I am greatly confused with Dedekind cuts... I am trying to prove that this is a Dedekind cut:

If $D$ and $E$ are in $\mathbb{Q}$ and are Dedekind cuts, then prove that $$D*E=(-\infty, 0] \cup \{r_1r_2\mid 0 < r_1 \in D, 0 < r_2 \in E\}$$ is a Dedekind cut as well.

My three propositions of a Dedekind cut are:

1.) If $r\in D$ and $s < r$, then $s \in D$.

2.) There is a number $x \in \mathbb{Q}$ so that $r\leq x$ for all $r \in D$.

3.) If $r \in D$, then there is a number $s \in D$ so that $r < s$.

After looking at many sources, my concept of a Dedekind cut is falling short... and so this proof is.

I would be greatly appreciative for a simple definition and example of a Dedekind cut, and/ or help on this proof. Thx!

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    $\begingroup$ Note that $D$ and $E$ cannot simultaneously "be in $\mathbb{Q}$" and be Dedekind cuts. $\endgroup$ – Arturo Magidin Jun 6 '12 at 4:15
  • $\begingroup$ I assume that should be read as "$D$ and $E$ are subsets of $\mathbb{Q}$"? $\endgroup$ – Nate Eldredge Jun 6 '12 at 4:26
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    $\begingroup$ Are $r_1$ and $r_2$ supposed to be positive? $\endgroup$ – Mark Bennet Jun 6 '12 at 4:29
  • $\begingroup$ While regarded as a set theoretical construction (although not entirely true) this question has absolutely nothing to do with [set-theory] (or even [elementary-set-theory]). $\endgroup$ – Asaf Karagila Jun 6 '12 at 7:13
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    $\begingroup$ Your definition says a Dedekind cut is a subset of $\mathbb Q$ that is (1) downward closed, (2) bounded above and (3) which has no maximum. What is missing is a counterpart to condition (2), namely (0) not empty. $\endgroup$ – Marc van Leeuwen Jun 6 '12 at 7:52

Let’s take a closer look at the conditions defining a Dedekind cut. A subset $D$ of $\Bbb Q$ is a Dedekind cut if:

  1. whenever $r\in D$ and $s<r$, then $s\in D$;
  2. there is a number $q\in\Bbb Q$ such that $r\le q$ for all $r\in D$; and
  3. for each $r\in D$ there is an $s\in D$ such that $r<s$.

Think of the rationals in the usual pictorial fashion, laid out as a line extending infinitely far in both directions, negative rationals to the left and positive rationals to the right. Condition (1) says that if $D$ contains some rational $r$, it contains every rational to the left of $r$ as well. One set that satisfies this condition is $\Bbb Q$ itself. Another is $(\leftarrow,0)$, the set of negative rationals, containing every rational strictly to the left of $0$. Yet another is $(\leftarrow,1]$, the set of rationals at or to the left of $1$. On the other hand, the set $(\leftarrow,0)\cup(2,4)$ does not satisfy condition (1): it contains $3$, but it doesn’t contain $1$ even though $1<3$. It’s not an initial segment of $\Bbb Q$.

Condition (2) is the simplest: it just says that $D$ cannot extend infinitely far to the right. There must be some rational number $q$ such that every member of $D$ is at or to the left of $q$. Almost all of the sets that satisfy condition (1) satisfy condition (2) as well; the only exception is $\Bbb Q$ itself, and condition (2) is designed specifically to rule out $\Bbb Q$.

Condition (3) is perhaps the hardest to get a grip on, but what it says is actually quite simple: it says that $D$ must not have a largest element. No matter what $r$ you choose in $D$, $D$ contains some bigger number $s$. This rules out sets like $(\leftarrow,1]=\{q\in\Bbb Q:q\le 1\}$, that have a maximum element.

A Dedekind cut, therefore, is a subset of the rational numbers that looks more or less like $(\leftarrow,2)$, say: it’s an initial segment of the rational number line, it’s not the whole line, and it has no largest element. In fact, every $r\in\Bbb Q$ defines a Dedekind cut in just this way, namely, the cut $$(\leftarrow,r)=\{q\in\Bbb Q:q<r\}$$ consisting of every rational to the left of (smaller than) $r$.

However, these aren’t the only Dedekind cuts. For example, let $$D=\{q\in\Bbb Q:q<0\text{ or }q^2<2\}\;.$$ It’s not hard to check that this $D$ satisfies conditions (1)-(3); if you’ve not already seen this, you probably will soon. But $D$ is not $(\leftarrow,r)$ for any rational number $r$, because if it were, we could show that $r^2=2$, when in fact we know that $\sqrt2$ is irrational. It’s these ‘extra’ Dedekind cuts, the ones not of the form $(\leftarrow,r)$ for any rational number $r$, that make Dedekind cuts useful and interesting: they ‘fill in’ the gaps in $\Bbb Q$ corresponding to the irrational numbers and allow us to construct the real numbers rigorously starting with just the rationals. In the end the cuts $(\leftarrow,r)$ for $r\in\Bbb Q$ are going to correspond to the rationals, and the ‘extra’ Dedekind cuts are going to correspond to the irrationals. But in order to make that work, we have to be able to define the various arithmetic operations on these Dedekind cuts in such a way that they behave the way they should. Your exercise here is part of showing how to define multiplication of Dedekind cuts.

