Proof that you can't have an Emulator (wrong)

This is wrong; you can have an Emulator (update 20/8/2025)

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This is an odd recent result that makes no sense.

(1) List all one argument functions so that each function has a number.

f1(x) is the function on the 1st row where n=1
f2(x) is the function on the 2nd row where n=2
etc

let #f be the notation that gives the row number of the function so that #fn = n

(2) Suppose there is an emulator function E(#f,x) which takes the row number from (1) and provides the result that f(x) would provide so that:

E(#f,x) == f(x)

(3) For each particular value of x, lets call it 'a', in (2) there is a function Ea(n) such that:

Ea(#f) == E(#f,a) == f(a)

In other words for every value of x there is a corresponding function fx(n) which provides the result of the function fn(x) on x.

(4) Each Ea(x) if it exists must have a row number #Ea. Substituting in 3 gives

Ea(#Ea) == E(#Ea, a) == Ea(a)

Therefore:

#Ea = a

This means that the row position of each Emulator of functions on value a is 'a' itself.

This is clearly a problem already. The following is simply looking for a contradiction in this statement:

But f5 is the name we have for the function at row 5, and 'fa' for the function at row 'a'

So:

fa(x) == Ea(x) == fx(a)

The function at row a applied to x gives the same result as the function at row x applied to a.

This is clearly nonsense as the order of function listing at the start was quite arbitrary... but how do we prove it.

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This is correct:

Ea(#Ea) == E(#Ea, a) == Ea(a)

And so:

Ea(#Ea) == Ea(a)

But what does this assure that #Ea = a.

if #Ea = b

Ea(b) == Ea(a)

For b = a there would need to be just one 'b' for which Ea(b) == Ea(a). However we cannot guarantee this. There nay be many numbers 'b' where Ea(b) == Ea(a).

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What we can do however is use this construction to prove Kleene's Fixed Point Theorem. #TODO