An uncountable countable set
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From: Dik T. Winter on 15 Aug 2006 00:00 In article <1155486885.745105.193680(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > Every union of finite segments is a finite segment. "finite" does not > > > apply here because there are *infinitely* many finite numbers. > > > > Just because there are infinitely many finite numbers, the word "finite" > > *does* apply. An infinite union of finite segments is *not* necessarily > > finite. Where do you find that an infinite union of finite segments is > > finite? Prove it. > > By definition each natural number is finite. As v. Neumann's model > gives every natural number but nothing else, every member of the set is > finite. Yes. So what? Where do you find that an infinite union of finite segments is finite? Are you not able to find that? > > > many. Hence we can unify infinitely many sequences. Nevertheless every > > > natural number is finite. > > > > Yes. But if we unify infinitely many sequences the result is *not* > > necessarily finite and *not* necessarily a finite number. Consider the > > sequence of finite sets {}, {0}, {0,1}, {0,1,2}, ... ; there are infinitely > > many of them. > > Relevance? > Hint: > You keep on intermingling number of numbers and sizes of numbers. > Every number of the infinitely many numbers is a finite number. That > alone is important. Says who? You state that an infinite union of finite sets is finite. I ask you for a quote or a proof, and you refrain to give some. Are you not able to either prove that or give a quote? -- 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 15 Aug 2006 07:15 Dik T. Winter schrieb: > > > Yes. But if we unify infinitely many sequences the result is *not* > > > necessarily finite and *not* necessarily a finite number. Consider the > > > sequence of finite sets {}, {0}, {0,1}, {0,1,2}, ... ; there are infinitely > > > many of them. > > > > Relevance? > > Hint: > > You keep on intermingling number of numbers and sizes of numbers. > > Every number of the infinitely many numbers is a finite number. That > > alone is important. > > Says who? You state that an infinite union of finite sets is finite. > I ask you for a quote or a proof, and you refrain to give some. Are you > not able to either prove that or give a quote? It is the definition of a natural number that it is a (positive) finite number. Regards, WM
From: mueckenh on 15 Aug 2006 07:18 Dik T. Winter schrieb: > 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? In the true list. That is the definition of the true list. > > > > 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. Yes. > 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. The largest index position required is given in the left column of this list 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. Regards, WM
From: mueckenh on 15 Aug 2006 07:20 Dik T. Winter schrieb: > > Nonsense. Cantor invented the list and constructed the first one. > > Quote, please, especially for the latter remark. Über eine elementare Frage der Mannigfaltigkeitslehre. [Jahresbericht der Deutsch. Math. Vereing. Bd. I, S. 75-78 (1890-91).] > > > > > 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. Regards, WM
From: mueckenh on 15 Aug 2006 07:23
Dik T. Winter schrieb: > > > > 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. Every natural number is finite. > > 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. > 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. > > > > > 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. Only if Cantor's list 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. Each segment of the number 0.111... terminates: 0.1, 0.11, 0.111,.... Do you conclude now that 0.111... terminates? > > > > > > 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. 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. Regards, WM |