An uncountable countable set
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From: Virgil on 19 Aug 2006 13:53 In article <1155998665.523128.168920(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > > That they > > > > are countable comes from other considerations. > > > > > > It doesn't matter where that comes from. In fact all numbers which can > > > be constructed form a countable set. > > > > Proof? And pray use the mathematical sense of constructed. Not what you > > think it should mean. > > There is no question that only countable sets of numbers can be > constructed in the mathematical sense. That depends on what construction methods are allowed. Inherent in the axioms of ZF and beyond are the existences of uncountable sets. > And as only a countable set of such sets can be constructed in > reality As no part of mathematics can be "constructed in reality", as soon as one delves into mathematics, one is outside the restrictions of being "constructed in reality".
From: Virgil on 19 Aug 2006 13:56 In article <1155999150.167954.72660(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > > I do not only *claim* that 0.111... can be indexed by the natural > > > > numbers; > > > > I can *prove* it, as I have done again and again. > > > > > > You supposed that all digits of 0.111... could be indexed, and than > > > you proved that they can be indexed. > > > > No. I *defined* 0.111... such that all digits could be indexed. > > I *defined* that all digit positions which can be indexed belong to > numbers of my list. The indexes are natural numbers and all natural > numbers are in my list. And that same set of naturals can index the list twice over and still have infinitely many naturals left to index other things. > > And I showed that your number is not in my list. > > Hence one of our definitions must be wrong. Yours, obviously. > > Either there are not all natural numbers in my list and nowhere else, > or your number cannot be indexed completely. Or, as usual, "Mueckenh" is wrong again.
From: Virgil on 19 Aug 2006 14:05 In article <1155999688.567981.312170(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > Why then did you want to allow to assign finite sequences of edges only? > > > > To simplify matters. Natural numbers have in any base notation a finite > > number of digits. In allowing finite sequences of digits I allow for > > the occurrence of (for instance) both 0 and 00 and 000. So I give you > > more freedom. > > But in fact that is less freedom because the sequence of edges per path > is infinite. The sequence of edges leading up to any edge is finite. So a finite sequence of 0's and 1's, 0 for left child, 1 for right child, uniquely identifies each edge, but it takes an endless sequence of 0's and 1's to identify any endless path. Thus in infinite binary trees, the set of edges is countable but the set of endless paths is not.
From: Virgil on 19 Aug 2006 14:13 In article <1156000079.633380.83620(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Exactly. Above formula holds for every finite number, as can be shown > by induction. But it doesn't hold for an infinite number of finite > numbers. That begs the question of whether there are infinitely many finite (natural) numbers. There re two standard definitions which distinguish between finite sets and non-finite sets, at least in ZFC and NBG, and assuming C, they are equivalent: (1) A set is finite if and only if it is bijectable with some natural number. (2) (Due to Dedekind) A set is finite if and only if there does not exist any injection from the set to any of its proper subsets. In either case, all sets which are not finite are called infinite.
From: Virgil on 19 Aug 2006 14:23
In article <1156000079.633380.83620(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > Oh. I must have missed something, because I have not seen a proof. > > Given an injection f: N -> P(N), why does the set > > M(f) = {n in N | n !in f(n)} > > not exist? > > Excuse me, the non-existing set is the triple {f, n, M_f(n)} > > Regards, WM What does "M_f(n)" mean? As far as I can see, M_f or M(f) is set, not a function so "M_f(n)" is not even defined, much less {f,n,M_f(n)). In any case, there are lots of injections, f, from N to P(N). Consider f(n) = {n}, for example. In which case one easily sees that M(f) = {}. Which exists quite nicely, thank you very much! |