It is well known that there are more irrational numbers than there are rational ones, and a typical prove for that is the Dedekind cut. Thinking about alternative proves I came up with the following idea:

  • Interpret any $n\in\mathbb N$ as a hexadecimal number and convert it to a string (using e.g. UTF-8 encoding).
  • Of all the strings obtained this way, only consider those that provide plain English instructions via which one or more real numbers can be obtained.
  • All numbers obtained this way are called "describable", the set of them shall be denoted $\mathbb D$.

Now my problem: Since $\mathbb R$ is uncountable, this means there must be an uncountable amount of "indescribable" real numbers $\mathbb I := \mathbb R\backslash \mathbb D$. But using the "describability" defined above, $\mathbb N$ can provide infinitely many descriptions of arbitrary lengths (thus $\mathbb D$ is countable as well), and I fail to see how the countability of $\mathbb N$ can imply the uncountability of $\mathbb I$ (other than already asserting the uncountability of $\mathbb R$). So can the uncountability of $\mathbb I$ be proven?

Since no requirement is made about finite evaluation of the description of any $d\in\mathbb D$, I think $\mathbb D$ is a superset of the computable numbers, though I'm not sure.

  • $\begingroup$ Even more confusing, since the Dedekind cut typically only proves that there is at least one irrational number, that one could be described as "the one number obtained by performing the following Dedekind cut: ...", thus the indescribable numbers are even different from "Dedekind numbers". $\endgroup$ Sep 27, 2017 at 7:36
  • $\begingroup$ Would the downvoter mind to explain what the question lacks? $\endgroup$ Sep 27, 2017 at 9:21
  • $\begingroup$ How can $\mathbb N$ provide infinitely many description of infinite length? Isn't that a typo? $\endgroup$
    – skyking
    Sep 27, 2017 at 9:25
  • $\begingroup$ @skyking That's a bit hand-waving, you're right. It should be more about "infitely many descritptions of arbitrary length" - now that I think about it, this suggests $\mathbb R$ is the closure of $\mathbb D$ since that arbitrary length approaches but never reaches infinity, right? $\endgroup$ Sep 27, 2017 at 9:31
  • $\begingroup$ $\mathbb R$ is the completion/closure of $\mathbb D$, not because arbitrary length aproaches infinity, but the fact that $\mathbb Q\subset\mathbb D$ and the completion/closure of $\mathbb Q$ is $\mathbb R\supset \mathbb D$. Infinite description does not make much sense anyway I think (especially if you can't describe them in a finite way in which case we would be back with a finite description). $\endgroup$
    – skyking
    Sep 27, 2017 at 9:37

2 Answers 2


By contradiction, I (think I) can merely prove that there are either uncountably many indescribable numbers or none at all: Assume $\mathbb I\neq \{\}$ and $\mathbb I$ is countable. Then there exists a mapping of $\mathbb N$ to $\mathbb I$. That in turn implies all $i\in\mathbb I$ can actually be described⁺, contradicting $\mathbb I = \mathbb R\backslash \mathbb D$.

My problem is I don't have a clue how to prove that there are indescribable numbers at all, since intuitively one can always describe at least one indescribable number as "a random number picked from the set of indescribable numbers".

⁺ Since $\mathbb I$ is asserted countable, it can be ordered. Then obviously each element of $\mathbb I$ could be described by the string "The $n^\text{th}$ element of $\mathbb I$".

  • 1
    $\begingroup$ I think that this is incorrect as you would if $\mathbb I\ne\emptyset$ have that there's at least a function from $\mathbb N$ to a subset $\mathbb I^\circ\subset \mathbb I$. Your construct would mean that the elements of $\mathbb I^\circ$ are describable and so theres describable numbers in $\mathbb I$. This is clearly wrong as we "know" there are undescribable numbers. There's a number of ways to dismiss the apparent paradox in the fact that one thinks one can describe numbers in $\mathbb I$. $\endgroup$
    – skyking
    Sep 27, 2017 at 9:32
  • $\begingroup$ @skyking That makes my mind spin in circles... doesn't the impossibility of a countable subset of $\mathbb I$ imply that $\mathbb I$ must be empty? $\endgroup$ Sep 27, 2017 at 9:36
  • $\begingroup$ No, it's just an apparent paradox. We know from Cantor's proof that $\mathbb R$ is uncountable and we know that $\mathbb D$ is countable - therefore $\mathbb I=\mathbb R\setminus\mathbb D$ must be uncountable. $\endgroup$
    – skyking
    Sep 27, 2017 at 9:38
  • $\begingroup$ @skyking Agreed (unless there's a mistake and $\mathbb D$ is uncountable despite its construction), but therefore there must be a fallacy here which I fail to correct... $\endgroup$ Sep 27, 2017 at 9:43

There's some "meta" character of this construct. Basically what you mean by describable is that the number has a formula for it in the formal language of mathematics (or in another language). The way the apparent paradox can be dismissed varies depending an the actual formal system that's in use. This can be done in various ways - disallowing your construct at some step.

From Cantor's proof we surely know that $\mathbb R$ is uncountable and one can certainly argue that $\mathbb D$ should be countable. The consequence of this is that $\mathbb I=\mathbb R\setminus\mathbb D$ should be uncountable.

Of course a proof not using Cantor's proof is quite hard to construct. Obviously a proof for the non-emptyness of $\mathbb I$ can't rely on a concrete description of any element of $\mathbb I$. It also can't rely on any other reference to a number in $\mathbb I$ since without them being describable and without $\mathbb I\ne\emptyset$ having already been established one cannot use such a construct. This means that one would probably need to use proof by contradiction at some point.


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