elaborating on trees

Author: rzach

content/sets-functions-relations/relations/orders.tex

@@ -60,12 +60,23 @@
 
 \begin{ex}
 The relation of \emph{divisibility without remainder} gives us a
-partial order which isn't a linear order. For integers $n$, $m$, we
-write $n\mid m$ to mean $n$ (evenly) divides $m$, i.e., iff there is
-some integer~$k$ so that $m=kn$. On $\Nat$, this is a partial order, but not a
-linear order: for instance, $2\nmid3$ and also $3\nmid2$. Considered
-as a relation on $\Int$, divisibility is only a preorder since
-it is not anti-symmetric: $1\mid-1$ and $-1\mid1$ but $1\neq-1$.
+partial order which isn't a linear order. For integers $n$ and~$m$, we
+write $n \mid m$ to mean $n$ (evenly) divides $m$, i.e., iff there is
+some integer~$k$ so that $m = kn$. On~$\Nat$, this is a partial order,
+but not a linear order: for instance, $2 \nmid 3$ and also $3 \nmid
+2$. Considered as a relation on $\Int$, divisibility is only a
+preorder since it is not anti-symmetric: $1 \mid -1$ and $-1 \mid 1$
+but $1 \neq -1$.
+\end{ex}
+
+\begin{ex}
+The \emph{extension} relation on a set of sequences~$A^*$ is the
+following: $s \sqsubseteq s'$ iff $s = \emptyseq$ (the empty
+sequence), $s = s'$, or $s = \tuple{s_1, \dots, s_n}$ and $s' =
+\tuple{s_1, \dots, s_n, s_{n+1}, \dots, s_m}$. If $s \sqsubseteq s'$
+we also say that $s$ is an \emph{initial segment} of~$s'$. The
+extension relation on $A^*$ is a partial order but not a linear order,
+e.g., if $a \neq b$, then $ab \not\sqsubseteq ba$ and $ba \not\sqsubseteq ab$.
 \end{ex}
 
 \begin{defn}[Strict order]

content/sets-functions-relations/relations/trees.tex

@@ -13,66 +13,117 @@
 all parts of logic is a \emph{tree}. Finite trees occur in elementary
 parts of logic: for example, !!{formula}s can be understood in terms
 of their decomposition into a syntax tree, while !!{derivation}s in
-natural deduction also take the form of a finite tree.
+many derivation systems also take the form of finite trees.
 %
 Infinite trees appear already in the proof of the completeness
 theorems for propositional and first-order logic, and are used
-throughout mathematical logic. For example, in descriptive set theory,
-many pointclasses of real numbers (such as Borel sets or analytic sets)
-have representations in terms of trees.
+throughout mathematical logic.
+
+The set-theoretic concept of a tree is closely related to the notion
+of a tree in graph theory. Here is a picture of a (finite) tree:
+
+\begin{center}
+\begin{tikzpicture}[nodes={draw, circle}, -]
+\node{r} [grow'=up]
+    child { node {a} 
+        child { node {c} }
+        child { node {d} }
+        child { node {e} }
+    }
+    child { node {b} };
+\end{tikzpicture}
+\end{center}
+
+The lowermost node~$r$ is the root. Every node other than $r$ has
+exactly one parent node immediately below it. We can think of the relation
+a node~$x$ stands in to a node~$y$ if $y$ can be reached from $x$ by
+following edges upwards as $x$ being an ancestor of~$y$. 
+
+The ancestor relation in a tree is a strict partial order. This
+motivates the set-theoretic definition. To state it we need two
+concepts. A \emph{minimal element} in a set~$A$ partially ordered
+by~$\le$ is !!a{element} $x \in A$ such that for all $y \in A$ we have
+that~$x \le y$. A set is \emph{well-ordered} by~$\le$ if every one of
+its subsets has a minimal element.
 
 \begin{defn}[Tree]
-A \emph{tree} is a pair $T = \tuple{X,\le}$ such that $X$ is a set
-and $\le$ is a partial order on $X$ with a unique minimal element
-$r \in X$ (called a \emph{root}) such that for all $t \in X$,
-the set $\Setabs{s}{s \le t}$ is well-ordered by $\le$.
+A \emph{tree} is a pair $T = \tuple{A, \le}$ such that $A$ is a set
+and $\le$ is a partial order on~$A$ with a unique minimal element
+$r \in A$ (called the \emph{root}) such that for all $x \in A$,
+the set $\Setabs{y}{y \le z}$ is well-ordered by~$\le$.
 \end{defn}
 
 \begin{defn}[Successors]
-Suppose $T = \tuple{X,\le}$ is a tree.
-%
-If $t,s \in X$, $t < s$, and there is no $s' \in X$ such that
-$t < s' < s$, then we say that $s$ is a \emph{successor} of $t$.
+Suppose $T = \tuple{A, \le}$ is a tree.
+If $x,y \in A$, $x < y$, and there is no $z \in A$ such that
+$x < z < y$, then we say that $y$ is a \emph{successor} of~$x$.
 \end{defn}
 
