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I'm studying a proof on this book I can't copy here the proof I'm studying because it is a little bit long. I only need help on a little statement. I don't understand in the page 180 of the book why we can choose $\alpha,\alpha^\prime\in\{\alpha_1,\alpha_2,\alpha_3\}$ distinct such that $x_\alpha-x_{\alpha^\prime}\not\in\{x_\beta-x_{\beta^\prime},x_{\beta^\prime}-x_\beta\}$.

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The set $\{x_{\beta} - x_{\beta'}, x_{\beta'} - x_{\beta}\}$ contains at most one positive number. Since there are at least two positive values of $x_{\alpha_{i}} - x_{\alpha_j}$ for $i,j \in \{ 1, 2, 3\}$, the claim follows. – JavaMan Jun 8 '11 at 17:48
"Since there are at least two positive values..." means that there are $(i,j)$ and $(\bar{i},\bar{j})$ such that $x_{\alpha_i}-x_{\alpha_j}$ and $x_{\alpha_{\bar{i}}}-x_{\alpha_{\bar{j}}}$ are positive. Why does this imply the claim? – Jacob Fox Jun 8 '11 at 18:02
If I misunderstand your question, please let me know (I have not indeed read the previous 179 pages!). The set $\{z , -z\}$ contains at most one positive value. We have two positive values, say $x$ and $y$. Therefore, at least one of $x$ or $y$ are NOT in the set $\{z, -z\}$. – JavaMan Jun 8 '11 at 18:14
you're right, sometimes my brain doesn't work. How can I prove that there are at least two $x_\alpha-x_{\alpha^\prime}$ distinct? – Jacob Fox Jun 8 '11 at 18:26
@Jacob; "Help on a proof" is a very unhelpful title. Worse, you've now edited another question into having this exact same title. Please make titles informative, don't edit them to make them less informative – Arturo Magidin Jun 26 '11 at 23:05
up vote 2 down vote accepted

If you accept that there are at least two positive (distinct) values for $x_\alpha - x_{\alpha'}$ then the claim follows since $\{x_\beta - x_{\beta'}, x_{\beta'} - x_\beta\}$ contains at most one positive value. Indeed, take one of the positive values for $x_\alpha - x_{\alpha'}$. If it is not in $\{x_\beta - x_{\beta'}, x_{\beta'} - x_\beta\}$ then chose it, otherwise chose the other one.

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how can I prove that there are at least two positive (distinct) values for $x_\alpha-x_{\alpha^\prime}$? – Jacob Fox Jun 8 '11 at 18:27
wlog assume that $x_{\alpha_1} < x_{\alpha_2} < x_{\alpha_3}$. Then $x_{\alpha_3} - x_{\alpha_1}$ and $x_{\alpha_3} - x_{\alpha_2}$ are two distinct values. – Levon Haykazyan Jun 8 '11 at 19:12

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