Recall A'= $\{x \mid \phi^A_x(x)$ halts and accepts $\}$

In this article, a proof that A' not computable in A is given: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2006/PAPERS/Flood.pdf

He constructs the function f(x)= $\phi_x^A(x)$ +1 if x in A and 0 if x not in A

Shouldn't it be f(x)= $\phi_x^A(x)$ +1 if x in A' and 0 if x not in A'

since we need $\phi^A_x(x)$ to halt and output a number provided it actually can?

then f will be computable in A and we can then go on to prove that f $\neq$ $\phi^A_x$ for the witness x


You are right. It seems you completely understand this already: in any case the result can be found many places, as it is a basic result about the Turing degrees.


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