Cantor Confusion
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From: Virgil on 28 Jan 2007 14:44 In article <1169986746.552791.97880(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Jan., 20:27, Virgil <vir...(a)comcast.net> wrote: > > In article <1169915856.541369.8...(a)p10g2000cwp.googlegroups.com>, > > > > > The name is not the thing named. > > > > > If there really is a *thing*, this may be true. > > > Even if there isn't, it is true. Because a name which does exist cannot > > be a thing which does not exist. > > But then the name is all that the name describes. "Santa Claus" is a name that does not describe a name. > > > > > Depends on how it is stated. If one's induction is of the form: > > There is a set S such that > > (1) The first natural is a member of S, and > > (2) The successor of every member of S is a member of S > > Then one's conclusion should be that N is a subset of S. > > In fact? If there is a path of lengths n then there is a path of > length n+1. And there is a path of length 1. What is the length of > the union of all these paths (which contains only finite paths)? If that union is to be regarded as a path at all then it cannot be a finite path as it would have to be be longer than every finite path (for every path of length n, the union wold have to be of length at least n+1 ). So its length is not finite. Those of us who do not have coniption fits over the use of the word call that /infinite/. > > > A set is finite if there does not exist any injection from it into > > any of its proper subsets. > > Now apply this to paths. The union of all finite paths allows the mapping of each finite path of length n to a path of length n+1 while giving the same union, so that that union is infinite (in the sense of being not-finite). > > The length of p is the union of the lengths of all finite paths. Which is the first limit ordinal. > > Do you complain? You are familiar with facts that aren't. Only with WM's claimed facts which it transpires aren't. > > > Equivalently, a path in a tree is a maximal sequence of linked nodes in > > that tree with each successive node being the child of its immediate > > predecessor node. > > And this definition, which is equivalent to mine, does not tolerate > different complete sets of paths in one and the same tree. It certainly tolerates infinite paths, and for every infinite binary trees requires them. > > > It is the maximality in its given tree that is essential for a path, as > > anything less is not a path in that tree. > > Correct. Think about it. Thinking about it leads to the inescapable conclusion that infinite binary trees must have infinitely long (endless) paths. > > > Any set which can be listed in its entirety is, by definition cuntable > > and any set which is not so listable is, by definition, uncountable. > > Anything else is irrelevant. > > Where can I see the set N be listed in its entirety? Listing, at least for the purpose of determining countability, is defined as providing a surjective function from the set of naturals to the set in question. The identity function on N, idn: N --> N : x |--> x, is such a surjection, Q.E.D. > > > Cantor showed with two entirely different proofs that the reals cannot > > be listed. So Andy is talking nonsense. > > Everybody not completely mindbended knows that the naturals cannot be > listed too. By the definition of 'listing' as a surjection from N to the set to be listed, the naturals are trivially listable. > > > Already False! There is no (finite) binary index for any of those > > uncountably many reals whose binary expansions require infinitely many > > 1's. For example 1/3. > > There is a finite binary index to any listed sequence of the form > 0.010101 [...] 01 which ever will be written. Every repeating binary IS 'written' as soon as the repetition is written, since it is then represented by a convergent infinite series. > > Regards, WM
From: Virgil on 28 Jan 2007 14:48 In article <1169986808.694831.100210(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Jan., 22:52, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > I am interested in the fact that every set of natural numbers has a > > > finite maximum.Then you should perhaps not talk to contemporary set > > > theorists who are > > accustomed to the > > fixed idea being far from being a > > > fact that there *are* sets of natural numbers which do > > not have maxima at all. In every standard set theory there is a set of all natural numbers, and in no standard set theory does that set have a maximal member. And WM can not presented us with any set theory in which either the set of all naturals cannot exist or in which it can exist but be finite, at least by any reasonable definition of finite.
From: Andy Smith on 28 Jan 2007 14:53 davidmarcus(a)alum.mit.edu writes >On Jan 28, 12:32 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: >> William Hughes <wpihug...(a)hotmail.com> writes > >> >No, the fact that one form of Cantor's diagonalization cannot be used >> >in one particular case does not mean that Cantor's diagonalization >> >is dead and buried. Cantor's diagonalization can be used in >> >many other cases and can be used to show that the reals >> >are not countable. > >> Thanks for that. Understood. But actually when Randy said "I do know >> that the set of infinite binary strings is uncountable, but I know that >> because it is easily proven" I assumed that he meant Cantor's >> diagonalisation - I had a mental equivalence set of real numbers <=> set >> of infinite binary strings, clearly too primitive a perspective. > >Cantor's diagonal argument works for binary strings because you don't >have the problem of two different strings representing the same >number. > Thanks. Yes (I had recognised that). So now suddenly William Hughes post suddenly comes into focus - you can associate a real number with each binary string, and hence the reals are uncountable vi the diagonalisation argument. -- Andy Smith
From: Andy Smith on 28 Jan 2007 14:59 davidmarcus(a)alum.mit.edu writes >On Jan 28, 10:11 am, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > >> Nobody has actually said that I can't attempt to enumerate the reals in >> the suggested fashion (on an assumption that they were countable) - >> although it is clear that doing so as I proposed can only ever index the >> (subset of) the rationals with a terminating sequence of zeroes in their >> expansion ... which (as I previously said) is prima facie evidence for >> the reals being uncountable. > >Of course, you can attempt to enumerate the reals any way you wish. >However, we don't see why the fact that some specific enumeration >fails is evidence that the reals are uncountable. > I had understood that countability or otherwise was independent of a specific arrangement . So if a specific arrangement clearly doesn't allow all the reals to be countable (if they were), then that seems like good evidence for the initial hypothesis that the reals were countable to be false. But I don't want to reiterate old discussions, and I see now (from your and WH's posts) what was wrong with mine. -- Andy Smith
From: Virgil on 28 Jan 2007 15:04
In article <1169986892.031377.245650(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > You claim that if something is true for every set L_n, > > then it is true for N. > > I do not talk about N. This symbol has become the aim of heaviest > abuse. All that abuse being authored by WM himself. > > > > We know something about the maximum that is true > > for every set L_n. > > > > So you do want to prove something about the maximum > > of the set N. > > I want to see whether the union of all finite numbers can be an > infinite number. In models in which natural numbers are sets, unions of sets are also sets, and the union of all such natural numbers, N, is a set. N is clearly not a finite set, nor is it a natural number, but there is no reason it cannot be a number of some other type. We already know of rational numbers and real numbers and imaginary numbers, none of which have to be and some of which cannot be, natural numbers, so what does WM claim prohibits N from being some other-than-natural kind of number? > This question was raised in the framework of the > infinite tree. ". That depends on what one's definition of path is. Set theorists, and others, should decide the issue according to their own definition of "path" If a path is required to have two end nodes then the union of infinitely many paths of different lengths cannot be a path and more the the union of infinitely many different natural numbers can be a natural number. On the other hand, if a path is merely a maximal chain of linked nodes and need not contain any last or end node, then that same union IS a path. Since my own definition is the "maximal chain" definition, for me the union of those infinitely many finite paths is an infinite path. And as WM classifies me among the set theorists, his claim that "Set theorists asserted that a union of finite paths cannot be / contain any infinite path" is clearly false. > > Regards, WM |