An uncountable countable set
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From: David R Tribble on 24 Aug 2006 16:44 Tony Orlow wrote: > The more I reflect on that wonderful > newbie's contribution, the question as to how many bits are required to > list all the naturals in binary, the more I see it as that gaping hole > in the hull of this wreck. If set theory claims to have a cardinality > which fits every set, then this set stands out as the counterexample. > > Any finite number of bit positions produces a finite set of strings. > > Any countably infinite set of bit positions produces an uncountable set of > strings. Which explains why the reals are uncountable (when represented as infinite binary fractions in [0,1)). But it does not explain your claim that the naturals are uncountable when represented as finite bitstrings. It's pretty straightfoward to show that a countable number of bits produces a countable number of bitstrings. There is no "in-between" cardinality. It also flies in the face of your statements about infinite binary trees. If each node is numbered with a natural (being the finite bitstring of the left/right paths traversed from the root to the node), then the nodes, and thus the naturals, are obviously countable. On the other hand, you don't understand that the infinite paths in the tree (the ones that don't have a terminating node) are uncountable and correspond directly to your infinite bitstrings and to the real binary fractions in [0,1).
From: mueckenh on 25 Aug 2006 06:10 Dik T. Winter schrieb: > > > You apparently do not know what the word "computable" does mean. To > > > be precise: all algebraic numbers are computable in the mathematical > > > sense. > > > > To be precise: Each algebraic number is computable. But not all. > > Because then the list of all algebraic numbers was computable. > > But it is. So you can compute all solutions of a polynomial equation even of higher than fourth degree in finite time? I doubt that. If however one computer needs infinite time for one polynomial equation, then finitely many computers (more don't exist) cannot compute all algebraic numbers in eternity. Therefore, in eternity, one can compute each algebraic number - but not all. (In any meaningful sense). Regards, WM
From: mueckenh on 25 Aug 2006 06:12 MoeBlee schrieb: > Albrecht wrote: > > We must conclude that we can only say: we can find infinite many points > > on a line. This is a potential infinity. > > Do you have axioms for your mathematical theory of potential but not > actual infinity? Five Peano axioms. Regards, WM
From: mueckenh on 25 Aug 2006 06:13 Virgil schrieb: > In article <1156148773.420187.122140(a)h48g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > You are dreaming. It does follow from the form of the numbers of my > > list. > > Try to come back to reality. Simply show how a digit position n can be > > indexed the complete number of which, i.e., the complete sequence of > > digit positions 1 to n is not in the list. > > The Hilbert Hotel method! Why does the Hilbert Hotel method not apply to my list, but only to Dik's number? Regards, WM
From: mueckenh on 25 Aug 2006 06:14
Virgil schrieb: > The set of edges of an infinite binary tree are certainly countable, and > it is not difficult to construct an explicit bijection between them and > the set of binary representations of the naturals. > > The set of paths, however is not countable, as may is shown by > constructing an explicit bijection between the set of all such paths and > the set of all subsets on the set of naturals. > > Both the bijections referred to above have been presented here with no > one able to show either to be anything other than as advertised. Ok. Let's accept that. But by rational relation we find the set of paths being *not* larger than the set of edges. How is that possible? Regards, WM |