# If $p\in\mathbb{Q}^+$, for any $x\in\mathbb{R}$, there is a rational $q\in x$ such that $p+q\not\in x$

I just read about the construction of the real numbers in Enderton's Elements of Set Theory, and am now trying to go through all the exercises. Enderton chooses to construct the reals with left sides of Dedekind cuts. His definition is $y$ is a Dedekind cut if $\emptyset\neq y\neq\mathbb{Q}$, $y$ has no largest member, and $y$ is downward closed, i.e., $q\in y\wedge r<q\implies r\in y$. The following, question 19 of chapter 5 has me a bit baffled.

Assume that $p$ is a positive rational number. Show that for any real number $x$ there is some rational $q$ in $x$ such that $p+q\not\in x$.

I have this intuitive argument in my head. I guess I want to take $q\in x$, preferably closer to the right in $x$ on the rational line than not, and add it to $p$. If $p+q\not\in x$, then I am done. Otherwise, since $x$ has no greatest element, I can find $q'\in x$ such that $q'>q$, and add it to $p$ until I finally get a $q$ that pushes $p+q$ out of $x$, supposing of course that this process eventually ends. But this is horribly informal, and I have very little experience with analysis, and can't see a way to formalize it (if this is even the correct approach.)

Could someone please explain how to make this argument more rigorous, or perhaps point out a better way? Thanks.

• yunone, You may want to add a line or two for people who do not have Enderton's book, explaining what construction of the reals you are using (left sides of Dedekind cuts, I assume from what you wrote). – Andrés E. Caicedo Jan 14 '11 at 6:59
• @Andres, thanks for pointing that out! I have added it just now. – yunone Jan 14 '11 at 7:05

There is an integer $n$ such that $np\in x$ but $(n+1)p\notin x$. The reason is that there is some rational $r\notin x$, and there is some natural $m$ such that $mp>r$, in particular $mp\notin x$. But there is some $s\in x$, and there is some integer $k$ such that $kp<s$.
It follows that there is a least $m$ such that $mp\notin x$ (since the set of such integers is bounded below by $k$), which means that $(m-1)p\in x$. Ok, then you can take $q=(m-1)p$.
• Thanks for your answer, Andres. I see your claim that $(\exists m)mp>r$ was actually the result of Exercise 18, but I didn't realize it since it was of the form $r<p\cdot E(E(m))$, where $E(E())$ was the corresponding isomorphic embedding from $\omega\to\mathbb{Z}\to\mathbb{Q}$. And then $E(k)p<s$ follows from that, for if $s\geq 0_Q$, then we can take any $E(k)<0_Q$. And if $s<0_Q$, $E(k)p<s\Leftrightarrow -s<p\cdot (-E(k))$, and $-E(k)=E(j)$ for some $j\in\mathbb{Z}$ since $-E(k)$ must be positive, so it is just the same as the earlier case, correct? I understand much better now. Thank you! – yunone Jan 14 '11 at 7:53