Problem statement:

Given an arbitrary positive Dedekind Cut $\alpha = A|B$, prove that the following

$\begin{align*}\alpha^{-1} &= C|D \\ &= \{r\in\mathbb{Q}: r\leq 0 \text{ or } \exists b\in B, \text{ not the smallest element of } B, r = \frac{1}{b} \}|\text{ rest of } \mathbb{Q}\end{align*}$

is a Dedekind Cut.

A Dedekind Cut $X|Y$ fulfills three conditions:

  1. $X\cup Y = \mathbb{Q}, X\neq\emptyset, Y\neq\emptyset, X\cap Y\neq\emptyset$
  2. If $x\in X$ and $y \in Y$ then $x<y$.
  3. $X$ has no largest element.

Proof attempt:

  1. $0\in C, \frac{1}{a}\in D$ since $\frac{1}{a}\not\in C$ for all $a\in A$ and trivially from the definition $C\cup D=\mathbb{Q}, C\cap D = \emptyset$.
  2. $c<\frac{1}{\inf(B)}\leq d\hspace{1mm}\forall c\in C \hspace{1mm}\forall d\in D$ (WRONG)
  3. Suppose $C$ has a maximal element $\gamma$. Then it is true that $\gamma < \frac{1}{b}$ from the definition. But then $\gamma = \frac{\gamma}{2} + \frac{\gamma}{2} < \frac{\gamma}{2} + \frac{1}{2b} < \frac{1}{2b} + \frac{1}{2b} = \frac{1}{b}$. So there is no maximal element for $C$. (WRONG)

So $\alpha^{-1}$ is a cut.

I suspect that showing $\alpha \cdot \alpha^{-1} = 1$ is non trivial but I'll leave that for another post.

  • $\begingroup$ Did you mean $Y$ instead of $B$ in condition 1.? $\endgroup$ – J. W. Tanner May 10 at 1:22
  • $\begingroup$ Oh yes! Thanks! $\endgroup$ – Darius May 10 at 1:23
  • 1
    $\begingroup$ $B$ is a subset of $\mathbb{Q}$. How do you know that $\inf(B)$ exists in $\mathbb{Q}$? Indeed, if $\alpha$ corresponds to the cut of an irrational, then $\inf B$ will not be in $\mathbb{Q}$. So you cannot say $\frac{1}{\inf(B)} \in D$. $\endgroup$ – ZeroXLR May 10 at 6:45
  • $\begingroup$ @ZeroXLR, Axiom of Completeness got me again :( $\endgroup$ – Darius May 10 at 6:47
  • $\begingroup$ Any positive member of $A$, say $a$, acts as a lower bound for members of $B$ and hence all the members of $C$ are less than $1/a$ so that both $C, D$ are non-empty. This establishes the first condition for being a Dedekind cut. The rest is proved easily. $\endgroup$ – Paramanand Singh May 10 at 7:19

Since $\alpha=A|B$ is a positive Dedekind cut there is a positive member $a\in A$. Further since $a\in A$ we have $a<b$ for all $b\in B$. And therefore $1/b<1/a$ for all $b\in B$.

The cut $\beta=C|D$ as described in question is such that $C$ contains $0$ and all negative rationals and further all positive rationals of the form $1/b$ where $b\in B$ (and $b$ is not the least member of $B$). All such numbers are less than $1/a$ as mentioned in previous paragraph. Hence the set $C$ is non-empty and bounded above so that it a proper subset of $\mathbb{Q} $. By definition $D=\mathbb {Q} - C$ and it is now clear that $D$ is also non-empty and proper subset of $\mathbb {Q} $ and $C\cup D=\mathbb {Q}, C\cap D=\emptyset$. The first condition for being a Dedekind cut is satisfied for $\beta$.

For second condition let $x\in C, y\in D$. Since $C, D$ are disjoint we can't have $x=y$. Note that $y>0$ as all negative numbers and $0$ lie in $C$. If $x>y$ then both $x, y$ are positive and since $x\in C$ we have $1/x\in B$ and $1/y>1/x$ implies that $1/y\in B$. And therefore $y\in C$ which is contrary to our assumption. Thus we can't have $x>y$ and we are thus forced to conclude that $x<y$. Thus the second condition for being a Dedekind cut is also verified for $\beta$.

Third condition is easy. If $C$ has a largest member $c$ then $c>0$ and $1/c\in B$ and $1/c$ is not the least member of $B$. So there is some positive $b\in B$ with $b<1/c$. Now choose a $b'$ with $b<b'<1/c$ so that $b'\in B$ and clearly $b'$ is not the least member of $B$. Hence $1/b'\in C$. But $1/b'>c$ and this contradicts that $c$ is the largest member of $C$. Therefore $C$ has no largest member. It follows that $\beta=C|D$ is a Dedekind cut.

I have used symbol $\beta$ instead of $\alpha^{-1}$ because we are yet to prove that $\alpha\beta=\beta\alpha=1$.

There is no need to deal with things like $\inf B$. The idea of a Dedekind cut requires one to know the basic operations on rationals like $+, -, \times, /, <, >$ and nothing more. Therefore it is the easiest route to a theory of real numbers. One should not try to think deeply while dealing with a Dedekind cut.

It is rather ironical that such a simple idea took a long time to come by and it is even more ironical that it is taught in undergraduate courses. Anyone who knows how to add/subtract/multiply/divide/compare rationals and has some knowledge of basic set theory can understand these ideas.


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