Cantor Confusion
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From: mueckenh on 12 Oct 2006 13:48 briggs(a)encompasserve.org schrieb: > Waving the magic wand, parameterizing by t and appealing to an intuitive > notion of "if all steps of a process are defined it follows > that the outcome is uniquely defined" is a poor substitute. That notion > turns out to be false. That notion is not at all different from the notion pi = a_n*10(-n). Only the consequences are clearer in this case. > If someone is willing to contemplate the one and not the other then > you're not dealing with the mathematical part of the problem. You're > dealing with delusions of physicality. The way out of that morass is > not to scale things so that our intuitions are satisfied. It is to > define things clearly enough that our implicit intuititive assumptions > need not be silently invoked. There is no intuition involved if we find that lim {t-->oo} t > 1. This can be obtained from lim {t-->oo} 1/t < 1. It is just pure mathematics. Regards, WM
From: mueckenh on 12 Oct 2006 13:50 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > It is not > > > > contradictory to say that in a finite set of numbers there need not be > > > > a largest. > > > > > > There is no such thing as a number with arbitrary size > > > (i.e. a number that has the property that its size is > > > arbitrary). > > > > The number I will write down here has that property, before I write it > > down. > > > > 7 > > > > Now it has no longer this property. > > > > > > The prime number to be discovered next has the property to be unknown. > > - Until it is discovered. > > You are confusing "unknown" with "arbitrary". They do > not mean the same thing. I understand by "arbitrary" just what you understand: to be determined at will. If I write down "7" so it was arbitrary, just my choice to do so. > > > > > > > > Take a finite set B. > > > > > > By definition, B must have a finite > > > number of elements. This > > > number does not have arbitrary size. > > > > > > Each element has a size This size is not > > > arbitrary. > > > > > > So we have a finite number (which does not > > > have arbitrary size) > > > of elements, each of which does not > > > have arbitrary size. > > > > > > Thus B has a largest element. > > > > What is the largest number of the set of numbers mentioned in the New > > York Times in 2007? > > "unknown" but not "arbitrary". correct, unless I pay an advertisment posting that number. > > > You see, your proof is rubbish. B will have a largest element. And the > > set of all numbers ever used in the universe in eternity also will have > > a largest element. But it has not yet. > > Therefore it is unknown. However, it is not arbitrary. The largest element possible with 100 bits can be very different, according to my arbitrary choice of representation. Regards, WM
From: mueckenh on 12 Oct 2006 13:54 Dik T. Winter schrieb: > > But this has > > been forgotten by Cantor whose diagonal proof attaches the same weight > > to every digit. > > When comparing for equality each digit has equal weight. Otherwise there > would be no trichonomy. (Or do you claim that some numbers are more equal > to a given number than others? Or that some numbers are more unequal to > a given number than others?) For infinite sequences we have undefined index positions. Why should we need the factors 10^(-n) otherwise? > > > Or it could also be applied to > > the left of the decimal point, constructing an infinite natural number > > and showing that every list of natural numbers is incomplete. > > It can be done if you use some metric where the sequence 1, 11, 111, ... > converges. Why does it not converge without special precautions? If every digit of an infinite sequence had the same definiteness, then convergence against a certain value would be guaranteed. No. Infinite sequences can only converge by some special measures which pull the weight of most digits to zero. > And indeed, it can be done in the n-adics, where such > sequences indeed do converge, and the result is 1/(1-n). And *indeed* > you can show (with the diagonal proof) that the n-adics are *not* countable > for any n. With the diagonal proof you cannot show anything for infinite sets. Regards, WM
From: mueckenh on 12 Oct 2006 13:57 Dik T. Winter schrieb: > In article <1160646886.830639.308620(a)c28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > If every digit position is well defined, then 0.111... is covered "up > > to every position" by the list numbers, which are simply the natural > > indizes. I claim that covering "up to every" implies covering "every". > > Yes, you claim. Without proof. 1) "Covering up to n" means 2)"covering n" and "covering the predecessors of n". Therefore we need not prove (2) if (1) is true. > You state it is true for each finite > sequence, so it is also true for the infinite sequence. It is true *for every finite position*. I do not at all care how many such positions there are. The obvious covering of (2) by (1) does not depend on frequency. > That conclusion > is simply wrong. Why should it? Have you some argument behind your words? Regards, WM
From: mueckenh on 12 Oct 2006 13:58
Dik T. Winter schrieb: > In article <1160647755.398538.36170(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > It is not > > contradictory to say that in a finite set of numbers there need not be > > a largest. > > It contradicts the definition of "finite set". But I know that you are > not interested in definitions. We know that a set of numbers consisting altogether of 100 bits cannot contain more than 100 numbers. Therefore the set is finite. The largest number of such a set cannot be determined, as far as I know. Could you determine it? Or would you prefer to define that such ideas do not belong to mathematics? Then I would not be interested in that definition. Regards, WM |