From 30e20714fa82c6d0d6b1c06b81ebcefdb72e1004 Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Sun, 10 Jul 2022 19:24:51 +0200
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 =?UTF-8?q?am=20zmiany?=
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---
 sections/conj_classes.tex | 83 ++++++++++++++++++++++++++++-------------------
 1 file changed, 49 insertions(+), 34 deletions(-)

(limited to 'sections/conj_classes.tex')

diff --git a/sections/conj_classes.tex b/sections/conj_classes.tex
index d380df6..b122058 100644
--- a/sections/conj_classes.tex
+++ b/sections/conj_classes.tex
@@ -1,7 +1,8 @@
 \documentclass[../lic_malinka.tex]{subfiles}
 
 \begin{document}
-  Let $M$ be a countable $L$-structure. We define a topology on the $G=\Aut(M)$:
+  Let $M$ be a countable $L$-structure. Recall, we define a topology on 
+  the $G=\Aut(M)$:
   for any finite function $f\colon M\to M$ we have a basic open set 
   $[f]_{G} = \{g\in G\mid f\subseteq g\}$.  
 
@@ -10,12 +11,16 @@
   In this section, $M=(M,=)$ is an infinite countable set (with no structure
   beyond equality).
   
-  \begin{proposition} 
-	\label{proposition:cojugate-classes}
+  \begin{remark} 
+	\label{remark:cojugate-classes}
 	If $f_1,f_2\in \Aut(M)$, then $f_1$ and $f_2$ are conjugate if and only 
     if for each $n\in \bN\cup \{\aleph_0\}$, $f_1$ and $f_2$ have the same 
     number of orbits of size $n$.
-  \end{proposition}
+  \end{remark}
+
+  \begin{proof}
+	It is easy to see.
+  \end{proof}
 
   \begin{theorem}
 	Let $\sigma\in \Aut(M)$ be an automorphism with no infinite orbit and with
@@ -30,7 +35,7 @@
 	Let $A_n = \{\alpha\in Aut(M)\mid \alpha\text{ has infinitely many orbits of size }n\}$.
 	This set is comeagre for every $n>0$. Indeed, we can represent this set
 	as an intersection of countable family of open dense sets. Let $B_{n,k}$
-	be the set of all finite functions $\beta\colon M\to M$ that consists
+	be the set of all finite functions $\beta\colon M\to M$ that consist
 	of exactly $k$ distinct $n$-cycles. Then:
 	\begin{align*}
 	  A_n &= \{\alpha\in\ \Aut(M) \mid \alpha\text{ has infinitely many orbits of size }n\} \\
@@ -49,10 +54,10 @@
 	thing left to show is that the set of functions with no infinite cycle is 
 	also comeagre. Indeed, for $m\in M$ let 
 	$B_m = \{\alpha\in\Aut(M)\mid m\text{ has finite orbit in }\alpha\}$. This 
-	is an open dense set. It is a sum over basic open sets generated by finite
+	is an open dense set. It is a union over basic open sets generated by finite
 	permutations with $m$ in their domain. Denseness is also easy to see.
 
-	Finally, by the proposition \ref{proposition:cojugate-classes}, we can say that 
+	Finally, by the remark \ref{remark:cojugate-classes}, we can say that 
 	$$\sigma^{\Aut(M)}=\bigcap_{n=1}^\infty A_n \cap \bigcap_{m\in M} B_m,$$
 	which concludes the proof.
   \end{proof}
@@ -63,22 +68,23 @@
 	\label{fact:conjugacy}
     Suppose $M$ is an arbitrary structure and $f_1,f_2\in \Aut(M)$. 
     Then $f_1$ and $f_2$ are conjugate if and only if $(M,f_1)\cong
-    (M,f_2)$ as structures with one additional unary relation that is
+    (M,f_2)$ as structures with one additional unary function that is
     an automorphism.  
   \end{fact}
 
