Because of diagonalization. If $(f_e: e \in \mathbb{N})$ was a computable enumeration of all total computable functions from $\mathbb{N}$ to $\mathbb{N}$, such that every $f_e$ was total, then $g(i) = f_i(i)+ 1$ would also be a total computable function, but it would not be in the enumeration. That would contradict the assumptions about the sequence. Thus no computable enumeration of functions can consist of exactly the total computable functions.
Suppose we think of a universal computable function $h(e,i)$, where "universal" means $h$ is a computable binary function and that for every total computable unary function $f(n)$ there is some $e$ such that $f(i) = h(e,i)$ for all $i$. Then there must also be some $e$ such that $g(n) = h(e,n)$ is not a total function, because of the previous paragraph. Otherwise $h$ would give a computable enumeration of total computable unary functions that includes all the total computable unary functions.
Thus the requirement that every function is a system of functions is total is incompatible with the existence of a universal function in that system. For some weak systems, such as the primitive recursive functions, every function is total but there are not universal functions. Stronger systems that have universal functions, such as Turing computability, simply must have partial functions in order to allow the universal function to exist.
I just wanted to add that somebody found what appears to be a loophole in diagonalization. If you use a typed representation for the program, you can use the type system to disallow diagonalization and create a total self-interpreter. See Breaking Through the Normalization Barrier: A Self-Interpreter for F-omega for details.
– hatch22 – 2016-05-19T21:03:28.717Of course, System F is not a Turing complete system. The paper you linked is interesting; it seems that they manage to leverage the non-Turing-completeness in an interesting way. – Carl Mummert – 2016-05-20T11:50:10.547
I don't get why "then $g(i) = f_i(i) + 1$ would also be a total computable function". If $g$ is a total computable function then $\exists k, f_k = g$, then evaluating $g(k)$ requires evaluating $g(k) = f_k(k) + 1 = g(k) + 1$: conttradiction. So it seems that if there is an enumeration of total computable functions, we cannot even build $g$, so that we cannot reach a contradiction to disprove the initial hypothesis (We can reach a contradiction, but it just disproves $g$ being total computable). – agemO – 2018-07-23T06:03:08.323
And even using a shifted diagonal to avoid this problem seems to lead to contradictions. – agemO – 2018-07-23T06:24:01.403