Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In the process of learning Real Analysis, I encountered a definition of an additive inverse of a cut $\alpha$ to be $$\text {add inv of } \alpha \colon= \{p:\exists r>0 \text{ s.t.} (-p-r)\notin \alpha\} $$ The definition does make sense but I can't seem to understand the motivation behind defining it this way.
Why is this definition significant? Isn't this a bit uncalled for? Why good does this definition do to us ? Does it make some things easier (how so?)?
Sorry for asking too many questions at once. The thing is, I am a High school student. I just started studying Real Analysis on my own and even though it has been quite a rough road for me, the experience has been truly rewarding and satisfying.
I am slowly getting used to the rigor in mathematics and so I can't grasp these kinds of concepts as easily. So could you please elaborate a bit more than necessary.

Any help is much appreciated!

Thanks in advance!

share|improve this question
    
I wouldn't view the construction of the reals as being indicative of the kind of thinking that you need in order to do real analysis. –  Dylan Moreland Jan 8 '12 at 19:05
    
@DylanMoreland. Come on, he said he's just started. Besides, he's still in high school. To begin real analysis by constructing $\mathbb{N}$, then $\mathbb{Q}$ and $\mathbb{R}$ is the right way to go. –  Samuel Tan Jan 9 '12 at 2:08
1  
@Samuel I didn't mean to sound critical. I just meant that the construction is somewhat technical and that one can happily solve problems in, say, Rudin without having a great feel for cuts. In other words, the OP should not feel discouraged. I agree that it's good to work this out once in your life. –  Dylan Moreland Jan 9 '12 at 2:31
add comment

1 Answer

up vote 2 down vote accepted

How are you defining cuts? For the usual definition (which I give below), $-\alpha$ turns out to be $\{p:\exists r>0\, (-p-r\in \alpha)\}$, not $\{p:\exists r>0\, (-p-r\notin \alpha)\}$. I’ll go through this derivation in some detail, in hopes that you can adapt it to your situation.

To understand where the definition of $-\alpha$ comes from, you have to go back to the definition of the sum of two cuts. I’m going to assume the following definitions of cut and of addition of cuts:

$\alpha\subseteq \mathbb{Q}$ is a cut iff
$\;$
$\qquad(1)\qquad\varnothing\ne \alpha\ne\mathbb{Q}$;
$\qquad(2)\qquad\text{if }p<q\in \alpha,\text{ then }p\in \alpha$; and
$\qquad(3)\qquad \alpha\text{ has no greatest element}$.

If $\alpha$ and $\beta$ are cuts, $\alpha+\beta\triangleq\{p\in\mathbb{Q}:\exists a\in \alpha\,\exists b\in \beta(p=a+b)\}$.

I’ll also assume that the embedding $^*$ of $\mathbb{Q}$ into the set of cuts has been defined so that $q^*=\{p\in\mathbb{Q}:p<q\}$, so that $0^*=\{q\in\mathbb{Q}:q<0\}$ is intended to be the additive identity.

Clearly we want $-\alpha$ to be defined so that $\alpha+(-\alpha)=0^*$. Thus, I want $-\alpha$ to satisfy the following condition:

$\qquad\qquad\qquad\qquad$ for all $p\in\mathbb{Q}$, $p<0$ iff $\exists a\in \alpha\,\exists b\in -\alpha(p=a+b)$.

