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In my Real Analysis class we just started talking about Dedekind cuts and I'm very confused on how to prove this one homework assignment:

Given the set $-A=\{-(a+b):a\in\mathbb{Q}^+, b\in\mathbb{Q}\setminus A\}$, how do I prove this is a cut? Please help, I'm so lost...

Thanks!!

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Write down the definition of a Dedekind cut. Now go over all the properties in the definition and show that $-A$ satisfies them. –  Asaf Karagila Sep 1 '12 at 18:08
    
Thanks, but I'm having trouble thinking how to prove that $-A\neq \mathbb{Q}$. How can I do that? –  roboguy12 Sep 1 '12 at 18:17
    
What is $A$ in the first place? –  J. Loreaux Sep 1 '12 at 18:20
    
As $A$ is a cut, $b$ is bounded from below and thus $-(a+b)$ is bounded from above. –  Karolis Juodelė Sep 1 '12 at 18:22
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1 Answer 1

up vote 7 down vote accepted

I assume that $A$ is supposed to be a Dedekind cut. Start by making a sketch:

    +++++++++++++++++++++++++++++++)----------------------------------  
                   A

The line as a whole represents $\Bbb Q$, and everything to the left of the parenthesis is in the cut $A$. Now look at the definition of $-A$:

$$-A=\{-(q+b):q\in\mathbb{Q}^+, b\in\mathbb{Q}\setminus A\}$$

It depends on two subsets of $\Bbb Q$, $\Bbb Q^+$, and $\Bbb Q\setminus A$, so we should figure out where those are in the picture. $\Bbb Q\setminus A$ is easy:

    +++++++++++++++++++++++++++++++)(---------------------------------  
                   A                                Q\A

it’s everything to the right of $A$. And we know just what $\Bbb Q^+$ is: it’s the set of positive rationals. What happens when you add a positive rational $q$ to every member of $\Bbb Q\setminus A$? You shift the set $\Bbb Q\setminus A$ to the right by $q$ units:

    -------------------------------)(+++++++++++++++++++++++++++++++++
                   A                                Q\A

    ooooooooooooooooooooooooooooooo)-----------(++++++++++++++++++++++  
                   A                                   q + Q\A

That last picture shows $\{q+b:b\in\Bbb Q\setminus A\}$ for a particular $q\in\Bbb Q^+$. Now what happens when you look at the negatives of these rational numbers, $\{-(q+b):b\in\Bbb Q\setminus A\}$? You simply flip the line $180$° around $0$ to get a picture more or less like this:

    ++++++++++++++++++++++)-----------(ooooooooooooooooooooooooooooooo

The plus signs mark the set $-(q+\Bbb Q\setminus A)$, and the o’s mark the set $\{-a:a\in A\}$. The gap in the middle has length $q$. $\{-a:a\in A\}$ is always in the same place, but the location of $-(q+\Bbb Q\setminus A)$ depends on $q$: when $q$ is large, it’s far to the left of $\{-a:a\in A\}$, and when $q$ is small, it’s very close to $\{-a:a\in A\}$.

The set $-A$ that you’re to prove is a Dedekind cut is just the union of all these sets $-(q+\Bbb Q\setminus A)$ as $q$ ranges over the positive rationals, so it’s the union of all possible sets like those marked with plus signs in the pictures below:

    +++++++++++++++++)----------------(ooooooooooooooooooooooooooooooo

    ++++++++++++++++++++++)-----------(ooooooooooooooooooooooooooooooo

    +++++++++++++++++++++++++)--------(ooooooooooooooooooooooooooooooo

    ++++++++++++++++++++++++++++++)---(ooooooooooooooooooooooooooooooo

What do you think that union will look like in a sketch of this kind? Won’t it look something like this?

    +++++++++++++++++++++++++++++++++)(ooooooooooooooooooooooooooooooo

That looks an awful lot like a Dedekind cut. Now you just have to prove it. For instance, you have to show that there is some rational that is not in the set. From the picture it appears that any $-a$ with $a\in A$ should work, so you should try to prove that this is the case. (Remember, the $q$’s that are being added are all strictly positive.)

You need to show that if $p$ is rational and $p<r\in-A$, then $p\in -A$. The picture certainly makes that look plausible, and it’s not hard to show. If $r\in-A$, then $r=-(q+a)$ for some $q\in\Bbb Q^+$ and $a\in A$. Can you find another positive rational $q'$ such that $p=-(q'+a)$? The number $r-p$ may be useful.

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