Editorial corrections

1 files changed, 45 insertions, 31 deletions

diff --git a/lic_malinka.tex b/lic_malinka.tex
index af116cd..d38f36e 100644
--- a/lic_malinka.tex
+++ b/lic_malinka.tex
@@ -128,7 +128,7 @@
 

       % \begin{example}

   Every countable set is meagre in any $T_1$ space, so, for example, $\bQ$

-  is meagre in $\bR$ (though being dense), which means that the set of

+  is meagre in $\bR$ (although it is dense), which means that the set of

   irrationals is comeagre. Another example is...

       % \end{example}

 

@@ -140,7 +140,7 @@
 

   \begin{definition}

     Suppose $X$ is a Baire space. We say that a property $P$ \emph{holds

-    generically} for a point in $x\in X$ if $\{x\in X\mid P\textrm{ holds for

+    generically} for a point $x\in X$ if $\{x\in X\mid P\textrm{ holds for

     }x\}$ is comeagre in $X$.  

   \end{definition}

 

@@ -162,15 +162,16 @@
     $T$ is \emph{the tree of all legal positions} in the Banach-Mazur game

     $G^{\star\star}(A)$ when $T$ consists of all finite sequences $(W_0,

     W_1,\ldots, W_n)$, where $W_i$ are nonempty open sets such that

-    $W_0\supseteq W_1\supseteq\ldots\supseteq W_n$. In another words, $T$ is

+    $W_0\supseteq W_1\supseteq\ldots\supseteq W_n$. In other words, $T$ is

     a pruned tree on $\{W\subseteq X\mid W \textrm{is open nonempty}\}$.

   \end{definition} 

 

   \begin{definition}

     We say that $\sigma$ is \emph{a pruned subtree} of the tree of all legal

-    positions $T$ if $\sigma\subseteq T$ and for any $(W_0, W_1, \ldots,

+    positions $T$ if $\sigma\subseteq T$, for any $(W_0, W_1, \ldots,

     W_n)\in\sigma, n\ge 0$ there is a $W$ such that $(W_0, W_1,\ldots, W_n,

-    W)\in\sigma$ (it simply means that there's no finite branch in~$\sigma$).  

+    W)\in\sigma$ (it simply means that there's no finite branch in~$\sigma$) and

+	$(W_0, W_1,\ldots W_{n-1})\in\sigma$ (every node on a branch is in $\sigma$).

   \end{definition}

 

   \begin{definition}

@@ -302,7 +303,7 @@
     the inductive hypothesis $x\in V_{n}$ for the unique $V_n\in S_n$. It

     must be that $V_{n+1}\in \cV_{p_{V_n}}$, because $V_n$ is the only

     superset of $V_{n+1}$ in $S_n$. But $\cV_{p_{V_n}}$ is disjoint, so

-    there is no other $V'_{n+1}\in \cV_{p_{V_n}}$ suc h that $x\in

+    there is no other $V'_{n+1}\in \cV_{p_{V_n}}$ such that $x\in

     V'_{n+1}$. Moreover, there is no such set in

     $S_{n+1}\setminus\cV_{p_{V_n}}$, because those sets are disjoint from

     $V_{n}$. Hence there is no $V'_{n+1}\in S_{n+1}$ other than $V_n$

@@ -461,8 +462,16 @@
 

   \subsection{Random graph} 

   

-  In this section we'll take a closer look on a class of finite graphs, which

-  form a Fraïssé class.

+  In this section we'll take a closer look on a class of finite unordered graphs, 

+  which is a Fraïssé class. 

+

+  The language of unordered graphs $L$ consists of a single binary

+  relational symbol $E$. If $G$ is an $L$-structure then we call it a

+  \emph{graph}, and its elements $\emph{vertices}$. If for some vertices 

+  $u, v\in G$ we have $G\models uEv$ then we say that there is an $\emph{edge}$ 

+  connecting $u$ and $v$. If $G\models \forall x\forall y (xEy\leftrightarrow yEx)$

+  then we say that $G$ is an \emph{unordered graph}. From now on we omit the word

+  \textit{unordered} and  graphs as unordered.

 

   \begin{proposition}

     Let $\cG$ be the class of all finite graphs. $\cG$ is a Fraïssé class.

