-1

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?

user76568
  • 4,542
  • @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'.. :) – user76568 Nov 06 '13 at 20:32
  • They dont like good answers too – Abdulh Khazzak Gustav ElFakiri Nov 06 '13 at 20:34
  • @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. – Asaf Karagila Nov 06 '13 at 20:48
  • 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. – Asaf Karagila Nov 06 '13 at 20:49
  • I think it is a duplicate, in that the other answer covers all the reals. Once you have the reals, you have the irrationals as part of them. – Ross Millikan Nov 06 '13 at 20:52
  • @AsafKaragila I don't quite follow. ZFC DOES have 'representations' of all rationals using ONLY sets, and no "number 'codings'"/sequences involved. The original questions is about existence of such, for ALL irrationals. Please correct me if I'm wrong. – user76568 Nov 06 '13 at 21:06
  • 1
    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. – Asaf Karagila Nov 06 '13 at 21:14
  • I'm sorry to say, but your edit shows that you don't understand the question either (without trying to sound arrogant or something like that). – Asaf Karagila Nov 06 '13 at 21:23
  • Maybe this thread will help you understand your question, and its answer better. – Asaf Karagila Nov 06 '13 at 21:24
  • @AsafKaragila thanks for the reference. I'll read it tomorrow when I'm focused. STILL, I don't understand why you invalidate my understanding.. Maybe my question is not formally correct or accurate, but I'm pretty sure you can understand it, and understand that I understand what I'm asking.. – user76568 Nov 06 '13 at 21:33
  • Dror, I'm not invalidating your understanding or anything. I just answered a lot of these questions for the past three years now. Both on this site, to people in real life, to students in class. With this experience you learn to sometimes identify from the question, that the asker had heard or read about a result, but really don't have the mathematical basis for understanding it. The arrogant, but truly correct, way to answer this is to send you to study logic and set theory for a while, and when you're understood how do we use $\sf ZFC$ as a foundation, you'll find the answer by yourself. – Asaf Karagila Nov 06 '13 at 21:38
  • @AsafKaragila I understand your point and agree fully. Meanwhile, should I delete this question in your opinion? – user76568 Nov 06 '13 at 21:43
  • let us continue this discussion in chat – user76568 Nov 06 '13 at 22:06
  • Dror, I'm sorry for not responding. I dislike the chat feature of this website, so I do my best to avoid it. As for deleting, as it stands the thread will be deleted sometime within 30 days (closed threads with a negative score, and no answer with a positive score get deleted within 30 days I believe, but maybe less?) So it's up to you, really. – Asaf Karagila Nov 08 '13 at 12:49
  • I don't know exactly what you're trying to ask. If you're asking something similar to this like I think you are, then this is not a duplicate. It's a real question that asks whether ZF can prove or disprove the existence of irrational numbers. Some people maybe just assume the complete ordered field axioms of real numbers and think it's impossible to prove whether they're well founded or not and find ZF a dubious model of set theory because it proves the axiom of regularity. According to – Timothy Jan 23 '19 at 06:30
  • https://meta.stackexchange.com/questions/217754/stance-on-repost-question-instead-of-editing-to-invalidate-answers/217757#217757, if you're still confused about something and the question this one is marked as a duplicate of doesn't answer your question, maybe you could ask another question as a fixed up version of this question linking this question explaining how the one this one is marked as a duplicate of doesn't answer your question, and clearly explain in detail how you think and ask how a result seems to contradict how you think. – Timothy Jan 23 '19 at 06:34

1 Answers1

2

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 $\subseteq$.

Malice Vidrine
  • 9,653