From 5e47a4972076f718dcb1f0766fa7bb8016f8056d Mon Sep 17 00:00:00 2001
From: Franciszek Malinka <franciszek.malinka@gmail.com>
Date: Thu, 21 Apr 2022 01:23:51 +0200
Subject: Aspell

---
 lic_malinka.tex | 86 ++++++++++++++++++++++++++++-----------------------------
 1 file changed, 43 insertions(+), 43 deletions(-)

(limited to 'lic_malinka.tex')

diff --git a/lic_malinka.tex b/lic_malinka.tex
index 85541cf..e1a45b3 100644
--- a/lic_malinka.tex
+++ b/lic_malinka.tex
@@ -122,7 +122,7 @@
 
   \begin{definition} 
     We say that $A$ is \emph{comeagre} in $X$ if it is
-    a complement of a meagre set. Equivalently, a set is comeagre iff it
+    a complement of a meagre set. Equivalently, a set is comeagre if and only if it
     contains a countable intersection of open dense sets.  
   \end{definition}
 
@@ -154,7 +154,7 @@
   \end{definition}
 
   There is an important theorem on the Banach-Mazur game: $A$ is comeagre
-  iff $\textit{II}$ can always choose sets $V_0, V_1, \ldots$ such that
+  if and only if $\textit{II}$ can always choose sets $V_0, V_1, \ldots$ such that
   it wins. Before we prove it we need to define notions necessary to
   formalise and prove the theorem.
 
@@ -196,7 +196,7 @@
   Intuitively, a strategy $\sigma$ works as follows: $I$ starts playing
   $U_0$ as any open subset of $X$, then $\textit{II}$ plays unique (by
   (iii)) $V_0$ such that $(U_0, V_0)\in\sigma$. Then $I$ responds by
-  playing any $U_1\subseteq V_0$ and $\textit{II}$ plays uniqe $V_1$
+  playing any $U_1\subseteq V_0$ and $\textit{II}$ plays unique $V_1$
   such that $(U_0, V_0, U_1, V_1)\in\sigma$, etc.
 
   \begin{definition} 
@@ -214,7 +214,7 @@
     $G^{\star\star}(A)$.
   \end{theorem}
 
-  In order to prove it we add an auxilary definition and lemma.
+  In order to prove it we add an auxiliary definition and lemma.
   \begin{definition}
     Let $S\subseteq\sigma$ be a pruned subtree of tree of all legal positions
     $T$ and let $p=(U_0, V_0,\ldots, V_n)\in S$.  We say that S is
@@ -227,7 +227,7 @@
   \end{definition}
 
   \begin{fact} 
-    If $\sigma$ is a winnig strategy for $\mathit{II}$ then
+    If $\sigma$ is a winning strategy for $\mathit{II}$ then
     there exists a nonempty comprehensive $S\subseteq\sigma$. 
   \end{fact}
 
@@ -360,7 +360,7 @@
   \subsection{Definitions} 
   \begin{definition}
     Let $L$ be a signature and $M$ be an $L$-structure. The \emph{age} of $M$ is
-    the class $\bK$ of all finitely generated structures that embedds into $M$.
+    the class $\bK$ of all finitely generated structures that embeds into $M$.
     The age of $M$ is also associated with class of all structures embeddable in
     $M$ \emph{up to isomorphism}.
   \end{definition}
@@ -411,8 +411,8 @@
   \end{definition}
 
   \begin{definition}
-    Let $M$ be an $L$-structure. $M$ is \emph{ultrahomogenous} if every
-    isomorphism between finitely generated substrucutres of $M$ extends to an
+    Let $M$ be an $L$-structure. $M$ is \emph{ultrahomogeneous} if every
+    isomorphism between finitely generated substructures of $M$ extends to an
     automorphism of $M$.
   \end{definition}
 
@@ -422,7 +422,7 @@
     \label{theorem:fraisse_thm}
     Let L be a countable language and let $\bK$ be a nonempty countable set of
     finitely generated $L$-structures which has HP, JEP and AP. Then $\bK$ is 
-    the age of a countable, ultrahomogenous $L$-structure $M$. Moreover, $M$ is
+    the age of a countable, ultrahomogeneous $L$-structure $M$. Moreover, $M$ is
     unique up to isomorphism. We say that $M$ is a \emph{Fraïssé limit} of $\bK$
     and denote this by $M = \Flim(\bK)$.
   \end{theorem}
@@ -432,7 +432,7 @@
   of this theorem appears another, equally important \ref{lemma:weak_ultrahom}.
 