@@ -475,8 +484,8 @@
     supergraphs of $A$, we can assume without loss of generality, that 

     $(B\setminus A) \cap (C\setminus A) = \emptyset$. Then 

     $A\sqcup (B\setminus A)\sqcup (C\setminus A)$ is the graph we're looking

-    for (with edges as in B and C and without any edges between the disjoint

-    parts).

+    for (with edges as in B and C and without any edges between $B\setminus A$

+	and $C\setminus A$).

   \end{proof}

 

   \begin{definition}

@@ -487,14 +496,14 @@
   The concept of the random graph emerges independently in many fields of

   mathematics. For example, one can construct the graph by choosing at random

   for each pair of vertices if they should be connected or not. It turns out

-  that the graph constructed this way is exactly the random graph we described

-  above.

+  that the graph constructed this way is isomorphic to the random graph with

+  probability 1.

 

   The random graph $\FrGr$ has one particular property that is unique to the

   random graph.

 

   \begin{fact}[random graph property]

-    For each finite disjoint $X, Y\subseteq \FrGr$ there exists $v\in\FrGr$

+	For each finite disjoint $X, Y\subseteq \FrGr$ there exists $v\in\FrGr\setminus (X\cup Y)$

     such that $\forall u\in X (vEu)$ and $\forall u\in Y (\neg vEu)$.

   \end{fact}

   \begin{proof}

@@ -521,8 +530,10 @@
     such that $\emptyset = f_0\subseteq f_1\subseteq f_2\subseteq\ldots$ and $a_n \in

     \dom(f_n)$ and $b_n\in\rng(f_n)$.

 

-    Suppose we have $f_n$. We seek $b\in G$ such that $f_n\cup \{\langle

-    a_{n+1}, b\rangle\}$ is a partial isomorphism. Let $X = \{a\in\FrGr\mid

+	Suppose we have $f_n$. We seek $b\in G$ such that $f_n\cup \{\langle

+    a_{n+1}, b\rangle\}$ is a partial isomorphism. 

+    If $a_{n+1}\in\dom{f_n}$, then simply $b = f_n(a_{n+1})$. Otherwise,

+	let $X = \{a\in\FrGr\mid

     aE_{\FrGr} a_{n+1}\}\cap \dom{f_n}, Y = X^{c}\cap \dom{f_n}$, i.e. $X$ are

     vertices of $\dom{f_n}$ that are connected with $a_{n+1}$ in $\FrGr$ and

     $Y$ are those vertices that are not connected with $a_{n+1}$. Let $b$ be

@@ -532,9 +543,9 @@
     $b_{n+1}$ in the similar manner, so that $f_{n+1} = f_n\cup \{\langle

     a_{n+1}, b\rangle, \langle a, b_{n+1}\rangle\}$ is a partial isomorphism.

 

-    $f = \bigcup^{\infty}_{n=0}f_n$ is an isomorphism between $\FrGr$

+    Finally, $f = \bigcup^{\infty}_{n=0}f_n$ is an isomorphism between $\FrGr$

     and $G$. Take any $a, b\in \FrGr$. Then for some big enough $n$ we have

-    that $aE_{\FrGr}b\Leftrightarrow f_n(a)E_{G}f_n(b) \Leftrightarrow f(a)E_{G}f(b)$ .

+    that $aE_{\FrGr}b\Leftrightarrow f_n(a)E_{G}f_n(b) \Leftrightarrow f(a)E_{G}f(b)$.

   \end{proof}

 

   Using this fact one can show that the graph constructed in the probabilistic

@@ -542,7 +553,8 @@
 

   \begin{definition} We say that a Fraïssé class $\bK$ has \emph{weak

     Hrushovski property} (\emph{WHP}) if for every $A\in \bK$ and an isomorphism

-    of substructures of $A$ $p\colon A\to A$, there is some $B\in \bK$ such 

+	of its substructures $p\colon A\to A$ (also called a partial automorphism of $A$),

+	there is some $B\in \bK$ such 

     that $p$ can be extended to an automorphism of $B$, i.e.\ there is an 

     embedding $i\colon A\to B$ and a $\bar p\in \Aut(B)$ such that the following

     diagram commutes: 

@@ -567,10 +579,10 @@
   The language and theory of unordered graphs is fairly simple. We extend the

   language by one unary symbol $\sigma$ and interpret it as an arbitrary

   automorphism on the graph structure. It turns out that the class of such

-  structures forms a Fraïssé class.

