I interpreted your algorithm for listing computables to be something like “enumerate the Turing machines that output ‘.’ then 0s and 1s and list what they print”, without worrying about the fact that some computables repeat. This technically lists all computables even though some machines will get stuck in some non-halting behavior that prints nothing after printing finitely many 0s and 1s. But you can’t figure out the nth digit of an arbitray thing in the list because you don’t always know if a machine will continue to print 0s or 1s or get indefinitely “stuck”.
It seems that you actually meant that you could have an algorithm that begins by enumerating the Turing machines that print computables and never get stuck in some non-halting configuration. This can’t be done.
If you have any surjection: N→S⊂(N→{0,1}) and diagonalize against it, you know the result is not in S. This fact doesn’t depend on the actual nature of the surjection N→S, just that S is the image. Here S is the computables.
I interpreted your algorithm for listing computables to be something like “enumerate the Turing machines that output ‘.’ then 0s and 1s and list what they print”, without worrying about the fact that some computables repeat.
I’m pretty sure my argument did not mention how computables are listed at all, rather proving that for any specific listing the inverse-diagonal is computable as well.
If you have any surjection: N→S⊂(N→{0,1}) and diagonalize against it, you know the result is not in S. This fact doesn’t depend on the actual nature of the surjection N→S, just that S is the image. Here S is the computables.
Yes. However, it’s the specific choice of set “computables” which creates the contradiction: I agree with “inverse-diagonal for rationals is an irrational number” and like.
Once again: for any “user-provided” computable table of computable digit sequences, I can, in finite time, get value for any specific position in table; therefore, each digit of inverse sequence is computable; therefore, I conclude that the inverse-diagonal sequence is itself computable (if I’m not mistaken in definitions).
My claim is that such a table does not exist because it leads to a contradiction. If you add it as an assumption, you can obtain a false conclusion because the assumption itself can never hold.
I interpreted your algorithm for listing computables to be something like “enumerate the Turing machines that output ‘.’ then 0s and 1s and list what they print”, without worrying about the fact that some computables repeat. This technically lists all computables even though some machines will get stuck in some non-halting behavior that prints nothing after printing finitely many 0s and 1s. But you can’t figure out the nth digit of an arbitray thing in the list because you don’t always know if a machine will continue to print 0s or 1s or get indefinitely “stuck”.
It seems that you actually meant that you could have an algorithm that begins by enumerating the Turing machines that print computables and never get stuck in some non-halting configuration. This can’t be done.
If you have any surjection: N→S⊂(N→{0,1}) and diagonalize against it, you know the result is not in S. This fact doesn’t depend on the actual nature of the surjection N→S, just that S is the image. Here S is the computables.
I’m pretty sure my argument did not mention how computables are listed at all, rather proving that for any specific listing the inverse-diagonal is computable as well.
Yes. However, it’s the specific choice of set “computables” which creates the contradiction: I agree with “inverse-diagonal for rationals is an irrational number” and like.
Once again: for any “user-provided” computable table of computable digit sequences, I can, in finite time, get value for any specific position in table; therefore, each digit of inverse sequence is computable; therefore, I conclude that the inverse-diagonal sequence is itself computable (if I’m not mistaken in definitions).
My claim is that such a table does not exist because it leads to a contradiction. If you add it as an assumption, you can obtain a false conclusion because the assumption itself can never hold.