1
$\begingroup$

My understanding behind motivation of additive inverse of a cut set is as follows :

For example, for the rational number 2 the inverse is -2. Now 2 is represented by the set of rational numbers less than it and -2 is represented by the set of rational numbers less than it. So, if the cut set $\alpha$ represents a rational number then the inverse of $\alpha$ is the set $\{-p-r : p\notin \alpha , r >0\}$. But if the cut set does not represent a rational number then is the above definition is correct ? I think we will miss the first rational number which does not belong to $\alpha$ intuitively. Should not the set $\{-p : p\notin \alpha \}$ be the inverse now ? Confused.

$\endgroup$
  • $\begingroup$ If the cut does not represent a rational number, there is no "first rational number which does not belong to $\alpha$". That is, for an irrational cut, $\{ p \colon p \notin \alpha\}$ and $\{p+r\colon p \notin\alpha, r > 0\}$ are the same set. $\endgroup$ – Daniel Fischer Jul 7 '13 at 10:29
  • $\begingroup$ @DanielFischer: why ? $\endgroup$ – RIchard Williams Jul 7 '13 at 10:41
  • $\begingroup$ If you cheat a bit, and take $\mathbb{R}$ as given for the moment, the cut set corresponding to $\alpha\in\mathbb{R}$ is $c(\alpha)=\{r\in\mathbb{Q}\colon r<\alpha\}$, and the complement $u(\alpha)=\{r\in\mathbb{Q}\colon\alpha\leqslant r\}$. So $\alpha\in\mathbb{Q}\iff\inf u(\alpha)\in u(\alpha)$, a cut is rational if and only if the complement contains a smallest element. If the complement does not contain a smallest element, for each $p$ in the complement, you can find a (small) $\varepsilon>0$ such that $p-\varepsilon$ is also in the complement. Then write $p=(p-\varepsilon)+\varepsilon$. $\endgroup$ – Daniel Fischer Jul 7 '13 at 10:49
2
$\begingroup$

There is no "first rational number that does not belong to $\alpha$," except when $\alpha$ represents that rational number.

If there is a "first rational," $r$, not in $\alpha$, that would mean that for any $r'<r$, $r'\in\alpha.$ Since $r\notin\alpha$, that means that $\alpha$ is the cut for $r$.

The entire motivation for the odd definition of $-\alpha$ is the case where the cut $\alpha$ does represent a rational number. When $\alpha$ represents an irrational, we can just write: $$-\alpha = \{r:-r\not\in\alpha\}$$

This doesn't work when $\alpha$ represents a rational number $q,$ because, under this (flawed) definition, we'd have $-q\in-\alpha$, and it is bigger than all other elements of $-\alpha$.

$\endgroup$
0
$\begingroup$

I think the confusion arises when we are trying to identify a rational number say $2$ with the cut $\{ x\mid x \in \mathbb{Q}, x < 2\}$. When using Dedekind cuts as a definition of real numbers it is important to stick to some convention and follow it properly. For example to represent a real number either we choose

1) set containing smaller rationals

2) or set container larger rationals

3) or both the sets

At the same time after choosing one of these alternatives it is important to clarify whether the sets contains an end point (like least member in option 2) and greatest member in option 1)) or not.

In this particular question I believe the definition uses option 1) with the criteria that there is no greatest member in the set. When this definition is adopted and you define the additive inverse of a real number then we must ensure that the set corresponding to the additive inverse does not have a greatest member. This needs to be taken care only when the cut represents a rational number.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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