An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Franziska Neugebauer on 16 Jul 2006 14:59 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > >> > Franziska Neugebauer schrieb: >> > >> >> Once again: There is no column without a 1. /Hence/ every column >> >> *-sums up to 1. >> > >> > Contradiction to A n : n has trailing zeros. >> >> There is no contradiction. For all j e N there is an n e N >> such that a_jn = 1 namely n = j, since a_jj = 1. > > Just that establishes the contradiction. forall n : n has trailing > zeros. A contradiction is a proposition having the shape p & ~p > And the linearity of the list numbers enforces the column without > zeros. A column having "all zeros" exists, if you can name it (by specifying its column number j e N) or if you can prove its existence. You haven't done so yet. Thus there is no evidence, that a column with "all zeros" exists. F. N. -- xyz
From: Dik T. Winter on 16 Jul 2006 18:59 In article <nyAH0KfRjhuEFwtL(a)212648.invalid> David Hartley <me9(a)privacy.net> writes: .... > >> Works, p. 214, translated: > >> The question, through which transformations the ordinal number of a > >> well-ordered set is changed, and through which it is not changed, is > >> easily answered, those, and only those transformations leave the > >> ordinal number unchanged that can be rewritten as a finite or infinite > >> set of transpositions, that is, of interchanges of two elements. .... > Question: Which transformations preserve ordinal number? > > Answer: Those which are realised by a finite or infinite set of > transpositions. > > "Transformation" is not defined here. For that matter, neither is the > action of an infinite set of transpositions, nor is there any proof. If > you (WM) want to use this proposition in your argument, you must fill > all these gaps. There is one thing more that bothers me about the quote. It speaks about "can be rewritten as ... set of transpositions". What is the precise meaning here? As transpositions in general do not commute, I would not use the word "set" unless the set contains only transpositions that mutually do commute. But as this is from letters by Cantor to Dedekind (and not published work, so probably research in progress) we likely never will know the exact meaning. But, whatever, the book with Cantor's letters to Dedekind is in our library, so I will look sometime next week whether more about that has been written. (I think it was Dedekind, but I can check that too.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 16 Jul 2006 19:09 In article <1153051163.295503.224950(a)s13g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > In article <1152978585.858028.263870(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > > In usual mathematics 0.999... is assumed to exist. > > > > > > > > It is not assumed. There is a definition floating around giving a meaning > > > > to that notation. > > > > Without the definition that notation makes no sense. > > > > > > And with that definition it makes no sense either, as the definition is > > > of he same kind as the squared circle. > > > > You do not believe in limits? Otherwise, why does the definition not make > > sense? Quote the definition for precisely that notation: > > 0.999... = lim{n -> oo} sum{k = 1..n} 9.10^(-k) > > what part of that definition makes no sense? > > "n --> oo" because there is no natural number oo. n cannot become oo > whether there are infinitely many natural numbers or not. Each is > finite. Thus, for n e |N, the result is always different from 1.000... > . n -> oo does *not* mean that n will become some oo, because n will not become oo; it only means that n grows without bound. And as you should know how the limit given above is defined, you ought to know that. For each (real) epsilon > 0 there is an n0 such that for n > n0 |1 - sum{k = 1..n} 9.10^(-k)| < epsilon at what place does n become oo? In the above, take n0 = log_10(-epsilon) when epsilon < 1.0, else take n0 = 0. > It is not that I fight against the identification of 0.999... with 1 or > against the identification of the limit of the series of > Gregory-Leibniz with pi/4 for computational purposes. But you fight against the identification for mathematical purposes? > But the assertion > that the digits of these limits could be used to construct a diagonal > number is simply nonsense. Only assertion and meaning. No content. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 16 Jul 2006 19:17 In article <1153052131.852951.273540(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1153039371.156834.8960(a)s13g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Virgil schrieb: > > ... > > > > 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. > > > > > > But the successor of 4 does not consist of the ingredients of 4 only. > > > > Eh? > > The successor of 4 is 5, and 5 = 0.11111 which cannot be indexed by 4 = > 0.1111. I think you are meaning that the digits of 5 can not be indexed by the numbers 1 to 4. That is right. > Hence, > 0.111... as the succesor of all naturals must consist of more 1's, than > any natural, if it is to be a number. (1): 0.111... is not *the* successor of anything, we may sloppily say that it is *a* successor of all naturals, just like 10 is *a* successor of 2 (2): we can safely call that also a number, but it is *not* a natural number, it depends entirely on how we define the concept of number > > > > So the successor of all natural numbers cannot consist of only all > > > natural numbers only. > > > > As stated this makes no sense. But the successor of every natural number > > is a natural number. > > The successor of all naturals is not a natural and, therefore, must be > larger (because it is not less). Yes, it is larger than all naturals, but I would not call it *the* successor, but *a* successor, or, if you wish, *the smallest* successor. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 16 Jul 2006 19:21
In article <+OQB7KjS6iuEFw9k(a)212648.invalid> David Hartley <me9(a)privacy.net> writes: > In message <1153051034.733810.190150(a)b28g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de writes .... > >Obvious is that, if infinite sets do exist and can be exhausted, the > >set of my transpositions does exist and can be exhausted. > > Obvious is, you haven't the faintest idea of how to prove your claims. Mueckenheim thinks that you can take out the elements of a set one by one and can exhaust it that way, that is, there is a point when you take out the last element. In fact, with this thinking he is contradicting the axiom of infinity, and that is the only thing he is fighting. When infinite sets exist, they can not be exhausted. And without exhaustion his proof does not work, so it is not a contradiction within set theory with the axiom of infinity. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |