Proving the Addition of Dedekind Cuts

[hitmeoff]

Homework Statement

Show that if \alpha and \beta are Dedekind cuts then so is \alpha + \beta={r₁+r₂: r₁ \epsilon \alpha and r₂\epsilon\beta

Homework Equations

The Attempt at a Solution

If a dedekind cut is a cut whereby the sets on the left side of the cut and the right side of the cut would contain all the rational numbers then the cut itself is an irrational number right? If you add to irrationals don't you get another irrational number, thereby having another cut where the LH U RH still contains all the rationals?

 

[VeeEight]

Have you checked the three properties (non-empty, closed below, and has no greatest rational element)?

 

[HallsofIvy]

How is addition of Dedekind cuts defined?

 

[hitmeoff]

Not sure, the question was stated as I wrote it.

And this part: \alpha + \beta={r₁+r₂: r₁ \epsilon \alpha and r₂\epsilon\beta

throws me off... is this to say that alpha is the left hand set and beta the right hand set (or vice-versa) else I am not sure what they mean by r1 in Alpha and r2 in Beta. I thought the alpha and beta's were the actual cuts?

 

[Hurkyl]

im not sure what they mean by r1 in Alpha and r2 in Beta

What is the definition of a Dedekind cut?

 

[hitmeoff]

Its a "cut" that separates all rationals into two sets and has the props:

1. The sets are non-empty
2. Every rational is in one set or the other
3. Every number in the Left set is less than every number in the Right set.

So is the question stating that Alpha is on set and beta is the other set? If so then I don't know what the mean by Alpha + Beta?

 

[Hurkyl]

If that is the definition of Dedekind cut your book uses, then something is very wrong in the statement of your homework problem. What exactly was asked?

 

[Hurkyl]

Oh, FYI, there is another commonly used definition of Dedekind cut. As compared to your definition, this other definition defines a Dedekind cut as being the left-hand set.

I.E. if (L,R) is a your-Dedekind cut, then L is an other-Dedekind cut. Conversely, if L is an other-Dedekind cut, then (L, Q-L) is a your-Dedekind cut.


Your originally stated problem makes sense with this definition.

 

[hitmeoff]

Oh, FYI, there is another commonly used definition of Dedekind cut. As compared to your definition, this other definition defines a Dedekind cut as being the left-hand set.

I.E. if (L,R) is a your-Dedekind cut, then L is an other-Dedekind cut. Conversely, if L is an other-Dedekind cut, then (L, Q-L) is a your-Dedekind cut.Your originally stated problem makes sense with this definition.

ok, now the problem makes all kinds of sense.

So let \alpha and \beta be dedekind cuts.

WLG \alpha \subseteq \beta

\forallr₁ \epsilon \alpha and r₂ \epsilon \beta r₁+r₂\epsilon\beta which, by definition has no largest rational?

Note: Damn it, this only works if we are talking about the righ-hand set. Then I don't get how r1 + r2 \epsilon beta and be guaranteed not to have a largest rational since r1 and r2 could be rational elements and this set is bounded above

 

[HallsofIvy]

ok, now the problem makes all kinds of sense.

So let \alpha and \beta be dedekind cuts.

WLG \alpha \subseteq \beta

\forallr₁ \epsilon \alpha and r₂ \epsilon \beta r₁+r₂\epsilon\beta which, by definition has no largest rational?

Note: Damn it, this only works if we are talking about the righ-hand set. Then I don't get how r1 + r2 \epsilon beta and be guaranteed not to have a largest rational since r1 and r2 could be rational elements and this set is bounded above

You don't need to get "r1 + r2 \epsilon beta " and that is not necessarily true. All you need to do is show that \{ r_1+ r_2| r_\in \alpha, r_2\in \beta\} is a cut.

You need to prove:
1) It is non-empty.
2) There is some rational number that is not in it.
3) If x is in the cut and y is not then x< y.
4) It contains no largest number.

The first three are pretty simple. To prove the fourth, by contradiction, suppose it were not true. That is, suppose there exist r which is the largest member of this set. Then there exist r_1\in \alpha and r_2\in \beta such that r_1+ r_2= r. Now use the fact that \alpha and \beta have no largest member.

(I apologize for my first response. I misread the question and didn't realize that you were defining "\alpha+ \beta".)