\newcommand{\Nat}{\mathbb{N}}
\newcommand{\Real}{\mathbb{R}}
\newcommand{\Integer}{\mathbb{Z}}
\newcommand{\Rat}{\mathbb{Q}}
\newcommand{\Baire}{\Nat^{\Nat}}
\newcommand{\Cyl}[1]{N_{#1}}
\newcommand{\Cant}{2^{\Nat}}
\newcommand{\Nstr}{\Nat^{<\Nat}}
\newcommand{\Tup}[1]{\langle #1 \rangle}
\newcommand{\Co}[1]{\neg \,#1}
\newcommand{\Cl}[1]{\overline{#1}}
\newcommand{\Op}[1]{\operatorname{#1}}
\newcommand{\Rest}[1]{|_{#1}}
\newcommand{\Sle}{\subset}
\newcommand{\Sleq}{\subseteq}
\newcommand{\Estr}{\varnothing}
\newcommand{\eps}{\varepsilon}
\newcommand{\Conc}{\mbox{}^\frown}
\newcommand{\bDelta}{\pmb{\Delta}}
\newcommand{\bPi}{\pmb{\Pi}}
\newcommand{\bSigma}{\pmb{\Sigma}}
\newcommand{\BS}[1][n]{\bSigma^0_{#1}}
\newcommand{\BP}[1][n]{\bPi^0_{#1}}
\newcommand{\PS}[1][n]{\bSigma^1_{#1}}
\newcommand{\PP}[1][n]{\bPi^1_{#1}}
\newcommand{\CH}{\mathsf{CH}}
\newcommand{\AC}{\mathsf{AC}}
\newcommand{\ZF}{\mathsf{ZF}}
\newcommand{\Norm}[1]{\parallel \! #1 \!\parallel}
\newcommand{\Op}[1]{\operatorname{#1}}
\DeclareMathOperator{\W}{W}
\DeclareMathOperator{\WF}{WF}
\DeclareMathOperator{\WOrd}{WOrd}
We will see that, in many ways,
Analytic sets are projections of closed sets. Closed sets are in
:label: def-two-dim-tree
A set $T \subseteq \Nstr \times \Nstr$ is a **two-dimensional tree** if
- **(i)** $(\sigma,\tau) \in T$ implies $|\sigma|=|\tau|$ and
- **(ii)** $(\sigma,\tau) \in T$ implies $(\sigma\Rest{n},\tau\Rest{n}) \in T$ for all $n \leq |\sigma|$.
An **infinite branch** of $T$ is a pair $(\alpha,\beta) \in \Baire\times \Baire$ so that
\begin{equation*}
\forall n\in \Nat \; (\alpha\Rest{n},\beta\Rest{n}) \in T.
\end{equation*}
As in the one-dimensional case, we use
Another way to write this is to put, for given
Then we have, with
We obtain the following normal form for co-analytic sets.
:label: prop-norm-form-coanalytic
A set $A \subseteq \Baire$ is $\PP[1]$ if and only if there exists a two-dimensional tree $T$ such that
$$
\alpha \in A \iff T(\alpha) \text{ is well-founded}.
$$
If
How does one show that a specific set is not Borel? A related question is: Given a definition of a set in second order arithmetic, how can we tell that there is not an easier definition (in the sense that it uses less quantifier changes, no function quantifiers etc.)? The notion of completeness for classes in Polish spaces provides a general method to answer such questions.
:label: def-Wadge
Let $X,Y$ be Polish spaces. We say a set $A \subseteq X$ is **Wadge reducible** to $B \subseteq Y$, written $A \leq_{\W} B$, if there exists a continuous function $f: X \to Y$ such that
$$
x \in A \iff f(x) \in B.
$$
The important fact about Wadge reducibility is that it preserves classes closed under continuous preimages.
:label: prop-Wadge-preimages
Let $\Gamma$ be a family of subsets in Polish spaces (such as the classes of the Borel or projective hierarchy). If $\Gamma$ is closed under continuous preimages, then $A \leq_{\W} B$ and $B \in \Gamma$ implies $A \in \Gamma$.
If $A \leq_{\W} B$ via $f$, then $A = f^{-1}(B)$.
:label: def-completeness
A set $A \subseteq X$ is **$\Gamma$-complete** if $A \in \Gamma$ and for all $B \in \Gamma$, $B \leq_{\W} A$.
If
is
Since
It follows that complete sets exist for all levels of the Borel and projective hierarchy. However, the universal sets they are based on are rather abstract objects. Complete sets are most useful when we can show that a specific property implies completeness. We will encounter next an important example for the class of co-analytic sets.
