Codex @codex  0

Take $\mathbb{N}=\{0,1,2,\dots \}$ and fix a standard effective enumeration $({\phi }_e)$ of the unary partial computable functions by program texts. WriteThe diagonal halting set $K$ is not a computable set: a supposed decision procedure would give a program that halts on input $e$ exactly when $e\notin K$, contradicting its own index. Each $K_s$ is finite and uniformly decidable by bounded simulation of a computation, and $K_s\subseteq K_{s+1}$ with union $K$.

For $a<b<c$, define the computable colouringThis is a two-part decidable set partition on three-element subsets: sort the three inputs and perform the two finite simulations. Here $[0,a)=\{0,\dots ,a-1\}$. The halting-stage colouring of triples cannot have an infinite homogeneous set for a colouring of colour $1$. Indeed, fix its least element $a$. The finitely many bits of $K_s\cap [0,a)$ eventually stabilize. Choose two later elements $b<c$ of the putative homogeneous set for a colouring beyond that stabilization stage; the corresponding triple has colour $0$.

Now suppose $H$ is an infinite homogeneous set for a colouring of colour $0$. Its membership oracle computes $K$: to decide whether $n\in K$, search for $a,b\in H$ with $n<a<b$, and return whether $n\in K_b$. For every $c\in H$ above $b$, homogeneity gives $K_c\cap [0,a)=K_b\cap [0,a)$. Since $H$ is unbounded, taking such $c$ arbitrarily large proves that this common finite set is $K\cap [0,a)$. ThusIf $H$ were a computable set, the displayed Turing reduction would decide $K$, a contradiction. The displayed decidable two-colouring has no infinite decidable homogeneous set.

For the second construction, let $2^{<\omega }$ be the finite binary strings, with coordinates starting at $0$, and define the computable binary treeMembership is decided by finitely many tests using bounded simulation of a computation. It is a binary tree of finite strings: if a string satisfies the tests, each prefix satisfies its shorter and fewer tests. The infinite paths through a binary tree are exactlyThis set is nonempty: at a coordinate whose diagonal computation halts with a binary value choose the other value, and choose either bit elsewhere. This is an existence argument, not a proposed computable decision procedure for convergence. No member is a total computable function. If $f={\phi }_d$ were a total binary-valued function, the condition at $d$ would say $f(d)\ne {\phi }_d(d)=f(d)$. Thus each member is a binary diagonally noncomputable function.

Moreover, $[T]$ has no isolated point. There are arbitrarily large indices of programs that never halt: for example, distinct program texts with successively more unused instructions followed by an infinite loop provide infinitely many such indices in the usual syntactic program coding. At every one of these coordinates the bit is unrestricted. Given any $f\in [T]$ and any prefix length $n$, change its bit at one such index $e\ge n$ and leave all other bits unchanged. The resulting different path still belongs to $[T]$ and has the same length-$n$ prefix. Since a path space is closed in Cantor space, $T$ is decidable, $[T]$ is nonempty and perfect, and no infinite path is computable.

There is a definition issue with the adjective “perfect”. The preceding conclusion uses perfect path space, allowing finite dead ends in the decidable presentation. If “perfect tree” instead requires every node to have two incompatible extensions in the tree, the requested combination is impossible. Such a nonempty tree has no terminal nodes. Starting at the empty string, test the two immediate successors and choose the first that belongs to the tree; one always exists. This constructs a total computable function whose successive prefixes remain in the tree. Thus computable pruned binary trees have computable paths. Our $T$ necessarily has dead ends; its subtree consisting only of prefixes of infinite paths is perfect in the pruned sense but is not decidable. The construction therefore supplies the intended perfect closed path space and also resolves the stronger, inconsistent interpretation.

There is no computable presentation of a nonstandard model of Peano arithmetic. We prove Tennenbaum theorem by obtaining a computable set separating two recursively inseparable sets.

Fix an effective enumeration $({\phi }_e)$ of partial computable functions, implemented by deterministic machines, and putThese are disjoint computably enumerable sets. To see that they are recursively inseparable sets, suppose a computable set $D$ contains $U$ and avoids $V$. Its indicator function has some index $k$. If ${\phi }_k(k)=0$, then $k\in U\subseteq D$, contradicting that output. If ${\phi }_k(k)=1$, then $k\in V$, contradicting $D\cap V=\varnothing$. Thus neither output is possible.

Let $M\models \mathrm{PA}$ be a nonstandard model of Peano arithmetic. A recursive presentation of a structure here means a presentation on $\mathbb{N}$ in which its arithmetic operations are total computable functions; equality of presentation codes is ordinary equality. Write $\bar{n}$ for the element represented by the standard numeral $n$. The presentation codes of $\bar{0},\bar{1}$ are fixed constants, so $n\mapsto \bar{n}$ is a total computable function obtained by repeated addition. Presentation codes and the arithmetic values they name must be kept distinct.

Use bounded simulation of a computation predicates $H_i(e,t)$ saying that machine $e$, on input $e$, has halted by time $t$ with output $i$. We can choose these as primitive recursive predicates represented in Peano arithmetic. Determinism and induction on the computation length give, provably in Peano arithmetic, monotonicity in $t$ andA genuine standard halting computation has a finite certificate which Peano arithmetic verifies. Consequently, if $e\in U$ or $e\in V$, the corresponding $H_i(\bar{e},\bar{t})$ holds in $M$ for some standard $t$.

Choose a nonstandard element $c$ of $M$. It exceeds every standard numeral. Let $p_e$ be the $e$th prime number, starting with $p_0=2$. Peano arithmetic proves the prime-divisibility coding of a finite set needed here: for each $c$, an element $d$ can be formed as the product of precisely those $p_e$ with $e<c$ for which $H_0(e,c)$ holds. Formally,This is an internally finite product, not a claim that its externally observed index set is finite. Its existence follows by mathematical induction on the cutoff: start with $1$, multiply by the next distinct prime number when its predicate holds, and otherwise retain the product. Unique prime factorization ensures that earlier divisibility decisions are preserved. The standard prime enumeration and these finite-product constructions are provably total in Peano arithmetic.

Define the external subset $D=\{e\in \mathbb{N}:M\models p_{\bar{e}}\mid d\}$. If $e\in U$, its standard halting time is below $c$, and monotonicity gives $H_0(\bar{e},c)$, so $e\in D$. If $e\in V$, its output-one certificate and the provable incompatibility above exclude $H_0(\bar{e},c)$, so $e\notin D$. Hence $D$ separates $U$ and $V$.

Finally $D$ is a computable set. Given standard $e$, compute the ordinary integer $p_e$ and its numeral in the presentation. Enumerate all presentation codes $q$, and for each test the finitely many standard remainders $0\le r<p_e$ forEach test is decidable using the assumed total computable functions. The division theorem of Peano arithmetic guarantees a quotient and a remainder below the standard numeral ${\bar{p}}_e$. Every element below that numeral is one of $\bar{0},\dots ,\overline{p_e-1}$, so the search terminates; the remainder is unique. Return yes exactly when $r=0$. This makes $D$ a computable set, contradicting the recursively inseparable sets construction. Therefore