The problem goes as follows. Let f: N --> N, such that f is partial, N is the natural numbers, and let m $\in$ N. Construct a non-computable function g such that g(x) = f(x) for x$\le$m.
Proof: By Cantor's theorem, there exist uncountably many functions on the naturals, including partial, non-computable ones. Call this class of all partial functions Z. Suppose we are given an f from Z. Either f will be computable or it will not be. We must show that there is an x, and a g, such that f(x) = g(x) with x$\le$m: m$\in$N and g non-computable.
Two cases eventuate:
f is non-computable. Then we can set g = f.
f is computable.
Need help with (what looks like) non-trivial case. Thanks!
If $h(x)$ is a noncomputable total function, $f(x)$ is any function, and $m$ is fixed, then $$ h'(x) = \begin{cases} h(x) : x > m \\ f(x) : x \leq m,f(x) \downarrow\\ 0 : \text{otherwise} \end{cases} $$ is a total, noncomputable function that agrees with $f$ (when $f$ is defined) for all $x \leq m$.