# Apparent contradiction in the definition of a Dedekind cuts

Taken from this Wikipedia article, are the following statements:

A Dedekind cut is a partition of the rational numbers into two sets A and B (...)

The set B may or may not have a smallest element among the rationals.

If $$B \subset \Bbb Q$$, then its smallest element is necessarily in the rationals. What am I missing here?

To clarify, I've understood a Dedekind cut to be a function $$F : \Bbb Q \mapsto A \land B$$, such that $$A = (- \infty, x), \ B =[x, \infty)$$, or alternatively, $$A$$ has some arbitrary lowest value $$y \ne x$$.

EDIT:

So, I think I understand. I assumed that $$B$$ had a smallest element, and that the sentence was saying that this smallest element could either be rational or irrational (which would be contradictory).

However, it is just a roundabout way of saying $$B$$ might have a smallest element, or not.

So, now I see. $$F : \Bbb Q \mapsto A \land B, \\ A =(- \infty, x) \land B = (x, \infty) \iff x \notin \Bbb Q \\ A = (- \infty, x) \land B = [x, \infty) \iff x \in \Bbb Q$$.

Realized I was using notation that implies $$A$$ and $$B$$ are intervals of the reals. That was an accident. I meant it as they were intervals of the rationals. Same properties as an interval, just excluding all irrationals.

• The set of rationals $>\sqrt 2$ does not have a least element. For that matter, the set of rationals $>0$ doesn't have a least element. The set of rationals $≥\sqrt 2$ doesn't have a least element, though the set of rationals $≥0$ does.
– lulu
Commented Jul 15, 2022 at 23:32
• What does $A \land B$ mean? Commented Jul 15, 2022 at 23:32
• $\mathbb{Q}$ doesn't have the least upper bound or greatest lower bound property (that's the problem this construction is trying to fix!). That is, there are sets $B\subset \mathbb{Q}$ bounded from below that have no smallest element (which is even true in $\mathbb{R}$; consider $(0, \infty)$) or infimum. Commented Jul 15, 2022 at 23:36
• $B$ need not have a smallest element at all. If it does have a smallest element it is a rational, you are correct about that, but if it's a Diedekind cut corresponding to an irrational it does not have a smallest element at all. Commented Jul 15, 2022 at 23:41

## 1 Answer

Intuitively a Dedikind cut cuts the rationals into two sets. $$A$$ which are all the rationals that are smaller than all the elements of $$B$$. And $$B$$ which are all the rationals that are bigger than all the elements of $$A$$. The cut will be so that $$A$$ has no biggest element and $$B$$ may or may not have a smallest element.

Another way of putting it is there is a SLICE that cuts all the rationals into two groups. $$A$$ is the set of all rationals to that are smaller than where the SLICE cut and $$B$$ is the set of all rationals that are bigger or equal to where the SLICE cut. If the SLICE cut at exactly a rational number that rational number will be the lowest element of $$B$$. If the slice cut did not occur at exactly a rational number then $$B$$ doesn't have a smallest element at all.

Now, you might say that if the SLICE didn't cut at a rational number it must have cut at an irrational number. SSSHHHH! No spoilers! The thing is we only know about rational numbers. We do not know anything else exists. But we can still SLICE the rationals into two sets. Sometimes we can slice at a rational number and that rational number goes into $$B$$. And sometimes we can slice between rational numbers and $$B$$ contains all the numbers bigger than than cut but there is no smallest rational number that is biggest than the cut.

An example is if we SLICE where the square of positive rational numbers are less than $$2$$ and where they are bigger than $$2$$. Then $$A=\{r\in \mathbb Q| r\le 0$$ or $$r^2 < 2\}$$ and $$B = \{r \in \mathbb Q| r > 0$$ AND $$r^2 > 2\}$$.

To clarify, I've understood a Dedekind cut to be a function F:Q↦A∧B, such that A=(−∞,x), B=[x,∞), or alternatively, A has some arbitrary lowest value y≠x.

ALmost but not quite.

Spoiler alert. Once we invent... I mean describe... the real numbers then each Dedikin cut is the pair of sets $$A = (-\infty, x)\cap \mathbb Q$$ and $$B = [x, \infty)\cap \mathbb Q$$ for a specific real number $$x$$. For every dedekind cut there is a real number, and for every real number there is an distinct Dedikind cut.

But if $$x \not \in \mathbb Q$$ then $$x \not \in [x,\infty)\cap \mathbb Q$$ and $$B= [x,\infty)\cap \mathbb Q= (x,\infty)\cap Q$$. Both $$[x,\infty)\cap \mathbb Q$$ and $$(x, \infty)\cap \mathbb Q$$ are the same thing and $$B$$ has no smallest element.

....or alternatively, A has some arbitrary lowest value y≠x.

You are misunderstanding that part. $$A$$ does not have a lowest element. But for any value $$w \in B$$ we have all $$y \in A$$ lead to $$y < w$$. Likewis if $$x \in \mathbb R$$ the the dedekind cut made by $$x$$ is $$A = \{y\in \mathbb Q| y < x\}$$ and $$B = \{w\in \mathbb Q| w\ge x\}$$.

• Very nice (and novel) explanation using SLICE. +1 Commented Aug 27, 2022 at 14:12