# Thread: sum of two Dedekind cuts

1. ## 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!

2. Originally Posted by zzzhhh
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 $0\in(A_x+A_y)\cup(B_x+B_y)$.

3. 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)).

4. Originally Posted by zzzhhh
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 $(A_x+A_y)\cup(B_x+B_y)$.

5. Thank you for the heuristic counterexample, another evidence that $\mathbb{Q}$ has gaps.