I’ll say nothing about the exercise itself, as breeden has already covered that in some detail.

Added: In view of the comments, I think that I should emphasize that in what I’ve written above, $(\leftarrow,q)$ and so forth are to be understood as subsets of $\Bbb Q$. That is, I’m writing $(\leftarrow,q)$ as an abbreviation for $\{r\in\Bbb Q:r<q\}$, not for $\{r\in\Bbb R:r<q\}$.

As another example to show how a set can be bounded above (as is required by (2)) and still have no largest element, let $D$ be the set of negative rationals. Every member of $D$ is less than $0$ (and hence less than every positive rational as well), but $D$ has no largest element. If $m$ and $n$ are positive integers, so that $-\frac{m}n\in D$, then $-\frac{m}n<-\frac{m}{2n}<0$, so $-\frac{m}{2n}$ is a member of $D$ that is bigger than $-\frac{m}n$. This shows that $D$ has no largest element: give me any element of $D$, and I’ve just shown you how to find a bigger one.

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    $\begingroup$ @Joshua: To see the difference between (2) and (3), let $D= (\leftarrow,2)$. Every member of $D$ is less than $2$ (or $3$, or $5/2$), but $D$ has no largest element: if $r\in D$, then $\frac12(r+2)$ is also in $D$, and it’s bigger than $r$. The sets $(\leftarrow,\sqrt2)$ and $(\leftarrow,\sqrt2]$ aren’t equal as sets of real numbers, because of course $\sqrt2$ is in the second but not the first. But those two sets contain exactly the same rational numbers: $\sqrt2$ is the only real number that’s in one but not the other, and it isn’t rational. $\endgroup$ – Brian M. Scott Jun 6 '12 at 21:52
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    $\begingroup$ @Joshua: You’re right: $(\leftarrow,2]$ violates (3), because it has $2$ as its largest element, but it satisfies (2), because it’s bounded above. In fact, it’s bounded above by $2$ (as well as by any larger rational): every member of $(\leftarrow,2]$ is $\le 2$. $\endgroup$ – Brian M. Scott Jun 6 '12 at 22:23
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    $\begingroup$ @Joshua: Every member of $(\leftarrow,2)$ is at or to the left of $2$ (or $3$, or $5$, ...): (2) does not require that the upper bound belong to the cut $-$ it just has to exist. $\endgroup$ – Brian M. Scott Jun 6 '12 at 22:34
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    $\begingroup$ @Joshua: Somehow you’ve got it into your head that (2) says that $D$ has to have an upper bound in $D$. It doesn’t: it just says that $D$ has to have an upper bound in $\Bbb Q$. $100$ is a perfectly good upper bound for $(\leftarrow,0)$: it isn’t in $(\leftarrow,0)$, but if $r\in(\leftarrow,0)$, then $r\le100$. (2) also doesn’t say that there has to be some member of $D$ equal to the upper bound. In fact, if you replaced (2) by (2'), ‘there is a number $q\in\Bbb Q$ such that $r<q$ for every $r\in D$’, (1),(2'), and (3) together would still define the same collection of Dedekind cuts. $\endgroup$ – Brian M. Scott Jun 7 '12 at 1:23
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    $\begingroup$ @Joshua: Great! $\endgroup$ – Brian M. Scott Jun 10 '12 at 1:13

You forgot to add that $D$ and $E$ are positive Dedekind cuts.

1) Let $r$ be in $D*E$, and let $s < r$. If $s \leq 0$, then $r$ is in $D*E$ trivially. Suppose that $s > 0$. So that $r > 0$. Then we must have $r = xy$, for some $x$ in $D$ and $y$ in $E$ where $x > 0$ and $y > 0$. Then $s = \frac{s}{r}(xy) = (\frac{sx}{r})y$. Now $\frac{s}{r}$ < 1, so $\frac{sx}{r} < x$, and it follows from (1) for $D$ that $\frac{sx}{r}$ is in $D$. So $s$ is in $D*E$.

2) By (2) There are $x$ and $y$ in $\mathbb{Q}$ such that for any $r$ in $D$ and $s$ in $E$ we have $r \leq x$ and $s \leq y$. Since $D$ and $E$ are positive, $xy$ is positive. Show that for any $z$ in $D*E$ that $z \leq xy$.

3) Let $r$ be in $D*E$. If $r < 0$, then there is nothing to show, since 0 is in $D*E$. Otherwise, use that $r = xy$ for $x$ in $D$ and $y$ in $E$ where $x$ and $y$ are greater than 0. Use that there exists $s$ and $t$ in $D$ and $E$ respectively where $x < s$ and $y < t$, to show that $r < st$.

  • $\begingroup$ Thank you! I am working on this... We are using 0 as a center number to show that s is in D? s< 0 because we see the interval (-infinite, 0]... and s> 0... because both r_1 and r_2 are more than zero? $\endgroup$ – user5262 Jun 7 '12 at 0:41

Since breeden has already given you fine hints for proving the result is a Dedekind cut (once the statement has been fixed), I'll address the issue of the intuition behind it.

