An uncountable countable set
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From: Dik T. Winter on 15 Aug 2006 19:45 In article <1155674032.400913.53980(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Again: all digits can be covered means, for each digit position there is an n such that the digits upto that digit position are covered by n (and this is the same as all digits can be indexed) but that does *not* mean: there is an n that covers all digits. > Hence > all digits cannot be covered. All digits can be covered, but there is no n that covers all digits. > > > 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. But *all* digit positions can only be indexed by natural numbers. There are no digit positions that can not be indexed by natural numbers. Again, K is defined as the (not-natural) number such that for each natural p the p-th digit of K is 1, and there are no other digit positions. But I think there is a misunderstanding here. In English (I think) the sentence "every digit of 0.111... can be covered" means for every digit you chose there is a covering. And the same is true (I think) for the sentence "all digits of 0.111... can be covered" (although the latter is a bit ambiguous). They do *not* mean, "there is a covering of all digits". > 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. Not so. But to be clear (and disambiguate the matter): 1. Define K as above. 2. For each digit there is a p in N such that that digit can be indexed. 3. For each digit there is a p in N such that that digit and the preceding ones can be covered. 4. There is no p in N such that K is covered. (2) and (3) are equivalent, but you continue stating that (2) is equivalent to: 5. There is a p in N such that K is covered. But that one is false and does not follow. It would be better if you started to define the terms you are using. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Aug 2006 20:10 In article <1155674284.887539.128610(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > Every natural number is finite. > > > > Still makes no sense. > > But it is a fact (by definition). You snipped context. It made no sense in the context (and was not an answer to my question). My question was: > How do I apply that to the "limits" you are using above? your answer was: > It is always the same: n takes the values of all finite natural numbers, > one ofter the other, without ever becoming infinite. And my question is unanswered. When I try to extract a definition from that, I find: lim{n -> oo} 1/n > 0. because 1/n never becomes 0. > > > > > 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? As I said (pray read), it is the word "because" I objected to. And I stated, quoted just below, that it *is* a well defined quantum. So why ask for something that I have already answered days ago? > > > > 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. Let us try it with: > > > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 let's say a_n = {2, 4, 6, ..., 2n} and b_n = {1, 2, 3, ..., n}. What is the subset equal to b_n in a_n? What are the remainings? > Do you see your formal objection as the only chance to escape? Nope. You use "because" in your terms, and *that* is the word I objected to. I ask you to justify that word. > > > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 it is much simpler than that. For each n the quotient is 1, and so, using standard terminology for limits, it is also 1 in the limit. This is not > > > > > (because omega is a fixed, well defined and well deteremined > > > > > quantum, according to Cantor) (and neither does it prove that aleph-0/aleph-0 can be defined and is 1, yes, I prefer aleph-0 when talking about cardinalities). > > > 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. Ok, but in that case you never have finite trees. > > > 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. This makes absolutely no sense. > You have just divided something (the list) in aleph_0 pieces and taken > one of them. I can also start with a pair of lines, a triple of lines, a quadruple of lines, or whatever fininite number of lines I wish to take. In all those cases I have split the list in aleph_0 pieces and taken one of them. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Aug 2006 20:32 In article <1155674426.389672.172370(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. That is not a specific list, but an arbitrary list. I hope you understand the difference. This is a standard start of a proof by contradiction. Assume an arbitrary object (not construct, but assume), and show that it does not fulfill certain requirements. This proves that no such object can have the "certain requirements". Anyhow, that assumed list can be enumerated by definition. > > > 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)? The lines are countable (by assumption). But pray assign natural numbers to each edge in your tree. I will even allow you to assign finite sequences of digits 0 (go left) and 1 (go right) to your tree (the mapping to natural numbers is quite standard in that case). What is the finite sequence of digits assigned to the edge that leads to 1/3? You claim that the edges are countable, so you should have an answer to my question. And before you mutter: I just assign 0 to it and change all other assignments so that it fits. Be prepared that I will ask you to give such a sequence for another number. I will not tell you which, but if your mapping does not map that other number to a finite sequence, your mapping is shown incomplete. And this works *regardless* the mapping you are using. Your mapping really should contain each and every real number before I consider further. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Aug 2006 21:48 In article <J42BHx.IzE(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <1155664930.866986.157410(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > 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. Let's have an easier example. The blocks are 1/2^n high and 1-1/2^n wide. When you make a stair of them, when you complete, you have a stair with height 1 and width 1. But there is neither a block that is 1 wide, nor a block that is at height 1. On the other hand, for every positive k, there are blocks that are beyond the height of 1-k and beyond the width of 1-k. Both in height and in width the blocks just do not reach the boundary line. And note that in both "infinity is reached". This is pretty similar to your stair where you have a stair with a height and width of aleph-0, but where neither are actually reached by any objects. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 16 Aug 2006 06:41
Dik T. Winter schrieb: > > > > 1 0.1 > > 2 0.11 > > 3 0.111 > > ... > > > > which contains all digit positions which can be indexed - by > > definition. > > > > 0.111... is not in the list. > > Indeed, it is not in the list. But K can be indexed. The problem you > appear to have is that K does not have a largest index position. The list does not have a largest number either. There is no largest index position which could be indexed by the numbers of the list. Nevertheless the number 0.111... has positions which are larger than all positions of numbers in the list. Otherwise 0.111... would be in the list. Regards, WM |