As I said earlier, I agree that the algorithm goes on forever.
At stage 1 it produces 0 and 1.
At stage 2 it produces 00, 01, 10, and 11.
At stage n it produces all the bitstrings of length n.
So if you claim a bitstring like 10101010101010… is produced by the algorithm, please tell me at what stage it’s produced.
Ok, 111… never gets counted. What about 0101010101…?
At what stage? How? Every string produced by the algorithm is a finite string. Their lengths are unbounded in the sense that there are arbitrarily large finite length strings. But there are no infinite length strings.
I’m in no hurry. After all we are discussing the infinite.
Last part, first part, second part? Not sure what you mean.
However I do realize I’ve been a little unclear about something. I don’t care about the digits to the left of the decimal point. That is, everything I can say about pi = 3.14159… is much easier to say about pi - 3 = .14159…
So when I think of a bitstring like 01010101010101… I am REALLY thinking about the number between 0 and 1 whose binary expansion is .0101010101…
So: All my bitstrings have an implied binary point in front of them. All the real numbers I care about are between 0 and 1. If we are using decimals then I’ll just talk about pi - 3 = .14159…
If you had a procedure to enumerate just the reals between 0 and 1, you could easily extend that to all the reals. Everything to the left of the decimal point is irrelevant.
Was that what you meant by first part and second part?
I’m willing to agree that any bitstring ending in an infinite sequence of 0’s can be truncated. So if 1 is on the list then 100000000… is on the list. It doesn’t affect my argument.
Also, any sequence ending in 1111… doesn’t affect the argument either, since any such number is rational. I’m more interested in irrationals that never have any pattern and that do not have trailing sequences of only 1’s or only 0’s.