that extends $\gamma$ and have at least
+	one can easily construct a permutation that extends $\gamma$ and has at least
 	$k$ many $n$-cycles.
 
 	Now we see that $A = \bigcap_{n=1}^{\infty} A_n$ is a comeagre set consisting
@@ -97,21 +97,21 @@
   generated by $\dom(g)$ and $\rng(g)$ respectively. Of course, in our
   case, $g$ is good
   if and only if $[g]_{\Aut(\Gamma)} \neq\emptyset$ (because of ultrahomogeneity
-  of $\Gamma$.
+  of $\Gamma$).
 
   Also it is important to mention that an isomorphism between two finitely
   generated structures is uniquely given by a map from generators of one structure
   to the other. This allow us to treat a finite function as an isomorphism
-  of finitely generated structure.
+  of finitely generated structures.
 
   \begin{proof}
 	Let $\Gamma = \Flim(\cC)$ and $(\Pi, \sigma) = \Flim(\cD)$. First, by the Theorem
 	\ref{theorem:isomorphic_fr_lims}, we may assume without the loss of generality
 	that $\Pi = \Gamma$. Let $G = \Aut(\Gamma)$,
 	i.e. $G$ is the automorphism group of $\Gamma$.
-	We will construct a strategy for the second player in the Banach-Mazur game
-	on the topological space $G$. This strategy will give us a subset 
-	$A\subseteq G$ and as we will see a subset of the $\sigma$'s conjugacy class.
+	We will construct a winning strategy for the second player in the Banach-Mazur game
+	(see \ref{definition:banach-mazur-game})
+	on the topological space $G$ with $A$ being $\sigma$'s conjugacy class. 
 	By the Banach-Mazur theorem (see \ref{theorem:banach_mazur_thm}) this will prove 
 	that this class is comeagre.
 
@@ -129,15 +129,6 @@
 	chooses $g_0, g_1,\ldots$ such that
 	$f_0\subseteq g_0\subseteq f_1\subseteq g_1\subseteq\ldots$, which identify
 	the corresponding basic open subsets $B_{f_0}\supseteq B_{g_0}\supseteq\ldots$.
-
-	% Our goal is to choose $g_i$ in such a manner that the resulting function
-	% $g = \bigcap^{\infty}_{i=0}g_i$ will be an automorphism of the Fraïssé limit 
-	% $\Gamma$ such that $(\Gamma, g) = \Flim{\cD}$. 
-	% Precisely, we will find $g_i$'s such that 
-	% $\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$ and by
-	% the Fraïssé theorem \ref{theorem:fraisse_thm}
-	% it will follow that $(\Gamma, g)\cong (\Pi, \sigma)$. Hence,
-	% by the fact	\ref{fact:conjugacy}, $g$ and $\sigma$ conjugate in $G$.
 	
 	Our goal is to choose $g_i$ in such a manner that 
 	$\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$ and $(\Gamma, g)$ is ultrahomogeneous
@@ -145,24 +136,14 @@
 	that $(\Gamma, \sigma)\cong (\Gamma, g)$, thus by the Fact \ref{fact:conjugacy}
 	we have that $\sigma$ and $g$ conjugate.
 
-	% Once again, by the Fraïssé theorem and by Lemma 
-	% \ref{lemma:weak_ultrahom} constructing $g_i$'s in a way such that
-	% age of $(\Gamma, g)$ is exactly $\cD$ and so that it is weakly ultrahomogeneous
-	% will produce expected result. 
-	First, let us enumerate all pairs of structures
-	$\{\langle(A_n, \alpha_n), (B_n, \beta_n)\rangle\}_{n\in\bN},\in\cD$
-	such that the first element of the pair embeds by inclusion in the second,
-	i.e. $(A_n, \alpha_n)\subseteq (B_n, \beta_n)$. Also, it may be that 
-	$A_n$ is an empty. We enumerate the elements of the Fraïssé limit 
-	$\Gamma = \{v_0, v_1, \ldots\}$.
-
 	Fix a bijection $\gamma\colon\bN\times\bN\to\bN$ such that for any
 	$n, m\in\bN$ we have $\gamma(n, m) \ge n$. This bijection naturally induces
 	a well ordering on $\bN\times\bN$. This will prove useful later, as the
 	main ingredient of the proof will be a bookkeeping argument.
 
