Can you help me with this?

Let $x=A \mid B$ and $x'=A' \mid B'$ be cuts in $\mathbb Q$. It is defined

$$x+x'=(A+A') \mid \text{ rest of } \mathbb Q$$

Show that although $B+B'$ is disjoint from $A+A'$, it may happen in degenerate cases that $\mathbb Q$ is not the union of $A+A'$ and $B+B'$.

EDIT: As the comment below asked, I'll include the definition of a cut in $\mathbb Q$.

A cut in $\mathbb Q$ is a pair of subsets $A,B$ of $\mathbb Q$ such that

(a) $A\cup B=\mathbb Q$, $A\neq \emptyset$, $B\neq \emptyset$, $A\cap B=\emptyset$

(b) if $a\in A$ and $b\in B$ then $a<b$.

(c) $A$ contains no largest element.

  • 3
    $\begingroup$ You might want to include the definition of "cuts" as well. Such low level details might depend crucially on the exact definitions (and could even vary from one book to another). $\endgroup$ – Srivatsan Jan 3 '12 at 15:46

You can simply take the cut corresponding to $\pm\sqrt2$, i.e.
$\newcommand{\Q}{\mathbb{Q}}A=\{x\in\Q; x<\sqrt{2}\}$ and $B=\{x\in\Q; x>\sqrt{2}\}$;
$A'=\{x\in\Q; x<-\sqrt{2}\}$ and $B'=\{x\in\Q; x>-\sqrt{2}\}$.

Clearly $A+A'=\{x\in\Q; x<0\}$ and $B+B'=\{x\in\Q; x>0\}$.

If you prefer to define the cut without any reference to real numbers you can get rid of $\sqrt{2}$ easily. E.g. $A=\{x\in\Q; x\le 0 \lor x^2<2\}$.


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