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My professor defined Dedekind cuts in the following way: "Given two nonempty sets E,F $\subset \mathbb{R}$, we say that the pair (E,F) is a Dedekind cut of $\mathbb{R}$ if

  1. $E \cap F = \emptyset $
  2. $E \cup F = \mathbb{R}$
  3. $x < y$ for all $x \in E$ and for all $y \in F$"

However, I was under the impression that Dedekind cuts could only be taken of the rational numbers, in order to construct the real numbers. Can you take a Dedekind cut of the Real line?

Edit: I should mention that he defined a cut this way in order for him to state the following Dedekind Axiom:

"For every Dedekind cut (E,F) of $\mathbb{R}$ there exists a unique $L \in \mathbb{R}$ such that $x \leq L \leq y$ for all $x \in E$ and for all $y \in F$"

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4  
You can, but you don't really get anything new. –  Michael Greinecker Sep 3 '12 at 17:53
    
Nothing prevents you from defining those sets and calling them Dedekind cuts, so it is certainly valid. But you are correct in your thinking that these sets will not be as interesting the the same idea with $\mathbb{Q}$, since Dedekind cuts are typically used to construct $\mathbb{R}$. –  nullUser Sep 3 '12 at 17:53
    
Perhaps a particular example of a set $E$ and $F$ will be illustrative: $E=(-\infty,0)$, and $F=[0,\infty)$. –  process91 Sep 3 '12 at 17:59
    
As Henning Makholm once commented: "One way to look at it is that if you do the Dedekind construction on $\bf R$ instead of $\bf Q$, you don't get any new elements out of it -- the completion-by-cuts is an idempotent construction." –  Pedro Tamaroff Sep 3 '12 at 20:02
    
For those who say "you get nothing new" - that is strictly true. But the details of definitions like this can help in proving the basic arithmetic properties of the real numbers without requiring discussion of lots of cases for positive and negative numbers. For or example, if - as is sometimes required - E must have a maximum, then the negative of a real number requires careful handling. The definition does have a role - look carefully at how it is used. –  Mark Bennet Sep 3 '12 at 21:32

4 Answers 4

up vote 13 down vote accepted

You can take Dedekind cuts of any linearly ordered set and use them to form the order completion of that set. If you start with a complete linear order, however, like $\langle \Bbb R,\le\rangle$, you get nothing new: the Dedekind completion is order-isomorphic to old one. (By the way, it’s usual to add one more condition, either that $E$ has no maximum element or that $F$ has no minimum element.)

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Given an ordered set $Q$ felt "incomplete" one can use Dedekind cuts to complete it: The Dedekind cuts of $Q$ are the pairs $(E,F)$ described in your question, with the additional condition that $E$ should not contain a maximal element. Denote the set of such pairs by $\bar Q$. Then (a) $\bar Q$ is again ordered: $(E,F)\leq(E',F')$ if $E\subset E'$; (b) the set $\bar Q$ is "order complete", (c) there is an embedding $\phi:\ Q\to\bar Q$ respecting the order: $$q\mapsto (E_q, F_q):=\bigl(\{x\in Q|x<q\}, \{x\in Q|x\geq q\}\bigr)\ ;$$ and (d) $\phi(Q)$ is dense in $\bar Q$. So $\bar Q$ is the "minimal" order complete set containing a copy of $Q$.

This idea of cuts is very handy also in existence proofs involving the "order complete" set ${\mathbb R}$, e.g., in the proof of the intermediate value theorem or similar statements. There the principle given by your professor is applied: If you cut ${\mathbb R}$ in two by means of an ax then the ax will touch exactly one real number $\xi$. When you cut ${\mathbb Q}$ in two by an ax at $\sqrt{2}$ then all points $x\in{\mathbb Q}$ have a positive distance from the ax.

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That is certainly well-defined, and yes, they are still called Dedekind cuts. Every cut is principal (i.e. either E has a maximum or F has a minimum), which expresses the fact the reals are complete.

If one so desired, one use cuts to give precise meaning to the idea of specifying places "infinitesimally close" to real numbers, or specifying ways to approach real numbers: e.g. the cut $0^+ := \left( (-\infty, 0], (0, +\infty) \right)$ represents $0$ approached from the positive side.

Interestingly (to me), if you remove the condition that $E$ and $F$ are nonempty, then Dedekind cuts are in one-to-one correspondence with orderings of $\mathbb{R}(x)$, the field of rational functions with real coefficients: the cut specifies where $x$ should be placed into the ordering. So relative to the ordering defined by $0^+$, $x$ is a positive infinitesimal element.

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One way of looking at Dedekind cuts is model-theoretic.

A Dedekind completion of a linear order $(X,\le)$ is a minimal linear order $X'$ containing $X$ such that every type of the form $p_{E,F}(x)=\{e\le x,x\le f\vert e\in E, f\in F \}$ is realized, where $E$, $F$ satisfy the conditions you specified. It can be shown that it is unique up to a $X$-preserving isomorphism.

But then you can notice that if $Y\subseteq X$ is dense, then $p_{E\cap Y,F\cap Y}\equiv p_{E,F}$, so the Dedekind completion of a linear order is isomorphic to the Dedekind completion of its any dense suborder, or, to put it differently, taking parameters from just $Y$ (or $\bf Q$ in your case) does yield anything less.

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