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\documentclass[../lic_malinka.tex]{subfiles}
\begin{document}
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
$[f]_{G} = \{g\in G\mid f\subseteq g\}$.
\subsection{Prototype: pure set}
In this section, $M=(M,=)$ is an infinite countable set (with no structure
beyond equality).
\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{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
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 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\} \\
&= \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 has 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 union over basic open sets generated by finite
permutations with $m$ in their domain. Denseness is also easy to see.
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}
\subsection{More general structures}
\begin{fact}
\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 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 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 finitely generated $L$-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 $\Flim(\cC)$.
\end{theorem}
Before we get to the proof, 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 structures (if it yields one) and \textit{vice versa}.
\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 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.
Recall, $G$ has a basis consisting of
sets $\{g\in G\mid g\upharpoonright_A = g_0\upharpoonright_A\}$ for some
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 partial
automorphism $g_0$ of finitely generated substructures of $\Gamma$.
By $B_{g}\subseteq G$ we denote a basic
open subset given by a partial isomorphism $g$. Again, Note that $B_g$
is nonempty because of ultrahomogeneity of $\Gamma$.
With the use of Corollary \ref{corollary:banach-mazur-basis} we can consider
only games where both players choose 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
$\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$ 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.
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
$X_{-1} = \emptyset$. 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 partial
automorphism $f_n$. We will construct a partial automorphism
$g_n\supseteq f_n$ together with a finitely generated substructure
$\Gamma_n \subseteq \Gamma$ and a set $X_n\subseteq\bN^2$
such that the following properties hold:
\begin{enumerate}[label=(\roman*)]
\item $g_n$ is is a partial automorphism of $\Gamma$ and an automorphism of
finitely generated substructure $\Gamma_n$,
\item $g_n(v_n)$ and $g_n^{-1}(v_n)$ are defined.
\end{enumerate}
Before we give the third point, suppose recursively that $g_{n-1}$ already
satisfy all those properties. Let us enumerate
$\{\langle (A_{n,k}, \alpha_{n, k}), (B_{n,k}, \beta_{n,k}), f_{n, k}\rangle\}_{k\in\bN}$
all pairs of finitely generated structures with automorphisms such
that the first substructure embed into the second by inclusion, i.e.
$(A_{n,k}, \alpha_{n,k})\subseteq (B_{n,k}, \beta_{n,k})$, and $f_{n,k}$
is an embedding of $(A_{n,k}, \alpha_{n,k})$ in the $(\FrGr_{n-1}, g_{n-1})$.
We allow $A_{n,k}$ to be empty. Although $g_{n-1}$ is a finite function,
we may treat it as a partial automorphism as we have said before.
\begin{enumerate}[resume, label=(\roman*)]
\item
Let
$(i, j) = \min\{(\{0, 1, \ldots, n\} \times \bN) \setminus X_{n-1}\}$ (with the
order induced by $\gamma$). Then $X_n = X_{n-1}\cup\{(i,j)\}$ and
$(B_{i,j}, \beta_{i,j})$ embeds in $(\FrGr_n, g_n)$ so that this diagram
commutes:
\begin{center}
\begin{tikzcd}
& (\Gamma_n, g_n) & \\
(B_{i,j}, \beta_{i,j}) \arrow[ur, dashed, "\hat{f}_{i,j}"] & & (\FrGr_{n-1}, g_{n-1}) \arrow[ul, dashed, "\subseteq"'] \\
& (A_{i,j}, \alpha_{i,j}) \arrow[ur, "f_{i,j}"'] \arrow[ul, "\subseteq"]
\end{tikzcd}
\end{center}
\end{enumerate}
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.
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
think it is simply a map from one generators of $\Gamma'_n$ to the
others. By the weak ultrahomogeneity of $\Gamma$, we may assume that
$\Gamma'_n\subseteq \Gamma$.
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)
without the loss of generality we
may assume that $\Gamma_n\subseteq\Gamma$. This way we've constructed $g_n$
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 $\cD$ 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.
Take any $(B, \beta)\in\cD$. Then, there are
$i, j$ such that $(B, \beta) = (B_{i, j}, \beta_{i,j})$ and $A_{i,j}=\emptyset$.
By the bookkeeping there was $n$ such that
$(i, j) = \min\{\{0,1,\ldots n\}\times\bN\setminus X_{n}\}$.
This means that $(B, \beta)$ embeds into $(\Gamma_n, g_n)$, hence it embeds
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 $n\in\bN$
such that $(i,j) = \min\{\{0,1,\ldots n-1\}\times X_{n-1}\}$ and
$(A,\alpha) = (A_{i,j},\alpha_{i,j}), (B,\beta)=(B_{i,j},\beta_{i,j})$ and
$f = f_{i,j}$. This means that there is a compatible embedding of $(B,\beta)$ into
$(\Gamma_n, g_n)$, which means we can also embed it into $(\Gamma, g)$.
Hence, $(\Gamma,g)\cong(\Gamma,\sigma)$.
By this we know that $g$ and $\sigma$ 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}
\label{theorem:key-theorem}
Let $\cC$ be a Fraïssé class of finitely generated $L$-structures with WHP
and canonical amalgamation. Then $\Flim(\cC)$ has a generic automorphism.
\end{theorem}
\begin{proof}
It follows trivially from Corollary \ref{corollary:whp+canonical-iso}
and the above Theorem \ref{theorem:generic_aut_general}.
\end{proof}
\subsection{Properties of the generic automorphism}
Let $\cC$ be a Fraïssé class of finitely generated $L$-structures with
weak Hrushovski property and canonical amalgamation.
Let $\cD$ be the Fraïssé class (by the Theorem \ref{theorem:key-theorem}
of the structures of $\cC$ with additional automorphism of the structure.
Let $\Gamma = \Flim(\cC)$.
\begin{proposition}
\label{proposition:fixed_points}
Let $\sigma$ be the generic automorphism of $\Gamma$. Then the set
of fixed points of $\sigma$ is isomorphic to $\Gamma$.
\end{proposition}
\begin{proof}
Let $S = \{x\in \Gamma\mid \sigma(x) = x\}$.
First we need to show that it is an infinite. By the theorem \ref{theorem:generic_aut_general}
we know that $(\Gamma, \sigma)$ is the Fraïssé limit of $\cD$, thus we
can embed finite $L$-structures of any size with identity as an
automorphism of the structure into $(\Gamma, \sigma)$. Thus $S$ has to be
infinite. Also, the same argument shows that the age of the structure is
exactly $\cC$. It is weakly ultrahomogeneous, also by the fact that
$(\Gamma, \sigma)$ is in $\cD$.
\end{proof}
\end{document}
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