From 753ff60029e269239a60f64cf036626d7ab7377e Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Thu, 21 Apr 2022 01:14:27 +0200
Subject: Main proof maybe finished

---
 lic_malinka.tex | 78 ++++++++++++++++++++++++++++++++++++++++++++++++---------
 1 file changed, 66 insertions(+), 12 deletions(-)

(limited to 'lic_malinka.tex')

diff --git a/lic_malinka.tex b/lic_malinka.tex
index 47dea88..85541cf 100644
--- a/lic_malinka.tex
+++ b/lic_malinka.tex
@@ -717,7 +717,8 @@
 	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 random graph
 	such that $(\FrGr, g) = \Flim{\cH}$, i.e. the Fraïssé limit of finite
-	graphs with automorphism. By the Fraïssé theorem \ref{theorem:fraisse_thm}
+	graphs with automorphism. Precisely, $\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$.
+	By the Fraïssé theorem \ref{theorem:fraisse_thm}
 	it will follow that $(\FrGr, g)\cong (\FrAut, \sigma)$. By the 
 	proposition \ref{proposition:graph-aut-is-normal} we can assume without
 	the loss of generatlity that $\FrAut = \FrGr$ as a plain graph. Hence,
@@ -732,11 +733,18 @@
 	i.e. $(A_n, \alpha_n)\subseteq (B_n, \beta_n)$. Also, it may be that 
 	$A_n$ is an empty graph. We enumerate the verticies of the random graph 
 	$\FrGr = \{v_0, v_1, \ldots\}$.
-	
-	Just for sake of fixing a technical problem, let $g_{-1} = \emptyset$. 
-	Suppose that player \textit{I} in the $n$-th move chose a finite partial
-	isomoprhism $f_n$. We will construct $g_n\supseteq f_n$ such that
-	following properties hold:
+
+	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 naturaly 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.
+
+	Just for sake of fixing a technical problem, let $g_{-1} = \emptyset$ and
+	$X_{-1} = \emptyset$. 
+	Suppose that player \textit{I} in the $n$-th move chooses a finite partial
+	isomoprhism $f_n$. We will construct $g_n\supseteq f_n$ and a set $X_n\subseteq\bN^2$
+	such that following properties hold:
+
 	\begin{enumerate}[label=(\roman*)]
 	  \item $g_n$ is an automorphism of the induced subgraph $\FrGr_n$,
 	  \item $g_n(v_n)$ and $g_n^{-1}(v_n)$ are defined,
@@ -747,14 +755,15 @@
 		$(A_{n,k}, \alpha_{n,k})\subseteq (B_{n,k}, \beta_{n,k})$, and $f_{n,k}$
 		is an embedding	of $(A_{n,k}, \alpha_{n,k})$ in the $\FrGr_{n-1}$ (which
 		is the graph induced by $g_{n-1}$). Let 
-		$(i, j) = \min\{\{0, 1, \ldots\} \times \bN \setminus X_{n-1}\}$. Then
+		$(i, j) = \min\{(\{0, 1, \ldots\} \times \bN) \setminus X_{n-1}\}$ (with the
+		order induced by $\gamma$). Then $X_n = X_{n-1}\cup\{(i,j)\}$ and
 		$(B_{n,k}, \beta_{n,k})$ embedds in $(\FrGr_n, g_n)$ so that this diagram
 		commutes:
 
