An uncountable countable set
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From: mueckenh on 6 Aug 2006 12:50 Dik T. Winter schrieb: > In article <1153948199.656603.23130(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > In article <1153829288.635491.96140(a)m73g2000cwd.googlegroups.com> muecken= > > h(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > ... > > > > I need not count to w. If a column contains one 1 then the *+sum is 1. > > > > This is a definition like that of Cantor's list. > > > > > > Depends. I have no idea (and it is very dissimilar), whatever way I > > > look at it, I find: > > > (*+){k = 1 .. n} Ak = An > > > and I can not substitute oo for n (so as to include all Ak), because there > > > is no Aoo. > > > > Therefore you cannot need and cannot use it. It is simply not > > available. > > Yes, so it is not defined for the *+ sum of all An... But the *+ sum is defined for all digit positions which can be indexed by a natural number. > > > > > (every digit of n can be indexed) ==> (n is in the list) > > > > > > Where is the proof? > > > > It is even an equivalence by *definition*. The list contains all > > natural numbers. All indexes *are* natural numbers and vice versa. > > I do not see the equivalence. The equivalence is: > (every digit of n can be indexed) ==> (every index is in the list) > I still see no proof that n is in the list, because I do not see a > proof that n is an index. n is a natural number. An index is a natural number. "Index" and "natural number" are synonymous. > > > > There is an easy proof of: > > > (n is in the list) ==> (every digit of n can be indexed) > > > but not of the converse. > > > > You cannot index other digits than those which are indexes. All indexes > > are all naturals. All naturals are by definition in he true list. > > Yes. This does *not* proof that the *+ sum of them all (once defined) is in > the list, because there is no proof that it is an index. Rather, there is an > easy proof that it is *not* an index. Then it cannot be indexed. Then it contains digit positions which are not indexed by natural numbers. Indexing is a symmetric operation: A digit position n can be indexed by a natural number 0.111...1 (with n digits) if and only if the natural number (in unary representation) can be indexed by the digit position of said number. > > > A class of numbers is a set of ordinals of sets with same cardinality. > > Possible, that is a proper definition. > > > Their cardinality in newer set theory (and already by the late Cantor) > > is expressed by the least ordinal. The cardinality of sets w , 2w, w^2, > > ... is expressed by aleph_0 or by omega. > > Yes, already by the late Cantor. > > > >are for sets that are in the same equivalence > > > class with respect to bijection. But perhaps I am wrong, but I think that > > > today the distinction between w and aleph-0 is made more clear than in > > > Cantor's later times. > > > > On the contrary! Cantor distinguished very clearly in all his written > > papers. > > Except in his later times, as you state so eloquently just above. Or > do you *not* see a contradiction between these two statements? Between which two statements should I see a contradiction? Fact is: 1) Cantor distinguished neatly between omega and aleph_0 in all his written work. 2) Modern set theorists (and the late Cantor) use omega to denote the order type of |N as well as the cardinality of countable sets. Regards, WM
From: mueckenh on 6 Aug 2006 13:04 Virgil schrieb: > > "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. > > Question: Which digit(s) of the decimal expansion of 1/9 cannot be > indexed by some finite natural? Those by which 1/9 differs from any number in the true list: i.e. those which have not a natural index (because every number in the true list has a natural index) and every natural index marks a position of a number which is in the list. > Answer: there is no digit in that expansion that cannot be indexed by a > finite natural. > > To maintain the existence of such a digit requires proof that it exists, > and "mueckenh" has no such proof because no such digit exists. It requires proof that 0.111.. does exist and can be indexed. But there is no proof. > > > > > > 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. > > What holds for every member of a list need not be true of the list > itself. But for every member. The list may be infinitely long. But *every* member is finite. Regards, WM
From: mueckenh on 6 Aug 2006 13:06 Virgil schrieb: > In article <1154722206.554768.237310(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > What you claim does not follow. > > What is true for each element of a set need not be true of the set > itself. Oracular statement. *All indexes are elements*. Index positions are 1's of the number 0.111... Regards, WM
From: mueckenh on 6 Aug 2006 13:10 Virgil schrieb: > If by w, "mueckenh" means the ordinal we call omega, it is not a > cardinal at all, though it has one, namely aleph_0, which IS the > cardinal of the set of all finite naturals, with or without 0. Try to learn some modern set theory. There omega is used as cardinal number too. Regards, WM
From: mueckenh on 6 Aug 2006 13:16
Virgil schrieb: > > But according to the axiom of infinity only (1) is true and (2) is > > false. Of course both statements are equivalent > > That claim about what the AoI says is itself false. It says that the complete set *does* exist with all of its elements. The number of elements cannot be finite, but the number of elements does exist. Hence it is an infinite number. The same must hold for the rows and the x's in the rows. Or the diagonal must bend. > > The AoI says that there is at least one set which contians {} as a > member and which for every member x also contains (x union {x}) as a > member. And that is all it says. > > > > That means: an actually infinitely long triangle is an actually > > infinitely broad triangle. > > is the limiting case a triangle at all? By what definition of triangle? By the definition of a straight line. > > > > > But in this case neither the paths, nor the edges are terminated. So while > > > in all cases the number of edges is less > > Actually in any tree all edges terminate in nodes. in an infinite tree > (with no terminal or leaf nodes) maximal paths will not terminate, but > every edge still terminates. But the number of edges does not terminate. > > > > > That is correct. But in no case the number of paths can be larger than > > that of edges (which is countable). > > That is provably false and has been proved false. > > The set of edges of an infinite binary tree has been shown to biject > with the infinite set of finite natural numbers. > > The set of paths of an infinite binary tree has been shown to biject > with the power set of the infinite set of finite natural numbers. > > So that "mueckenh" is claiming a bijection between a set and its power > set. which is a no-no. This is not an arument. I am claiming that these theorems lead to contradictions. Cantor's theorem may prove Card N < Card R. The binary tree shows the opposite result. > > Infinitely many naturals have infinitely many differences of 1 which > > accumulate to an infinite number. > > That doesn't prove anything. At least not in ZF or NBG. Set theory cranks are like other cranks: No argument or evidence can ever be sufficient to make a crank abandon his belief. Regards, WM |