Here's a quick sketch to show that there is no Turing machine to decide whether an arbitrary class of problems is decidable.

I should clarify what I mean by class of problems: a class of problems $T$ is a Turing machine which enumerates the elements (natural numbers, say) of a recursively enumerable set one after the other, such that each element in the set is eventually printed. The problem intuitively captured by $T(n)$ is: "is the number $n$ in this set?". This captures the usual problems in the field of computability, such as "is i the index of a Turing machine that halts on empty input?".

Suppose there was machine $M$ which, given as input a class of problems $T$ answered $\mathit{true}$ if that class is decidable and $\mathit{false}$ otherwise.

Now take an arbitrary Turing machine $T$. We build the following class of problems $T'$ in the following manner:

1. Simulate $T$.
2. If $T$, halts, enumerate the indices of the Turing machines that halt on empty input.

Now it is clear that if $T$ halts, then $M(T')$ returns $\mathit{false}$, as the set of indices halting Turing machines is not a decidable (recursive) set.

If $T$ does not halt, then it $T'$ does not enumerate any numbers, which makes it exactly the class of problems containing no indices! Therefore $M(T')$ answers $\mathit{true}$, since that class is decidable (by the machine that always rejects).

Therefore, $M(T')$ returns $\mathit{true}$ iff $T$ does not halt, and $\mathit{false}$ otherwise. Thus the existence of $M$ allows us to solve the halting problem for an arbitrary machine $T$, which is a contradiction.

Here's a quick sketch to show that there is no Turing machine to decide whether an arbitrary class of problems is decidable.

I should clarify what I mean by class of problems: a class of problems $T$ is a Turing machine which enumerates the elements (natural numbers, say) of a recursively enumerable set one after the other, such that each element in the set is eventually printed. The problem intuitively captured by $T(n)$ is: "is the number $n$ in this set?". This captures the usual problems in the field of computability, such as "is i the index of a Turing machine that halts on empty input?".

Suppose there was machine $M$ which, given as input a class of problems $T$ answered $\mathit{true}$ if that class is decidable and $\mathit{false}$ otherwise.

Now take an arbitrary Turing machine $T$. We build the following class of problems $T'$ in the following manner:

1. Simulate $T$.
2. If $T$ halts, enumerate the indices of the Turing machines that halt on empty input.

Now it is clear that if $T$ halts, then $M(T')$ returns $\mathit{false}$, as the set of indices halting Turing machines is not a decidable (recursive) set.

If $T$ does not halt, then $T'$ does not enumerate any numbers, which makes it exactly the class of problems containing no indices! Therefore $M(T')$ answers $\mathit{true}$, since that class is decidable (by the machine that always rejects).

Therefore, $M(T')$ returns $\mathit{true}$ iff $T$ does not halt, and $\mathit{false}$ otherwise. Thus the existence of $M$ allows us to solve the halting problem for an arbitrary machine $T$, which is a contradiction.

Here's a quick sketch to show that there is no Turing machine to decide whether an arbitrary class of problems is decidable.

I should clarify what I mean by class of problems: a class of problems $T$ is a Turing machine which enumerates the elements (natural numbers, say) of a recursively enumerable set one after the other, such that each element in the set is eventually printed. The intuitively captured by $T$ is: "is the number $n$ in this set?". This captures the usual problems in the field of computability, such as "is i the index of a Turing machine that halts on empty input?".

Suppose there was machine $M$ which, given as input a class of problems $T$ answered $\mathit{true}$ if that class is decidable and $\mathit{false}$ otherwise.

Now take an arbitrary Turing machine $T$. We build the following class of problems $T'$ in the following manner:

1. Simulate $T$.
2. If $T$, halts, enumerate the indices of the Turing machines that halt on empty input.

Now it is clear that if $T$ halts, then $M(T')$ returns $\mathit{false}$, as the set of indices halting Turing machines is not a decidable (recursive) set.

If $T$ does not halt, then it $T'$ does not enumerate any numbers, which makes it exactly the class of problems containing no indices! Therefore $M(T')$ answers $\mathit{true}$, since that class is decidable (by the machine that always rejects).

Therefore, $M(T')$ returns $\mathit{true}$ iff $T$ does not halt, and $\mathit{false}$ otherwise. Thus the existence of $M$ allows us to solve the halting problem for an arbitrary machine $T$, which is a contradiction.

Here's a quick sketch to show that there is no Turing machine to decide whether an arbitrary class of problems is decidable.

I should clarify what I mean by class of problems: a class of problems $T$ is a Turing machine which enumerates the elements (natural numbers, say) of a recursively enumerable set one after the other, such that each element in the set is eventually printed. The problem intuitively captured by $T(n)$ is: "is the number $n$ in this set?". This captures the usual problems in the field of computability, such as "is i the index of a Turing machine that halts on empty input?".

Suppose there was machine $M$ which, given as input a class of problems $T$ answered $\mathit{true}$ if that class is decidable and $\mathit{false}$ otherwise.

Now take an arbitrary Turing machine $T$. We build the following class of problems $T'$ in the following manner:

1. Simulate $T$.
2. If $T$, halts, enumerate the indices of the Turing machines that halt on empty input.

Now it is clear that if $T$ halts, then $M(T')$ returns $\mathit{false}$, as the set of indices halting Turing machines is not a decidable (recursive) set.

If $T$ does not halt, then it $T'$ does not enumerate any numbers, which makes it exactly the class of problems containing no indices! Therefore $M(T')$ answers $\mathit{true}$, since that class is decidable (by the machine that always rejects).

Therefore, $M(T')$ returns $\mathit{true}$ iff $T$ does not halt, and $\mathit{false}$ otherwise. Thus the existence of $M$ allows us to solve the halting problem for an arbitrary machine $T$, which is a contradiction.