(chap-trees)=
\newcommand{\N}{\mathbb{N}}
\newcommand{\Nat}{\mathbb{N}}
\newcommand{\Real}{\mathbb{R}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\eps}{\varepsilon}
\newcommand{\Cant}{2^{\Nat}}
\newcommand{\Baire}{\Nat^{\Nat}}
\newcommand{\Rest}[1]{|_{#1}}
\newcommand{\Cl}[1]{\overline{#1}}
\newcommand{\Str}[1][2]{{#1}^{<\Nat}}
\newcommand{\Sle}{\subset}
\newcommand{\Sleq}{\subseteq}
\newcommand{\Sgr}{\supset}
\newcommand{\Sgeq}{\supseteq}
\newcommand{\Conc}{\mbox{}^\frown}
\newcommand{\Estr}{\varnothing}
\newcommand{\Nstr}{\Nat^{<\Nat}}
\newcommand{\Cyl}[1]{N_{#1}}
\DeclareMathOperator{\KB}{KB}
\DeclareMathOperator{\lex}{lex}
Let
:label: def-tree
A **tree** on $A$ is a set $T \subseteq \Str[A]$ that is **closed under prefixes**, that is
$$
\forall \sigma, \tau \; [\tau \in T \: \& \: \sigma \Sleq \tau \; \Rightarrow \; \sigma \in T ]
$$
We call the elements of
A sequence
An important criterion for a tree to have infinite paths is the following.
:label: thm-Koenigs-lemma
Any tree $T$ with infinitely many nodes that is **finite branching** (i.e. each node has at most finitely many immediate extensions) has an infinite path.
We construct an infinite path inductively.
Let $T_\sigma$ denote the tree "above" $\sigma$, i.e. $T_\sigma = \{ \tau \in \Str[A] \colon \sigma\Conc\tau \in T\}$. If $T$ is finite branching, by the **pigeonhole principle**, at least one of the sets $T_\sigma$ for $|\sigma| = 1$ must be infinite. Pick such a $\sigma$ and let $\alpha\Rest{1} = \sigma$.
Repeat the argument for $T = T_\sigma$ and continue inductively. This yields a sequence $\alpha \in [T]$.
If
is well-founded, i.e. it does not have an infinite descending chain.
If
-
If
$\sigma$ is a terminal node, i.e.$\sigma$ has no extensions in$T$ , then let$\rho_T(\sigma) = 0$ . -
If
$\sigma$ is not terminal, and$\rho_T(\tau)$ has been defined for all$\tau \Sgr \sigma$ , we set$\rho_T(\sigma) = \sup {\rho_T(\tau)+1 \colon \tau \in T, \tau \Sgr \sigma }$ . -
Finally, set
$\rho(T) = \sup{\rho_T(\sigma) + 1 \colon \sigma\in T } = \rho_T(\Estr)+1$ , where$\Estr$ denotes the empty string.
Now suppose
\begin{equation*}
\sigma \leq_{\lex} \tau \quad :\Leftrightarrow \quad \
\sigma = \tau ; \text{ or } ; \exists i <|\sigma|,|\tau| : \left( \sigma_i <_A \tau_i \text{ and } \forall j < i ; \sigma_j = \tau_j \right )
\end{equation*}
This ordering extends to
:label: prop-leftmost-branch
If $\leq_A$ is a well-ordering of $A$ and $T$ is a tree on $A$ with $[T] \neq \emptyset$, then $[T]$ has a $\leq_{\lex}$-minimal element, the **leftmost branch**.
We **prune** the tree $T$ by deleting any node that is not on an infinite branch. This yields a subtree $T' \subseteq T$ with $[T'] = [T]$.
Let $T'_n = \{\sigma \in T' \colon |\sigma| = n \}$. Since $\leq_A$ is a well-ordering on $A$, $T'_1$ must have a $\leq_{\lex}$-least element. Denote it by $\alpha\Rest{1}$. Since $T'$ is pruned, $\alpha\Rest{1}$ must have an extension in $T$, and we can repeat the argument to obtain $\alpha\Rest{2}$.
Continuing inductively, we define an infinite path $\alpha$ through $T'$, and it is straightforward to check that $\alpha$ is a $\leq_{\lex}$-minimal element of $[T']$ and hence of $[T]$.
We can combine the
:label: def-Kleene-Brouwer
The **Kleene-Brouwer ordering** $\leq_{\KB}$ of $\Str[A]$ is defined as follows.
$\qquad \sigma \leq_{\KB} \tau$ $\quad :\Leftrightarrow \quad$ $\sigma \Sgeq \tau \;\;$ or $\; \; \sigma \leq_{\lex} \tau$.
