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August 22nd, 2017, 04:23 AM  #11 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
The Dedekind cut is one way to construct the reals. Here is another: Consider the set of all "Cauchy sequences" of rational numbers. That is, the set of all infinite sequences of rational numbers, $\displaystyle a_0, a_1, a_2, \cdot\cdot\cdot, a_n, \cdot\cdot\cdot$ such that $\displaystyle a_n a_m$ goes to 0 as m and n go to infinity independently. We say that two such sequences $\displaystyle \{a_n\}$ and $\displaystyle \{b_n\}$ are equivalent if and only if the sequence $\displaystyle (a_0 b_0), (a_1 b_1), \cdot\cdot\cdot, (a_n b_n), \cdot\cdot\cdot$ goes to 0. It is easy to show that this is an equivalence relation and then we define the set of all real numbers to be the set of all equivalence classes. Last edited by greg1313; August 24th, 2017 at 01:41 AM. 
August 22nd, 2017, 05:55 AM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra 
Most numbers are not describable/computable$*$. Doesn't that mean that we can't describe an appropriate Cut without referring to the number itself. The only way out of this that I can see is to use the limit of a Cauchy sequence to define the Cut. But this approach effectively uses Cauchy sequences to define the reals, making the Cuts redundant except for the way they impose an ordering relation on the reals. $*$ I'm aware that "describability" is a slippery concept, but the flavour of the idea is what I wish to invoke. 
August 22nd, 2017, 01:15 PM  #13  
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  Quote:
Likewise the set of all Cauchy sequences of rationals is also a set by the rules of set theory, once we believe in the rationals. (Then we have the set of all functions from the naturals to the rationals that satisfy the definition of a Cauchy sequence). Definability/computability never come into this discussion in any way that I can see. Can you explain why you see a connection? Last edited by Maschke; August 22nd, 2017 at 01:25 PM.  
August 22nd, 2017, 03:28 PM  #14  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra  Quote:
$$L = \{x \in \mathbb Q: x^2 \lt 2\} \quad R = \{x \in \mathbb Q: x^2 \ge 2\}$$ I'd already spotted a possible inconsistency in considering this to be less well defined than the set of all Cauchy sequences, although I haven't worked it all through in my mind. But while I know that we can't describe a condition for every real without using an infinite number of characters, the set of binary sequences (not ending in an infinite string of 1s) is well defined and can be easily used to define a Cauchy sequence, giving us the reals in the halfopen interval $[0,1)$ and thus all reals. I'm aware that I'm using infinite binary sequences while at the same time not being comfortable with infinite sequences of symbols in the set definition, but haven't thought the conclusion of that dichotomy through. But I also have a feeling that the Cauchy route uses a sequence that I know makes sense to generate a number. The Dedekind Cut seems to require more from its infinite sequence of symbols unless it relies on the Cauchy sequence itself. I guess this is a question about what it means to construct the reals. Do we just create an amorphous blob of objects that must contain them all several times over, or do we create each one separately. Last edited by skipjack; August 22nd, 2017 at 10:56 PM.  
August 22nd, 2017, 06:05 PM  #15  
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  Quote:
We see expressions like that as examples of cuts; but that's not how cuts are defined. Rather than go into all the details I'll just refer readers to Rudin, PMA, 3rd edition, page 17ff. Earlier he states without proof theorem 1.19, that there exists a complete ordered field. Now on page 17 he proves that theorem. He defines a cut as any subset of the rationals that's nonempty, not all the rationals, that's downward closed (if x is in the cut and y < x then y is in the cut); and that has no largest element. From that definition, he shows that the set of cuts, with <, +, and * defined appropriately, is a complete ordered field that satisfies all of the axioms for the real numbers given earlier in the chapter. There is no reference to any specific cut. Any concrete example we give is of course of a very special form, namely describable by a formula, or computable, or some such characterization. Of course there aren't enough of those to make up all the reals and the construction never discusses any particular cuts. Note that the set of cuts exists. First, the power set of the rationals exists by the powerset axiom. Among all those subsets, some are cuts. Namely they are nonempty, downward closed, etc.; and each condition is a predicate that can be applied to the power set via the axiom of specification. So the set of cuts exists. Last edited by Maschke; August 22nd, 2017 at 06:11 PM.  
August 23rd, 2017, 10:29 PM  #16  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
$L = \{x \in \mathbb Q: x^2 \lt 2 \ \ or \ \ x \lt 0 \}$ $R = \{x \in \mathbb Q: x^2 \ge 2 \ \ or \ \ x \gt 0 \}$ With this extra condition a number like $ \ \ 5 \ \ $ surely belongs to $ \ \ L $ Last edited by agentredlum; August 23rd, 2017 at 10:33 PM.  
August 24th, 2017, 01:45 AM  #17 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  Country Boy, [tex]...[/tex] and [latex]...[/latex] no longer work here. Please use [math]...[/math]. Thanks.

October 27th, 2017, 05:53 AM  #18  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
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