Part of the problem is that TMs are allowed to go backwards through their tapes. This makes concocting a Pumping Lemma difficult for a basic reason.
Recall that the basic proof of the Pumping Lemma for regular languages was to take a DFA $D$ for $L$, and then count the number of local configurations (the numbers of states multiplied by the size of the alphabet). If $D$ reads any string in $L$ of length greater than this, then there must be a (state,symbol) pair which is reached two times. By "pumping" the substring that appears between the times $D$ in in these two states, the definition of a DFA makes it clear that $D$ must do the exact same thing each time through each of these pumps.
For PDAs the reasoning is similar.
Note that if you attempted to do the same for a TM $M$ you would run into a problem since it could have been that the steps between reaching the same (state,symbol) pair you read a symbol to the left of that substring. Then the pumping may not work because moving to the left at a pump may result in a different action being taken (because there is no guarantee that the last symbol of this pumped substring is the same as the last symbol before this pump, and you might have to take into account even earlier symbols, just compounding the problem). So not only would a proof of a Pumping Lemma have to take into account the local (state,symbol) configuration, but actually the full configuration would be needed. In this way the set of configurations that need to be distinguished becomes unbounded, and so the pigeonhole argument fails.
Hint: It may be clearer to state that the set of internal states of the Turing machine $T$ is either $\{0,1\}$ for the infinitely (or finitely many strings that it may produce on its tape), especially if the language $\Sigma$ is recursively enumerable (or equivalently, in the range of a partial recursive function $\psi$).
I assume, then, that the language $\Sigma$ is recursively enumerable. We say that $\Sigma$ is decidable if $\Sigma$ lays in the range of a total recursive function $\psi_e(n)$ where $e$ is the code number of the machine $T^e$ that halts on input $n$, and where $n \in \mathbb N$. Intuitively, total recursion holds if $\psi_e$ is defined for every input.
Rice's theorem states that any nontrivial property $P$ of a partial recursive function $\psi_e$ is undecidable (computably unsolvable by $T^e$). We take $P$ to be decidable (either $\psi_e(n)=0$ $\vee$ $\psi_e(n)=1$ if $P$ holds or does not hold, respectively). More directly, let $0'$ be the degree of unsolvability of the problem for deciding if $T^e$ halts on arbitrary input $n$. If the solution set for deciding if $P$ holds is unsolvable, then $P$ is undecidable and the solution set is $0'$.
Therefore, Enumerating the pairs $(n,i)$ if $P$ holds for a string $s_n$ in $\Sigma$ is simply taking $P$ as computable and showing that any $P^i, i \in \mathbb N$ holds or does not hold for a finite (or infinite) $s_n$. Supposing that it does hold, then the solution set it produces should be decidable, as the characteristic $\chi_\Sigma(x) = 0$ if $P(x)$ holds for $x \in \Sigma$ or $\chi_\Sigma(x) = 1$ if $P(x)$ does not hold for $x \in \Sigma$. If this pair is decidable, then the solution set is decidable, since a set is computably enumerable iff its characteristic function is, and not if otherwise. A property of $\psi$ could taken as any property that the sequence of partial functions $\psi_1(n),...,\psi_e(n)$ terminates on input $n$, or if each member of the solution set is a proof that $P(x)$ holds.
Best Answer
Hint: A set is decidable iff it and its complement are recursively enumerable.