Quote:
|
Originally posted by Lebell If we represent infinity as "N", and call your hotel N1, we can call my hotel N2, to signify the increased holding power.
|
Close, except that you simply described N^2 rather than 2^N (where ^ is the power function and N is positive integers). By the axiom of choice we decide that 2^N=R, the real line. To get to R we'd need a list of all the infinitely long hotels (not just an infinite list of infinite hotels). The easiest mathematically correct proof is to extend the Power Function to infinite series, although this is still non-trivial. Heuristically, the power function (in topological terms) is the set of all possible subsets of a set. We know by induction that for any finite set there is no mapping from the set to its power set. {By contradiction} Consider some set (S) and its power set (P). Let A be an onto, 1-1 map from S to P. Consider the set A^-1(P) = S (the reverse mapping from an element of the power set to the original set), then we can simply "swap" orderings such that for each A^-1i (the ith element of the inverse map) matches Pi such that Si = Pi. But this means there is some Pj such that Pj != Sj, because we could simply consider the element Aj = Pj = (Si,Sk), Si!=Sk. This means that there is no onto, 1-1 function from S to P, which means that |S|<|P|, where |*| is the "size of" operations. Trivially we can map Pj->Sj, for Pj = {Sj} for an onto, 1-1 map, meaning that |S|!>|P|, thus |P|>|S|.
This obviously has flaws, but you get the idea.