[proofplan] Assume that the halting set has a total computable characteristic function. Use it to construct a partial computable function $g$ which intentionally disagrees with the diagonal computation $\varphi_e(e)$: it halts exactly when the characteristic function says that $e$ is not in the halting set. Since the numbering is acceptable, $g$ has some index $d$. Evaluating the construction at its own index $d$ forces $d \in K$ iff $d \notin K$, a contradiction. [/proofplan] [step:Assume a computable decision function for the halting set] Suppose, toward a contradiction, that $K$ is computable. Then there exists a total computable function \begin{align*} h: \mathbb{N} &\to \{0,1\} \end{align*} such that, for every $e \in \mathbb{N}$, \begin{align*} h(e) = 1 \iff e \in K. \end{align*} Equivalently, for every $e \in \mathbb{N}$, \begin{align*} h(e) = 1 \iff \varphi_e(e) \text{ is defined}. \end{align*} [/step] [step:Construct a partial computable diagonal function that reverses the decision] Define a partial function \begin{align*} g: \mathbb{N} &\rightharpoonup \mathbb{N} \end{align*} as follows. On input $e \in \mathbb{N}$, first compute $h(e)$. If $h(e) = 0$, halt and output $1$. If $h(e) = 1$, do not halt. Thus \begin{align*} g(e) \text{ is defined} \iff h(e) = 0. \end{align*} Since $h$ is total computable, this procedure defines a partial computable function $g: \mathbb{N} \rightharpoonup \mathbb{N}$. [guided] We use the supposed decision function $h$ to build a new partial computable function which behaves opposite to the diagonal halting prediction. Define \begin{align*} g: \mathbb{N} &\rightharpoonup \mathbb{N} \end{align*} by the following algorithm on input $e \in \mathbb{N}$: compute the value $h(e)$, which is possible because $h$ is total computable. If the computation gives $h(e) = 0$, then halt and output $1$. If the computation gives $h(e) = 1$, then enter a nonterminating computation. This gives the exact equivalence \begin{align*} g(e) \text{ is defined} \iff h(e) = 0. \end{align*} The construction is partial computable because it consists of a total computable preliminary computation, followed by a computable branch: one branch halts with the fixed value $1$, and the other branch intentionally runs forever. The point of the definition is that $g$ halts precisely on those inputs which $h$ declares not to be in the diagonal halting set. [/guided] [/step] [step:Choose an index for the diagonal function] Because $(\varphi_e)_{e \in \mathbb{N}}$ is an acceptable numbering of the partial computable functions, the partial computable function $g: \mathbb{N} \rightharpoonup \mathbb{N}$ has an index. Hence there exists $d \in \mathbb{N}$ such that \begin{align*} \varphi_d = g. \end{align*} [/step] [step:Evaluate the constructed function at its own index] By the definition of $K$ and the equality $\varphi_d = g$, \begin{align*} d \in K &\iff \varphi_d(d) \text{ is defined} \\ &\iff g(d) \text{ is defined}. \end{align*} By the construction of $g$, \begin{align*} g(d) \text{ is defined} \iff h(d) = 0. \end{align*} Since $h$ is the characteristic function of $K$, \begin{align*} h(d) = 0 \iff d \notin K. \end{align*} Combining these equivalences gives \begin{align*} d \in K \iff d \notin K. \end{align*} This is impossible. Therefore the assumed total computable characteristic function $h$ does not exist, and $K$ is not computable. [/step]