My Math Forum Dedekind's Theory of Irrationals

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 August 22nd, 2017, 05:23 AM #11 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 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 02:41 AM.
 August 22nd, 2017, 06:55 AM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 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, 02:15 PM   #13
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 Originally Posted by v8archie 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.
I don't see how either the Dedekind or the Cauchy constructions relate to computability. In the former case we're considering the set of all pairs of Lower/Upper-type partitions of the rationals. That's a valid set that we can build given that we believe in the rationals.

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 02:25 PM.

August 22nd, 2017, 04:28 PM   #14
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 Originally Posted by Maschke In the former case we're considering the set of all pairs of Lower/Upper-type partitions of the rationals.
That's not a visualisation I've seen before. I'd always seen a cut defined with an explicit condition such as
$$L = \{x \in \mathbb Q: x^2 \lt 2\} \quad R = \{x \in \mathbb Q: x^2 \ge 2\}$$
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 half-open 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 11:56 PM.

August 22nd, 2017, 07:05 PM   #15
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 Originally Posted by v8archie That's not a visualisation I've seen before. I'd always seen a cut defined with an explicit condition such as $$L = \{x \in \mathbb Q: x^2 \lt 2\} \quad R = \{x \in \mathbb Q: x^2 \ge 2\}$$
Oh I see. That couldn't possibly work. There aren't enough explicit expressions (no matter how you define them) to construct a complete ordered field, which the reals are required to be. (Defined to be, in fact).

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 07:11 PM.

August 23rd, 2017, 11:29 PM   #16
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Quote:
 Originally Posted by v8archie That's not a visualisation I've seen before. I'd always seen a cut defined with an explicit condition such as $$L = \{x \in \mathbb Q: x^2 \lt 2\} \quad R = \{x \in \mathbb Q: x^2 \ge 2\}$$
For this to work properly as Dedekind intended , you have to add an extra condition otherwise the partition to L and R sets gets jumbled up

$L = \{x \in \mathbb Q: x^2 \lt 2 \ \ or \ \ x \lt 0 \}$

$R = \{x \in \mathbb Q: x^2 \ge 2 \ \ or \ \ x \gt 0 \}$

With this extra condition a number like $\ \ -5 \ \$ surely belongs to $\ \ L$

Last edited by agentredlum; August 23rd, 2017 at 11:33 PM.

 August 24th, 2017, 02:45 AM #17 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond Country Boy, $$...$$ and $...$ no longer work here. Please use $...$. Thanks.
October 27th, 2017, 06:53 AM   #18
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 Originally Posted by agentredlum For this to work properly as Dedekind intended , you have to add an extra condition otherwise the partition to L and R sets gets jumbled up $L = \{x \in \mathbb Q: x^2 \lt 2 \ \ or \ \ x \lt 0 \}$ $R = \{x \in \mathbb Q: x^2 \ge 2 \ \ or \ \ x \gt 0 \}$
That last should be "and $x\gt 0$", not "or". With "or", 1/2, 1, 1.4 would be in this set which you do not intend.

Quote:
 With this extra condition a number like $\ \ -5 \ \$ surely belongs to $\ \ L$

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