An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 12 Jul 2006 14:35 In article <1152718533.811441.322410(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152639759.882706.252490(a)s13g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > The first edge is mapped on the first path. > > > > > > > > Which of the infinitely many paths through that edge is the "first" > > > > one? > > > > > > > That at the root. > > > > All paths start at the root. > > There these all are one path. You may call it bunch of paths. What I call a path is a set of edges with (1) one and only one edge from the root node and (2) one and only one edge from each child node of one of its edges which is also a parent node of the tree. In a binary tree in which each path ends at a leaf node, the number of paths equals the number of leaf nodes. > > > > > > > > > If this splits in two, 1/2 of the first edge in addition to the new > > > > > edge is mapped on the new paths. > > > > > > > > You are not allowed to split edges. Besides which, the first edge has > > > > already been entirely used up. > > > > > > It is inherited by the childs of the first path. > > > > Paths do not have children, only nodes have children. > > Call it as you like. The first bunch splits in two bunches. These two > inherit half of the first edge each. On the contrary, each path "inherits" all of its branches. > > > > In the set of naturals {} is the first. > > What does half of {} mean? Or any other fractional part of {}? > > In the set of naturals 1 is the first. The first edge is one edge, not > {} edge. > However, one edge is the first one, and this first one is divided in > two halves. But half an edge is not an edge, so one does not have any paths with "half" of an edge.
From: Virgil on 12 Jul 2006 14:36 In article <1152718605.477772.83560(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152639642.547075.258360(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > All positions which can be enumerated are in the list, by definition. > > > > > Hence, the decimal representation of 1/9 does not exist. > > > > > > > > Then how is it that so many people use it? > > > > > > Because they do only believe they used it. > > > > For numbers, such things only exist by belief in them anyway, so if > > enough people believe in 0.111..., then it does exist. > > And presently enough people believe in infinity. Too many don't.
From: Virgil on 12 Jul 2006 14:40 In article <1152718724.627520.95560(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152638895.480011.23630(a)h48g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > 0,111... is a sequence which does not represent a natural. > > > > > > And it cannot be completely indexed by natural numbers. > > > > If such an obviously well ordered set has digits not indexed by members > > of N it must have a first such digit > > > > Which digit is that first non-indexed digit, "mueckenh"? > > > > Can't answer? > > It is the first 1 behind the zeros at the rights hand side. > 0.111... - 0.111...1 = 0.000...0111... But given your 0.111...1, there is always a larger one, say 0.111...11. So that the first 1 in 0.000...0111... WAS indexed after all. And "mueckenh" is wrong again!
From: Virgil on 12 Jul 2006 14:44 In article <1152718835.603216.109240(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152638595.246620.165700(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > > > It can only be proved > > > > > after having counted the digits from 1 to n without leaving out a > > > > > single one. > > > > > > > > Nonsense. The Cantor rule generates a digit to go in the nth decimal > > > > place of the number being created without any reference to any other > > > > decimal place. > > > > > > > You cannot identify any place without counting from 1 to that place. > > > > The axiom of infinity guarantees us all those "places" without counting > > anything. > > But it does not guarantee what is existing there. In order to find out > you must count. I do not know what "axiom of infinity", 'mueckenh" is referring to , but the ones in ZF, or ZFC or NBG guarantee existence of all of the "naturals" ( as finite ordinals) without having to count anything.
From: David Hartley on 12 Jul 2006 14:44
In message <virgil-612DD6.17462711072006(a)news.usenetmonster.com>, Virgil <virgil(a)comcast.net> writes >> Given a sequence of total-orderings <_n of a set S we can define a >> partial ordering <_L on S by >> >> For any x,y in S, x <_L y iff there exists n in N st. >> for every m > n, x <_m y >> >> This ordering will not necessarily be total, but if it is, it seems >> reasonable to say that the sequence has a limit, and that it is <_L. >> >> Using this definition, Daryl McCullough, Dik Winter and I have all given >> examples of sequences of well-orderings whose limit is not a >> well-ordering. WM's original proposal was rather muddled, but his >> present version works. He *has* described how to construct a sequence of >> orderings of the positive rationals, starting with that induced by an >> arbitary enumeration, whose limit is the usual ordering. (In fact, the >> same technique will give a sequence whose limit is any given total >> ordering, and a similar, although a bit more complicated, argument shows >> that there are sequences starting with any given total ordering whose >> limit is any given ordering isomorphic to N.) > >How do you define this alleged sequence of orderings? What is the >process by which one goes from one ordering to the next in this >sequence? > >What is your measure of how close to the limit state you have >progressed? E.g., what are your analogues to the deltas and epsilons of >more prosaic limit processes? > See definition at top of quoted section. (There are no epsilons and deltas, it's purely set-theoretic. If you consider the orderings as subsets of S x S, <_L = union (n=1 to oo) [intersection (k=n to oo) <_k ] which I believe is a semi-standard idea of the limit of a sequence of sets.) As you frequently point out to WM, limits are not reached step-by-step, they are just defined in terms of the whole sequence. In WM's construction, each ordering can be constructed from the previous one by using finitely many transpositions, but that is irrelevant. Just define the nth ordering as the original with its first n elements reordered by size. The transpositions are relevant in so far as WM claims that Cantor wrote that a sequence of transpositions can not change the order-type. Either Cantor was wrong (or using a different idea of the limit), or WM is misinterpreting him, my German is not good enough to tell. In any case, it's a proposition WM needs to prove to complete his argument, and he hasn't done so. -- David Hartley |