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Is each Dedekind cut a unique real number? or when we apply the process(Dedekind cut), do we get a bunch of real numbers instead of a unique one.

If we get a unique real number, is the unique real number then plotted as a line segment between rationals on the number line? Or is it plotted as a single point (just like 0 and 1)?

If not so, Is the bunch of numbers that we get infinite?

novice
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There are several ways to construct (or if you prefer, define) the real numbers. The two most familiar are as Dedekind cuts in $\Bbb Q$, the ordered set of rational numbers, and as equivalence classes of Cauchy sequences of rational numbers with respect to a certain equivalence relation. If one constructs them using Dedekind cuts, then each Dedekind cut is by definition a unique real number. If one constructs them in some other way, it’s no longer the case that each real number is a Dedekind cut in $\Bbb Q$, but it is still true that there is a nice bijection between the set of Dedekind cuts in $\Bbb Q$ and the reals as constructed.

Saying that the reals are the numbers on the real number line is not a definition: in order to make it one, you’d need to define the real number line independently of the notion of real number. As it stands, you’re pretty much just saying that a real number is a member of the set of real numbers.

Brian M. Scott
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  • Yes I knew it was a circular definition, which is why I asked what it is the standard definition. Thanks. – novice Jan 22 '15 at 01:55
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    @novice: You're welcome. There really isn't one. The mathematically important fact is that all of the constructions produce isomorphic objects: up to isomorphism there is only one Dedekind-complete ordered field. – Brian M. Scott Jan 22 '15 at 02:03
  • So it's by definition a unique real number - But it is used to plugin the gap between two rational points and is plotted as a line segment between them? – novice Jan 22 '15 at 02:07
  • @novice: No, it's a single point, not a segment. No two rational numbers are adjacent: the average of any two rational numbers is a rational number between them. The gaps in $\Bbb Q$ are a bit more subtle than that. For example, if $R$ is the set of positive rational numbers $x$ such that $x^2>2$, and $L$ is the rest of the rationals, then every member of $L$ is less than each member of $R$, but there is no rational number between $L$ and $R$. In $\Bbb R$ the number $\sqrt 2$ fills that gap. – Brian M. Scott Jan 22 '15 at 02:25
  • That gap can only be filled by a line segment right? How can a point fill that gap. – novice Jan 22 '15 at 02:34
  • ok. I guess the gap is assumed to be a point. Could you please also help explain - How to obtain a set with no greatest element – novice Jan 22 '15 at 12:46
  • @novice: The set of integers is a familiar one with no largest element. Another is the set of real numbers. Yet another is the interval $(0,1)$: if $0<x<1$, $\frac12(x+1)$ lies strictly between $x$ and $1$, so it’s an element of $(0,1)$ bigger than $x$. Thus, no element of $(0,1)$ can be the largest. – Brian M. Scott Jan 22 '15 at 20:59
  • But then its an infinite set, using an abstract quantity like infinity. How do we prove that the set has no greatest element. I agree though its a way to getting infinitely close to square root of two. – novice Jan 23 '15 at 03:22
  • @novice: Let $r\in L$. If $r<0$, then $0$ is a larger element of $L$. If $r\ge 0$, then $r^2<2$, so $s-r^2>0$. Let $n$ be any integer greater than $\frac{2r+1}{2-r^2}$. Then $$\frac{2r}n+\frac1{n^2}=\frac1n\left(2r+\frac1n\right)\le\frac1n(2r+1)<2-r^2;,$$ so $$\left(r+\frac1n\right)^2=r^2+\frac{2r}n+\frac1n^2<2;,$$ and therefore $r+\frac1n\in L$. Clearly $r+\frac1n>r$, so $L$ has no largest element: this procedure always finds a larger one. – Brian M. Scott Jan 23 '15 at 14:19
  • Yes, but dedekind cut is a way of obtaining all the reals from rationals. After this process is addition defined on all the reals. So Im not sure if you can perform arithmetic on it before defining it. – novice Jan 23 '15 at 16:12
  • nvm.. I got it thanks – novice Jan 23 '15 at 16:42