An uncountable countable set
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From: Dik T. Winter on 15 Aug 2006 09:52 In article <1155640559.355146.166090(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > 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. What is the relation with your statement that "an infinite union of finite sets is finit"? Where is the *proof* of that statement? 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: Dik T. Winter on 15 Aug 2006 09:56 In article <1155640712.772073.24850(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. The definition of the true list is that it can enumerated. Some days ago you agreed that that is something different. Still I ask you, For each p in N, K[p] = 1 and there are no other digits. So *every* digit of K can be indexed, and so by your definition K can be indexed. What is false about this statement? > > 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. 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. -- 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 10:03 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. > > > > > 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. -- 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 10:18 In article <1155641022.427146.293090(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Still makes no sense. > > > 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. > > 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. > > > ??? > > > 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? > > 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? No. But you do not get non-terminating paths if you add edges, or levels of edges, one by one. > > 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? -- 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 13:46
Virgil schrieb: > > An infinite sum of 1's is not infinite? > > "Mueckenh" again proves that he cannot read. It is false and stupid to > say that an infinite sum being infinite prohibits infiniteness. Don't misquote me. It prohibits infiniteness of the set of natural numbers. > But > saying that in no way implies that an infinite sum of 1's is not > infinite. But it is not a natural number. Regards, WM |