An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Dik T. Winter on 27 Jul 2006 19:54 In article <1153948709.115674.15190(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > An arbitrary union of finite sequences is not necessarily a finite > > > > sequence. This is similar to: the union of finite sets is a finite > > > > set. But this argues that the union of {1}, {1, 2}, {1, 2, 3}, ... > > > > is a finite set. Consider the elements of those sets as indices to > > > > the elements of the sequences. This is only true when N is finite, > > > > but that is the negation of the axiom of infinity. .... > I meant unions of finite segments like {1}, {1, 2}, > {1, 2, 3}, ... remain a finite segment. Because we can identify tem > with natural numbers which always remain finite. Pray keep in the system where you are trying to prove an inconsistency. When we take a finite sequence of sets you are right. But I take the sequence of all such sets, which I can do by the axiom of infinity. That you can identify them with natural numbers is a red herring. In the first place: The union of a collection of sets is defined as the set consisting of all elements that occur in at least one of the collection of sets. So we have the (infinite) collection of sets {1}, {1, 2}, ..., and the union of them consists of all elements that occur in at least one of the sets of that collection. As for each natural number there is a set in that collection that it occurs in, all natural numbers do occur in at least one set from the collection, so it occurs in their union. Also there is no non-natural in any of the collection of sets, so their union does not contain a non-natural. Hence the union is the set of all naturals, which is infinite. This all because of the axioms used, in which the axiom of infinity plays a dominant role. And if you want to refute the definition, I may note that you did something very similar with your *+ sum if you look at it at each column, which I did allow (for each column). And note also my article where I did show how you could do it if you regarded your things just as strings. -- 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 4 Aug 2006 16:06 Dik T. Winter schrieb: > Let's state that more proper. The index of every digit that can be > indexed is in the true list (of natural numbers). But later you use > the word "index" to mean something completely different. You are > stating there that if every digit of a number can be indexed the number > *itself* is in the list. But that does not follow. I does. Indexing, or let's say "indexibility", is a symmetric relation: If a digit position of a number like 1/9 = 0.111... can be indexed by a list number, then also the list number can be indexed by the corresponding digit position of 1/9. If 1/9 is not in the list but can be indexed completely by list numbers, this symmetry is broken, because: What means "not in the list"? It means that 1/9 has more 1's than every list number. > You need a proof > for that. And there is no such proof. As I stated earlier, every > *terminating* number of which all digits can be indexed through numbers > in the list is itself in the list. That is trivial. The last digit > index position is the number itself. But that is *not* true for a > non-terminating list, as it does not have a last index position. But it is true for every number which can be indexed by list numbers, because all of them are finite. And indexing up to position n means covering up to position n. Indexing up to every position means covering up to every position. That is why a chose this structure of numbers. > > > > My logic is very clear: > > But unproven. It is given by the fact that indexing up to n means covering up to n. Indexing all = every means covering all = every. > > > (every digit of n can be indexed) ==> (n is in the list) > > Where is the *proof* of this? > > The left hand states: > for all n in N, K[p] = 1 > your conclusion reads: > there is an n such that An = K > As I said before, this is true *only* if N is finite. "Every digit of n can be indexed" This property is true for every number which is in the list, i.e. for every natural number in binary representation. It is true with no doubt because every list number does index (and cover) itself. This is a *symmetric* relation. Now you claim that a number with more 1's than every list number can also be indexed by the list numbers. This would destroy the symmetry. It is wrong. > But N is not finite > by the axiom of infinity. Every index number = natural number is finite. Hence what you claim for finite numbers of the list, holds for every number of the list. Regards, WM
From: mueckenh on 4 Aug 2006 16:10 Dik T. Winter schrieb: > > > The last conclusion requires proof (and it is false). What is the case > > > is that for every Ai the indexing of digits will terminate at some point, > > > and the indexing of digits of 0.111... will never terminate when we start > > > at the first digit and go on. Because there will always be a further > > > digit. (As there will always be a further An.) > > > > But each one will be finite. > > You mean the index of the digit and the An. Yes. All you agree with for finite numbers applies to all index numbers, because each one is finite. > > > (every digit of n can be indexed) ==> (n is in the list) > > Where is the proof? There is an easy proof of: > (n is in the list) ==> (every digit of n can be indexed) > but not of the converse. Indexing is a symmetric relation. A position n of number N can index a position m of number M, if and only if n = m. As 0.111... has more digit positions than the binary representation of any natural number, your claim implies that indexing is not a symmetric relation. Or you must claim that infinity is not larger than finity. Regards, WM
From: mueckenh on 4 Aug 2006 16:14 Dik T. Winter schrieb: > > > > > > > The proof is given by the list: > > > > > > > > > > > > 1 0.1 > > > > > > 2 0.11 > > > > > > 3 0.111 > > > > > > ... ... > > > > > > w 0.111... > > > > > > > > > > That is neither a proof, nor a list. > > > > > > > > The list is given in the upper part. Call the whole an extended list. > > > > It shows that w is not the number of the naturals, because w appears in > > > > the sequence of ordinal numbers only once. > > > > > > That is only a statement, *not* a proof. > > > > Why is the ordinality of the left column of the extended list another > > than the ordinality of the right column? > > I do not state so. The ordinality of both "lists" is w+1. Why do you think > they are different? > > In general, the ordinal number of any initial segment of ordinal numbers > is the smallest ordinal number that is larger than all ordinal numbers in > that segment. (This can be done because the ordinal numbers form a > well-ordered set.) And note that 0 is also an ordinal number (of the empty > set). So the ordinal number of {} is 0, of {0, 1, 2} is 3, etc. > Finally, the ordinal number of {0, 1, 2, 3, 4, ..., w} is w+1. And so > is the ordinal number of {1, 2, 3, 4, 5, ..., w}, because there is an > order preserving bijection between the two. > > You apparently did not appreciate that 0 is an ordinal number. (And that > is independent of Bourbaki.) I find that in the *true list* every number n gives the cardinality of the sequence of *positive* ordinal numbers from 1 to n inclusive and the cardinality of the 1's of the corresponding unary representation. 0 = 0.0 1 = 0.1 2 = 0.11 3 = 0.111 .... n = 0.111...1 (with n 1's) I find that this is no longer true for positive ordinal w of the extended list 0 = 0.0 1 = 0.1 2 = 0.11 3 = 0.111 .... w = 0.111... This shows that w is not the cardinal number of the natural numbers alone. It is independent of the question whether or not 0 is an ordinal. Regards, WM
From: mueckenh on 4 Aug 2006 16:24
Franziska Neugebauer schrieb: > >> 2. There is also no use of "actual" vs. "potential" infinity after > >> the advent of set theory either. > > > > You continuously contradict yourself. Compare Hilbert's statement, > > which he uttered *after* the advent of set theory. > > I neither will compare Hilbert's statements nor will I discuss the > timeline of set theory. What, then, is "after the advent" of set theory? Why do you use this phrase if you can't distinguish historical dates? Regards, WM |