Addition/Multiplication for Dedekind cuts?

[DrWillVKN]

I don't really understand the properties for adding/multiplying dedekind cuts. I get that they're closed, commutative and associative because that follows from the rational numbers (and the cut just partitions a rational number into 2 classes of rationals, plus the "cut" that only contains one number), but the other properties are confusing.

First, I'm not really clear on how to add/multiply cuts- which happens due to them containing only rationals. I thought you take the limit of cuts (which converge to a precise limit) and this limit is the precise number that you are able to add/multiply, but it is not an actual pinpointed number and just a limit, yet limits behave just like numbers. But then I have trouble tying all of this together.

EDIT: Okay I think I'm mixing up two ways of constructing the real numbers.

For the identity property for addition, 0* can be chosen as all the negative rational. Say p is a point in a dedekind cut A, and q is another point such that p < q. Then p = q + [-(q-p)], where -(q-p) is a point in 0*. Thus A \subset A + 0*, and A + 0* \subset A due to the property of dedekind cuts to be closed downward. This makes A + 0* = A, or am i misunderstanding a concept?

The inverse is really hard for me to imagine. A + B = 0*, so B is supposed to be a cut that gives a cut of all the negative rationals?

Multiplication seems to be similar to addition, but the cases are divided into positive and negative. 1* should be part of the identity property.

 

[micromass]

I don't really understand the properties for adding/multiplying dedekind cuts. I get that they're closed, commutative and associative because that follows from the rational numbers (and the cut just partitions a rational number into 2 classes of rationals, plus the "cut" that only contains one number), but the other properties are confusing.

First, I'm not really clear on how to add/multiply cuts- which happens due to them containing only rationals. I thought you take the limit of cuts (which converge to a precise limit) and this limit is the precise number that you are able to add/multiply, but it is not an actual pinpointed number and just a limit, yet limits behave just like numbers. But then I have trouble tying all of this together.

Cuts have nothing to do with limits (at least not immediately). What a cut represents is all the numbers smaller than a given number. Thus the number 2 is represented by the cut

\{x\in \mathbb{Q}~\vert~x&lt;2\}

So the supremum of a cut is the number it represents. However, the supremum doesn't always exist, because we're working in the rationals here. For example, the cut

\{x\in \mathbb{Q}~\vert~x^2&lt;2\}

doesn't have a supremum. Indeed, that supremum would be \sqrt{2} but this isn't rational!

For the identity property for addition, 0* can be chosen as all the negative rational. Say p is a point in a dedekind cut A, and q is another point such that p < q. Then p = q + [-(q-p)], where -(q-p) is a point in 0*. Thus A \subset A + 0*, and A + 0* \subset A due to the property of dedekind cuts to be closed downward. This makes A + 0* = A, or am i misunderstanding a concept?

This is ok.

The inverse is really hard for me to imagine. A + B = 0*, so B is supposed to be a cut that gives a cut of all the negative rationals?

Do some easy examples first. You know the inverse of 2 in -2. The cut of 2 is

A=\{x\in \mathbb{Q}~\vert~x&lt;2\}

You must transform this into the cut of -2, that is

B=\{x\in \mathbb{Q}~\vert~x&lt;-2\}

The thing to look at is -B, that is

-B=\{-x\in \mathbb{Q}~\vert~x&lt;-2\}

Then you obtain all numbers larger than 2. So given a cut A, a good choice for the inverse would be

\{-x\in \mathbb{Q}~\vert~x\notin A\}

However, this cut might have a largest element (which is not allowed), so we need to remove that!

 

[HallsofIvy]

To multiply two cuts, start with "positive" cuts. A cut is positive if and only if it contains at least one rational number. If a and b are positive cuts, then ab is the set of all rational numbers, xy, where x\in a, y\in b. Show that this is a cut. Show that if a is any cut, then a1= a. Remember that "1" is the set of all rational numbers less than 1.

Then define: if a< 0, b> 0, ab= -(-a)(b), if a> 0, b< 0, ab= -(a)(-b), and if a< 0, b< 0, ab= (-a)(-b).

 

[mycroft]

For the identity property for addition, 0* can be chosen as all the negative rational. Say p is a point in a dedekind cut A, and q is another point such that p < q. Then p = q + [-(q-p)], where -(q-p) is a point in 0*. Thus A \subset A + 0*, and A + 0* \subset A due to the property of dedekind cuts to be closed downward. This makes A + 0* = A, or am i misunderstanding a concept?.

Sorry to jump into this but actually I'm going through the same proof and I don't get this.

-(q-p) in 0* I understand as some negative rational in the set of all negative rationals but the jump from there to saying A \subset A + 0* I just can't follow and the more I think about it the worse it gets.

I imagine A + 0* as all rationals in A \geq 0 and A \subset A + 0* just baffles me :(

 

[HallsofIvy]

I imagine A + 0* as all rationals in A \geq 0 and A \subset A + 0* just baffles me :(

Then you are being too imaginative! "All rationals in A\geq 0" is not a cut.
(Do you see why?)

0* is the set of all rational number less than 0- in other words the negative rationals. A+ 0* consists of all sums of rationals in A and rationals in 0*. But since all rationals in 0* are negative, we always get numbers less than the one in A.

That is, if a\in A, o\in O*, then a- o< a. Since one of the properties of a cut is "if a\in A and b< a, then b\in A, it follows that every member of A+ O* is in A. That is, A+ O* is a subset of A.

Of course, it is also one of the properties of a cut that there is no largest member. If a\in A, there exist b\in A with b> a. Then c= (a- b)< 0 so c is in O*. It follows that b+ c= b+ (a- b)= a is in A+ O*. That is, A is a subset of A+ O*: A= A+ O*.

 

[mycroft]

Then you are being too imaginative! "All rationals in A\geq 0" is not a cut.
(Do you see why?)

Yes, it's not closed downward.

0* is the set of all rational number less than 0- in other words the negative rationals. A+ 0* consists of all sums of rationals in A and rationals in 0*. But since all rationals in 0* are negative, we always get numbers less than the one in A.

I think my mistake was imagining adding x \in A and y \in B as x_n + y_n thus causing O* to cancel the elements of A < 0. Am I right in assuming now that it is the set of each element x \in A added to all y \in B.

That is, if a\in A, o\in O*, then a- o< a. Since one of the properties of a cut is "if a\in A and b< a, then b\in A, it follows that every member of A+ O* is in A. That is, A+ O* is a subset of A.

Of course, it is also one of the properties of a cut that there is no largest member. If a\in A, there exist b\in A with b> a. Then c= (a- b)< 0 so c is in O*. It follows that b+ c= b+ (a- b)= a is in A+ O*. That is, A is a subset of A+ O*: A= A+ O*.

Is it because Q is dense (and we are working with a cut) that A + O^* \subseteq A and A \subseteq A + O^* ? I.e., you are not losing the larger elements of A by subtracting a set of negative numbers.

Many books (Rudin) use \subset so I'm not sure my interpretation is correct at all.



Thanks a lot for your help. Once I get myself muddled it's very hard to get unmuddled!