# sum of two Dedekind cuts

• August 13th 2009, 01:10 AM
zzzhhh
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!
• August 14th 2009, 12:23 AM
Opalg
Quote:

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 $0\in(A_x+A_y)\cup(B_x+B_y)$.
• August 14th 2009, 09:19 AM
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)).
• August 14th 2009, 02:10 PM
Opalg
Quote:

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 $(A_x+A_y)\cup(B_x+B_y)$.
• August 15th 2009, 02:13 AM
zzzhhh
Thank you for the heuristic counterexample, another evidence that $\mathbb{Q}$ has gaps.