An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 14 Jul 2006 15:43 In article <1152882972.884911.319600(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > In the binary tree everyone can see that no path can split without two > > > additional edges. So it is by no means possible that there were more > > > paths than edges. > > > > As long as paths all end. > > And if thy do not end, they run without edges? Flying? If no path ends then more paths than edges is not only possible, it is necessary. Does "mueckenh" deny that for infinite binary trees there is a bijection between the set of paths and the power set of the set of edges?
From: Virgil on 14 Jul 2006 15:50 In article <1152883061.851461.27130(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > The set of naturals shold be the same set as the set of indices of > > > 0.111... . But it cannot. > > > > Maybe not in "mueckenh"'s world but it works fine in mine. > > It cannot. It does. > You see it by the following argument. > Define 0 *+ 0 = 0, 0 *+ 1 = 1 *+ 0 = 1 *+ 1 = 1. > Do the following addition *+ > > 0.1 > 0.11 > 0.111 > ... > > What is the result? 0.111... Not until you define 1 *+ <blank space> and infinite 'sums' And, as far as I can see, your "addition" is equivalent to "union" , in which case it only means the union of all finite naturals (as sets) is the set of all naturals, which is naturally infinite.
From: Virgil on 14 Jul 2006 16:00 In article <1152883183.067840.307620(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > This is a set and it exists as well as any other infinite set. Not more > > > and not less. > > > > What set? > > All the transpositions required form an infinite set. As such it should > be possible to be exhausted. Then it must be a sequence, well ordered, and the ordering cannot be ignored without compromising the alleged effects of the transpositions. And then there is no definition of a "limit" possible. After each transpostion there will still be only a finite initial segment of the sequence of rationals correctly ordered by magnitude followed by an infinite tail segment containing infinitely many descending (in magnitude) sequences of rationals.
From: Virgil on 14 Jul 2006 16:31 In article <1152883379.086210.127070(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > For infinite binary trees each such list of branchings is an infinite > > > > sequence. And there are uncountably many possible such infinite > > > > sequences. > > > > > > But we know that there are nearly twice as much edges. > > > > We only know that for finite trees. "Twice as much" doesn't work for > > infinite sets. > > But less than twice also does not work. But infinitely many less (edges than paths) does nicely. For a binary tree in which every pathe is infinite, there is a bijection from the set of all infinite paths to the power set of the set of all edges. > > > > > There are > > > infinitely many of them too. And the the set of edges is countable. > > > > Then show me an ennumeration of them. > > I wait until you will have learned to calculate with fractions. Then > you can see that every path carries two edges. In any binary tree in which every path has a terminal edge, there is an easy bijection between terminal edges and paths, but this does not happen in trees for which paths do not have terminal edges. In an infinite binary tree one can still show that the set of edges is countable: {0,1} for "first" edges (from the root node), {2,3,4,5} for "second" edges, having with one preceding edge, {6,7,8,9,10,11,12,13} for those with two preceding edges, and so on. One could, if pressed, work out an exact formula for the natural corresponding to each edge based on the left versus right branchings from the root node through itself, establishing a bijection from N to the set of edges. In an infinite tree one can show that the endless paths are uncountable" Each endless path is an end less sequence of left or right branchings, so can be represented by an endless string of characters from {"L", "R"}. Each such endless string corresponds uniquely to that subset of N which contians an n if and only if the n'th character in the string is an "L". This clearly establishes bijections from the set of paths to the set of endless strings of {"L","R"} characters to the set of all subsets of N. "Mueckenh" chooses to ignore these incontrovertible proofs of the fact that for an infinite binary tree there are only countably many edges but uncountably many paths. "Mueckenh"'s only option is to deny existence of infinite binary trees. Once he allows them, as he has, all is lost.
From: Virgil on 14 Jul 2006 16:36
In article <1152883498.624854.7380(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > 0.111... cannot be indexed by the members of any natural. Does 0.111... > > > have more 1's than can be indexed by any natural? Or what prevents it > > > from being indexed by any of the naturals? > > > > What cannot be indexed by any member of N is N itself, but N itself can > > be "indexed" by the members of Succ(N) = N \union {N}. > > Doesn't N consists of naturals only? Can't each natural index itself? > What remains, if each natural has indexed itself? The set of all naturals to index itself, and all subsequent ordinals to index themselves. > > > > Each ordinal has a successor, even limit ordinals like N, and any > > ordinal can be indexed by the members of its successor. > > Each ordinal can be indexed by itself. 1 is indexed by 1. 1,2 is > indexed by 1 and 2. 1,2,3 is indexed by 1,2 and 3 and so on. But all > naturals are not indexed by all naturals? Extremely ridiculous! > Extremely! Then why is no one but "mueckenh" laughing? Every ordinal has successors, whether in ZF or NBG or most other set theories, and to index any of them it is sufficient to use at the members of any successor. |