# Cardinality of set of Dedekind cuts (elementary)

Under the Dedekind construction the irrationals are defined as those cuts $(A,B)$ where $B$ has no least element ($A$ not having a greatest element by definition), for example the $q^2=2$ case. I can see how I can construct a countable number of irrationals that way but I can't see how to get an uncountable number of irrationals without an uncountable number of symbols. If I need an uncountable number of symbols then I don't need the Dedekind construction to start with. What am I missing?

• There are far more sets of rationals than there are rationals. Dedekind cuts correspond to certain sets of rationals, so there's no a priori reason why there can't be more of them than there are rationals. Note that a cut cannot, in general, be written with any finite string of symbols. – MJD Nov 3 '14 at 23:32
• – MJD Nov 3 '14 at 23:34
• What precisely do you mean by a symbol? – Andrés E. Caicedo Nov 3 '14 at 23:37
• @MJD: ...more sets of rationals: I have no issue with that for any finite set of whatevers. That allows only an enumberable number of cuts, which added to the rationals still leaves me with a enumerable set. Thanks for the link. I will check out Cauchy sequences. – user533933 Nov 9 '14 at 21:07
• Thanks for the link. I will check out Cauchy sequences. Your comment on that page starting "Assuming that there are uncountably infinitely many" I guess expresses the lump I am struggling to swallow. – user533933 Nov 9 '14 at 21:19

In our enlightened day and age we know that you don't have to name something for it to exist. Did atoms came to existence just when they were named? Was everything afloat until Newton wrote down the laws of gravity?

In mathematics, especially abstract mathematics that deals with infinite sets, we don't have to name something in order for it to exist. Sometimes we can just show that something with the certain property exists, or we can show that it is impossible for something to exist.

The point is that we don't have to define the Dedekind cuts using arithmetic, or some other well-known function. We can prove that there are uncountably many of them. And this means that many of the numbers defined using Dedekind cuts have no definition using algebraic operations on $\Bbb Q$. But why is that a problem?

• Atoms have not all been named. The class has been named, but its members have not. I happen to believe that atoms are countable so I would not expect any problems there anyway. Your point would be pertinent if you could say a non-enumerable number of atoms had been named. – user533933 Nov 9 '14 at 21:29
• I'm not sure what you mean. – Asaf Karagila Nov 9 '14 at 21:32
• Okay, I still don't follow your comment. – Asaf Karagila Nov 9 '14 at 22:07
• It might do you some good to read and learn about basic first-order logic and basic model theory. In particular every first-order theory which has an infinite model, has one of each infinite cardinality. And you can wreck the definability pretty badly like that. If you have an issue with undefinable objects, then perhaps the route for you goes through intuitionistic logic and constructive mathematics. Although I think it's preferable to approach that part of mathematics only after having some reasonable grip of classical mathematics. – Asaf Karagila Nov 9 '14 at 22:25
• Struggling with the stackoverflow ui here. sorry for mess. > Sometimes we can just show that something with the certain property exists Whatever description you use will do as a name. Even if the thing in question does not exist. > we don't have to define the Dedekind cuts. We can prove that there are uncountably many of them. I'm not happy with an uncountable number of undefined things. That might as well be a seminary school argument about the will of god. Undefined puts it outside mathematics. – user533933 Nov 9 '14 at 22:28

There is an alternative, geometric interpretation of Dedekind cuts which may be helpful.

View each positive, rational $\frac{n}{m}$ as the integer point in the first quadrant of the plane $\langle n,m \rangle$.

Then each line through the origin determines a unique cut of the plane into a lower set and an upper set - i.e., points below the line and points above the line.

There are uncountably many such cuts since there are uncountably many such lines. Those lines are $y=rx$ for each real number $r$.

• Your response fails on 2 counts:My question is about Dedekind cuts. I could phrase the same question about Cantor's diagonalization argument but I am here talking of Dedekind cuts. – user533933 Nov 9 '14 at 21:32
• @user533933 You need to view the points in the plane simple as a countable collection of points - ignoring their "value" as rational numbers. The cuts then divide this countable set of points (e.g. Q) into two sets and they do it uncountably many times. I could add this visualization is given in a number of text books. – Epsilon Nov 9 '14 at 21:35
• 1) My question is about Dedekind cuts specifically. An explanation in terms of some other rationalization does not resolve the issue for Dedekind cuts. 2) It seems you are indexing an uncountable infinity using the countable rational. There is in fact only a countable number of pairs of points of the form ((m,n), (0,0)). The value of r would be n/m which is rational. – user533933 Nov 9 '14 at 21:43
• How do "they do it uncountably many times"? Can you point me to links that don't require me to spend money? – user533933 Nov 9 '14 at 21:46
• @user533933 Fair enough. I can understand you objections on the grounds that it does rephrase the problem by making it more abstract. However, the point of this visualization is that the uncountablility arises from the geometry of the line in the plane. A straight line through the origin will partition the set uncountably many times, as can be seen using simple Euclidean geometry. – Epsilon Nov 9 '14 at 21:48