TITLE: Why did nobody prove undecidability by the "too many problems" argument?
QUESTION [19 upvotes]: To the best of my knowledge, the two major breakthroughs in regarding negative
results in computation/recursion theory came in 1936 by Church and Turing.
They both prove that some variant of the Halting problem was undecidable.
However, a very simple proof of the statement "There is an undecidable problem"
already existed at that time, namely the fact that there are simply too many
problems and too few algorithms.
Assuming that any algorithm has a finite description, and that they take natural
numbers as input, we are trying to make a bijection between the natural numbers
$\mathbb{N}$ and its powerset $2^{\mathbb{N}}$.
Adopting Cantor's diagonal argument to Machines, we can observe that if
$\mathcal{M} = \{M_i \mid i \in \mathbb{N}\}$ is the set of machines, and for
every machine $M \in \mathcal{M}$, that $M(n)$ is either yes (halts) or no (does
not halt), for $n \in \mathbb{N}$.
Then we can construct the set
$$U = \{i \in \mathbb{N} \mid M_i(i) \text{ is no} \}$$
like Cantor would, and pick the $i$ s.t. the domain of $M_i$ is $U$, and ask
whether $i \in U$, in both cases reaching a contradiction.
Question: Why wasn't this discovered between 1875/90 and 1935?  (Or was it?)

REPLY [19 votes]: This is similar to Rene Schipperus's answer, but instead of working with recursive enumerability, I'll use your notion of "finite description". You say that every algorithm must have a finite description, and I agree. But I'd also say that a problem must have a finite description.  A random, indescribable set of natural numbers should not be considered a problem, because it can't be "asked".  
The essential point of undecidability theorems is not the triviality that some subsets of $\mathbb N$ have no decision procedure, but that some definable subsets of $\mathbb N$ have no decision procedure.  (Rene's answer concerns the stronger result that the subsets in question can be taken to be not only definable but recursively enumerable.)