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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.

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    $\begingroup$ 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 $\aleph_0$ $\endgroup$ – Mauro ALLEGRANZA Jan 25 at 11:38
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    $\begingroup$ 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. $\endgroup$ – koralgooll Jan 25 at 11:52
  • $\begingroup$ 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. $\endgroup$ – Mauro ALLEGRANZA Jan 25 at 11:54
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jan 25 at 12:10
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    $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen Jan 28 at 21:52
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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).

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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 $x$.

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    $\begingroup$ Isn't this circular? Dedekind cuts are uncountable because they correspond to real numbers which are uncountable. But why are real numbers uncountable? Because they correspond to Dedekind cuts which are uncountable... $\endgroup$ – user253751 Jan 25 at 19:49
  • $\begingroup$ @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. $\endgroup$ – José Carlos Santos Jan 25 at 19:50

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