This is a bit hard to sort out, so let’s pretend for a moment that we’ve actually constructed $\mathbb{R}$ and look at a concrete example. Suppose that $\alpha$ is intended to be $\sqrt 2$: $\alpha=\{p\in\mathbb{Q}:p<\sqrt 2\}$. Clearly $-\alpha$ should be $-\sqrt 2$, or $\{p\in\mathbb{Q}:p<-\sqrt 2\}$. But $p<-\sqrt 2$ iff $-p>\sqrt 2$, so $-\alpha$ ought to be $\{p\in\mathbb{Q}:-p>\sqrt 2\}=\{p\in\mathbb{Q}:-p\notin\alpha\}$. This suggests that perhaps we should define $-\alpha$ in general to be $\{p\in\mathbb{Q}:-p\notin \alpha\}$. This almost works, but there’s a small problem if $\alpha$ is a rational cut $q^*$: in that case $\{p\in\mathbb{Q}:-p\notin q^*\}=\{p\in\mathbb{Q}:-p\ge q\}=\{p\in\mathbb{Q}:p\le -q\}$, which is not a cut, since it has a greatest element. We want $-(q^*)$ to be simply $(-q)^*$, i.e., $\{p\in\mathbb{Q}:p<-q\}$. There’s a simple (if slightly clumsy-looking) way around the problem: we simply set $$-\alpha\triangleq\{p\in\mathbb{Q}:-p\notin\alpha\text{ AND }-p\text{ is not the least element of }\mathbb{Q}\setminus\alpha\}\;.\tag{1}$$ It’s not hard to check that this really does define a cut. To check that it defines the right cut, suppose first that $a\in\alpha$ and $b\in-\alpha$. Then $-b\notin\alpha$, so $a<-b$, and therefore $a+b<0$, as desired.

Now suppose that $p$ is any negative rational; we need to show that there are $a\in\alpha$ and $b\in-\alpha$ such that $p=a+b$. This amounts to finding $a\in\alpha$ and $-b\in\mathbb{Q}\setminus\alpha$ such that $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$ and $a+b=p$ or, equivalently, $a-p=-b$.

Let $r=-p>0$; then we want to find $a\in\alpha$ and $-b\in\mathbb{Q}\setminus\alpha$ such that $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$ and $a+r=-b$. To give this some intuitive content, we’re trying to find rationals $a$ and $-b$ that are exactly $r$ apart and that ‘straddle’ the cut $\alpha$.

To do this, let $q\in\alpha$ and $s\in\mathbb{Q}\setminus\alpha$ be arbitrary. Since $\mathbb{Q}$ is non-Archimedean, there is an $n\in\mathbb{Z}^+$ such that $n>(s-q)/r$ and hence $q+nr>s$, so $\{n\in\mathbb{Z}^+:q+nr\notin\alpha\}\ne\varnothing$. Let $m=\min\{n\in\mathbb{Z}^+:q+nr\notin\alpha\}$; then $q+(m-1)r\in\alpha$ and $q+mr\notin\alpha$. Thus, if $q+mr$ is not the least element of $\mathbb{Q}\setminus\alpha$, we can simply take $a=q+(m-1)r$ and $-b=q+mr$ to get $a+r=-b$ with $a$ and $-b$ as desired.

If $q+mr$ is the least element of $\mathbb{Q}\setminus\alpha$, we have to be a little cleverer: in this case we let $a=q+\left(m-\frac12\right)r$ and $-b=q+\left(m+\frac12\right)r$. Clearly $a+r=-b\in\mathbb{Q}\setminus\alpha$, $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$, and $a<q+mr=\min(\mathbb{Q}\setminus\alpha)$, so $a\in\alpha$, again exactly as desired.

Notice that in the indented part of the argument we were actually showing that if $r$ is any positive rational, there are $a\in\alpha$ and $p\in-\alpha$ such that $-r=a+p$ or, equivalently, such that $a=-p-r$. In other words, we were showing that $\{p:\exists r>0\;(-p-r\in\alpha)\}\subseteq-\alpha$. On the other hand, if $p\in-\alpha$ by definition $(1)$, then $-p\notin\alpha$; let $a\in\alpha$ be arbitrary, set $r=-p-a$, and note that $r>0$ and $-p-r=a\in\alpha$, showing that $-\alpha\subseteq\{p:\exists r>0\;(-p-r\in\alpha)\}$. Thus, definition $(1)$ can be replaced by: $$-\alpha\triangleq\{p\in\mathbb{Q}:\exists r>0\;(-p-r\in\alpha)\}\;.\tag{2}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.