An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 10 Jul 2006 14:39 In article <1152524049.197170.304760(a)35g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152450210.343596.317970(a)m79g2000cwm.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Cantor's fallacy is the claim the indexation of list numbers and > > > diagonal digits carries over to the limit. > > > > What limit is that? > > n --> oo On the contrary , Cantor has a simple rule that applies for all n in N, entirely and at once. No limit processes requires. > > > > All Cantor assumes is that if two reals differ in some place in > > restricted certain ways (to avoid the double representation problem of > > certain reals) then they are not equal as reals. > > But this double representation problem shows that differing by digits > is not sufficient a criterion to differ by size in the limit n --> oo. The differences may be small, but never zero. And any difference greater than zero is enough. > > > > That does not require any "carry over to the limit" assumption. > > All I assume is that if two rationals differ in the well-order then > they are not equal by size and can be ordered by size. But, as I have shown in another post, "mueckenh"'s algorithm does not sort the rationals into their natural order. In fact, it does not sort them by size at all, it merely shuffles them around without regard to size.
From: Virgil on 10 Jul 2006 14:47 In article <1152524415.524476.303070(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > According to ZFC there is an infinite set of natural numbers which can > > > be exhausted. This set is the set of indices of my transpositions. > > > > But it cannot be "exhausted" by sequential operation. > > Then the set of the algebraic numbers cannot be exhausted, i.e., they > cannot be well-ordered? An infinite set can only be well ordered by establishing a finite set of rules which define the ordering unambiguously. This has been done for algebraic numbers. > > Give me a rational number of the well-order which cannot be put in the > order by magnitude with its predecessors. If you can, then you are > right. I you cannot, then you are wrong, as usual. Give me an algorithm which will, for an arbitary positive natural n, put the first n elements of a list of numbers into increasing numerical order, regardless of their original ordering. Until you can do this, it is nonsense to suppose you can do it for an infinite list.
From: Virgil on 10 Jul 2006 14:50 In article <1152524704.821204.5090(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152450354.585870.85210(a)b28g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > But it reaches past every finite value!!! > > > > > > > > > > There is always a finite value past every finite value. So we are and > > > > > remain sufficiently save within the domain of finit values. > > > > > > > > So one never reaches an end. That is what "infinite" for a sequence like > > > > the naturals means, "without end". > > > > > > Correct. And that is valid for Cantor's list too. > > > > Except that with Cantor's method, one need not proceed through the list > > sequencially, but can deal with them all simultaneously, which with > > WM's transpositions is impossible. > > Wrong. In order to determine the n-th line, one must count from 1 to n. Cantor does not claim to be able to provide any such list. And it is the provider of the list who must do that. Cantor only says that when someone provides him with that list, that he can find a number not listed.
From: Virgil on 10 Jul 2006 14:56 In article <1152524952.938645.11070(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > And, in fact, all of the Cantor comparisons are proved simultaneously. > > How many parallel calculators in how many parallel universes are run by > the fairies under your bed? When a rule is stated in a form which applies equally well for all natural numbers, then it is either valid for all or invalid for all. When one writes that n + n = 2*n for all n in N, one is not required to prove it for any individual values of n when one can prove it for an arbitrary n. That is one of the many facets of mathematics that "mueckenh" apparently does not grasp: general rules.
From: Virgil on 10 Jul 2006 15:09
In article <1152525125.467445.43240(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152450704.938408.69690(a)h48g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > What we have said, and what is quite true, is that for every natural > > > > there is a larger natural. > > > > > > Of course. And therefore it is impossible to exhaust all of them or to > > > find a set which is larger than all naturals together. > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > > > > Except in such systems as ZF, ZFC and NBG where there is an axiom > > requiring that such sets exist. > > > > And until WM or someone else, can find some internal > > contradiction within one of these systems, they remain eminently usable, > > and will be the most common standards on which set theories are based. > > There is no axiom that uncountable sets exist. But as a consequence of the axioms that do exist, one can deduce their existence. > In the contrary, it can > be shown that, given ZFC is free of contradictions, it must have a > countable model. This is a result of Skolem, who, therefore, argued > that all the evidence that had been given for the existence of > uncountable sets was inconclusive. But in countable models of ZF, there are sets, S, for which no bijection from N to S exists --> within that model <-- !!! http://www.ltn.lv/~podnieks/gta.html One has to go outside the model. |