## Dedekind Cut Addition Degenerate Cases

The addition of two dedekind cuts is defined to be $\displaystyle A|B + C|D = \{a+c | a\in A\text{ and } c\in C \} | \{\text{Rest of \mathbb{Q}} \}$

Now let $\displaystyle A|B$ and $\displaystyle C|D$ be rational cuts. I am asked as to why it is not defined as

$\displaystyle A|B + C|D = \{a+c | a\in A\text{ and }c\in C \} | \{b+d | b\in B \text{ and } d\in D \}$

and am told that in some degenerate cases that $\displaystyle \{a+c | a\in A\text{ and }c\in C \}\cup \{b+d | b\in B \text{ and } d\in D \}\neq\mathbb{Q}$

I cannot come up with any case in which the union fails to be all of the rational numbers. In fact, since both cuts are rational cuts it seems to me that

$\displaystyle A|B + C|D = \{a+c | a\in A\text{ and }c\in C \} | \{b+d | b\in B \text{ and } d\in D \}=$

$\displaystyle \{a+c | a<r_1\text{ and }c<r_2} | \{b+d | b\geq r_1\text{ and } d\geq r_2 \}$ for some $\displaystyle r_1,r_2\in\mathbb{Q}$

And of course this means that we have $\displaystyle \{x\in\mathbb{Q} | x<r_1+r_2\} | \{x\in\mathbb{Q} | x\geq r_1+r_2\}$ and clearly every rational number is in one of these two sets

Any help is much appreciated