+  structures is a Fraïssé class.

 

   \begin{proposition}

-    Let $\cH$ be the class of all finite graphs with automorphism, i.e.

+    Let $\cH$ be the class of all finite graphs with an automorphism, i.e.

     structures in the language $(E, \sigma)$ such that $E$ is a symmetric

     relation and $\sigma$ is an automorphism on the structure. $\cH$ is

     a Fraïssé class.

@@ -579,21 +591,23 @@
     Countability and HP are obvious, JEP follows by the same argument as in

     plain graphs. We need to show that the class has the amalgamation property.

 

-    Take any graphs $(A, \alpha), (B, \beta), (C,\gamma)$ such that $A$ embeds

-    into $B$ and $C$. Let $D$ be the amalgamation of $B$ and $C$ over $A$ as in

+    Take any $(A, \alpha), (B, \beta), (C,\gamma)\in\cH$ such that $A$ embeds

+    into $B$ and $C$. Without the loss of generality we may assume that 

+	$B\cap C = A$ and $\alpha\subseteq\beta,\gamma$. 

+	Let $D$ be the amalgamation of $B$ and $C$ over $A$ as in

     the proof for the plain graphs. We will define the automorphism

     $\delta\in\Aut(D)$ so it extends $\beta$ and $\gamma$. 

-	We let $\delta_{\upharpoonright X} = \id_X$ for $X\in \{A,

-    B\setminus A, C\setminus B\}$. Let's check the definition is correct. In

-    order to do that, we have to show that for any $u, v\in

-    D\quad(uE_Dv\leftrightarrow \delta(u)E_D\delta(v))$. We have two cases:

+	We let $\delta\upharpoonright_B = \beta, \delta\upharpoonright_C = \gamma$.

+	Let's check the definition is correct. We have to show that 

+	$(uE_Dv\leftrightarrow \delta(u)E_D\delta(v))$ holds for any $u, v\in

+	D$. We have two cases:

     \begin{itemize}

       \item $u, v\in X$, where $X$ is either $B$ or $C$. This case is trivial.

       \item $u\in B\setminus A, v\in C\setminus A$. Then

         $\delta(u)=\beta(u)\in B\setminus A$, similarly $\delta(v)=\gamma(v)\in

         C\setminus A$. This follows from the fact, that $\beta\upharpoonright_A

         = \alpha$, so for any $w\in A\quad\beta^{-1}(w)=\alpha^{-1}(w)\in A$,

-        similarly for $\FrGr$. Thus, from the construction of $D$, $\neg uE_Dv$

+        similarly for $\gamma$. Thus, from the construction of $D$, $\neg uE_Dv$

         and $\neg \delta(u)E_D\delta(v)$.

     \end{itemize}

   \end{proof}

@@ -604,15 +618,15 @@
 

   \begin{proposition}

 	\label{proposition:graph-aut-is-normal}

-    The Fraïssé limit of $\cH$ interpreted as a plain graph is isomorphic to

-    the random graph $\FrGr$. 

+	The Fraïssé limit of $\cH$ interpreted as a plain graph (i.e. as a reduct

+	to the language of graphs) is isomorphic to the random graph $\FrGr$. 

   \end{proposition}

 

   \begin{proof}

     It is enough to show that $\FrAut = \Flim(\cH)$ has the random graph

     property. Take any finite disjoint $X, Y\subseteq \FrAut$. Without the loss

-    of generality assume that $X\cup Y$ is invariant to $\sigma$, i.e.

-    $\sigma(v)\in X\cup Y$ for $v\in X\cup Y$. This assumption can be done

+    of generality assume that $X\cup Y$ is $\sigma$-invariant, i.e.

+    $\sigma(v)\in X\cup Y$ for $v\in X\cup Y$. This assumption can be made 

     because there are no infinite orbits in $\sigma$, which in turn is due to

     the fact that $\cH$ is the age of $\FrAut$.