   \begin{proof}
     Suppose that $f_1 = g^{-1}f_2g$ for some $g\in \Aut(M)$. 
-    Then $g$ is the automorphism we're looking for. On the other hand if 
+	Then $g$ is the isomorphism between $(M,f_1)$ and $(M,f_2)$. 
+	On the other hand if 
     $g\colon (M, f_1)\to (M, f_2)$ is an isomorphism, then 
     $g\circ f_1 = f_2 \circ g$ which exactly means that $f_1, f_2$ conjugate.
   \end{proof} 
 
   \begin{theorem}
 	\label{theorem:generic_aut_general}
-	Let $\cC$ be a Fraïssé class of finite structures of a theory $T$ in a 
-	relational language $L$. Let $\cD$ be the class of the finite structures of
-	$T$ in the language $L$ with additional unary function symbol interpreted
+	Let $\cC$ be a Fraïssé class of finite $L$-structures. 
+	Let $\cD$ be the class of structures from $\cC$ with additional unary 
+	function symbol interpreted
 	as an automorphism of the structure. If $\cC$ has the weak Hrushovski property
 	and $\cD$ is a Fraïssé class, then there is a comeagre conjugacy class in the
 	automorphism group of the $\Flim(\cC)$.
@@ -86,13 +92,13 @@
 
   \begin{proof}
 	Let $\Gamma = \Flim(\cC)$ and $(\Pi, \sigma) = \Flim(\cD)$. Let $G = \Aut(\Gamma)$,
-	i.e. $G$ is the automorphism group of $\Gamma$. First, by the theorem
+	i.e. $G$ is the automorphism group of $\Gamma$. First, by the Theorem
 	\ref{theorem:isomorphic_fr_lims}, we may assume without the loss of generality
 	that $\Pi = \Gamma$.
 	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 a conjugacy class in $G$.
-	By the Banach-Mazur theorem \ref{theorem:banach_mazur_thm} this will prove 
+	$A\subseteq G$ and as we will see a subset of the $\sigma$'s conjugacy class.
+	By the Banach-Mazur theorem (see \ref{theorem:banach_mazur_thm}) this will prove 
 	that this class is comeagre.
 
 	Recall, $G$ has a basis consisting of open
@@ -100,28 +106,36 @@
 	finite set $A\subseteq \Gamma$ and some automorphism $g_0\in G$. In other
 	words, a basic open set is a set of all extensions of some finite partial
 	isomorphism $g_0$ of $\Gamma$. By $B_{g}\subseteq G$ we denote a basic 
-	open subset given by a finite partial isomorphism $g$. From now on we will
-	consider only finite partial isomorphism $g$ such that $B_g$ is nonempty.
+	open subset given by a finite partial isomorphism $g$. Note that $B_g$
+	is nonemty because of ultrahomogeneity of $\Gamma$.
 
-	With the use of corollary \ref{corollary:banach-mazur-basis} we can consider 
-	only games, where both players choose finite partial isomorphisms. Namely,
+	With the use of Corollary \ref{corollary:banach-mazur-basis} we can consider 
+	only games where both players choose finite partial isomorphisms. Namely,
 	player \textit{I} picks functions $f_0, f_1,\ldots$ and player \textit{II}
 	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, $\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$, 
-	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 $G$.
-
-	Once again, by the Fraïssé theorem \ref{theorem:fraisse_thm} and the
-	\ref{lemma:weak_ultrahom} lemma 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
+	% 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
+	with age $\cD$. By the Fraïssé theorem (see \ref{theorem:fraisse_thm}) it will follow
+	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 
@@ -131,12 +145,13 @@
 	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 argument of the proof will be constructed as a bookkeeping argument.
+	main ingredient of the proof will be a bookkeeping argument.
 
-	Just for sake of fixing a technical problem, let $g_{-1} = \emptyset$ and
+	For technical reasons, let $g_{-1} = \emptyset$ and
 	$X_{-1} = \emptyset$. 
 	Suppose that player \textit{I} in the $n$-th move chooses a finite partial
-	isomorphism $f_n$. We will construct $g_n\supseteq f_n$ and a set $X_n\subseteq\bN^2$
+	isomorphism $f_n$. We will construct a finite partial isomorphism $g_n\supseteq f_n$ 
+	and a set $X_n\subseteq\bN^2$
 	such that following properties hold:
 
 	\begin{enumerate}[label=(\roman*)]
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