Is there any "real" number that may not actually be a real because we haven't found its Dedekind cut?

Question

I just watched a video that shows how real numbers are constructed using Dedekind cuts, and what I understood was that a real number is a subset of Q which, among a few other conditions, contains no greatest element. The video also shows an example of how this property can be proven for $\sqrt 2$. I also found these proofs for $pi$ and $e$.

I was wondering if there was another way to construct the reals using Turing machines, since the Dedekind cut feels a bit like a "by negation" kind of definition, and a Turing machine would be more of a procedural, actual way to construct them. But it turns out Turing machines are countable and reals are not, so there must be real numbers that can't be computed by Turing machines.

Which then led me to this other question, in which they give a few samples of non-computable real numbers. But now I wonder, given any of these numbers (say Chaitin's constant), can we prove that the Dedekind cut that corresponds to this number (and to every other number that we intuitively think of as a real number) has no greatest element? Or could it be the case that Chaitin's constant turns out to be a non-real number (which I have no idea what it'd even mean, since it definitely seems to be a point on the real line)?

I know there are other ways to construct real numbers that I haven't learned about yet, so perhaps one of them can be used to prove it?

Answer 1

Your reasoning about counting Turing machines versus counting real numbers has a certain weakness. You might imagine to make a list of all Turing machines; this can be done automatically, since the description of a TM is syntactically clear, and they would be numbered from 1,2,3,... and onwards. Having done that, you might imagine to examine each Turing machine, starting from number 1, then number 2, and so on, to isolate those that happen to compute real numbers, which obviously only some of them would do. If you determine that TM numbered n computes a real number, you would give that real number a label i(n) to show that it is the i(n)'th real number computable by a TM.

It leaves the small obstacle to have to determine whether a given TM computes a real number or not. You probably see where this is going. The solution to the halting problem says that you cannot even determine whether the given TM ever halts, much less whether it halts and outputs a real number description that you desire.

It follows that you cannot argue in this way that there is any "set of computable real numbers", nor that such a set would be countable.

Something like Chaitin's constant is obviously a well-defined real number, since it is given by a description of the individual decimals in its decimal expansion. It is easy in principle to construct a corresponding Dedekind cut or a Cauchy sequence. Not so much in practice.

Answer 2

The Dedekind cut $A$ representing a certain real number $\alpha$ consists of all rational numbers that are $<\alpha$. Among the numbers in $A$ there are arbitrarily precise rational approximations to $\alpha$, but also totally uninteresting numbers in reference to the value of $\alpha$. But the latter play no rôle in the subsequent construction. The reason why we are carrying them along is that we want a single and simple mathematical object $A$ representing the number $\alpha$, without having to bother about extra data like error bounds.

The condition that such a cut should have no maximal element is not a "definition by negation", but of a purely technical nature. A cut defining an irrational number, like $\pi$, automatically has no maximal element. But we want to have the rational numbers as ordinary real numbers as well. Without taking care a rational number $\rho$ could be represented by all rational numbers $<\rho$ as well as by all rational numbers $\leq\rho$. In order to avoid ambiguities one requires once and for all that the second kind of "cuts" is excluded.