Is anyone interested in an algorithm that builds a well ordered, uncountable, subset of the real numbers (irrational > 1) without the axiom of choice? I think I have found such an algorithm, but am not 100% sure that the set it builds is not countable. If the set it builds is countable and a mapping from the integers to that set can be stated (as opposed to knowing such a mapping exists) then the algorithm continues. That would indicate that the set generated is uncountable. Of course, such an algorithm would need an uncountable number of iterations to build such a set. The algorithm is not practical in that sense, but it would still show that such a well ordering is possible without invoking the axiom of choice. I do not think it would be easy to decide if the set has the cardinality of the real numbers or not. However, someone very good with continued fractions might be able to figure it out. The algorithm uses continued fraction notation to represent the numbers used. That allows one to write the numbers as an infinite vector of natural numbers and uses that notation for the algorithm.

  • 3
    $\begingroup$ I don't this is possible $\endgroup$ May 14 at 3:47
  • 2
    $\begingroup$ Sorry, *think is missing... here's a reference math.stackexchange.com/questions/44559/… $\endgroup$ May 14 at 4:23
  • $\begingroup$ I think you can make uncountable choice if you can make uncountable number of iterations, so you are implicitly assuming axiom of choice. $\endgroup$
    – Anixx
    May 14 at 10:43


Browse other questions tagged or ask your own question.