By a Dedekind cut, we mean, an ordered pair $(L,U)$ of subsets of $\mathbb{Q}$ such that they are disjoint, their union is $\mathbb{Q}$, and

  • Each member of $L$ is smaller than each member of $U$

  • $L$ contains no largest rational number.

Let $(L,U)$ be a Dedekind cut for which $L$ contains some positive rationals.

Let $L'$ be the collection of non-positive rationals, along with those positive rationals $x$ whose product with all positive rationals of $L$ is $<1$. Let $U'$ be the complement of $L'$ in $\mathbb{Q}$.

Can we say that $(L',U')$ defined in this way is the multiplicative inverse of $(L,U)$?

(One may see this wiki-link for product of Dedekind cuts).


2 Answers 2


Yes. One needs to check whether $(L,U)\times(L',U')=(L_1,U_1)$, where $L_1=(-\infty,1)$, $U_1=[1,\infty)$. Note also that only the multiplication of the lower sets needs be considered, since the uppers sets are their complements.

By the definition of multiplication of Dedekind cuts, since $L$ and $L'$ contain positive rationals, the lower set of their product is defined to contain

  • all non-positive rationals
  • all positive rationals $ab$ where $a\in L$, $b\in L'$.

Take any rational $c<1$. Pick a rational $a\in L$ such that $a>cL$; this is possible since $c<1$. So for any $b\in L$, $(c/a)b<1$, which implies $c/a\in L'$ and so $c=(c/a)a\in L'\times L$. This shows that $(-\infty,1)\subseteq L$.

Take any rational $c\ge1$. If $c=ab$ with $a\in L$, $b\in L'$, it would contradict the way $L'$ is defined. So $L=(-\infty,1)$.


Actually, you need to be slightly more careful than that.

Suppose the cut (L,U) is divided by a rational value $f$ (i.e U has $f$ as a minimum value). Then for every positive rational $r$ in L, $f^{-1} r$ < 1, so $f^{-1}$ is in L' by definition. It is easy to see that $f^{-1}$ is the maximum value in L', so (L',u') doesn't quite meet the given definition of a cut.


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