 	For technical reasons, let $g_{-1} = \emptyset$ and
-	$X_{-1} = \emptyset$. 
+	$X_{-1} = \emptyset$. Enumerate the elements of the Fraïssé limit 
+	$\Gamma = \{v_0, v_1, \ldots\}$.
 	Suppose that player \textit{I} in the $n$-th move chooses a finite partial
 	automorphism $f_n$. We will construct a finite partial automorphism 
 	$g_n\supseteq f_n$ together with a finitely generated substructure
@@ -204,26 +185,13 @@
 		\end{center}
 	\end{enumerate}
 
-	% First item makes sure that no infinite orbit will be present in $g$. The 
-	% second item together with the first one are necessary for $g$ to be an 
-	% automorphism of $\Gamma$. The third item is the one that gives weak
-	% ultrahomogeneity. Now we will show that indeed such $g_n$ may be constructed.
-	
 	First, we will satisfy the item (iii). Namely, we will construct $\Gamma'_n, g'_n$
 	such that $g_{n-1}\subseteq g'_n$, $\Gamma_{n-1}\subseteq\Gamma'_n$,
 	$g'_n$ gives an automorphism of $\Gamma'_n$
 	and $f_{i,j}$ extends to an embedding of
 	$(B_{i,j}, \beta_{i,j})$ to $(\Gamma'_n, g'_n)$. But this can be easily
-	done by the fact, that $\cD$ has the amalgamation property. Moreover, without
-	the loss of generality we can assume that all embeddings are inclusions.
+	done by the fact, that $\cD$ has the amalgamation property. 
 
-    \begin{center}
-      \begin{tikzcd}
-                                  & (\Gamma'_n, g'_n) & \\
-        (B_{i,j}, \beta_{i,j}) \arrow[ur, dashed, "\subseteq"] &   & (\Gamma_{n-1}, g_{n-1}) \arrow[ul, dashed, "\subseteq"'] \\
-								  & (A_{i,j}, \alpha_{i,j}) \arrow[ur, "\subseteq"'] \arrow[ul, "\subseteq"]
-      \end{tikzcd}
-    \end{center}
 
 	It is important to note that $g'_n$ should be a finite function and once
 	again, as it is an automorphism of a finitely generated structure, we may
@@ -231,14 +199,7 @@
 	others.	By the weak ultrahomogeneity of $\Gamma$, we may assume that 
 	$\Gamma'_n\subseteq \Gamma$.
 
-	% \begin{center}
-	%   \begin{tikzcd}
-	% 	B_{i,j}\cup\Gamma_{n-1} \arrow[d, "\subseteq"'] \arrow[r, "\subseteq"] & \Gamma \\
-	% 	\Gamma'_{n}\arrow[ur, dashed, "f"'] 
-	%   \end{tikzcd}
-	% \end{center}
-
-	Now, by the WHP of $\cK$ we can extend $\langle\Gamma'_n\cup\{v_n\}\rangle$ together
+	Now, by the WHP of $\cC$ we can extend $\langle\Gamma'_n\cup\{v_n\}\rangle$ together
 	with its partial isomorphism $g'_n$ to a finitely generated structure $\Gamma_n$ 
 	together with its
 	automorphism $g_n\supseteq g'_n$ and (again by weak ultrahomogeneity)
@@ -247,7 +208,7 @@
 	that has all desired properties.
 
 	Now we need to see that $g = \bigcap^{\infty}_{n=0}g_n$ is indeed an automorphism
-	of $\Gamma$ such that $(\Gamma, g)$ has the age $\cH$ and is weakly ultrahomogeneous.
+	of $\Gamma$ such that $(\Gamma, g)$ has the age $\cD$ and is weakly ultrahomogeneous.
 	It is of course an automorphism of $\Gamma$ as it is defined for every $v\in\Gamma$
 	and is an union of an increasing chain of automorphisms of finitely generated
 	substructures.
@@ -257,7 +218,8 @@
 	By the bookkeeping there was $n$ such that 
 	$(i, j) = \min\{\{0,1,\ldots n-1\}\times\bN\setminus X_{n-1}\}$.
 	This means that $(B, \beta)$ embeds into $(\Gamma_n, g_n)$, hence it embeds
-	into $(\Gamma, g)$. Hence, the age of $(\Gamma, g)$ is $\cH$. 
+	into $(\Gamma, g)$. Thus, $\cD$ is a subclass of the age of $(\Gamma, g)$.
+	The other inclusion is obvious.	Hence, the age of $(\Gamma, g)$ is $\cH$. 
 
 	It is also weakly ultrahomogeneous. Having $(A,\alpha)\subseteq(B,\beta)$,
 	and an embedding $f\colon(A,\alpha)\to(\Gamma,g)$, we may find $n\in\bN$
@@ -267,11 +229,9 @@
 	$(\Gamma_n, g_n)$, which means we can also embed it into $(\Gamma, g)$.
 
 	Hence, $(\Gamma,g)\cong(\Gamma,\sigma)$.
-	By this we know that $g$ and $\sigma$ conjugate in $G$. As we stated in the
-	beginning of the proof, the set $A$ of possible outcomes of the game (i.e.
-	possible $g$'s we end up with) is comeagre in $G$, thus $\sigma^G$ is also
-	comeagre and $\sigma$ is a generic automorphism, as it contains a comeagre
-	set $A$.
+	By this we know that $g$ and $\sigma$ are conjugate in $G$, thus player \textit{II}
+	have a winning strategy in the Banach-Mazur game with $A=\sigma^G$,
+	thus $\sigma^G$ is comeagre in $G$ and $\sigma$ is a generic automorphism.
   \end{proof}
 
   \begin{theorem}
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