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For a given cut set $\alpha$, what is the intuition behind considering the set of all such $p$'s such that some number less than $-p$ does not belong $\alpha$ as the inverse of $\alpha$ ? i.e. $\beta=\{p: \exists r>0 \text{such that} -p-r \notin \alpha\}$

BTW, a cut set or a Dedekind cut set $\alpha$ is a subset of $Q$ with the following properties :

1) $\alpha \neq \phi$ and $\alpha \neq Q$

2) if $p \in \alpha$ and $q < p$ then $q \in \alpha$

3) $\alpha$ should not have any greatest element

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.

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  • $\begingroup$ Could you please define what exactly a cut set is ? A quick google search only gives a definition in the context of graphs. I have a feeling you want to talk about Dedekind cuts in case of which it would be better if we knew what a 'cut set' of a Dedekind cut is. $\endgroup$ – nsoum Jun 28 '13 at 13:41
  • $\begingroup$ In 1), the "or" should be an "and". $\endgroup$ – Michael Greinecker Jul 3 '13 at 2:31
  • $\begingroup$ Could you be more explicit in your explanation of the inverse of $\alpha$? In particular, could you write it as a set? $\endgroup$ – Michael Albanese Jul 4 '13 at 6:07
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If we have $\alpha$ and simply take $\beta = \{-x\mid x\in \alpha\}$ then this is not a cut set - it is "the wrong half". So we think that $\gamma:=\mathbb Q\setminus \beta$ is a better idea, but that also poses problem, namely when $\alpha$ happens to represent a rational number. The somewhat indirect sounding formulation you quote is used to circumvent this las t specific problem.

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  • $\begingroup$ Thanks for your reply. But, I think there should be some better motivation for this. $\endgroup$ – RIchard Williams Jun 28 '13 at 13:55

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