[stymaar]

> Consider that great big set. It's still countable.

You didn't prove that statement, and actually Cantor's diagonal arguments [1] proves you wrong:

Consider the set of «all real numbers which you can actually specify IN ANY FASHION AT ALL». If you can count it, you can order them in a certain fashion: n0, n1, n2, n3 … It's easy to specify a number X which is not part of this set (which is in contradiction with the definition of this set, and demonstrates ad absurdum that this set is not countable). To build such X, consider the n-th decimal of the n-th number: if it's zero, the n-th decimal of X is 1, otherwise the n-th decimal is 0. Then by construction, X is different from any number of this set, which mean X is not a part of the set. □

[1]: https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

[moefh]

I don't think that argument works.

How do you guarantee that you can compute the n-th decimal of the n-th number in the list? In fact, from what I understand, this paper[1] shows how to specify exactly a number can't be computed like that.

[1] https://arxiv.org/pdf/1003.0480.pdf

[stymaar]

Why do you need to compute the n-th decimal of a number for this proof to work ? let Ω be a Chaitin Omega number it's decimal are non-computable, yet for every x, Ω + x is a perfectly valid number.

Then `floor(Ω.10^(n))-10.floor(Ω.10^n-1)` is also a number (a natural one) and actually it's the n-th decimal of Ω. It cannot be computed, but it still exist no matter what.

[gertef]

The number you are paraphrasing cannot be precisely identified in finite time.

[stymaar]

Oh, you're right. For that I would need to explicit the construction of the sequence n0, n1, … but that's impossible : if I can find such sequence, my proof holds but at the same time such sequence shows that the set is countable => contradiction, hence there is no such sequence.

But well, now that we've proven that such sequence doesn't exist, we've proven that the given set is not countable !

Not the most elegant proof ever, but I think it works.

[stymaar]

Actually it doesn't: what if the given sequence exists but cannot be expressed with words ? (This is exactly the same thing than «a real number that can't ne expressed with words» which is what I'm trying to demonstrate not to exist. I'm going nowhere)