I’m reading a book on axiomatic set theory, classic Set Theory: For Guided Independent Study, and at the beginning of chapter 4 it says:

So far in this book we have given the impression that sets are needed to help explain the important number systems on which so much of mathematics (and the science that exploits mathematics) is based. Dedekind’s construction of the real numbers, along with the associated axioms for the reals, completes the process of putting the calculus (and much more) on a rigorous footing.

and then it says:

It is important to realize that there are schools of mathematics that would reject ‘standard’ real analysis and, along with it, Dedekind’s work.

How is it possible that “schools of mathematics” reject standard real analysis and Dedekind’s work? I don’t know if I’m misinterpreting things but, how can people reject a whole branch of mathematics if everything has to be proved to be called a theorem and cannot be disproved unless a logical mistake is found?

I’ve even watched this video in the past: https://www.youtube.com/watch?reload=9&v=jlnBo3APRlU and this guy, who’s supposed to be a teacher, says that real numbers don’t exist and that they are only rational numbers. I don’t know if this is a related problem but how is this possible?

**Answer**

Although the possibility of different axioms is a concern, I think the major objection the author is speaking of is largely about *constructivism* (i.e. intuitionistic logic). There really is a big gap between rational numbers and real ones: with enough memory and time, a computer can represent any rational number and can do arithmetic on these numbers and compare them. This is not true for real numbers.

To be specific, but not too technical: let’s start by agreeing that the rational numbers $\mathbb Q$ are a sensible concept – the only controversial bit of that involving infinite sets. A Dedekind cut is really just a function $f:\mathbb Q\rightarrow \{0,1\}$ such that (a) $f$ is surjective, (b) if $x<y$ and $f(y)=0$ then $f(x)=0$, and (c) for all $x$ such that $f(x)=0$ there exists a $y$ such that $x<y$ and $f(y)=0$.

Immediately we’re into trouble with this definition – it is common that constructivists view a function $f:\mathbb Q\rightarrow\{0,1\}$ as some object or oracle that, given a rational number, yields either $0$ or $1$. So, I can ask about $f(0)$ or $f(1)$ or $f(1/2)$ and get answers – and maybe from these queries I could conclude $f$ was not a Dedekind cut (for instance, if $f(0)=1$ and $f(1)=0$). However, no matter how long I spend inquiring about $f$, I’m never going to even be able to verify that $f$ is a Dedekind cut. Even if I had two $f$ and $g$ that I knew to be Dedekind cuts, it would not be possible for me to, by asking for finitely many values, determine whether $f=g$ or not – and, in constructivism, there is no recourse to the law of the excluded middle, so we cannot say “either $f=g$ or it doesn’t” and then have no path to discussing equality in the terms of “given two values, are they equal?”*.

The same trouble comes up when I try to add two cuts – if I had the Dedekind cut for $\sqrt{2}$ and the cut for $2-\sqrt{2}$ and wanted $g$ to be the Dedekind cut of the sum, I would never, by querying the given cuts, be able to determine $g(2)$ – I would never find two elements of the lower cut of the summands that added to at least $2$ nor two elements of the upper cut of the summands that added to no more than $2$.

There are some constructive ways around this obstacle – you can certainly say “real numbers are these functions alongside proofs that they are Dedekind cuts” and then you can define what a proof that $x<y$ or $x=y$ or $x=y+z$ looks like – and even then prove some theorems, but you never get to the typical axiomatizations where you get to say “an ordered ring is a set $S$ alongside functions $+,\times :S\times S \rightarrow S$ and $<:S\times S \rightarrow \{0,1\}$ such that…” because you can’t define these functions constructively on $\mathbb R$.

(*To be more concrete – type theory discusses equality in the sense of “a proof that two functions $f,g$ are equal is a function that, for each input $x$, gives a proof that $f(x)=g(x)$” – and the fact that we can’t figure this out by querying doesn’t mean that we can’t show specific functions to be equal by other means. However, it’s a *huge* leap to go from “I can compare two rational numbers” – which is to say, I can always produce, from two rational numbers, a proof of equality or inequality – to “a proof that two real numbers is equal consists of…” understanding that the latter definition does not let us always produce a proof of equality or inequality for any pair of real numbers)

**Attribution***Source : Link , Question Author : Andrea Burgio , Answer Author : Milo Brandt*