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I m not at all a specialist so im gonna say how i understand the terms i use even if it's basic for some so we are in the same page.

Notation: let $P$ be a turing machine $P(t)$ is the word writen after $t$ steps. let $M$ be a word $M[n]$ is the nth caractere of the word. Let M be a word then $P_{M}(t)$ is the word after $P$ act on $M$ after $t$ step.

Let $M\in \{0;1\}^\mathbb{N}$, $M$ is said calculable if there exist a turing machine $P$ such that $\forall n, \exists t$ such that $\forall t'>t: P(t')[n] = M[n]$.

Let $M \in \{0;1\}^\mathbb{N}$, $P$ turing machine $M$ is said accepted by $P$ if $\exists t$ such that $\forall t'>t: P_{M}(t')[1] = 1$. $M$ is said rejected by $P$ if $\exists t$ such that $\forall t'>t: P_{M}(t')[1] = 0$.

  1. Does it exists a turing machine which accept only and all the calculable words ?
  2. Does it exists a turing machine which reject only and all the non calculable words ?

i dont ask a turing machine doing 1+2 as im fairly sure it doesnt exists and it should not be hard to prove.

edit: i preshot, 1) and 2) are not obviously equivalent as a word can be neither accepted or rejected

Guill Guill
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  • Under your definition, every word is calculable: just let $P$ be the Turing machine that disregards any input and simply prints out $M$, say. – David Gao Jan 28 '24 at 10:56
  • oh oups i ve to edit. Done – Guill Guill Jan 28 '24 at 10:59
  • Again, your edit doesn’t solve the problem. Every word is still calculable, just because you can just choose the Turing machine that simple prints out the desired word. (There’s a reason why decidability is always about a set of words instead of just a word - a singleton set of just one word is always decidable, per the argument I just gave.) – David Gao Jan 28 '24 at 11:07
  • oh damn im very not focused, its ofc N not n – Guill Guill Jan 28 '24 at 11:08
  • done, and also the questions were not good ofc it has to accept all the one if else it too easy :) , i hope it's good this time – Guill Guill Jan 28 '24 at 11:11
  • does the question makes sens now David Gao ? – Guill Guill Jan 28 '24 at 11:30
  • It makes sense now - the definitions are very non-standard though. It’s likely neither 1 nor 2 can be achieved. Just thinking about the easiest way to prove it. – David Gao Jan 28 '24 at 11:39
  • I can’t quite prove this, so I’ll leave a comment here to explain my thoughts. As the Turing machines in question allow infinitely long inputs, it’s better to regard inputs as an element of the Cantor set. The set of all calculable elements, in the OP’s definition, is in $\Sigma_4^0$. However, for any fixed Turing machine $P$, the set of elements of the Cantor space for which $P$ accepts, in the OP’s definition, seems to be in $\Sigma_2^0$ - and the same holds for the set of elements $P$ rejected. As such, it would seem this is impossible. – David Gao Jan 28 '24 at 21:22
  • I’m not used to work with arithmetic hierarchy, especially as applied to subsets of the Cantor space, though, so someone else would need to figure out whether this line of reasoning actually works. – David Gao Jan 28 '24 at 21:26

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I still don't have the answer but i've an argument for negative answer, or at least it prevent one simple way to answer positively to the question.

Statement: For all $i$, every fonction which at a turing machine $P$ and an integer $n$ associate a number greater than: $max\{t, P(t)[i] \neq P(\infty)[i]\}$ if $P[i]$ converge and $\infty$ if $P[i]$ does not converge, is not calculable (in the classical sens).

proof: a simple diagonal argument, by absurd. Let's suppose $f$ which at $i, P$ associate $f(i,P)$ the function stated in the theorem is calculable. Let's consider $u_k$ a calculable sequence which enumerate all turing machine. Then we can construct a convergent turing machine $Q$ such that $\forall i, f(i,Q) > fin \{f(i,u_k), k\leq i\}$ where $fin$ designe the finites values in the set, leading to a contradiction when Q begin to show in the $u_k$.

Guill Guill
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