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Math Help - sum of two Dedekind cuts

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    sum of two Dedekind cuts

    Let x=(A_x,B_x), y=(A_y,B_y), x,y\in\mathbb{R} be two Dedekind cuts. Is it true that (A_x+A_y)\cup(B_x+B_y)=\mathbb{Q}? where A_x+A_y=\{r+s|r\in A_x\;{\rm{and}}\;s\in A_y\}, the meaning of B_x+B_y is similar.
    I met with this problem while trying to prove the equivalence of two formulations of addition of real numbers, therefore, to avoid iterative proof please restrict addition to rationals only. Thanks!
    Last edited by zzzhhh; August 13th 2009 at 07:37 AM.
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    Quote Originally Posted by zzzhhh View Post
    Let x=(A_x,B_x), y=(A_y,B_y), x,y\in\mathbb{R} be two Dedekind cuts. Is it true that (A_x+A_y)\cup(B_x+B_y)=\mathbb{Q}? where A_x+A_y=\{r+s|r\in A_x\;{\rm{and}}\;s\in A_y\}, the meaning of B_x+B_y is similar.
    I met with this problem while trying to prove the equivalence of two formulations of addition of real numbers, therefore, to avoid iterative proof please restrict addition to rationals only. Thanks!
    Look carefully at this example. Let A_x = \{s\in\mathbb{Q}:s<0\}, B_x = \{s\in\mathbb{Q}:s\geqslant0\}, A_y = \{s\in\mathbb{Q}:s\leqslant0\}, B_y = \{s\in\mathbb{Q}:s>0\}. Then ask yourself whether 0\in(A_x+A_y)\cup(B_x+B_y).
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    Thanks but the definition of Dedekind cut prevents this A_y= \{s\in\mathbb{Q}|s\leq 0\} from being part of a Dedekind cut, since the left part of a dedekind cut should not have greatest element. I am using Karel Hrbacek's "Introduction to Set Theory"(P88 Def5.4) and Baby Rudin(P17 (III)).
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    Quote Originally Posted by zzzhhh View Post
    Thanks but the definition of Dedekind cut prevents this A_y= \{s\in\mathbb{Q}|s\leq 0\} from being part of a Dedekind cut, since the left part of a dedekind cut should not have greatest element. I am using Karel Hrbacek's "Introduction to Set Theory"(P88 Def5.4) and Baby Rudin(P17 (III)).
    In that case, the counterexample needs to be a bit more complicated (but not much more). Define A_x = \{s\in\mathbb{Q}:s<0\text{ and }s^2>2\} and A_y = \{s\in\mathbb{Q}:s<0\text{ or }s^2<2\}, with B_x and B_y their complements in \mathbb{Q}. (In more familiar notation, A_x = (-\infty,-\sqrt2)\cap\mathbb{Q}, A_y = (-\infty,\sqrt2)\cap\mathbb{Q}, B_x = (-\sqrt2,\infty)\cap\mathbb{Q}, B_y = (\sqrt2,\infty)\cap\mathbb{Q}.) Again, look at 0 to see that it is not in (A_x+A_y)\cup(B_x+B_y).
    Last edited by Opalg; August 15th 2009 at 12:34 AM.
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    Thank you for the heuristic counterexample, another evidence that \mathbb{Q} has gaps.
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