I'm studying Dedekind's Cuts and his construction of Real numbers from the Rational ones. Here we are allowed to use $\Bbb{Q}$ as an ordered field and all all its properties (Archimedean Property, his usual operations, etc), but it's not allowed to mention real numbers.

Definition We say that $\alpha\subset\Bbb{Q}$ is a cut (known as Dedekind's Cut) if
1) $\alpha \neq \emptyset$ and $\alpha \neq \Bbb{Q}$;
2) If $r\in \alpha$ and $r'<r$, then $r'\in \alpha$;
3) If $r\in \alpha$, then $\exists r'\in \alpha$ such that $r'>r$.

Then, in the middle of the way, I came across this lemma, which I coudn't prove:

Given $0<a<1$ and a cut $\alpha$, there exists $0<l_1<l$ such that $la^2\in \alpha$ but $l_1a\notin \alpha$.

Obviously, until here, we have developed some properties concerning cuts, but I'd rather not quote them here... It really seems to be not so difficult to do it, but I wasn't able to.
Need some help! Some hint? Thanks


It is tacitly assumed that the cut $\alpha$ contains some positive rationals.

There is an $a'$ such that $a<a'<1$. We now need to localize the cut $\alpha$ quite precisely. Starting from an arbitrary positive $x\in\alpha$ and an $y\notin\alpha$ you have to construct two numbers $c\in\alpha$ and $d\notin\alpha$ with $c\geq x$ and $$a' d<c<d\ .$$ (A hint: Divide the interval $[x,y]$ into $N\gg1$ equal pieces. When $N$ is large enough there will be two successive partition points that can serve as $c$ and $d$.)

Now put $$\ell_1:={d\over a},\qquad\ell:={a'\>d\over a^2}\ .$$ It is easy to verify that $\ell_1$, $\ell$ satisfy the requirements.

  • $\begingroup$ Oh, sure, $\alpha$ contais some positive rationals. I have forgotten to say it... Sorry! $\endgroup$ – Ders Jan 7 '15 at 15:59
  • $\begingroup$ And thanks for the help! $\endgroup$ – Ders Jan 7 '15 at 15:59
  • $\begingroup$ But I didn't understand what guarantees the existence of such $d$... @Christian Blatter $\endgroup$ – Ders Jan 7 '15 at 16:31

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