TITLE: Relationship between non-standard computation and TM(oracle)?
QUESTION [6 upvotes]: We know that there are non-standard models of arithmetic, and in such models there are non-standard proofs of standardly unprovable sentences. Now, we can imagine a version of representability relative to some non-standard model of some arithmetical theory such that the said function would be what I would call a "non-standard computation". At the same time we can imagine the said computation otherwise correct, as in it would be the same as adding some real/oracle to a TM. Is there a nice way to connect these concepts? Is it a trivial connection? Can it give us insights about the base theory X or the sentence S such that X + "sentence S" $\vdash$ TM(oracle) halts?

REPLY [6 votes]: Let address the version of the question posed by Gro-Tsen in his
comments.

Sets (a) and (c) are the same, and these are exactly the standard system of the model $M$, as mentioned in the answer of Patrick
Lutz. The standard system is equivalently defined as the set of traces on the standard
numbers of any sequence that is definable from parameters in $M$, whether or not
it is computable in $M$, since every (possibly nonstandard) finite initial
segment of a definable sequence is an element of $M$. Any sequence as in (a) or (c) will be the standard part of a definable
sequence in $M$ and thus in the standard system, and conversely,
any set in the standard system is the standard part of a pseudo
finite sequence in $M$, which can be hard-coded into a
(nonstandard) program $e$ in $M$, fulfilling (a) or (c).

Set (b) consists of the sets of natural numbers that are the
trace of a set that is computably enumerable in $M$ by a standard
program $e$, such that this program gives output on all standard
input. This is equivalent to the sets of natural numbers such that
both it and its complement are the standard parts of sets that $M$ thinks are
computably enumerable using a standard program. In a model of true
arithmetic, this would be exactly the computably decidable sets,
since the operation of a standard program
$e$  would be the same in $\mathbb{N}$ as in $M$ in this case. Thus, this set can be strictly smaller than the sets in (a) and (c). Meanwhile, the universal
algorithm
(see also Arithmetic potentialism and the universal
algorithm
(arXiv:1801.04599) shows that
any given set of natural numbers can be placed into set (b) for a
suitable choice of $M$, since the universal algorithm $e$, which is
a standard program, can compute any set in a suitable model $M$ of
PA, and by arranging that the program enumerate the set in
increasing order, one will also be able to compute the complement
with a standard program. Nevertheless, set (b) is never equal to
the sets in (a) and (c), because the model $M$ can unify the
sets of (b) into a single subset of $\mathbb{N}\times\mathbb{N}$,
using program $e$ on slice $e$. This will be a set in the standard
system, from which all sets in (b) are computable, but the standard
system must have sets not computable from any given set.

Set (d) consists of the standard parts of the sets that $M$
thinks are computably decidable by a standard program. In a model
of true arithmetic, these are exactly the computably decidable
sets, but by using the universal algorithm as above, we can place
any given set into set (d). The difference between set (d) and set
(b) is that the programs in (b) must halt only on all standard
input, but in (d) they must halt on all input. So set (d) is
contained in set (b). I would have to think harder whether the
sets can differ. But the argument I gave above shows that (d) is
never equal to (a) or (c).

Sets (e) and (f) consists exactly of the finite sets of natural numbers.
Any finite set of natural numbers has the property in (e) and (f), and
conversely, if a set is infinite and definable in a nonstandard
model $M$ of PA, then it must have nonstandard elements, since
otherwise it would stand as a counterexample to induction in $M$.


A modified version of (e) would be the property (e'), asserting that $X\subset\mathbb{N}$ and there is some program $e\in M$ that halts on all input in $M$, such that $\forall i\in \mathbb{N} (i\in X\iff \{e\}(i)\downarrow=1)$.  This would be equivalent to (a) and (c).