Proof that total computable functions are not enumerable

Question

In an answer to this question, a sketch of the proof that total computable functions are not enumerable is made:

Because of diagonalization. If $(f_e:e \in N)$ was a computable enumeration of all total computable functions from N to 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.

I really don't understand why

then $g(i)=f_i(i)+1$ would also be a total computable function

Indeed if $g$ was a total computable function, then $\exists k, f_k=g$, so that $g(k) = f_k(k) + 1 = g(k) + 1$, which is impossible.

So it seems to me that building $g$ is impossible, so that we cannot reach a contradiction by saying that we have built another total computable function which is not in our enumeration.

I tried shifting the diagonal to avoid this effect, but other loops of the same kind can't be ruled out.

Answer 1

$g$ is total computable by definition.

• By assumption $f : \mathbb{N}^2 \to \mathbb{N}$ is total computable.
• $1$ certainly is.
• $+$ certainly is.
• The concatenation of $+$ and $f$ is computable by construction, and that's $g$.

The details depend on the exact definition of "computable" (i.e. the machine model) you use; see e.g. here.

So under the assumption that such $f$ existed, $g$ is computable; therefore, it must have an index in $f$, which is impossible (as you note); therefore, the assumption must be false.

Answer 2

The problem really is in how you're approaching the proof by contradiction. You're objecting to one conclusion ("$g$ is computable and total") by drawing a different conclusion ("$g$ isn't total computable, since it's not on the list" or "... since there is a way to interpret it as referring to itself") contradicting it. But this is exactly the point: that from our hypothesis we can conclude contradictory things, so our hypothesis must be wrong. That is, within a proof by contradiction you have to sort of give yourself "tunnel vision:" when trying to prove Thing A inside a proof by contradiction, don't let its clear falseness stop you!

It may be helpful in this case to rewrite the usual proof as a direct proof, and I've done this below. Incidentally, exactly the same thing can be done with Cantor's diagonal argument, and it might help to get comfortable with that first if this proof is still confusing.

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First, convince yourself of the following:

If $f$ is a total computable function of two variables, then the function $g_f: x\mapsto f(x,x)+1$ is also total computable.

Exactly how you do this will depend on which definition of "computable" you're using, but they'll all make it fairly easy. For example, in $\lambda$-calculus this is immediate: $g_f$ is just $\lambda x.S(f(x,x)).$ With Turing machines, this will take a bit more work but really only a bit.

We can now prove:

If $f$ is a total computable function of two variables, then $g_f$ is not any of the "rows" of $f$: for each $x$ we have $g_f\not=f(x,-)$ (since in particular $g_f(x)\not=f(x,x)$).

Now an effective enumeration of (some, not necessarily all) total computable functions of one variable is really just a single total computable function of two variables: "$(f_e)_{e\in\mathbb{N}}$" is really the same as the two-ary function $f: (x,y)\mapsto f_x(y)$. So phrasing things differently, what we have is:

If $(f_e)_{e\in\mathbb{N}}$ is an effective enumeration of total computable functions, then there is a total computable function $g$ which is not equal to any of the $f_e$s.

Namely, let $g(x)=f_x(x)+1$.

Note that everything here is direct: I haven't used a proof by contradiction anywhere. But now it should be obvious that there is no effective enumeration of all the total computable functions, since that would contradict the point above.

The basic flow of ideas in the proof-by-contradiction-version of this argument is:

• We start with a hypothetical object: an effective enumeration $E$ of all the total computable functions.

• Well, in particular it's an effective enumeration of some total computable functions (it just happens to hit them all) ...

• ... And for any effective enumeration of total computable functions, we have a total computable function not so enumerated.

• So in particular there is a total computable function not in $E$, and this contradicts the assumption on $E$, so we're done.

Note, though, that this is really just adding an unnecessary level of confusion: it's really better to proceed as above, where we prove directly a fact ("For every effective enumeration of total computable functions, we can find (in fact, effectively in an index for the enumeration!) a total computable function not so enumerated") which implies the thing we want ("No effective enumeration of total computable functions gets all of them").

Answer 3

The answer is that you have found a legitimate problem with these naive proofs by diagonalisation. In practice, diagonalisation cannot be applied to computational theory. Diagonalising is only possible in the fantastical land of actually infinite sets. There are no such things in the world of computation.

The problem is this: It is impossible to build a machine g case by case, because we would require an infinite number of cases, and the description of g would grow in length forever. However, to define g differently to any total computable function requires that g itself is not a total computable function, otherwise it would be different to itself! But we cannot begin by assuming what has to be proven.

$g(i)=f_i(i)+1$ is not a definition. It is a property. It means "define g of i such that ..."

Unless a procedure can be legally described which computes g, then this property is meaningless in the world of computation. The reason that functions with such properties can be assumed to exist in the fantastical world of actually infinite sets is because the Axiom of Choice allows them to choose elements from every set in an infinite set all at once. Obviously in the finite world of computation, this is impossible.

Finally, I would like to point out that Turing, in his original 1936 paper does not give such a stupidly naive argument for the existence of uncomputable real numbers. His proof is much more complex, precisely because he was aware that direct self-reference such as this must be avoided.