This means
We now have
:label: prop-KB-wellorder
Assume $(A,\leq)$ is a well-ordered set. Then for any tree $T$ on $A$,
$\qquad$ $T$ is well-founded $\quad \Leftrightarrow \quad$ $\leq_{\KB}$ restricted to $T$ is a well-ordering.
Suppose $T$ is not well-founded. Let $\alpha \in [T]$. Then $\alpha\Rest{0}, \alpha\Rest{1}, \dots$ is an infinite descending sequence with respect to $\leq_{\KB}$.
Conversely, suppose $\sigma_0 >_{\KB} \sigma_1 >_{\KB} \dots$ is an infinite descending sequence in $T$. By the definition of $>_{\KB}$, this implies $\sigma_1(0) \geq_A \sigma_2(0) \geq_A \dots$ as a sequence in $A$. Since $A$ is well-ordered, this sequence must eventually be constant, say $\sigma_n(0) = a_0$ for all $n \geq n_0$.
Since the $\sigma_n$ are descending, by the definition of $\leq_{\KB}$ it follows that $|\sigma_n| \geq 2$ for $n > n_0$. Hence we can consider the sequence $\sigma_{n_0+1}(1) \geq_A \sigma_{n_0+2}(1) \geq_A \dots$ in $A$. Again, this must be constant $= a_1$ eventually. Inductively, we obtain a sequence $\alpha = (a_0, a_1, a_2, \dots) \in [T]$, that is, $T$ is not well-founded.
The order type of a well-founded tree under $\leq_{\KB}$ usually is not equal to its rank $\rho(T)$.
We can also define an ordering on
For
For
These two mappings allows us henceforth to see trees as subsets of the natural numbers. This will be an important component in exploring the relation between topological and arithmetical complexity.
Let
:label: prop-trees-closed
A set $F \subseteq A^\Nat$ is closed if and only if there exists a tree $T$ on $A$ such that $F = [T]$.
Suppose $F$ is closed. Let
\begin{equation*}
T_F = \{\sigma \in \Str[A] \colon \sigma \Sle \alpha \text{ for some }\alpha \in F\}.
\end{equation*}
Then clearly $F \subset [T_F]$. Suppose $\alpha \in [T_F]$. This means for any $n$, $\alpha\Rest{n} \in T_F$, which implies that there exists $\beta_n \in F$ such that $\alpha_n \Sle \beta_n$. The sequence $(\beta_n)$ converges to $\alpha$, and since $F$ is closed, $\alpha \in F$.
For the other direction, suppose $F = [T]$. Let $\alpha \in A^\Nat \setminus F$. Then there exists an $n$ such that $\alpha\Rest{n} \not\in T$. Since a tree is closed under prefixes, no extension of $\alpha\Rest{n}$ can be in $T$. This implies $\Cyl{\alpha\Rest{n}} \subseteq A^\Nat \setminus F$, and hence $A^\Nat \setminus F$ is open.
Let
This mapping has the following properties:
- It is monotone, i.e.
$\sigma \Sleq \tau$ implies$\phi(\sigma) \Sleq \phi(\tau)$ . - For any
$\alpha \in A^\Nat$ we have$\lim_n |\phi(\alpha\Rest{n})| = \infty$ . This follows directly from the continuity of$f$ : For any neighborhood$\Cyl{\tau}$ of$f(\alpha)$ there exists a neighborhood$\Cyl{\sigma}$ of$\alpha$ such that$f(\Cyl{\sigma}) \subseteq \Cyl{\tau}$ . But$\tau$ has to be of the form$\tau = f(\alpha)\Rest{m}$ , and$\sigma$ of the form$\alpha\Rest{n}$ . Hence for any$m$ there must exist an$n$ such that$\phi(\alpha\Rest{n}) \Sgeq f(\alpha)\Rest{m}$ .
On the other hand, if a function
This $\phi^$ is indeed continuous: The preimage of $\Cyl{\tau}$ under $\phi^$ is given by \begin{equation*} (\phi^)^{-1}(\Cyl{\tau}) = \bigcup {\Cyl{\sigma} \colon \phi(\sigma) \Sgeq \tau }, \end{equation} which is an open set.
We have shown the following.
:label: prop-product-continuous
A mapping $f:A^\Nat \to A^\Nat$ is continuous if and only if there exists a mapping $\phi$ satisfying (1) and (2) such that $f = \phi^*$.
Note that we can completely describe a topological concept, continuity, through a relation between finite strings.