An uncountable countable set
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From: Dik T. Winter on 14 Aug 2006 23:04 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 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. You have first to *define* what you mean with 10% of an infinite set. As long as I have not seen a definition, I have no idea what it means. Moreover, you used something additional "the first 10%", pray provide a definition. Are the evens one half of the natural numbers? In some sense they are, in another sense they are not. As always you utter terminology without ever giving proper definitions. -- 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 14 Aug 2006 23:25 In article <1155486119.103774.163800(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > 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 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". What you write is simply false and does not follow. > 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? -- 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 14 Aug 2006 23:48 In article <1155486471.373626.282060(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > But an omega-th line, because there is a line omega + 1. > > > > Neither are true. > > > Look here: > > {1, 3, 5,... , 2, 4, 6, ...} > > There is an element number omega and an element number omega + 1 etc. Insidious. What is the relevance to the discussion? The discussion was about {1, 2, 3, 4, ...}. > > No, of course not. It is not the list of Cantor, but the list provided > > to Cantor by somebody else. I hope that the person that provided the > > list did know how the list was enumerated. > > Nonsense. Cantor invented the list and constructed the first one. Quote, please, especially for the latter remark. > > > 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? -- 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 14 Aug 2006 23:41 In article <1155486317.573725.325460(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Yes, so what? Mathematics is there to give meaning to terminology, so that we can talk about things without confusion. > > 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. Makes no sense. What is lim{n -> oo} sum{i = 1 .. n} 1/i^2 ? What is lim{h -> oo} (f(x + 1/h) - f(x)) * h ? When does a limit exist? When does a limit not exist? How do you prove that lim{n -> oo} sum{i = 1 .. n} 1/i does not exist? > 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. 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. > > > 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. Yes, but the each and every path terminates. And there is no terminating path that leads to 1/3. > > > But no edge and no node is terminating a path. > > > > Eh? > > No path terminates. But you stated that a path consists of edges, and that each edge terminates at a node. So I can only conclude that each path terminates at a node. > > > > 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. Makes no sense. As this whole discussion about paths and whatever does not make much sense to me. -- 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 14 Aug 2006 23:56
In article <1155486556.890586.315500(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > > > Every n in N is in the list. K is not in the list. > > > > That is not an answer. Can the number K be indexed or not? > Let me quote (as you did diligently snip): > > EVERY INDEX IS THERE. Every index position is there. It is the set of > > all indexes and of all index positions. But 0.111... is not there. > > Because this number cannot be indexed. > > But when I state "for every n in N, K[n] = 1" can the number K not be > indexed? end quote. > No. If it could, then it was in the list. But "EVERY INDEx IS THERE. Every index position is there.". This holds for K, so why can it not be indexed? > > Yes, so can K be indexed or not? And if not, why not? Please use your > > definition of indexing. I specifically stated that K was that > > representation where each natural number points to a 1, and there are > > no other digit positions. So by your *own* definition it can be indexed. > > But indeed, it is not in the list (as you admit). > > Therefore it cannot be indexed. Every digit position which can be > indexed is in the true list. Pray explain. If I understand you well, a number can be indexed if every digit position of it is in the true list. This is true for K: K[p] = 1 for every p in N. So K can be indexed. But it is not in the true list because it is not a natural number. I still do not see a problem. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |