Rudin write that the members of $\mathbb R$ will be certain subsets of $\mathbb Q$, called cuts.

A cut is, by definition, any set $\alpha\subset \mathbb Q$ with the following three properties.

(I) $\alpha$ is not empty, and $\alpha\neq \mathbb Q$

(II) If $p\in \alpha$ and $q<p$ then $q\in\alpha$

(III) If $p\in \alpha$ then $p<r$ for some $r\in \alpha$

If $\alpha\in \mathbb R^{+}$ and $\beta \in \mathbb R^{+}$ we define $\alpha\beta$ to be the set of all $p$ such that $p\leqslant rs$ for some choices of $r\in \alpha$, $s\in \beta$, $r>0$,$s>0$.

Affirms that for any $\alpha\in \mathbb R^{+}$ and $\alpha\neq 0^{*}$ there is $\beta\in \mathbb R^{+}$ such that $\alpha\beta=1^{*}$.

How to define this cut $\beta$?


Answer. $$ \beta=\big\{p\in\mathbb Q : pq\le 1\,\,\text{for all}\,\,q\in\alpha\cap\mathbb Q^+\big\}. $$

  • $\begingroup$ You know the two first properties are implies but the third i can't prove. Can you help me? If $p\in \beta$ then $\beta<r$ for some $r\in \beta$ $\endgroup$ – ZFR Jun 1 '15 at 12:31
  • $\begingroup$ If $p\in \beta$ then $p\leqslant \frac{1}{q}$ for all $q\in \alpha, q>0$. So $\frac{1}{p}\geqslant q$ for all q. Then $1/p\notin \alpha$. What's next? $\endgroup$ – ZFR Jun 1 '15 at 12:38
  • $\begingroup$ Is it true that $\beta=0^{*}$? $\endgroup$ – ZFR Jun 1 '15 at 14:36

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