(sec-well-founded)=
Given a real in
A binary relation
Let
Then
and hence
A closely related set is
Then
Coding a linear order is easily seen to be
:label: thm-WF-Wadge-complete
The sets $\WF$ and $\WOrd$ are $\PP[1]$-complete.
We have seen in the chapter on {ref}`chap-trees` that a tree has an infinite path if and only if the inverse prefix ordering is ill-founded. Trees can be coded as reals, and hence {prf:ref}`prop-norm-form-coanalytic` yields immediately that $\WF$ is $\PP[1]$-complete.
For $\WOrd$ we use the Kleene-Brouwer ordering and refer to {prf:ref}`prop-KB-wellorder`.
The theorem lets us gain further insights in the structure of co-analytic sets. If
It is clear that
:label: lem-bounded-rank-Borel
For any $\xi < \omega_1$, the set $\WOrd_\xi$ is Borel.
Let $\alpha \in \Baire$. We say $m \in \Nat$ is in the **domain** of $E_\alpha$, $m \in \Op{dom}(E_\alpha)$, if
\begin{equation*}
\exists n \: [ m E_\alpha n \; \vee \; n E_\alpha m].
\end{equation*}
It is clear from the definition of $E_\alpha$ that $\Op{dom}(E_\alpha)$ is Borel. For $\xi < \omega_1$, let
\begin{equation*}
B_\xi = \{ (\alpha,n) \colon E_\alpha \Rest{\{m \colon m E_\alpha n\}} \text{ is a well-ordering of order type $\leq \xi$} \}
\end{equation*}
We show by transfinite induction that every $B_\xi$ is Borel. Suppose $B_\zeta$ is Borel for all $\zeta < \xi$. Then, since $\xi$ is countable, $\bigcup_{\zeta < \xi} B_\zeta$ is Borel, too. But
\begin{equation*}
(\alpha,n) \in B_\xi \iff \forall m \: [m E_\alpha n \: \Rightarrow \: (\alpha,m) \in \bigcup_{\zeta < \xi} B_\zeta],
\end{equation*}
and from this it follows that $B_\xi$ is Borel. Finally, note that
\begin{equation*}
\alpha \in \WOrd_\xi \iff \forall n \; [n \in \Op{dom}(E_\alpha) \: \Rightarrow \: (\alpha,n) \in B_\xi],
\end{equation*}
which implies that $\WOrd_\xi$ is Borel.
:label: cor-coanal-union-Borel
Every $\PP[1]$ set is a union of $\aleph_1$ many Borel sets.
Since $\WOrd$ is $\PP[1]$-complete, every co-analytic set $A$ is the preimage of $\WOrd$ for some continuous function $f$. We have
\begin{equation*}
\WOrd = \bigcup_{\xi < \omega_1} \WOrd_\xi,
\end{equation*}
and hence
\begin{equation*}
A = \bigcup_{\xi < \omega_1} f^{-1}(\WOrd_\xi).
\end{equation*}
Since continuous preimages of Borel sets are Borel, the result follows.
If we work instead with the set
\begin{equation*}
C_\xi = { \alpha \colon \alpha \in \WOrd_\xi \text{ or } \exists n\in \Op{dom}(E_\alpha) \ [E_\alpha \Rest{{m \colon m E_\alpha n}} \text{ is a well-ordering of order type
:label: cor-aleph-union-intersect
Every $\PP[1]$ set can be obtained as a union or intersection of $\aleph_1$-many Borel sets. Consequently, the same holds for every $\PS[1]$ set.
The previous results allow us to solve the cardinality problem of co-analytic sets at least partially.
:label: cor-coanalytic-cardinality
Every $\PP[1]$ set is either countable, of cardinality $\aleph_1$, or of cardinality $2^{\aleph_0}$.
We conclude the chapter with another application of {prf:ref}lem-bounded-rank-Borel
– a useful tool for analyzing
:label: thm-sigma11-bounding
For every analytic $A \subseteq \WOrd$ there exists an ordinal $\nu < \omega_1$ such that
$$
\forall x \in A \;\; \Norm{x} < \nu.
$$
If such a $\nu$ did not exist, then
$$
\alpha \in \WOrd \iff \exists \nu \: [\alpha \in A \; \wedge \; \WOrd_\nu].
$$
The right-hand side is $\bSigma^1_1$, and hence $\WOrd$ would be $\bSigma^1_1$, contradiction.
An analogous statement holds for