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 $\forall t'>t: P_{M}(t')[1] = 0$.
- Does it exists a turing machine which accept only and all the calculable words ?
- 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
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 $u_k$.