# Understanding of real numbers based on Dedekind cut?

I can not understand the construction of real numbers by Dedekind cuts. Can somebody help me with understanding? The problem which I have is that the cardinality of rationals is $$\aleph_0$$. Base on that, my assumption is that I can cut this only in $$\aleph_0$$ places in such a way that $$\forall a \in A, b \in B : a \lt b$$ and $$A \cup B = \mathbb{Q}$$. If it is true then it could find only $$\aleph_0$$ real numbers from $$\mathbb{R}$$. Please point me where is a logic mistake.

• No; a cut is a subset of $\mathbb Q$ and we have that the number of subset of an "$\aleph_0$ set" is greater than $\aleph_0$ – Mauro ALLEGRANZA Jan 25 at 11:38
• I understand, but each next $A$ i just $A \cup next \{q\}$ where $q \in \mathbb{Q}\$ assuming that I have such an ordering on $\mathbb{Q}$. Doesn't it imply that I have only $\aleph_0$ sets like that. – koralgooll Jan 25 at 11:52
• No; the typical example is $D = \{ q \in \mathbb Q \mid q^2 < 2 \}$ that defines $\sqrt 2$. This cut does not coincides with a rational. Thus, there is at least one more cut than rationals. – Mauro ALLEGRANZA Jan 25 at 11:54
• The problem is exactly that in $\Bbb{Q}$ there is no "next($q$)". Between any two rationals there are infinitely many others. And you can keep on bisecting either left or right half ad infinitum, leading to uncountably many cuts. – Jyrki Lahtonen Jan 25 at 12:10
• @koralgooll That is, indeed, a bit baffling at first. It is actually possible for a countably infinite set to have an uncountable chain of subsets, strictly ordered by inclusion. That is exactly what we see here. When I was in grad school on of the professors posted this as a challenge on his office door. I came up with a convoluted construction, and explained it to him at the next coffee break. He replied, This is nice. I like it. But have you heard of Dedekind cuts. I nearly punched him (in jest) because I was disgusted with myself. – Jyrki Lahtonen Jan 28 at 21:52

The mistake may be that you are trying to extrapolate from simpler linearly ordered sets. If, instead of $$\Bbb{Q}$$ you had a finite linearly ordered set in which every element has an immediate successor (and all but the first element is a successor of a single element) then the number of cutting points would, indeed, be more or less equal to the number of elements ($$\pm1$$ depending on whether you include the ends). Every possible cut occurs between an element $$x$$ and its successor $$s(x)$$. The same holds for infinite sets that have a similar successor function. For example, you can cut $$\Bbb{N}$$ at only countably infinitely many points – the two "halves" being $$[0,n]$$ and $$[n+1,\infty)$$ for all $$n$$.
However, the ordering of $$\Bbb{Q}$$ is more complicated. For one, it is dense. Between any two rational numbers there are infinitely many others. This means that there cannot be a successor function. Implying that you can keep on adding more cuts between any two earlier cutting points. This is in sharp contrast to cutting $$\Bbb{N}$$. It is easy to see that this leads to uncountably many cuts, but the details of that argument may be better explained by other means (such as Cantor's diagonal argument).
The mistake lies in your assertion that you can cut only in $$\aleph_0$$ points. Why? The set $$\Bbb R$$ is uncountable and, for each $$x\in\Bbb R$$, if you consider the cut$$\Bbb R=\bigl((-\infty,x)\cap\Bbb Q\bigr)\cup\bigl([x,\infty)\cap\Bbb Q\bigr),$$you get different cuts for distinct values of $$x$$.
• @user253751 It is not circular because this was written only to explain to the OP why it is not true that the fact that $\Bbb Q$ is countable implies that there are only contably many cuts. – José Carlos Santos Jan 25 at 19:50