1. Jul 12, 2011

### 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.

Last edited: Jul 12, 2011
2. Jul 12, 2011

### micromass

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

$$\{x\in \mathbb{Q}~\vert~x<2\}$$

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

$$\{x\in \mathbb{Q}~\vert~x^2<2\}$$

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

This is ok.

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

$$A=\{x\in \mathbb{Q}~\vert~x<2\}$$

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

$$B=\{x\in \mathbb{Q}~\vert~x<-2\}$$

The thing to look at is -B, that is

$$-B=\{-x\in \mathbb{Q}~\vert~x<-2\}$$

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!

3. Jul 14, 2011

### 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).

4. Aug 3, 2011

### mycroft

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 :(

5. Aug 4, 2011

### HallsofIvy

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*.

6. Aug 5, 2011

### mycroft

Yes, it's not closed downward.

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 $y \in B$.

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!