Permutacje komizernosc

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   \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...