[sharpener]

> Define a function z(n) = 1 - f(n)(n).

I don't understand the notation f(n)(n). Is it related to f_{nn} in LaTeX notation? Your later text suggests maybe it was aiming at f(n,n) so I will assume that.

I recognise a form of this argument and I might have tackled it in the supplementary materials I created that are referenced in the article. Let me know.

> However, z(k) = 1 - f(k)(k). Yet f(k) = z, so z(k) = 1 - z(k).

I'm assuming this was intended to be: z(k) = 1 - f(k). Yet f(k) = z, so z(k) = 1 - z(k).

For some k, z(k) = 0.5. f(k) = 0.5. Seems Ok.

[myWindoonn]

For future readers: z cannot return 0.5. z can only return 0 or 1. This is because z is closed over a function which returns 0 or 1, and inverts it to return 1 or 0.

This should remind folks of both Turing's Halting problem and Russell's paradox. z takes some f which claims to be a bijection (claims to Halt, claims to be a set of all sets) and finds a way to call f against a witness constructed from f.

[drdeca]

For each natural number k, f(k) is itself a function, and f(k)(k) means the value of that function at k.

Yes, you could basically think of it as f_{k,k} if you wanted to.

No, this is __not__ meant to be 1 - f(k) . f(k) is a function (or a sequence, if you prefer), not a particular value in {0,1}, f(k)(k) is a particular value in {0,1}.

0.5 is not in the set {0,1}, and therefore if z(k)=0.5 then z is not in {0,1}* .