A Dedekind cut is just a way of breaking up the rationals into two complementary sets. Imagine the rationals arrayed in the real line: a Dedekind cut is like picking a point on the line, and looking at all the rationals that are smaller than the point, excluding the point if the point happens to be a rational.

The intuition is that we want to use Dedekind cuts to represent the real numbers. If all we know are the rational numbers, how do we describe a real number? A real number $r$ is uniquely determined by the collection of rationals $q$ with $q\lt r$, in the sense that $$\{q\in\mathbb{Q}\mid q\lt r\} = \{q\in\mathbb{Q} \mid q\lt s\}$$ if and only if $r=s$. Moreover, $$\{q\in\mathbb{Q}\mid q\lt r\} \subseteq \{q\in\mathbb{Q}\mid q\lt s\}$$ if and only if $r\lt s$.

But that's cheating: we are actually using the reals, and we are supposed to be trying to construct them with only rationals as reference. So we try to identify which intrinsic properties these sets of "all rationals less than or equal to a particular real number" have, and use them to define these sets. Then we can use the sets as proxies for the real numbers.

The three properties you give are an intrinsic way of describing what "the set of all rationals less than or equal to a particular real" would be: if they contains a particular rational $r$, then they contain all rationals smaller than $r$ (property 1); they are bounded above in the rationals (property 2); and they have no largest element (property 3).

This allows us to use these sets as stand-ins for the reals, without having to actually define the reals: we can just work with sets of rationals. This makes the order properties of the reals very easy (as I noted above): it is easy to compare these sets of rationals in a way that maintains our intuitive notion of "order" for the reals they are proxies for. But it makes the additive and multiplicative structures a little more difficult: we need to find a way of "adding" and "multiplying" these sets in a way that will correspond to what we want "addition" and "multiplication" of reals to be.

Addition is pretty easy: to get all rationals less than $r+s$ from all rationals less than $r$ and all rationals less than $s$ you just add them pairwise. So we can define the sum of two Dedekind cuts $D$ and $E$ as $$D+E = \{ q_1+q_2 \mid q_1\in D\text{ and }q_2\in E\}.$$ And this works.

What you have here is part of the method to define "multiplication." If we just try to do the same thing we are going to fail, because $D$ and $E$ will have arbitrarily large negative elements, and when we take their product we will get arbitrarily large positive rationals, so we will not get a Dedekind cut (it will not be bounded above). So we need to do something more clever.

Intuitively, how do you get all rationals that are less than $rs$, if you have all rationals less than $r$ and all rationals less than $s$? Well, if $r$ and $s$ are both positive, you can look at all rationals $q_1$ and $q_2$ with $0\leq q_1\lt r$ and $0\leq q_2\lt s$, and consider $q_1q_2$. Every such product will be a positive rational less than $rs$; and every positive rational less than $rs$ will be such a product (though this is harder to establish). So we would like to define the product of positive Dedekind cuts that way: take all products $q_1q_2$ with $0\leq q_1\lt r$ and $0\leq q_2\lt s$; and then throw in all negative rationals to complete the Dedekind cut.

This is your definition here, but we need to check that it is indeed a Dedekind cut. breeden has given you good hints for doing this.

Now, what are examples of Dedekind cuts? Well, for any real number $a$, take $(-\infty,a)\cap\mathbb{Q}$. That's exactly what a Dedekind cut looks like (if you already know what the real numbers are).

If you don't know what the real numbers are, then for every rational $q$ the set $$\{ r\in\mathbb{Q}\mid r\lt q\}$$ is a Dedekind cut. There are other cuts; the canonical example of a cut that is not of the form "all rationals less than the rational $q$" is: $$\Bigl((-\infty,0)\cap \mathbb{Q}\Bigr)\cup \{r\in\mathbb{Q}\mid r\geq 0\text{ and }r^2\lt 2\}.$$

  • $\begingroup$ Thank you! I've come to understand the intuition behind Dedekind cuts, especially as a way to construct the real numbers by using the properties of a rational numbers, and not them, themselves. The mention of how to obtain rationals less than rs, if r and s are both positive is currently a great help on this proof for class. In this last dedekind cut, does it simplify to [0, root 2) and if so, wouldn't it violate property (1)? $\endgroup$ – user5262 Jun 6 '12 at 22:00
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    $\begingroup$ @Joshua: As a subset of the real numbers, the last should be written as $$\Bigl( (-\infty,0)\cap\mathbb{Q}\Bigr)\cup \{r\in\mathbb{Q}\mid r\geq 0\text{ and }r^2\lt 2\}$$ and corresponds to $(-\infty,\sqrt{2})\cap\mathbb{Q}$. $\endgroup$ – Arturo Magidin Jun 6 '12 at 23:57
  • $\begingroup$ Oh, I see now, I lopped off the infinite bound for 0, when x more than or at 0 is included in the -infinite to root 2 bound. Thank you! $\endgroup$ – user5262 Jun 7 '12 at 0:34

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