%! TEX root = LC.tex % vim: tw=50 % 02/12/2024 11AM \textbf{Strategy:} \begin{enumerate}[(1)] \item Given a countable model $\mathcal{M}$ of $\PA$, we note that we encode subsets of $\Nbb$ as elements of $\mathcal{M}$; \item If $\mathcal{M}$ is non-standard, then there is an element that codes a non-\gls{rec} set; \item If $\mathcal{M}$ also has \gls{rec} $\oplus$, then there is a membership \glsref[dec]{decision} procedure for any subset that it codes. \end{enumerate} \newcommand\pr{\operatorname{pr}}% Note that there is a $\Sigma_1$-formula $\pr(x, y)$ that captures $y$ being the $x$-th prime, and $\PA \syn \forall x . \exists ! y . \pr(x, y)$. So if $\Nbb$ thinks that $k$ is the $n$-th prime, then any model of $\PA$ thinks so too. Write $\pi_n$ for the $n$-th prime. \begin{fclemma}[Overspill] \label{lemma:2.2.19} % Lemma 2.2.19 Assuming: - $\mathcal{M}$ a non-standard model of $\PA$ - $\varphi(x)$ an $L_{\PA}$-formula - $\mathcal{M} \models \varphi(n)$ for all standard natural numbers $n$ Then: there is a nonstandard natural number $e$ such that $\mathcal{M} \models \varphi(e)$. \end{fclemma} \begin{proof} Say $\mathcal{M} \models \varphi(n)$ for all standard $n$, but only them. Then $\mathcal{M} \models \varphi(0)$ and $\mathcal{M} \models \forall n . (\varphi(n) \to \varphi(n + 1))$ holds (if $\varphi(n)$ holds, then $n$ and hence $n + 1$ are standard). By $I\varphi$ (induction), we conclude that $\mathcal{M} \models \forall n . \varphi(n)$. But $\mathcal{M}$ is non-standard, so there is non-standard $e \in \mathcal{M}$ with $\varphi(e)$, contradiction. \end{proof} Fix some $m \in \Nbb$, and a property $\varphi(x)$ of the natural numbers. \begin{itemize} \item There is a number $c$ such that $\forall k < m . (\varphi(k) \leftrightarrow \pi_k \mid c)$, namely the product of all primes $\pi_k$ with $k < m$ and $\varphi(k)$. \item We perceive $c$ as a code for the numbers with the property $\varphi$ below $m$, which we can decode by prime factorisation. \end{itemize} \begin{fcdefn}[Canonically coded] \glsadjdefn{cc}{canonically coded}{subset}% \label{defn:2.2.20}% % Definition 2.2.20 A subset $S \subseteq \Nbb$ is \emph{canonically coded} in a model $\mathcal{M}$ of $\PA$ if there is $c \in \mathcal{M}$ such that \[ S = \{n \in \Nbb : \exists y . (\pi_{\ul{n}} \times y = c)\} \] where $\ul{n}$ denotes the standard number $n$ in the model. \end{fcdefn} We could use other formulas to code subsets. Th subsets of $\Nbb$ coded in $\mathcal{M}$ are those $S \subseteq \Nbb$ for which there is a $\PA$-formula $\varphi(x, y)$ and $c \in \mathcal{M}$ such that $S = \{n \in \Nbb : \mathcal{M} \models \varphi(\ul{n}, c)\}$. As it turns out, coding via $\sigone$-formulae gives nothing new: \begin{fcprop}[] \label{prop:2.2.21} % Proposition 2.2.21 Assuming: - $C(u, x)$ be a $\delz$-formula - $\mathcal{M}$ a non-standard model of $\PA$ Then: given any $\tilde{b} \in \mathcal{M}$, there is $c \in \mathcal{M}$ such that, for any $n \in \Nbb$: \[ \mathcal{M} \models \exists k < \tilde{b} . C(k, n) \leftrightarrow \exists y . (\pi_{\ul{n}} \times y) = c .\] \end{fcprop} \begin{proof}[Proof (sketch*)] The following formula holds in $\Nbb$ for any $n$: \[ \forall b . \exists a . \forall u < n . (\exists k < b . C(k, u) \leftrightarrow \exists y . (\pi_u \times y) = a) .\] This is by the reasoning we gave when introducing codes, which works due to the bound on $k$ and $u$. This can be proved in $\PA$*. Thus \[ \mathcal{M} \models \forall b . \exists a . \forall u < \ul{n} . (\exists k < b . C(k, u) \leftrightarrow \exists y . (\pi_u \times y = a)) \] for any $n \in \Nbb$. So by \cref{lemma:2.2.19} there is a non-standard $w \in \mathcal{M}$ such that \[ \mathcal{M} \models \forall b . \forall a . \forall u < w . (\exists k < b . C(k, u) \leftrightarrow \exists y . (\pi_u \times y = a)) .\] So for any $\tilde{b} \in \mathcal{M}$, there must be $c \in \mathcal{M}$ such that \[ \mathcal{M} \models \forall u < w . (\exists k < \tilde{b} . C(k, u) \leftrightarrow \exists y . (\pi_u \times y = c)) .\] Now $w$ is non-standard, so $\mathcal{M} \models \ul{n} < w$ for all $n \in \Nbb$. So for any $\tilde{b} \in M$ there is $c \in \mathcal{M}$ with \[ \mathcal{M} \models \exists k < \tilde{b} . C(k, n) \leftrightarrow \exists y . (\pi_{\ul{n}} \times y = c) \] for all $n \in \Nbb$. \end{proof} \begin{fcdefn}[Recursively inseparable] \glsadjdefn{ri}{recursively inseparable}{sets}% \label{defn:2.2.22} % Definition 2.2.22 We say that subsets $A, B \subset \Nbb$ are \emph{recursively inseparable} if they are disjoint and there is no \gls{rec} $C \subseteq \Nbb$ with $B \cap C = \emptyset$ and $A \subseteq C$. \end{fcdefn} \begin{fcprop}[] There are \gls{renum} subsets $A, B \subseteq \Nbb$ that are \gls{ri}. \end{fcprop} \begin{proof} Fix an effective enumeration $\{\varphi_n : n < \omega\}$ of the \glspl{prf}. Define $A = \{n \in \Nbb : \varphi_n(n) = 0\}$ and $B = \{n \in \Nbb : \varphi_n(n) = 1\}$, which are clearly disjoint and are clearly \gls{renum}. Suppose there is a \gls{rec} $C$ with $A \subseteq C$ and $B \cap C = \emptyset$, and write $\chi_C$ for its (\gls{tr}) characteristic function. There must be $u \in \Nbb$ such that $\chi_C = \varphi_u$, as $\chi_C$ is \gls{tr}. Since $\chi_C(u)\downarrow$ and is either $0$ or $1$, we have either $u \in A$ or $u \in B$. If $u \in A$, then $\chi_C(u) = \varphi_u(u) = 0$, so $u \notin C$, contradicting $A \subseteq C$; so $u \in B$. But then $\chi_C(u) = \varphi_u(u) = 1$, so $u \in C$, contradicting $B \cap C = \emptyset$. Thus $A$ and $B$ are \gls{ri}. \end{proof}