-\begin{defn}[Infinite and finitely branching trees]
-Suppose that $T = \tuple{X,\le}$ is a tree.
-%
-$T$ is said to be \emph{infinite} if $X$ is an infinite set,
-\emph{finite} otherwise.
-%
-If $T$ is such that every $t \in X$ has only finitely many
-successors, then we say that $T$ is \emph{finitely branching}.
+The successors of $x \in A$ are also called its \emph{children}. If
+$y$ is a successor of~$x$, then we call $x$ the \emph{predecessor} or
+\emph{parent} of~$y$.
+
+\begin{prop}
+If $\tuple{A,\le}$ is a tree, then every $x \in A$ other than the root
+has at most one predecessor.
+\end{prop}
+
+\begin{proof}
+  Suppose $y < x$ and $y' < x$ and $y \neq y$. Then both $\{y,
+  y'\} \subseteq \Setabs{z}{z<x}$. Since $\Setabs{z}{z<x}$ is
+  well-ordered by~$\le$, it has a minimal element, which obviously
+  must be either $y$ or~$y'$. So either $y \le y'$ or $y' \le y$. We
+  assumed that $y \neq y'$, so actually either $y < y'$ or $y' < y$.
+  Since we assumed that $y < x$ and $y' < x$, we furthermore have that
+  either $y < y' < x$ or $y' < y < x$. So $y$ and $y'$ cannot both be
+  predecessors of~$x$.
+\end{proof}
+
+\begin{defn}
+A tree $T = \tuple{A, \le}$ is said to be \emph{infinite} if $A$ is an
+infinite set, and \emph{finite} otherwise. If $T$ is such that every
+$x \in A$ has only finitely many successors, then we say that $T$ is
+\emph{finitely branching}.
 \end{defn}
 
 \begin{defn}[Branches]
-Given a tree $T = \tuple{X,\le}$, a \emph{branch} of $T$ is a
-maximal chain in $T$, i.e.\ a set $B \subseteq X$ such that
-for any $a,b \in B$ either $a \le b$ or $b \le a$, and for any
-$c \in X \setminus B$ there exists $d \in B$ such that neither
-$c \le d$ nor $d \le c$.
+Given a tree $T = \tuple{A, \le}$, a \emph{branch} of~$T$ is a
+maximal chain in~$T$, i.e., a set $B \subseteq A$ such that
+for any $x, y \in B$ either $x \le y$ or $y \le x$, and for any
+$z \in X \setminus B$ there exists $u \in B$ such that neither
+$z \le u$ nor $u \le z$.
 %
 We use $[T]$ to denote the set of all branches of $T$.
 \end{defn}
 
 \begin{ex}
 A classic example of a finitely branching tree is the
-\emph{binary tree} of finite sequences of $0$s and $1$s,
-sometimes denoted $\{0,1\}^*$, ordered by the extension
+\emph{infinite binary tree} of finite sequences of $0$s and~$1$s,
+sometimes denoted $\{0,1\}^*$ or~$\Bin^*$, ordered by the extension
 relation $\sqsubseteq$ (e.g., $101 \sqsubseteq 101101$).
 Since any binary string can always be extended by adding
 a $0$ or a $1$ on the end, this tree contains infinitely
-many elements. Its root is the empty sequence $\emptyseq$.
+many elements: every element~$s$ has exactly two successors, $s0$ and~$s1$. Its root is the empty sequence $\emptyseq$.
+\end{ex}
+
+\begin{ex}
+Slightly more generally, the set of finite sequences of natural
+numbers~$\Nat^*$ with the extension relation~$\sqsubseteq$ is also a
+tree. It is obviously not finitely branching: every $s \in \Nat^*$ has
+infinitely many successurs~$sn$, one for every $n \in \Nat$. Every $A
+\subseteq \Nat^*$ which is closed under~$\sqsubseteq$ is a
+\emph{subtree} of~$\Nat^*$. (That is, $A$ is such that if $s \in A$
+and $s' \sqsubseteq s$, then also $s' \in A$.) All finite trees can be
+represented as finite subtrees of~$\Nat^*$.
 \end{ex}
 
 \begin{prop}[K\H{o}nig's lemma]
-If $T = \tuple{X,\le}$ is a finitely branching infinite tree,
+If $T = \tuple{A,\le}$ is a finitely branching infinite tree,
 then $T$ has an infinite branch.
 \end{prop}
 
-A special case of K\H{o}nig's lemma widely used in
-computability theory, known as \emph{weak K\H{o}nig's lemma},
-is the following:
-any infinite subtree of $\{0,1\}^*$ has an infinite branch.
+A special case of K\H{o}nig's lemma widely used in computability
+theory, known as \emph{weak K\H{o}nig's lemma}, is the following: any
+infinite subtree of $\{0,1\}^*$ has an infinite branch.
 
 \end{document}