Once again about Dedekind cuts

Question

Assume I have a Dedekind cut: $(-\infty,a)\cup(a,+\infty)$. This Dedekind cut defines irrational number $a$. And $(-\infty,a]\cup(a,+\infty)$ defines rational. I have a few questions about this:

1. How to prove that $\Bbb R$ in which every number defined this way is continuous?
2. How to prove that Dedekind cut defines only one number?
3. How to prove that there are no more excluded points on axis?

What if Dedeking cut actualy defines some weird interval like $\{\sqrt{2}-\epsilon\;;\sqrt{2}+\epsilon\}$ that does not contain rational numbers? And in that terms irrational number is not a point on axis, but that kind of weird microcontinuum interval.

Answer 1

First of all, you shouldn't describe a Dedekind cut as $(-\infty,a)\cup(a,+\infty)$, since this needs the irrational number $a$ to be defined, while the whole goal of Dedekind cuts is of course to define the irrational numbers.

---

The way I like to see Dedekind cuts, is as a set $X\subset \Bbb Q$ of rationals, such that:

• $X$ is not empty and $X$ is not all of $\Bbb Q$,
• $X$ is downward closed (i.e.: for any $x\in X$ and any $y<x$ also $y\in X$),
• $X$ has no maximal element (i.e.: for any $x\in X$ there is some $y\in X$ such that $y>x$).

Often Dedekind cuts are described as two sets, the other set being $\Bbb Q\setminus X$, but including this second set is not really necessary.

There are two important cases: either $X$ has a least upper bound (inside $\Bbb Q$), or it does not. For example, the set $\{q\in \Bbb Q\mid q<2\}$ is a Dedekind cut that has least upper bound $2$, while the set $\{q\in\Bbb Q\mid q<0\lor q^2<2\}$ is also a Dedekind cut, but does not have a least upper bound in $\Bbb Q$.

The real numbers $\Bbb R$ are defined to be the set of all Dedekind cuts on $\Bbb Q$. If $X,Y$ are two real numbers, we say that $X\leq Y$ if $X\subseteq Y$. So the ordering of real numbers comes straight from the inclusion of the Dedekind cuts that define them.

We call a real number $X$ rational if $X$ has a least upper bound in $\Bbb Q$, and call it irrational if $X$ does not have a least upper bound in $\Bbb Q$. To avoid confusion, I will say "rational number" whenever I mean an element $q\in\Bbb Q$ and I will say "rational real" whenever I mean a Dedekind cut with a least upper bound in $\Bbb Q$.

In practice and intuitively, we often see a real number as some kind of point on a line, instead of a set of rational numbers. However, this line, and this point, are defined by Dedekind cuts. Hence we cannot prove that a real number is a point on a line, since we define both the point and the line using Dedekind cuts.

---

Let's look at your three questions:

1. A set cannot be continuous. Continuity is a property of functions between topological spaces, not a property of sets. However, there is such a thing as the property of being a continuum, which is probably what you meant. A continuum is any set that is dense and complete.

   Denseness means that if $x<y$, then there is some $z$ such that $x<z<y$. The real numbers are dense, since $\Bbb Q$ is dense: if $X$ and $Y$ are Dedekind cuts, and $X\subsetneq Y$, then there is some $q\in Y$ with $q\notin X$. Because $Y$ has no maximal element, there is some $r>q$ with $r\in Y$. Then the set $Z=\{p\in \Bbb Q\mid p<r\}$ is a Dedekind cut that is in between $X$ and $Y$. Note that $Z$ is not just any real number, it is a rational real.

   Completeness means that for any set of reals $R\subset \Bbb R$ that has an upper bound, there is a least upper bound of $R$. This follows quite easily from the definition of Dedekind cuts, since we can just take the union of the Dedekind cuts in $R$. This union is a Dedekind cut itself, since $R$ has an upper bound (and thus there is some $Z\in\Bbb R$ such that $X\subsetneq Z$ for all $X\in R$, meaning $Z$ contains some rational number $q\in\Bbb Q$ with $q\notin X$ for any $X\in R$).

2. As said before, we cannot prove that a Dedekind cut defines one number, since we define real numbers to be exactly the same as Dedekind cuts. Using the definition, this question asks for proving that every Dedekind cut defines one Dedekind cut.

3. Similarly here, we cannot prove this, since both points and axis are defined to be exactly all of the Dedekind cuts. There are no excluded points, is saying that no Dedekind cuts are excluded. But this is obvious by definition, since "the set of all Dedekind cuts" contains "all Dedekind cuts".

Finally your last remark, does there exists some weird interval $[\sqrt 2-\epsilon,\sqrt 2+\epsilon]$ not containing any rational reals? You are defining this interval using two real numbers $X=\sqrt 2-\epsilon$ and $Y=\sqrt 2+\epsilon$.

Since real numbers are Dedekind cuts, suppose that $X\subsetneq Y$. Then, as we saw in the proof of denseness, there exists some rational real $X\leq Z\leq Y$. However, we asked for an interval $[X,Y]$ not containing any rational reals. Therefore $X\not\subsetneq Y$, and thus $X\geq Y$.

This shows it is impossible to have an interval containing more than one real that does not contain any rational reals.