From 80d1346f17a3ae138cf84cb25f27a6ff6ccf4696 Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Tue, 3 May 2022 12:45:25 +0200
Subject: Permutacje komizernosc

---
 lic_malinka.tex | 58 +++++++++++++++++++++++++++++++++++++++++++++++++++------
 1 file changed, 52 insertions(+), 6 deletions(-)

(limited to 'lic_malinka.tex')

diff --git a/lic_malinka.tex b/lic_malinka.tex
index d38f36e..fe992a1 100644
--- a/lic_malinka.tex
+++ b/lic_malinka.tex
@@ -648,20 +648,66 @@
 
   \section{Conjugacy classes in automorphism groups}
 
+  TODO:
+  \begin{itemize}
+	\item w głównym dowodzie mogę użyć wprost AP, nie muszę tego uzasadniać
+	  jeszcze raz.
+  \end{itemize}
+
+  Let $M$ be a countable $L$-structure. We define a topology on the $G=\Aut(M)$:
+  for any finite function $f\colon M\to M$ we have a basic open set 
+  $[f]_{G} = \{g\in G\mid f\subseteq g\}$.  
+
   \subsection{Prototype: pure set} 
   In this section, $M=(M,=)$ is an infinite countable set (with no structure
   beyond equality).
+  
   \begin{proposition} 
-    If $f_1,f_2\in \Aut(M)$, then $f_1$ and $f_2$ are conjugate if and only 
+	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}
 
-  \begin{proposition} The conjugacy class of $f\in \Aut(M)$ is dense if
-    and only if...  \end{proposition} \begin{proposition} If $f\in
-  \Aut(M)$ has an infinite orbit, then the conjugacy class of $f$ is
-  meagre.  
-  \end{proposition}
+  \begin{theorem}
+	Let $\sigma\in \Aut(M)$ be an automorphism with no infinite orbit and with
+	infinitely many orbits of size $n$ for every $n>0$. Then the conjugacy
+	class of $\sigma$ is comeagre in $\Aut(M)$.
+  \end{theorem}
+
+  \begin{proof}
+	We will show that the conjugacy class of $\sigma$ is an intersection of countably
+	many comeagre sets. 
+
+	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
+	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\} \\
+		  &= \bigcap_{k=1}^{\infty} \{\alpha\in \Aut(M)\mid \alpha\text{ has at least }k\text{ orbits of size }n\} \\ 
+		  &= \bigcap_{k=1}^{\infty} \bigcup_{\beta\in B_{n,k}} [\beta]_{\Aut(M)},
+	\end{align*}
+	where indeed, $\bigcup_{\beta\in B_{n,k}} [\beta]_{\Aut(M)}$ is dense in
+	$\Aut(M)$: 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
+	$k$ many $n$-cycles.
+
+	Now we see that $A = \bigcap_{n=1}^{\infty} A_n$ is a comeagre set consisting
+	of all functions that have infinitely many $n$-cycles for each $n$. The only
+	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
+	permutations with $m$ in their domain. Denseness is also easy to see.
+
+	Finally, 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}
 
   % \begin{proposition} 
     % An automorphism $f$ of $M$ is generic if and only if...
-- 
cgit v1.2.3