From af08209bb6f9fc9f3c293aed0b35649160174b34 Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Tue, 23 Aug 2022 22:03:59 +0200
Subject: More examples

---
 sections/examples.tex | 73 ++++++++++++++++++++++++++++++++++++++++++---------
 1 file changed, 61 insertions(+), 12 deletions(-)

(limited to 'sections/examples.tex')

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}
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