Dedekind cuts: proof every interval of $\mathbb{R}$ contains both rational and irrational numbers

I'm trying to follow the proof that every interval $(a,b)$ of $\mathbb{R}$ contains both rational and irrational numbers given on page 19 of Real Mathematical Analysis by Pugh. Word for word, he begins the proof:

This is certainly true of the interval $(0,1)$ since it contains the numbers $1/2$ and $1/\sqrt{2}$. For the general interval $(a,b)$, think of $a$, $b$ as [Dedekind] cuts $a=A|A'$, $b=B|B'$. The fact that $a<b$ implies the set $B \, \backslash \, A$ contains two distinct rational numbers, say $r$, $s$. Thus $a \leq r < s \leq b$.

The last sentence is where he loses me. Why is it that if $r \in B \, \backslash \, A$ then $r \geq a$ and if $s \in B \, \backslash \, A$ then $s \leq b$? These statements are intuitive, but I haven't been able to prove them. Let $r=R|R'$. I have tried to deduce a contradiction from $R \subset A$, $R \neq A$, but the lines of reasoning I come up with seem to depend on the truth of the statement I'm trying to prove. I checked the discussions of cuts in Enderton's Elements of Set Theory and Spivak's Calculus, but neither seems to offer any insight.

\begin{align} r > a & \iff R^\prime \subset A^\prime \\ & \iff ([r,\infty)\cap \mathbb Q) \subset ([a,\infty)\cap \mathbb Q) \\ & \iff r \in A^\prime \end{align}
Note, that $r \in B\setminus A$ implies $r\in A^\prime$.
• I understand $r>a \Leftrightarrow R' \subset A'$ and $r \in B \, \backslash \, A \Rightarrow r \in A'$. I don't understand why $r \in A' \Rightarrow R' \subset A'$ though -- can you give me more detail? – flakmonkey May 18 '15 at 21:17
• Note that $R^\prime = [r,\infty)\cap \mathbb Q$ and thus $r\in R^\prime$. You also have $A^\prime = [a,\infty)\cap \mathbb Q$. – Stephan Kulla May 18 '15 at 21:41