An uncountable countable set
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From: mueckenh on 13 Aug 2006 12:02 David R Tribble schrieb: > Mueckenh wrote: > >> And there is no definite number aleph_0 of lines. > >> > >> But if there were actually infinitely many, namely aleph_0 lines, then > >> already the first 10 % of lines were infinitely many. And 90 % of the > >> lines were infinitely long. > > > > 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? > > > > mueckenh wrote: > > At what point do you know that you have all aleph_0 lines? > > When your list (set, sequence, list) has a line for every natural. When do you have a line for every natural? > When you can show that every line is denumerated by a natural, How can you show that? I am sure you can only show that for a finite number and, therefore, for a finite number of numbers. > then you know you've got Aleph_0 lines. How do you know that? > > So tell me, where in this list is the 10% point? It is far beyond all your efforts described above can reach. Regards, WM
From: mueckenh on 13 Aug 2006 12:15 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. > 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. Therefore your assertion is correct but void of power to prove that any digit of 0.111... can be indexed. > > 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. > > > 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. > > > What counts (in the true sense of the word) is how many digits a number > > has. And every natural number has only finitely many digits, regardless > > of how many numbers there are. > > I never contradicted that. Why are you always saying that I claim > something which I do not claim? You claimed just above that this fact is not relevant. > > > Please stop your current intermingling of infinity in sizes of numbers > > and number of numbers. > > There is no ifinite size. from hat I conclude that there is no infinite > > set. > > Please stop claiming you found an inconsistency with the axiom of infinity > when your basic assumption already is in contradiction with it. My basic assumption is the existence of this axiom. I *derive* a contradiction. > > > > 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. > . > > > > 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. > Can't you understand that in mathematics it is on occasion very interesting > to compare infinite sets? Of course. For this reason I constructed the binary tree. Regards, WM
From: mueckenh on 13 Aug 2006 12:18 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. >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? Regards, WM
From: mueckenh on 13 Aug 2006 12:21 Dik T. Winter schrieb: > In article <J3uzrx.Apr(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > > In article <1155315217.378759.114420(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > Also you state that indexing is the same as covering. This is the > > same as telling: > > for all p there is an n such that ... > > is equivalent to: > > there is an n such that for all p ... > > which is false. > > > > Back again to quantifier dyslexia. > > For a simple (finite) example where quantifier dyslexia tells the wrong > things can be found at: > <http://mathworld.wolfram.com/EfronsDice.html> > let p and n be (different) dice numbers (from 1 to 4) the first statement > would be: > for all p there is an n such that ... > the second would be: > there is an n such that for all p ... > where ... stands for the statement "the probability of throwing higher > with die n than with die p is larger than the reverse". The first is true, > the second is false. You pick one first, next I pick one, and I have a 2:1 > chance of winning when we throw, regardles of the one you pick. > > When you look around a bit more there are many more interesting examples. > There is even one where the situation is clear with a set of three dice > of different colours, but where it reverses if you duplicate the set > and start throwing with two dice of the same colour. That is all very interesting and very ------------------- misleading. For a linear problem like the unary numbers of the list, there is no quantifier gambling possible. If position n is indexed by a unary number, then also all positions m < n are covered by this number n. There is no outcome in the set of numbers of the form 0.111...1. 1) If a digit is indexed then the sequence up to that digit is covered. 2) If there are no digits which cannot be indexed, then there are no digits, which cannot be covered. 3) That is equivalent to the statement, that all digits can be indexed and all digits are covered. You assert that every digit of 0.111... can be covered, but that 0.111... cannot be covered. That assertion is void of meaning. It makes no sense. And examples of dice or of all women who have men and of a woman who has all men do not apply to this case. Regards, WM
From: mueckenh on 13 Aug 2006 12:25
Dik T. Winter schrieb: > > > > > |{2, 4, 6, ..., 2n}| < 2n > > > > > > > > lim [n --> oo] {2, 4, 6, ..., 2n} = G = set of all positive even > > > > numbers > > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| = aleph_0 > > > > lim [n --> oo] n < aleph_0 (because n is finite) > > > > > > Now, again, you fail to define what you *mean* with n --> oo. > > > > The same is meant as in sequences and series like: > > lim [n --> oo] (1/n) = 0 > > n becoming arbitrarily large, running through all natural numbers, but > > always remaining a finite number < aleph_0. > > Well, that is not a definition at all. The limit of a sequence like 1/n > is defined as: > lim{n -> oo} 1/n = L if for each epsilon there is an n0 such that > for n > n0, |1/n - L| < epsilon. That is a special formalization which was given long, long after the limit had been in use. One of the first limit-users was Archimedes when exhausting the area of the parable. > How do I apply that to the "limits" you are using above? It is always the same: n takes the values of all finite natural numbers, one after the other, without ever becoming infinite. lim [n --> oo] 2n/n = 2 lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 (because omega is a fixed, well defined and well deteremined quantum, according to Cantor) > > I did not change. All the paths are infinite. Every edge starts from a > > node and ends in a node. > > In that case all paths are terminating paths. The number of paths is > countable, and each path is the representation of a rational number for > which the binary expansion terminates. 1/3 is not in the set of paths. ??? Here you see two nodes, a and c, and one edge, b: o a \ b o c These are elements of an infinite path. > > > > > > > The number of edges is twice the number of paths. > > > > > > > > > > Yup, and each path is terminated. > > > > > > > > ? I thought you had understood: No path is terminated. There is no edge > > > > without appending node and no node without appending edge. > > > > > > In the first picture, there is no edge without terminating node. In the > > > second there is no edge with terminating node. > > > > In the naturals there is no even number without a following odd number > > and there is no odd number without a following even number. In the tree > > there is no edge without a node from which it starts and to which it > > leads. > > > > But no edge and no node is terminating a path. > > Eh? No path terminates. Therefore there is no termination of a path, neither an edge nor a node is a terminating element of a path. > > > > What is 1/2 edge? > > > > Think about it. What is a half cake? What s a half horse? When our > > ancestors tried the fractions they may have asked those questions. Do > > you really have problems? > > Oh, I understand it in that sense. 1/2 edge either starts at a node and > does not terminate at a node, or it terminates at a node and does not > start at a node. Do you have that in mind? No. Two half edges give one whole edge as two half euros give one whole euro. If you have 2 euros for every edge then you have not less euros than edges and not less euros than paths. And you are very, very rich. Regards, WM |