This question already has an answer here:

I understand how integers and rationals are expressed/derived in ZFC. But what about the irrational numbers? Can they also be expressed? If not, are there other axiomatic set theories able to express them? As for Dedekind cuts, from my understanding (maybe wrong), any irrational in question must have a 1-1 explicit function to a natural number, in order for the method to work (As an example the square root of 2 has such a function).

It boils down to this: is there an isomorphism between the the set of irrationals (as "just" numbers) and a a set of pure sets?


marked as duplicate by Zhen Lin, Trevor Wilson, Asaf Karagila, Ross Millikan, Davide Giraudo Nov 6 '13 at 20:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @ZhenLin I don't know if It's a duplicate, because my questions is specifically about irrational numbers... If you still consider it as such I'll close it, as I'm starting to notice (-1) people don't like 'duplicates'.. :) $\endgroup$ – user76568 Nov 6 '13 at 20:32
  • $\begingroup$ They dont like good answers too $\endgroup$ – Abdulh Khazzak Gustav ElFakiri Nov 6 '13 at 20:34
  • $\begingroup$ @Dror: How do you represent the rational numbers? In $\sf ZFC$ context, for set theorists, real numbers are usually just sequences of integers, or sometimes binary sequences. The rational numbers would correspond to eventually zero (or eventually constant binary sequences) and irrational numbers correspond to the rest. $\endgroup$ – Asaf Karagila Nov 6 '13 at 20:48
  • $\begingroup$ Moreover, note that the constructions of $\Bbb Z$ from $\Bbb N$; of $\Bbb Q$ from $\Bbb Z$; or $\Bbb R$ from $\Bbb Q$... usually don't extend properly, but rather there is a "canonical copy" of the previous system in the new one, which we identify with the previous system. So when constructing the real numbers as sets, you end up with a whole other copy of the rationals or the integers or the natural numbers. $\endgroup$ – Asaf Karagila Nov 6 '13 at 20:49
  • 1
    $\begingroup$ Dror, once you have a representation for the rationals, you can use any of the standard constructions for the real numbers from the rationals, and that would end up as a set. If you can represent the real numbers as a set, in the universe of set theory, then all the real numbers were also represented by sets. In particular ALL the irrational numbers. $\endgroup$ – Asaf Karagila Nov 6 '13 at 21:14

Dedekind cuts are the most common way. The idea is that, given the rationals, you take the reals to be all the downward closed sets of the rationals (including all the ones with no greatest element). For a concrete example, $\sqrt{2}$ is $\{x\in\mathbb{Q}:x^2<2\}$.

The version I'm chiefly familiar with is Quine's from "Set Theory and its Logic". He defines a rational $a/b$ as $\{x+(x+y)^2:x\cdot b < a\cdot y\}\subset\mathbb{N}$. Then given a collection of such rationals $A$, their upper bound is $\bigcup A$. Most of these upper bounds fail to be other rationals themselves. In this treatment, rationals are a subset of the reals as they are intuitively, and "$\leq$" reduces to $\subseteq$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.