diff options
author | Franciszek Malinka <franciszek.malinka@gmail.com> | 2022-04-21 01:14:27 +0200 |
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committer | Franciszek Malinka <franciszek.malinka@gmail.com> | 2022-04-21 01:14:27 +0200 |
commit | 753ff60029e269239a60f64cf036626d7ab7377e (patch) | |
tree | 3d93731e79109efd456178412d21131d3f3f9ca5 /lic_malinka.tex | |
parent | d457bc06427c6e70c0d6cb19b1b6203bdd57ffc9 (diff) |
Main proof maybe finished
Diffstat (limited to 'lic_malinka.tex')
-rw-r--r-- | lic_malinka.tex | 78 |
1 files changed, 66 insertions, 12 deletions
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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