 		\begin{center}
 		  \begin{tikzcd}
-			(A_{i,j}, \alpha_{i,j}) \arrow[d, "\subseteq"'] \arrow[r, "f_{i,j}"] & (\FrGr_n, g_n) \\
-			(B_{i,j}, \beta_{i,j}) \arrow[ur, dashed, "\hat{f}_{i,j}"']
+			(A_{i,j}, \alpha_{i,j}) \arrow[d, "\subseteq"'] \arrow[r, "f_{i,j}"] & (\FrGr_{n-1}, g_{n-1}) \arrow[d, "\subseteq"] \\
+			(B_{i,j}, \beta_{i,j}) \arrow[r, dashed, "\hat{f}_{i,j}"'] & (\FrGr_n, g_n)
 		  \end{tikzcd}
 		\end{center}
 	\end{enumerate}
@@ -762,10 +771,55 @@
 	First item makes sure that no inifite orbit will not be present in $g$. The 
 	second item together with the first one are necessary for $g$ to be an 
 	automorphism of $\FrGr$. The third item is the one that gives weak
-	ultrahomogeneity. 
+	ultrahomogeneity. Now we will show that indeed such $g_n$ may be constructed.
+	
+	First, we will suffice the (iii) item. Namely, we will construct $\FrGr'_n, g'_n$
+	such that $g_{n-1}\subseteq g'_n$ and $f_{i,j}$ extends to an embedding of
+	$(B_{i,j}, \beta_{i,j})$ to $(\FrGr'_n, g'_n)$. Without the loss of generality
+	we may assume that $f_{i,j}$ is an inclusion and that $A_{i,j} = B_{i,j}\cap\FrGr_{n-1}$.
+	Then let $\FrGr'_n = B_{i,j}\cup\FrGr_{n-1}$ and $g'_n = g_{n-1}\cup \beta_{i,j}$.
+	Then $(B_{i,j}, \beta_{i,j})$ simply embedds by inclusion in $(\FrGr'_n, g'_n)$,
+	i.e. this diagram commutes:
+
+	\begin{center}
+	  \begin{tikzcd}
+		(A_{i,j}, \alpha_{i,j}) \arrow[d, "\subseteq"'] \arrow[r, "\subseteq"] & (\FrGr_{n-1}, g_{n-1}) \arrow[d, "\subseteq"] \\
+		(B_{i,j}, \beta_{i,j}) \arrow[r, dashed, "\subseteq"'] & (\FrGr'_n, g'_n)
+	  \end{tikzcd}
+	\end{center}
+
+	The argument that those are actually embeddings is almost the same as in
+	proof of the amalgamation property of $\cH$.	
+
+	It may be also assumed without the loss of generality that $\FrGr'_n\subseteq \FrGr$.
+	Of course by the recurisve assumption $\FrGr_{n-1}\subseteq\FrGr$. The 
+	$\FrGr'_n \setminus\FrGr_{n-1} = B_{i,j}\setminus A_{i,j}$ can be found in
+	$\FrGr$ by the random graph property -- we can find verticies of the remaining
+	part of $B_{i,j}$ each at a time so that all edges are correct.
+
+	Now, by the WHP of $\bK$ we can extend the graph $\FrGr'_n\cup\{v_n\}$ together
+	with its partial isomorphism $g'_n$ to a graph $\FrGr_n$ together with its
+	automorphism $g_n\supseteq g'_n$, where without the loss of generality we
+	may assume that $\FrGr_n\subseteq\FrGr$. 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 $\FrGr$ such that $(\FrGr, g)$ has the age $\cH$ and has weak ultrahomogeneity.
+	It is of course an automorphism of $\FrGr$ as it is defined for every $v\in\FrGr$
+	and is a sum of increasing chain of finite automorphisms.
+
+	Take any finite graph with automorphism $(B, \beta)$. 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 $(i, j) = \min\{\{0,1,\ldots\}\times\bN\setminus X_{n-1}\}$.
+	This means that $(B, \beta)$ embedds into $(\FrGr_n, g_n)$, hence it embedds
+	into $(\FrGr, g)$. With a similar argument we can see that $(\FrGr, g)$ is
+	weakly ultrahomogenous.
+
+	By this we know that $g$ and $\sigma$ conjugate in $G$. As we stated in the
+	beginning of the proof, the set 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.
   \end{proof}
 
-
-
   \printbibliography
 \end{document}
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