An uncountable countable set
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From: mueckenh on 15 Aug 2006 16:38 Dik T. Winter schrieb: > > Every natural number is finite. > > Still makes no sense. But it is a fact (by definition). > > > > > 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) > > > > > > You think so. > > > > No, I do not think so. Cantor did - and you do. > > Not in the way you think. It is the "because" I am objecting to. Is omega a fixed, well defined and well determined quantum or not? > > > > It is a well defined quantum, but that does not mean that > > > you can apply the standard arithmetic operations on it. In order to > > > have division, you need an integral domain. > > > > From a < b you can derive that a/b < 1 without any further definitions. > > How do you define definition on the cardinal numbers? Pray learn a bit > about mathematics. (Sorry, above I said "integeral domain", that should > have been "field".) You need a division ring to have division in general. > The ordinal numbers do not form a field, nor do the cardinal numbers. So > division is not defined on them. Next we can look how division is actually > defined. a / b = c if there is a unique c such that b * c = a. But there > is no such unique c when a and b are aleph-0, so, division is not defined. If a > b, then a has a sequence or subset equal to b. Now you can define a/b = 1 + remainings. ==> a/b > 1. Comparability (trichotomy) of cardinals is sufficient. Do you see your formal objection as the only chance to escape? > > > > > ??? > > > > Here you see two nodes, a and c, and one edge, b: > > > > o a > > > > \ b > > > > o c > > > > These are elements of an infinite path. > > > > > > Yes, but the each and every path terminates. > > > > Only if Cantor's list terminates. > > Why? Because my paths are just as long as Cantor's list. > > Each segment of the number 0.111... terminates: 0.1, 0.11, 0.111,.... > > Do you conclude now that 0.111... terminates? > > No. But you do not get non-terminating paths if you add edges, or levels > of edges, one by one. I do not add but define that the complete tree is there, just as Cantor's list. > > > > Makes no sense. As this whole discussion about paths and whatever does not > > > make much sense to me. > > > > You can divide a euro in 100 cent. You can give one cent to a beggar. > > And if 99 other people do the same, the beggar will carry 1 euro. > > > > No exercise the same with edges and paths and oo instead of 1oo. > > Pray explain. How do I divide something in oo pieces? Consider a line of Cantor's list. Have you? You have just divided something (the list) in aleph_0 pieces and taken one of them. Regards, WM
From: mueckenh on 15 Aug 2006 16:40 Dik T. Winter schrieb: > In article <1155640812.322879.187040(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > Nonsense. Cantor invented the list and constructed the first one. > > > > > > Quote, please, especially for the latter remark. > > > > =DCber eine elementare Frage der Mannigfaltigkeitslehre. > > [Jahresbericht der Deutsch. Math. Vereing. Bd. I, S. 75-78 (1890-91).] > > Do you have an URL? Is it in his "Werke"? Otherwise I have no access > to it. Werke, page 278. In particular the list is on p. 279. > > > > > > > Do you know how the infinite set of algebraic numbers can be > > > > > > enumerated? > > > > > > > > > > Yes. > > > > > > > > And why did you believe that the set of edges of my tree was > > > > uncountable? > > > > > > Because there is an easy proof? > > > > Are you really believing that the number of edges was uncountable? > > Take any edge you like. You can attach a natural number to it. That and > > nothing else is the definition of countability. > > That is not the definition of countability. The definition of countability > is that you can attach a (different) natural number to each and every > edge. Yes, of course. And what is your problem with the edges (how do they differ from the lines of Cantor's list)? Regards, WM
From: Virgil on 15 Aug 2006 17:57 In article <1155674032.400913.53980(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > 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. > > > > (3) is indeed also true if you reformulate to "all digits can be indexed > > and all digits can be covered". > > OK. All digits can be covered. But all digits are never covered. Hence > all digits cannot be covered. Again a new equivalence of set theoretic > logic: > All digits can be covered <==> All digits cannot be covered. That logic does not hold in ZF or ZFC or NBG. One wonders just what axiom system "Mueckenh" is using in which it does hold. Or is he, as we suspect, flying blind? > > > > > 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. > > > > To you, apparently. I did show a proof, what was wrong with the proof? > > You showed it for digit positions only which can be indexed by natural > numbers. > You claimed that all digit positions which are in the true list suffer > to index all positions of 0.111... which is not in the true list. You > constructed an "equivalence" like that one I wrote down above. > > Regards, WM
From: Dik T. Winter on 15 Aug 2006 19:20 In article <1155664930.866986.157410(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Nope. You claim the last, but it is false. Each finite segment can be covered, and each finite index can be covered. But the total is not finite, but contains only finite digit positions. > > > > 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. Again, you keep asserting such things without *ever* providing a proof. > > > > 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. My assertion is that K can be completely indexed but not completely covered. Again you misquote my assertion. > > > > 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. But it is not a natural number. > That is what set theory and the definition of natural number exclude. So? It is not a natural number. Those two do indeed tell that it is not a natural number. So what? > > 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. Why not? Each and every digit position is finite, so can be indexed, and, by *your* definition the number can be indexed. But there are infinitely many digit positions, as there are infinitely many natural numbers. > > > 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. Yes. When you can add all those numbers. But the result is not a natural number. > > > 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)? A set that is not finite is infinite, that is the definition of the word infinite. So yuor question is actually, why are there cardinal number? Well, they come in handy when comparing infinite sets. And (by definition) a cardinal number is the equivalence class of sets that can be put in bijection with each other. A set being in bijection with another set is easily shown to be an equivalence relation, and so we can split the sets in equivalence classes. That is all elementary mathematics. And those equivalence classes are called cardinal numbers. Just as the equivalence classes of Cauchy sequences are called real numbers and the equivalence classes of ordered pairs of integers are called rational numbers (with some particular equivalence relation in mind). All extremely elementary. > > > > 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. You keep asserting that. > 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. Just because there is no last line. If there were a last line, it would have width aleph-0, but there is no last line. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +3120592413
From: Dik T. Winter on 15 Aug 2006 19:26
In article <1155665028.991156.223740(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. There are many real numbers that can not be completed. But by the definition of "real number", each sequence of decimal digits defines a real number, and it is easily shown, by the above, that that real number is not in the list because it is different from each number in the list. (There are indeed also many real numbers that are not computable..., but they are nevertheless real numbers, by the definition of "real number".) > > > > 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. What is the inconsitency? Can you show what axiom of set theory or what proposition from set theory it violates? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |