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.
Understanding of real numbers based on Dedekind cut?
178 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
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 $x$.