Cantor Confusion
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From: Ross A. Finlayson on 22 Nov 2006 03:42 Han de Bruijn wrote. Hi Han, Some time ago, we were in a discussion about sampling the natural integers at random. I wonder about this: biject the unit interval of real numbers, of which it is said a uniform distribution exists, to the elements of NxNxNx..., N^N, the sequences of natural numbers. Sample, discarding samples that don't have only and exactly one of each (finite) element in the naturals. Then, of one of those, select the first element and it's a natural integer at uniform random. What do you think about that? Basically it says that if the reals and set of choice functions of the naturals is equivalent, then the natural integers can be uniformly sampled. Basically talk about well-ordering the reals, cardinals between N and P(N), the continuum hypothesis, functions on the naturals that are not real functions, and staccato pointillism introduce some complexities to the consideration. Thanks, Ross
From: Ross A. Finlayson on 22 Nov 2006 04:10 MoeBlee wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > The countability of the set W of all the words of a finite alphabet is > > proved by other means than the usual countability proof. The latter > > consists in constructing a bijection with N, i.e., in constructing a > > list. But it is impossible to construct a list of all words. It is > > impossible to construct a bijection between W and N. > > WRONG. It is a theorem that from the set of finite sequences of members > of a finite alphabet there is an injection into N. > > > Why should it be > > possible to construct a bijection between N and N? > > Because ANY x is 1-1 with x, by the identity function on x. > > > > > Note however that in order to attach a definite cardinal number to > > > > every set, well-ordering must be possible for every set. That is why > > > > Cantor insisted on well-ordering of all sets. > > > > > > Piffle. Well-ordering has nothing to do with the existence or not of > > > at least one bijection. > > > > > You are wrong. If a set cannot be well-ordered, it cannot be assigned a > > definite cardinal number. In order to assign a cardinal, at least one > > bijection to an ordinal is required. This requires at least one > > well-ordering. > > WRONG. By using Scott's method (which requires the axiom of > reqularity), we don't need the well ordering theorem to prove that > every set has a cardinality. > > MoeBlee Hey Moe, It does, then. There are reasons well-ordering is called a principle. So, if every set has a cardinal then it has an initial ordinal and every set is well-ordered? I can see why you would prefer that to non-trichotomy in the cardinals, because the lack of trichotomy would erase the meaning of those transfinite cardinals, they could not be distinct or transitive loops would exist and they could not be distinct. I have some suggestions how you can argue against that. I also know why. You might say collections with cardinals aren't sets, they generally are, collections defined by their elements. So, is choice not a dependent axiom? If cardinals are trichotomous they're ordinals. Fraenkel, Bar-Hillel, and Levy have a few interesting paragraphs to the effect of considering the _existential character_ of the "axiom." Wolfgang, I thought it was funny when you were translating Georg into English and described his ruminations of counting BACKWARDS from infinity. Today, you see, omega's a limit ordinal and can't, there is no mechanism for the status quo to do that. Another thing I thought was interesting was a quote about sets of numbers being basically ignorant of the soi-disant numbers they contain, where the translation in the English edition instead put it in terms that they were "abstracted." Translations definitely show the translater. The universe is infinite, and that's an example that infinite sets are equivalent. A counterexample to the powerset result is everywhere. Look at infinitesimals, like iota or dx, with checks and balances, sum them to integrate. With warm regards, Ross
From: Eckard Blumschein on 22 Nov 2006 05:31 On 11/22/2006 9:22 AM, Han de Bruijn wrote: > Eckard Blumschein wrote: > >> To what extent criticism by WM might be justified? >> Let me tell you my position: WM is most likely wrong if he sees the >> finished infinities the fundament of modern mathematics. > > Quite on the contrary. > > Eckard, please learn more about Wolfgang Mueckenheim in the first place: > > http://www.fh-augsburg.de/~mueckenh/ > > Wild guess: his ideas will turn out to be not so uncomfortable with you. Hallo Han, First of all I would like to urge you for restricting the discussion to sci.mat by removal of de.sci.mathematik. Nice to meet you again. I am also glad that you are trying to defend WM. I benefitted a lot from his booklet "Die Geschichte des Unendlichen" which was sent to me by him personally. Presumably you got me wrong concerning what I consider the fundament of modern mathematics. I consider set theory a toxic coat of modern mathematics rather than its genuine fundament. Modern mathematics is inconceivable without the contributions e.g. by Thales, Eudoxos, Pythagoras, Archimedes, Euclid, Fermat, Vieta, Descartes, Galilei, Newton, Leibniz, Euler, Bernoulli, Laplace, Lagrange, Fourier, Hermite, Lindemann, Gauss, Cauchy, Kronecker, Poincar?, Shannon, Kolgomoroff, and even Bill Gates. Dedekind and others contributed an illusion necessarily leading to endless quarrels. WM has to adapt too much to present mathematical terminology. When he confused Paul with Emile this was an excusable mistake to me. However I cannot follow him completely. Kind regards, Eckard
From: mueckenh on 22 Nov 2006 05:46 Randy Poe schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Randy Poe schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Randy Poe schrieb: > > > > > > > > > > > > > > I know that at most 10^100 digits of sqrt(2) can be determined, in > > > > > > principle. > > > > > > > > > > In principle, if a is the sqrt(2) to 10^100 digits, then > > > > > 0.5*(a + 2/a) is the sqrt(2) to 2*10^100 digits. > > > > > > > > > > What "in principle" prevents me from calculating 2/a, > > > > > or adding it to 1, or taking 0.5*(a + 2/a)? > > > > > > > > Lack of bits to represent 2/a if a already requires all bits available. > > > > > > That means it's hard in practice. > > > > > > But what prevents 2/a from existing in principle? > > > > It is not "hard in practice", but it is impossible. There is no chance. > > That means "in principle". > > No, that is not the meaning of "in principle". It is. What you mean is "in the magical realm of transfinity-worshippers" Regards, WM
From: mueckenh on 22 Nov 2006 05:47
William Hughes schrieb: > No. You can develop "potential set theory". Just define > an element of a potential set to be an element of any one > of the arbitrary sets that can be produced. Now go through > set theory and add the word "potential" in front of each > occurence of the word set. There is no difference between > saying "a potentially infinite set exists" and saying "an actually > infinite set exists". A potentially infinite set has no cardinal number. It cannot be in bijection with another infinite set because it does not exist completely. Regards, WM |