Hello everyone just like the title says I want to prove that $H(x) = \mu y T(x,x,y)$ has no total computable extension such that if we had a function $BIG(x)$ that is both total and agrees with $H(x)$ whenever $H(x)$ is defined, then $BIG(x)$ is not computable. This is a homework question so I don't want a full solution just some help!
The function $H(x)$ I believe returns the smallest computational history $y$ such that a program $\{x\}(x)$ (program takes input of its own configuration and runs) runs and halts. Maybe I am not quite clear what it means for $BIG$ to agree with $H(x)$ or what it's own input. I think I need to create a diagonal function that uses $BIG$ if $BIG$ was computable and show that if I had some program $e$ then it must agree with $BIG$ but based on my definition of that diagonal function it isn't. If you are reading this you might see the mess of my thinking, any help would be greatly appreciated!
This is very easy, because you already have your diagonal function. Hence let
$$\delta:x\mapsto BIG(x)+1$$
If $BIG$ is total and computable, so must be $\delta$. Let $y$ be the number of $\delta$. Then
$$\delta(y)=BIG(y)+1=\{y\}(y)+1=\delta(y)+1$$
Do you see the contradiction ?