I have studied the halting problem and the concepts of decidability and computability, however, I am stuck with the transfer of Turing machines to sets of natural numbers. Specifically, using the diagonalization approach given a function ($f(x)$ that outputs $1$ if $x \in \mathbb{N}$ and $0$ otherwise.
How can I demonstrate that there exists at least one set of natural numbers that is not computable? I studied the concept of computable sets but I somehow can't make the connection from the Halting problem and diagonalization approach of it to show that a set of natural numbers is not computable.
I started setting up the table by assuming all possible sets of natural numbers on the vertical axis and an enumeration of natural numbers that represent the inputs $x$. The table will be filled with zeroes and ones depending on whether the $x$ value on the x-axis is in a given set. I then assume the existence of some sort of negator function such as $g(x) = 1 - x$. Constructing now a set of natural numbers as function of the diagonal zeroes and ones but negated, I get to the intersect and that is where I am stuck:
What does this diagonal value mean? It is supposed to represent the contradiction and hence non-computability of that particular set but how do I describe this mathematically?
Lets assume I have \begin{array}{|l|l|l|l|l|} \hline \text{set} & 0 & 1 & 2 & 3 \\ \hline \text{set 1} & (0) & 1 & 0 & 0 \\ \hline \text{set 2} & 1 & (1) & 0 & 1 \\ \hline \text{set 3} & 0 & 0 & (0) & 0 \\ \hline \text{set-}i & 1 & 0 & 1 & (?) \\ \hline \end{array} Can someone please point me into the right direction, please?