An uncountable countable set
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From: mueckenh on 15 Aug 2006 13:55 Virgil schrieb: > > "Every digit of n can be indexed" > And can be 'indexed" by using only those digits whose position number > is a prime, leaving most of them unused as indices but still indexing > all of them. > > > > This is a *symmetric* relation. > > > Not at all. when one turns it about, only those digits in prime > positions are indexed leaving infinitely many un-indexed. You talk about enumerating the elements of a set and, as usual, you have not grasped the essential point of indexing. Regards, WM
From: mueckenh on 15 Aug 2006 13:58 Virgil schrieb: > In article <1155067333.259873.193700(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > Again, you are arguing with a particular model in mind. But when you > > > *do* argue in a particular model, pray do it correctly. Especially your > > > sentence: "Each union of finite segments is a natural number" is false. > > > > The v. Neumann model of natural numbers is just the model of segments. > > n = {0, 1, 2, ..., n-1} > > Every natural number is finite. > > Hence every union of finite segments is a finite segment. > > Nor quite. Every FINITE union of finite segments is finite, but infinite > unions need not be. In fact the union of all naturals in NBG equals the > set of all naturals, and by the axiom of infinity does not have any > largest member. But, by definition, EVERY member is finite. > > > You don't like set theoretic models if they are uncomfortable for your > > current arguing? > > Every model of NBG contradicts "Mueckenh"'s arguing, so he must be > infinitely uncomfortable with it.. > > > > Again, a model, with an unfounded statement. I was talking about sets, > > > not about specific models of natural numbers. > > > > The v. Neumann model of natural numbers is just the model of segments. > > n = {0, 1, 2, ..., n-1} > > Every natural number is finite. > > Hence every union of finite segments is a finite segment. > > "Mueckenh" repeats what is false in NBG (the von Neuman-Bernays-Goedel > model). The union of all von Neumann naturals is not finite. Which is the first infinite union? Regards, WM
From: mueckenh on 15 Aug 2006 14:02 Dik T. Winter schrieb: > In article <1155485742.529974.141050(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > Further: Indexing the digit number n is equivalent to covering the > > > > string up to digit number n. A finite string can never cover an > > > > infinite string. > > > > > > But *that* is irrelevant. > > > > Wrong. > > Why is it wrong Because it is very relevant. You state that an infinite string could be indexed by finite strings. Complete indexing includes covering. > > > > Consider K = 0.111... . For each n we can > > > index digit number n of K and so cover K up to digit number n. > > > > Of course you can index each n. But your "each n" stems from the true > > list. And we know that 0.111... is not in the true list, because it is > > distinguished from any element of the true list. > > Pray read what I state. Again, what you state is irrelevant. > > > Therefore your assertion is correct but void of power to prove that any > > digit of 0.111... can be indexed. > > Eh? This tells me exactly nothing. My assertion "each digit of K can be > indexed, but K can not be covered" is void of power to prove that any > (why any here?) digit of 0.111... can be indexed? If each digit of K > can be indexed, any digit of K can be indexed. Then any digit of K can be covered by a member of the true list. K is nothing else but all its digits. Then K is in the true list. > > > > > Don't forget: The infinite set of natural numbers contans only natural > > > > numbers, i.e., finite strings of 1's in unary representation. > > > > > > Yes, but there is no bound on the index positions, and hence the number > > > of index positions is infinite. > > > > The number of digit positions is irrelevant. The digit positions are > > all finite - in the true list. All numbers with merely finite digit > > positions are in the true list. 0.111... is not there. > > Yes, I never said otherwise. Why do you always come back with this > statement that I do not contradict? Because your assertion that K could be completely indexed and covered is equivalent to K being in the true list. > > > > > Why can't you learn that the > > > > asserted infinity of the number of numbers is completely irrelevant. > > > > > > That is just opinion. > > > > That is not opinon. In oder to index or to cover, only the digit > > postions of the numbers are relevant. It is completely irrelevant how > > many othe numbers are there. Because we always consider only one > > special number. > > Again, makes no sense to me. *If* there are infinitely many natural > numbers, there are also infinitely many digit positions. An infinite many of digit positions would result in an infinite number. That is what set theory and the definition of natural number exclude. > To cover a > number by a natural it is required that the number to be covered has > only finitely many digit positions. So a number with infinitely many > digit positions can not be covered by a natural number. Likewise a number with infinitely many digit positions can not be indexed by a natural number. > > > > My basic assumption is the existence of this axiom. I *derive* a > > contradiction. > > You do not. In all your derivations somewhere you assume that that axiom > is false. The axiom of infinity states that there are infinitely many numbers, hence infinitely many differences of 1. It is obvious that an infinite number of differences of 1 leads to an infinite number. I need not do more than show this implication. > > > > > > But if you claim that the set if finite numbers is finite, there is > > > > > a largest number. > > > > > > > > There is no largest number. The set of of natural numbers is infinite, > > > > > > Eh? Above you wrote that there is *not* an infinite set. Contradicting > > > yourself? > > > > Important: There is no largest number. The axiom says only that the set > > is infinite. > > I said, there is no largest number. The axiom says the set does exist. In > and of itself the axiom does *not* state that the set is infinite. It is > a theorem that the set is not finite, and so must be infinite. Why should a set which is not finite be actually infinite (i.e. have a cardinal number)? > > > > > There is no infinite natural number. Because there is no infinite > > > > finite number - and every natural number is finite. Even the axiom of > > > > infinity does not state the contrary. (But if the axiom of infinity had > > > > to be satisfied, then a finite infinite number would be required. > > > > Compare the staircase.) > > > > > > Again an assertion, without proof. > > > > The staircase is a proof. > > No. You may deny this, nevertheless it is true. If there is an actual number of steps, then the width is equal to the lenghts is aleph_0. What I do not understand is how you can believe that someone (except, perhaps, Virgil) could share your opinon that only in one dimension the infinity was actually realized, but not in the other. Regards, WM
From: mueckenh on 15 Aug 2006 14:03 Dik T. Winter schrieb: > In article <1155485913.329014.192160(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1155242105.069297.90260(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > David R Tribble schrieb: > > > ... > > > > > What is 10% of Aleph_0? If you start counting your lines, at what > > > > > point do you know that you've counted the first 10% of them? > > > > > > > > At what point do you know that you have all aleph_0 lines? > > > > > > When you start at 1, at no point. > > > > So Cantor's diagonal is never completed. We are never sure that it is > > not in the list. > > Nobody ever has said that the diagonal is ever completed. There is a > definition of that number, and it is easy to show, with that definition, > that that number (when completed) would be different from all numbers on > the list. But if it is impossible to complete it, then the result is void. > > > > But the axiom of infinity asserts that > > > the set of all aleph-0 lines does exist. I think you are asserting a new > > > axiom: 10% of the lines does exist. > > > > I think that 10 % must exist if the whole set = 100 % does exist? No? > > No. Here you see the inconsistence of set theory. The set shall exist actually = completely, but the first 10 % shall not exist. Regards, WM
From: Virgil on 15 Aug 2006 15:29
In article <1155484846.128106.136770(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > At what point do you know that you have all aleph_0 lines? > > > > When you have them all. > > That means never, because at least the last one does not exist. In > fact, there are some more missing, because not more than 10^100 can be > enumerated. > > Regards, WM "Mueckenh" believes that his imagination is constrained by the possibilities of the physical world. Mathematicians minds are not so constrained. |