An uncountable countable set
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From: mueckenh on 29 Jun 2006 09:55 Virgil schrieb: > In article <1151505096.329945.87930(a)i40g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Daryl McCullough schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de says... > > > > > > >Daryl McCullough schrieb: > > > > > > >> So the answer is no? There is no function f enumerating > > > >> all the real numbers? So the reals are uncountable. > > > > > > > >There is a mathematical form containing all real numbers of [0, 1]. In > > > >binary representation it is given here. Why not call it a function in > > > >an extended sense? > > > > > > The question is this: does it map the naturals onto the real numbers > > > in [0,1] such that every real number is in its image? > > > > > > In particular, with your representation, which natural number maps > > > to the real 1/3? > > > > It is not necessary to know this natural number (which remains unknown > > or does not exist) in oder to prove that the infinite branches of the > > tree are countable. > > > If no natural number maps onto 1/3,, then you do not have a surjection > from N ONTO R at all, and your mapping does not show the set of reals to > be countable after all. If the paths are not more than the nodes, and if the nodes are countable, then the set of paths cannot have cardinality larger than aleph_0. > > It is equally unneccessary to know which natural might map to 1/3 in > order to prove that very same set of branches uncountable, and > futhermore, such proofs of uncountability actually exist, whereas no > valid "proofs" of countability can exist. That is easy to explain. Proofs with the "wrong" result are declared invalid. Regards, WM
From: mueckenh on 29 Jun 2006 09:57 Virgil schrieb: > In article <1151505625.358363.7610(a)m73g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Daryl McCullough schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de says... > > > > > > >Virgil schrieb: > > > > > > > > > > > >> It is not "Cantor's list" but ANY list. > > > >> You are challenged to find a list which lists all reals, then Cantor > > > >> presented a rule for showing any list you provide is incomplete. > > > > > > > >I have already shown that any list of all or even some rationals and > > > >its diagonal number is uncountable. > > > > > > No, you didn't, since it is false. > > > > > > 1. Start with a function f that is a surjection from > > > N to Q (Q is the rationals). > > > 2. Compute the diagonal d(f). This is a real number that is > > > guaranteed to not be in the image f. > > But d(f) is NOT guaranteed to be a rational number, and without such > guarantee, your "proof" crashes. I did not say that d was a rational number. > > > > 3. Form a new set M(f) = Q union { d(f) }. > > > 4. This new set is *countable* > > So that now "mueckenh" is claiming that Q is NOT countable but that Q > union with some other set IS countable. Read again. Regards, WM
From: Daryl McCullough on 29 Jun 2006 11:12 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> If you are claiming that your tree includes all real numbers in [0,1] >> then yes, it is necessary to know that it contains the real number 1/3. > >Of course it contains the number 0.010101... Which natural number is mapped to that number? >> Cantor's claim is that no countable set contains all real numbers. >> To prove him wrong, you need to demonstrate both (1) that your tree >> has countably many branches, and (2) that it contains every real number. > >It contains every real number o [0,1] by its construction. It has not >more branches than nodes. The set of nodes is countable. To prove that a collection C is countable, you need to prove that there is a mapping from N to C such that every element of C is in the image. What is the mapping? What natural number is mapped to 1/3? -- Daryl McCullough Ithaca, NY
From: mueckenh on 29 Jun 2006 12:43 Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >Daryl McCullough schrieb: > > >> If you are claiming that your tree includes all real numbers in [0,1] > >> then yes, it is necessary to know that it contains the real number 1/3. > > > >Of course it contains the number 0.010101... > > Which natural number is mapped to that number? > > >> Cantor's claim is that no countable set contains all real numbers. > >> To prove him wrong, you need to demonstrate both (1) that your tree > >> has countably many branches, and (2) that it contains every real number. > > > >It contains every real number o [0,1] by its construction. It has not > >more branches than nodes. The set of nodes is countable. > > To prove that a collection C is countable, you need to prove that there > is a mapping from N to C such that every element of C is in the image. Wrong. The proof that a set contains less elements than a countable set is sufficient to show that the first one is countable. Regards, WM
From: Gc on 29 Jun 2006 12:44
mueckenh(a)rz.fh-augsburg.de kirjoitti: > An uncountable countable set > > There is no bijective mapping f : |N --> M, > where M contains the set of all finite subsets of |N > and, in addition, the set K = {k e |N : k /e f(k)} of all natural > numbers k which are mapped on subsets not containing k. > > This shows M to be uncountable. > > Regards, WM Think each element n of N as n:th prime. For every finite subset you have corresponding unique "Gödel number". So it`s clear that there is countable many finite subsets. The set K = {k e |N : k /e f(k)} is then N and you may choose for example 1 to be it`s pair. |