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--- a/sections/preliminaries.tex
+++ b/sections/preliminaries.tex
@@ -1,7 +1,27 @@
\documentclass[../lic_malinka.tex]{subfiles}
\begin{document}
+ Before we get to the main work of the paper, we need to establish basic
+ notions, known facts and theorems. This section provides a brief
+ introduction to the theory of Baire spaces and category theory.
+ Most of the notions are well known, interested reader may look at
+ \cite{descriptive_set_theory}, \cite{maclane_1978}
+
\subsection{Descriptive set theory}
+ In this section we provide an important definition of a \emph{comeagre} set.
+ It is purely topological notion, the intuition may come from the measure
+ theory though. For example, in a standard Lebesuge measure on the
+ real interval $[0,1]$, the set of rationals is of measure $0$, although
+ being a dense subset of the $[0,1]$. So, in a sense, the set of rationals
+ is \emph{meagre} in the interval $[0,1]$. On the other hand, the set
+ of irrational numbers is also dense, but have measure $1$, so it is
+ \emph{comeagre}.
+
+ This is only a rough approximation of the topological
+ definition. The definitions are based on the Kechris' book \textit{Classical
+ Descriptive Set Theory} \cite{descriptive_set_theory}. One should look into
+ it for more details and examples.
+
\begin{definition}
Suppose $X$ is a topological space and $A\subseteq X$.
We say that $A$ is \emph{meagre} in $X$ if $A = \bigcup_{n\in\bN}A_n$,
@@ -15,11 +35,10 @@
contains a countable intersection of open dense sets.
\end{definition}
- % \begin{example}
- Every countable set is meagre in any $T_1$ space, so, for example, $\bQ$
+ Every countable set is meagre in any $T_1$ space. So, $\bQ$
is meagre in $\bR$ (although it is dense), which means that the set of
- irrationals is comeagre. Another example is...
- % \end{example}
+ irrationals is comeagre. The Cantor set is nowhere dense, hence meagre
+ in the $[0,1]$ interval.
\begin{definition}
We say that a topological space $X$ is a \emph{Baire space} if every
@@ -43,10 +62,10 @@
$\bigcap_{n}V_n \subseteq A$.
\end{definition}
- There is an important theorem on the Banach-Mazur game: $A$ is comeagre
- 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.
+ There is an important theorem \ref{theorem:banach_mazur_thm} on the
+ Banach-Mazur game: $A$ is comeagre 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.
\begin{definition}
$T$ is \emph{the tree of all legal positions} in the Banach-Mazur game
@@ -96,16 +115,18 @@
$\bigcap_{n}V_n \subseteq A$.
\end{definition}
- Now we can state the key theorem.
-
\begin{theorem}[Banach-Mazur, Oxtoby]
\label{theorem:banach_mazur_thm}
Let $X$ be a nonempty topological space and let $A\subseteq X$. Then A is
comeagre $\Leftrightarrow$ $\textit{II}$ has a winning strategy in
$G^{\star\star}(A)$.
\end{theorem}
+
+ The statement of the theorem is once again taken from Kechris
+ \cite{descriptive_set_theory} 8.33. However, the proof given in the book is
+ brief, thus we present a detailed version. In order to prove the
+ theorem we add an auxiliary 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
@@ -336,7 +357,7 @@
\end{definition}
\begin{definition}
- The \emph{cospan category} of category $\cC$, refered to as $\Cospan(\cC)$,
+ The \emph{cospan category} of category $\cC$, referred to as $\Cospan(\cC)$,
is the category of cospan diagrams of $\cC$, where morphisms between
two cospans are normal transformations of the underlying functors.