Sir, I just came across your paper on "Cantor's Theorem" that there is no bijection from a set to its power set. I think you are right about the set M of "non-generators" being paradoxical. [...] I have been troubled about set theory since they told me in school that there is a rational between every two irrationals, yet more irrationals than rationals. It is obvious that "between every two irrationals there is a rational" implies that there are as many rationals as irrationals. However, I am frustrated that this could be so hard to prove while being immediately obvious to the intuition. I am also convinced that there cannot be more distinct Dedekind cuts than distinct rational numbers. Just drawing a sketch of some Dedekind cuts convinces me. The Dedekind cuts are 1) nonempty and 2) totally ordered by the relation "is a proper subset of". For finite sets it is easy to see, and prove by induction, that for such a collection of sets there are no more sets than elements. But I do not know how to make this a transfinite induction. Thank you for reading my long email. I hope that people are listening to your arguments!
Dear NN You were right. The reason is: There are at most countably many finite definitions like e = SUM1/n!. That is undisputed. So if there should be uncountably many reals, most of them cannot be defined - or can only be defined by infinite sequences. But that means the same as being undefined, because none of those sequences defines a number unless you know the last digit - which is impossible. So those "reals" cannot be used in mathematics (which means communication) because they cannot be communicated. They are not really real. And here comes a simple proof that the notion of uncountablility is in fact nonsense: Construct all real numbers of the unit interval as infinite paths of the complete infinite Binary Tree. It contains all real numbers between 0 and 1 as infinite paths i. e. infinite sequences of bits. [...] The complete tree contains all infinite paths. The structure of the Binary Tree excludes that are any two initial segments, B_k and B_(k +1), which differ by more than one infinite path. (In fact no B_k does contain any infinite path - but that is not important for the argument.) Hence either there are only countably many infinite paths. Or uncountably many infinite paths come into the tree after all finite steps of the sequence have been done. But if so, then it is by far more probable to assume that the single Cantor-diagonal comes into the Cantor-list after all lines at finite places have been searched. And then we have no reason to assume the existence of uncountable sets.