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-rw-r--r--sections/conj_classes.tex2
-rw-r--r--sections/examples.tex73
2 files changed, 62 insertions, 13 deletions
diff --git a/sections/conj_classes.tex b/sections/conj_classes.tex
index d37afd5..87fa416 100644
--- a/sections/conj_classes.tex
+++ b/sections/conj_classes.tex
@@ -144,7 +144,7 @@
such that the following properties hold:
\begin{enumerate}[label=(\roman*)]
- \item $g_n$ is is a partial automorphism of $\Gamma$ and an automorphism of
+ \item $g_n$ 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}
diff --git a/sections/examples.tex b/sections/examples.tex
index 0b45b32..f65ebce 100644
--- a/sections/examples.tex
+++ b/sections/examples.tex
@@ -79,7 +79,8 @@
On the other hand $\cL$ cannot have $WHP$. This follows from the fact that
that only automoprhism of a finite linear ordering is identity, so
we cannot extend a partial automoprhism sending exactly one element to
- some distinct element.
+ some distinct element. However, in this case, generic automoprhism exists
+ which was shown by Truss \cite{truss_gen_aut}.
\begin{definition}
Let $X$ be a set. A ternarny relation $\le^C \subseteq X^3$ is a
@@ -95,22 +96,70 @@
\end{itemize}
\end{definition}
- \begin{example}
- Let $\cC$ be the class of all finite cyclic orderings. Then:
- \begin{enumerate}
- \item $\cC$ is a Fraïssé class.
- \item $\cC$ does not have WHP.
- \item $\cC$ does not have CAP.
- \end{enumerate}
- \end{example}
+ It is easy to visualise a cyclic ordering as a directed (\textit{nomen omen})
+ cycle. For example, a 11-element cyclic order could be drawn like this:
+
+ \begin{figure}[h]
+ \centering
+ \begin{tikzpicture}
+ \def \n {11}
+ \def \radius {3cm}
+ \def \margin {10}
+
+ \foreach \s in {1,...,\n}
+ {
+ \node[draw, circle] at ({360/\n * (\s - 1)}:\radius) {};
+ \draw[->, >=latex] ({360/\n * (\s - 1)+\margin}:\radius)
+ arc ({360/\n * (\s - 1)+\margin}:{360/\n * (\s)-\margin}:\radius);
+ }
+ \node[draw, circle, fill=green] at ({360/\n * (1 - 1)}:\radius) {};
+ \node[draw, circle, fill=blue] at ({360/\n * (5 - 1)}:\radius) {};
+ \node[draw, circle, fill=red] at ({360/\n * (9 - 1)}:\radius) {};
+ \end{tikzpicture}
+ \end{figure}
+
+ For three elements $a, b, c$ we say that $a\le_bc$ if after ''cutting'' the
+ cycle at $b$ we have a path from $a$ to $c$.
+ In this particular example we can say that the green element is
+ red-element-smaller than the blue one.
\begin{example}
- The class of all finitely generated vector spaces $\cV$ is a Fraïssé class
- with WHP and CAP.
+ The class $\cC$ of all finite cyclic orders is a Fraïssé class, but does
+ not have WHP or CAP.
\end{example}
+
+ It is not hard to show that $\cC$ is indeed the Fraïssé class. As usual,
+ the hardest part is showing AP, which in this case is done analogously to
+ the linear orders. The Fraïssé limit of $\cC$ is a countable unit circle.
+
+ $\cC$ hasn't WHP by the similar argument to this for linear
+ orderings. Imagine a cycling order of three elements and a partial automoprhism
+ with one fixed point and moving second element to the third. This cannot be
+ extended to automoprhism of any finite cyclic order.
+
+ Also, $\cC$ cannot have CAP. A reason to that is that it do not admit
+ canonical amalgamation over the empty structure see this by taking
+ 1-element cyclic order and 3-element cyclic order with automoprhism other
+ than identity).
+
+ % \begin{example}
+ % The class of all finitely generated vector spaces over a countable field
+ % $\cV$ is a Fraïssé class with WHP and CAP.
+ % \end{example}
+ %
+ % The prove of this is relatively easy knowing that there is essentially one
+ % vector space of finite dimension and that every linear independent subset
+ % of a vector space can be extended to a basis of this space. The Fraïssé limit
+ % of $\cV$ is the $\omega$-dimensional vector space.
+
\begin{example}
- The class of all finite graphs $\cG$ is a
+ The class of all finite graphs $\cG$ is a Fraïssé class with WHP and free
+ amalgamation.
\end{example}
+
+ We have already shown this fact. Thus get that the random graph has a generic
+ automoprhism.
+
\begin{example}
Graphs without triangles.
\end{example}