# Show $\alpha^{-1}$ is a Dedekind Cut

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 $$x.
3. $$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.

• Did you mean $Y$ instead of $B$ in condition 1.? – J. W. Tanner May 10 at 1:22
• Oh yes! Thanks! – Darius May 10 at 1:23
• $B$ 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 $\frac{1}{\inf(B)} \in D$. – ZeroXLR May 10 at 6:45
• @ZeroXLR, Axiom of Completeness got me again :( – Darius May 10 at 6:47
• 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. – 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 $$a for all $$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 $$x. Thus the second condition for being a Dedekind cut is also verified for $$\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 $$b so that $$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.