An uncountable countable set
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From: mueckenh on 11 Aug 2006 12:38 Virgil schrieb: > The "Hilbert Hotel" method allows one always to insert one more and > still have rooms for all. Just this argument shows that the diagonal number of Cantor's list has always reserved a place in the list. there is no uncountable set. Why don't you argue in this direction? Because arguing in set theory is arbitrary and misleading? Regards, WM
From: mueckenh on 11 Aug 2006 12:44 Virgil schrieb: > In article <1155069769.640195.127530(a)75g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > So the edges leading to final nodes > > > are countable. What is the natural number that can be mapped to the > > > final edge in 0.111... ? > > > > There is no final edge. Nevertheless all edges can be enumerated. And > > there are two edges per infinite path. > > So we have infinitely many finite triangles, but without that "final > edge", no infinite triangle. Cantor's list has no final line. Is it finite? Regards, WM
From: mueckenh on 11 Aug 2006 12:46 David R Tribble schrieb: > >> Yup, and aleph-0 is *not* in the inductive set of natural numbers. > > > > Mueckenh wrote: > >> It is not in the set, but the number of the elements of the set is > >> aleph_0. Under this aspect aleph_0 should have some existence in the > >> set, namely as the number of elements. Of course it doesn't. Therefore > >> it cannot be at all. > > > > That statement leads to a contradiction. > > If Aleph_0 is a member of the set of naturals, then it is a natural. I never said that it is a natural. But according to Cantor it is a number which stands in trichotomy with other numbers, in particular with the natural numbers. > But then Aleph_0+1 must then also be a natural. And if Aleph_0 is the > cardinality of N, then Aleph_0+1 cannot be a member of N. > But Aleph_0+1 must be a member of N because it is a natural. > Contradiction: Aleph_0+1 cannot be both a member and not a member > of N. > > Since your statement results in a logical contradiction, your statement > must be false. So Aleph_0 cannot be a natural nor a member of N. I never said that it is a natural. Regards, WM
From: mueckenh on 11 Aug 2006 12:53 Dik T. Winter schrieb: > > There is no natural number oo. Therefore {k = 1 .. oo} is nonsense. > > There is, as you say, a definition for (*+){k = 1 .. n} Ak. More is not > > possible. > > More is possible, when you define what the notion means. No more is possible if naturals are required. > > Because: index = natural number = "number that can be indexed" > > Again you are using a term in a single sentence to mean two different things. No. It is exactly the same. > > > You state so, and remain stating so, without any proof. > > > > It is a definition. > > *What* definition? There is, as far as I know, no definition that states: > A number can be indexed if all its digit positions can be indexed. > so what definition are you alluding to. Pray, for once, be clear. Indexing is equivalent to covering. A number can be covered if all its digit positions can be covered. You state "all digit positions can be covered but the whole number cannot be covered". That is nonsense. > > It is impossible, because index number = number which can be indexed. > > Again, you fail to provide a definition for "can be indexed". In normal > terminology, a number can be indexed if we can assign a natural number > with it in the indexing process. That would be expressed by "enumerating" this number. I chose the word "indexing" in order to focus on the digit positions. > You appear to be meaning something > completely different, namely: a number can be indexed if all its digit > positions can be assigned a natural number to it. Correct. > > > Or he recognized that the first ordinal of a number class may well > > serve as the cardinal of this class. But this topic is not very > > interesting to me because neither of these numbers are numbers. > > So you know acknowledge that your statement: > > On the contrary! Cantor distinguished very clearly in all his written > > papers. > was false? No. In his written papers he distinguished very clearly. His later habit is only told us by G. KOWALEWSKI: Bestand und Wandel, Oldenbourg, München (1950) 202. Regards, WM
From: mueckenh on 11 Aug 2006 12:54
Dik T. Winter schrieb: > In article <1155067888.355942.225340(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > You are claiming that K is in the list of indices. That makes no > > > > > sense at all. > > > > > > > > I am claiming that every index is in the list of indexes. This list is > > > > infinite. There are infinitely many indexes. And, yes, 0.111... is not > > > > among them and therefore cannot be indexed. > > > > > > No proof, yet. K = 0.111... can also be written as "for every n in N, > > > K[n] = 1". > > > > Every number which can be written in this manner is contained in the > > list, because there is every n and every digit which can be indexed. > > Makes no sense at all. Proof please. > > > EVERY INDEX IS THERE. Every index position is there. It is the set of > > all indexes and o 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? Every n in N is in the list. K is not in the list. > To be clear, what do you mean with "can be indexed". You > are using two different, conflicting, definitions in your articles. I mean that a natural number can be attributed to every digit position. Regards, WM |