   \begin{definition}
-    We say that an $L$-structure $M$ is \emph{weakly ultrahomogenous} if for any
+    We say that an $L$-structure $M$ is \emph{weakly ultrahomogeneous} if for any
     $A, B$ finitely generated substructures of $M$ such that $A\subseteq B$ and
     an embedding $f\colon A\to M$ there is an embedding $g\colon B\to M$ which
     extends $f$.
@@ -446,8 +446,8 @@
 
   \begin{lemma}
     \label{lemma:weak_ultrahom}
-    A countable structure is ultrahomogenous if and only if it is weakly
-    ultrahomogenous.
+    A countable structure is ultrahomogeneous if and only if it is weakly
+    ultrahomogeneous.
   \end{lemma}
 
   This lemma will play a major role in the later parts of the paper. Weak
@@ -486,7 +486,7 @@
 
   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 verticies if they should be connected or not. It turns out
+  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.
 
@@ -501,38 +501,38 @@
     Take any finite disjoint $X, Y\subseteq\FrGr$. Let $G_{XY}$ be the
     subgraph of $\FrGr$ induced by the $X\cup Y$. Let $H = G_{XY}\cup \{w\}$,
     where $w$ is a new vertex that does not appear in $G_{XY}$. Also, $w$ is connected to 
-    all verticies of $G_{XY}$ that come from $X$ and to none of those that come
+    all vertices of $G_{XY}$ that come from $X$ and to none of those that come
     from $Y$. This graph is of course finite, so it is embeddable in $\FrGr$.
-    Wihtout loss of generality assume that this embedding is simply inclusion. 
+    Without loss of generality assume that this embedding is simply inclusion. 
     Let $f$ be the partial isomorphism from $X\sqcup Y$ to $H$, with $X$ and
     $Y$ projected to the part of $H$ that come from $X$ and $Y$ respectively. 
-    By the ultrahomogeneity of $\FrGr$ this isomoprhism extends to an automorphism
+    By the ultrahomogeneity of $\FrGr$ this isomorphism extends to an automorphism
     $\sigma\in\Aut(\FrGr)$. Then $v = \sigma^{-1}(w)$ is the vertex we sought. 
   \end{proof}
 
   \begin{fact}
     If a countable graph $G$ has the random graph property, then it is
-    isomorhpic to the random graph $\FrGr$.
+    isomorphic to the random graph $\FrGr$.
   \end{fact}
   \begin{proof}
-    Enumerate verticies of both graphs: $\FrGr = \{a_1, a_2\ldots\}$ and $G
+    Enumerate vertices of both graphs: $\FrGr = \{a_1, a_2\ldots\}$ and $G
     = \{b_1, b_2\ldots\}$.
     We will construct a chain of partial isomorphisms $f_n\colon \FrGr\to G$
     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 isomoprhism. Let $X = \{a\in\FrGr\mid
+    a_{n+1}, b\rangle\}$ is a partial isomorphism. 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
-    verticies of $\dom{f_n}$ that are connected with $a_{n+1}$ in $\FrGr$ and
-    $Y$ are those verticies that are not connected with $a_{n+1}$. Let $b$ be
-    a vertex of $G$ that is connected to all verticies of $f_n[X]$ and to none
+    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
+    a vertex of $G$ that is connected to all vertices of $f_n[X]$ and to none
     $f_n[Y]$ (it exists by the random graph property). Then $f_n\cup \{\langle
-    a_{n+1}, b\rangle\}$ is a partial isomoprhism. We find $a$ for the
+    a_{n+1}, b\rangle\}$ is a partial isomorphism. We find $a$ for the
     $b_{n+1}$ 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 isomoprhism between $\FrGr$
+    $f = \bigcup^{\infty}_{n=0}f_n$ is an isomorphism between $\FrGr$
     and $G$. Take any $a, b\in \FrGr$. Then for some big enough $n$ we have
     that $aE_{\FrGr}b\Leftrightarrow f_n(a)E_{G}f_n(b) \Leftrightarrow f(a)E_{G}f(b)$ .
   \end{proof}
@@ -576,10 +576,10 @@
     a Fraïssé class.
   \end{proposition}
   \begin{proof}
-    Countability and HP are obivous, JEP follows by the same argument as in
+    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 $A$ embedds
+    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
     the proof for the plain graphs. We will define the automorphism
     $\delta\in\Aut(D)$ so it extends $\beta$ and $\gamma$. (TODO: chyba nie
@@ -619,9 +619,9 @@
 
