Cantor Confusion
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From: Virgil on 2 Jan 2007 16:32 In article <1167737840.595327.127430(a)42g2000cwt.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > In the union of all finite trees, there are no infinite paths at all, > > > > > > No? Where do they end? You should be able to answer this question and > > > to prove your answer. > > > > In any union of finite trees, every path is a member of some finite > > tree, and it ends where the paths in that tree end. QED. > > Where is the ending of the union of all finite paths? Isn't it smilar > to the "ending" of all finite natural numbers? it is quite similar in that one has in both cases an infinite set of all whose members are finite. > > > > > Hint: In the union of all finite initial segments of N there is an > > > infinite initial segment, namely N (so some people claim at least). > > > > According to that definition there must be a member of a union which is > > not a member of any set in the union. That is definitely non-standard > > set theory. > > It is not standard that the object N = {1,2,3,...} is not in the union > of all initial segments {1,2,3,...,n}? It is not a member of that union because it is not a member of itself. WM again conflates being a member of a set (the uniion) with being a subset of that set. N is a subset of the union of all initial segments ( actually equals the union) but is not a member of that union. > > Regards, WM
From: Virgil on 2 Jan 2007 16:35 In article <1167738133.535489.164800(a)n51g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I have been knowing for several years already that the decimal or > binary representation of 1/3 does not exist. WM has then 'been knowing for several years' things contrary to fact.
From: Virgil on 2 Jan 2007 16:45 In article <1167738410.654375.175860(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1167514645.819727.261070(a)h40g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > Why don't you simply give the formula for the definable well > > > > > ordering? > > > > > The formula could be evaluated, and, step by step, we would get a > > > > > list > > > > > of all reals. > > > > > > > > Wrong. It would not give a list. It would give a well-ordering. Do > > > > you > > > > not know the difference? > > > > > > Every non empty set of a well-ordering has to have a first element. If > > > all reals are well-ordered, then all reals are first elements in some > > > sets. > > > > What is the relevance? If I order the naturals as: > > {1, 3, 5, 7, ..., 2, 4, 6, 8, ...} > > I have a well-ordering but not a list. A well-ordering is *not* a list. > > It is not a list in the technical meaning of "list", but it is a linear > ordering. The real are linearly ordered in their usual ordering, but not well-ordered. > > You keep to state that, without proof. For a well-ordering it is not > > necessary to be able to state the elements one by one. > > As your above example shows, one must be able to find for every x the > set where x is the first element Actually, one need only show that for every non-empty set there is a first element to establish a well-order. > > > > > > > Because non-distinguishable sets are not different sets. You just > > > > > proved that there is no element of a path which distinguished it > > > > > from > > > > > every other path. > > > > > > > Right. But that does *not* mean that it can not be distinguished from > > > > every other path. > > > > > > If every edge of a path is shared by another path, how would you > > > distinguish them? > > > > What is "them" here? Give me two paths and I can give you an edge that > > distinguishes them (or at least an algorithm that shows it). > > "Them" is two paths. If one path has no edge of itself, then there is > at least one other path which shares all edges with the former. Not in an infinite tree. In such a tree, no path need have any edge all to itself, and one can still find for eery two distinct paths an edge in one but not the other. (In any two infinite binary strings of {L,R} symbols, either they agree at every position and are the same tree or they differ at some finite position, locating an edge in each not in the other. > > Regards, WM
From: Virgil on 2 Jan 2007 16:48 In article <1167738761.822054.120400(a)k21g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > "3/2 < 2" and "|N| < |R|" have /completely different meanings/. The > > first is a linear order applied to a field. The second is an ordering > > based on surjections / injections. > > > Both are based upon linear orderings. > > > Sums of rational numbers are nothing else but sums > > > of integers, given in the unit of the greatest denominator. In case of > > > limits this kind of arithmetic is extended, but the basis is > > > calculating with natural numbers which are to be compared by mappings, > > > i.e., the arithmetic of ordinal numbers. > > > > 3 is not identical to 6. 7 is not identical to 14. And yet we say 3/7 > > is identical to 6/14. In fact, there are at least as many names for > > the number "3/7" as there are natural numbers. This is the same as the > > number of names for the number "2/9"; which is to say there is a > > bijection between the names for "3/7" and the names of "2/9". > > > > Your thoughts? > > Sorry, I have not the time to develop here the arithmetic of rational > and irrational numbers. There is nothing in CBrown's analysis that is not obvious to a schoolchild of reasonable intelligence, so why has WM not been able to grasp it?
From: Virgil on 2 Jan 2007 17:05
In article <1167738995.894212.208170(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > Why? How should we know which can possibly exist? > > > > > > > > L_D must contain the diagonal. > > > > Thus, L_D must contain any element that > > > > can be shown to exist in the diagonal. > > > > > > It does, after the existence has been shown. > > > > However, before existence has been shown it does > > not. > > Correct. Otherwise the actual existence of all natural numbers would be > implied. It is implied by all sorts of things. In particular, it is required in ZFC and NBG, and WM has produced no system of axioms in which it is not implied. > > > > If L_D contains the diagonal then it contains > > any element that can be shown to be in the diagonal > > Yes. L_D contains any element, that can be shown in the diagonal. Then it contains an element that it can be shown not to contain. > > > whether or not existence has already been shown. > > It must contain the element before existence has > > been shown. > > How should that be possible? With L_D containing all these elements, > their exisence would be deternmined. If the existence of an element not in L_D follows from assuming that L_D exists, then L_D cannot exist. And the existence of an L_D requires the existence of an element not in L_D, so L_D cannot exist. > L_D contains any element that can exist, from the first to the last > one. And there must be in addition an element not in L_D, namely, Max(L_D) + 1. > L_D cannot contain an element which does not yet exist, because the > elements (digits) of L_D are supposed to exist, but they cannot exist, > if they don't exist. Isn't that a triviality? For every L_D, there is a Max(L_D) + 1 in the diagonal but not in L_D. > > I think it is of little value to continue this discussion. You always > implicitly assume the actual existence of all natural numbers. And you always implicitly assume the existence of a largest natural number. > This > idea bears a contradiction. But this contradiction becomes more obvious > when considering the binary tree, which, according to some > correspondents here, is different from the union of all finite trees. To those who do not deliberately blind themselves, the distinction is clear. Every union of finite trees, each regarded as a set of finite paths, will end up as a set of finite paths. To assume that the union of any family of sets will contain any object not in any of the member sets directly contradicts the definition of unions. According to standard definitions of the union of a family of sets, an object is in that union if and only if it is in at least one of the sets in the family, and is not in the union if it is not in any of the set of the family. Does WM dispute this definition? If so, he should provide us with the definition he is using to produce his fantastical results. And unless WM has some definition of union different from everyone else's which he can produce, he is wrong again. |