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> 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 w
by sharpener 5y ago
> 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 5y agoFor 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 5y agoFor 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}* .