     Let $G_{XY}$ be the graph induced by $X\cup Y$. Take $H=G_{XY}\sqcup {v}$ 
     as a supergraph of $G_{XY}$ with one new vertex $v$ connected to all 
-    verticies of $X$ and to none of $Y$. By the proposition
+    vertices of $X$ and to none of $Y$. By the proposition
     \ref{proposition:finite-graphs-whp} we can extend $H$ together with its
-    partial isomoprhism $\sigma\upharpoonright_{X\cup Y}$ to a graph $R$ with
+    partial isomorphism $\sigma\upharpoonright_{X\cup Y}$ to a graph $R$ with
     automorphism $\tau$. Once again, without the loss of generality we can
     assume that $R\subseteq\FrAut$, because $\cH$ is the age of $\FrAut$. But
     $R\upharpoonright_{G_{XY}}$ together with $\tau\upharpoonright_{G_{XY}}$
@@ -629,7 +629,7 @@
 
     Thus, by ultrahomogeneity of $\FrAut$ this isomorphism extends to an 
     automorphism $\theta$ of $(\FrAut, \sigma)$. Then $\theta(v)$ is the vertex
-    that is connected to all verticies of $X$ and none of $Y$, because
+    that is connected to all vertices of $X$ and none of $Y$, because
     $\theta[R\upharpoonright_X] = X, \theta[R\upharpoonright_Y] = Y$.
   \end{proof}
 
@@ -712,7 +712,7 @@
 	player \textit{I} picks functions $f_0, f_1,\ldots$ and player \textit{II}
 	chooses $g_0, g_1,\ldots$ such that
 	$f_0\subseteq g_0\subseteq f_1\subseteq g_1\subseteq\ldots$, which identify
-	the coresponding basic open subsets $B_{f_0}\supseteq B_{g_0}\supseteq\ldots$.
+	the corresponding basic open subsets $B_{f_0}\supseteq B_{g_0}\supseteq\ldots$.
 
 	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
@@ -721,28 +721,28 @@
 	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,
+	the loss of generality that $\FrAut = \FrGr$ as a plain graph. Hence,
 	by the fact	\ref{fact:conjugacy}, $g$ and $\sigma$ conjugate in $G$.
 
 	Once again, by the Fraïssé theorem \ref{theorem:fraisse_thm} and the
 	\ref{lemma:weak_ultrahom} lemma constructing $g_i$'s in a way such that
-	age of $(\FrGr, g)$ is exactly $\cH$ and so that it is weakly ultrahomogenous
+	age of $(\FrGr, g)$ is exactly $\cH$ and so that it is weakly ultrahomogeneous
 	will produce expected result. First, let us enumerate all pairs of finite
 	graphs with automorphism $\{\langle(A_n, \alpha_n), (B_n, \beta_n)\rangle\}_{n\in\bN}$
-	such that the first element of the pair embedds by inclusion in the second,
+	such that the first element of the pair embeds by inclusion in the second,
 	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 
+	$A_n$ is an empty graph. We enumerate the vertices of the random graph 
 	$\FrGr = \{v_0, v_1, \ldots\}$.
 
 	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
+	$n, m\in\bN$ we have $\gamma(n, m) \ge n$. This bijection naturally 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 $X_n\subseteq\bN^2$
+	isomorphism $f_n$. 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*)]
@@ -757,7 +757,7 @@
 		is the graph induced by $g_{n-1}$). Let 
 		$(i, j) = \min\{(\{0, 1, \ldots\} \times \bN) \setminus X_{n-1}\}$ (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
+		$(B_{n,k}, \beta_{n,k})$ embeds in $(\FrGr_n, g_n)$ so that this diagram
 		commutes:
 
 		\begin{center}
@@ -768,7 +768,7 @@
 		\end{center}
 	\end{enumerate}
 
-	First item makes sure that no inifite orbit will not be present in $g$. The 
+	First item makes sure that no infinite 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. Now we will show that indeed such $g_n$ may be constructed.
@@ -778,7 +778,7 @@
 	$(B_{i,j}, \beta_{i,j})$ 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 $(\FrGr'_n, g'_n)$,
+	Then $(B_{i,j}, \beta_{i,j})$ simply embeds by inclusion in $(\FrGr'_n, g'_n)$,
 	i.e. this diagram commutes:
 
 	\begin{center}
@@ -792,9 +792,9 @@
 	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 
+	Of course by the recursive 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
+	$\FrGr$ by the random graph property -- we can find vertices 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
@@ -811,9 +811,9 @@
 	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
+	This means that $(B, \beta)$ embeds into $(\FrGr_n, g_n)$, hence it embeds
 	into $(\FrGr, g)$. With a similar argument we can see that $(\FrGr, g)$ is
-	weakly ultrahomogenous.
+	weakly ultrahomogeneous.
 
 	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.
-- 
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