Another AC anomaly?
in [Logic]
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From: Ross A. Finlayson on 30 Nov 2009 10:35 On Nov 29, 11:53 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <3bc52c73-77ee-479b-9a28-824fd9a4a...(a)z10g2000prh.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > > > > > > > (Binary and ternary (trinary) anti-diagonal cases require refinement.) > > > > > > But as neither I nor Cantor were not dealing with numbers in any base, > > > > > your objections are, as usual, irrelevant. > > > > > No, it was just noted a specific constructive counterexample to that > > > > lists of (expansions representing) real numbers don't contain their > > > > antidiagonals. > > > > It wasn't even that. > > > > Marshall > > > In binary or ternary an everywhere-non-diagonal isn't not on the list. > > What does "an everywhere-non-diagonal" mean? > Given a matrix, there is a main diagonal, called "the" diagonal. In binary, there's one anti-diagonal. In ternary, base three, an everywhere-non-diagonal is different at each place than the diagonal. > And does "isn't not on the list" mean the same as "is on the list"? > No. It means "isn't necessarily not on the list." > In any integer base, from 2 on up, and any list, there are constructably > as many non-members of the list as members of it. > No, that is in integer bases from 4 on up. Collecting across moduli reinterprets the input list in a higher radix, in terms of accesses to the elements of the "main diagonal". > > > > > Using AC, in ZFC, given a well-ordering of the reals > > Since no one has yet been able to give an explicit well ordering of the > reals, we won't give it to you. > I don't need it from you, the theory guarantees one exists (if ZFC were consistent). As well, in a suitable nonstandard construction of the real numbers, the domain and range of EF are naturally well-ordered. > > I described a > > symmetry based construction of a distribution of the natural integers > > at uniform random. > > Is that supposed to mean something in English? > > [Further garbage DELETED] Yeah it means just what it says. It means that in the universe of mathematical objects there are features of these sets of numbers and their products that allow a notation and writing of, consistently applied, a consistent constant infinitesimal probability, (i.e., iota), the sum of those over that support space being as expected, unity. Bishop and Cheng formulate a measure theory in a constructible universe that's countable: don't need the trans-finite. The observation that your quote was simply mistaken stands for itself. Ross F.
From: WM on 30 Nov 2009 13:18 On 30 Nov., 14:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > I think you are confusing the limit of a sequence of sets (which is a > set) and the limit of the sequence of the cardinalities of sets ( which > is a cardinality). In general: the limit of the cardinalities is not > necessarily the cardinality of the limit, however much you would like > that to be the case. If the limit of cardinalities is 1, then the limit set has 1 element. In my case the minimum cardinality is one and the minimum set has 1 element. This covers the limits of cardinality and set. Set theory say that the limit set is empty. Therefore set theory is wrong. Regards, WM
From: Virgil on 30 Nov 2009 15:36 In article <8daa8710-2bb7-4c99-b2b1-f4d5e64ffdb3(a)o10g2000yqa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 29 Nov., 18:35, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Nov 29, 8:05�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > WM has conceded that you can use induction > > to show that every element of the list has > > a final 1, and that there is a constructive > > proof that the diagonal number does not have a > > final 1. > > > > WM has a new argument. > > > > > Use induction to show that the diagonal number cannot have more digits > > > than every entry of the list. > > > > This cannot be done. �All you do is show that > > every one of an infinite number of different > > numbers, �none of which is the diagonal number, > > cannot have more digits than every entry of the list. > > There is a simple proof by contradiction: > Assume that the diagonal (in the example-list > > 0.0 > 0.1 > 0.11 > 0.111 > ...) > > has a digit that is not in an entry of the list. This would mean that > the list has an end. That is wrong by definition. > Therefore every seqeunce of 1's in the diagonal is in an entry of the > list. > > Regards, WM Your "sample list" must actually be 0.000... 0.1000... 0.11000... 0.111000... .... Whence an acceptable "anti-diagonal" is 1.111... which differs from every entry to the list. So that Muekenheim mucked up again.
From: Virgil on 30 Nov 2009 15:46 In article <8b0d1469-3424-4a9d-8115-7535aadc656a(a)a10g2000pre.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson(a)gmail.com> wrote: > On Nov 29, 11:53�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <3bc52c73-77ee-479b-9a28-824fd9a4a...(a)z10g2000prh.googlegroups.com>, > > �"Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > > > > > > > > > (Binary and ternary (trinary) anti-diagonal cases require > > > > > > > refinement.) > > > > > > > > But as neither I nor Cantor were not dealing with numbers in any > > > > > > base, > > > > > > your objections are, as usual, irrelevant. > > > > > > > No, it was just noted a specific constructive counterexample to that > > > > > lists of (expansions representing) real numbers don't contain their > > > > > antidiagonals. > > > > > > It wasn't even that. > > > > > > Marshall > > > > > In binary or ternary an everywhere-non-diagonal isn't not on the list. > > > > What does "an everywhere-non-diagonal" mean? > > > > Given a matrix, there is a main diagonal, called "the" diagonal. In > binary, there's one anti-diagonal. In ternary, base three, an > everywhere-non-diagonal is different at each place than the diagonal. If there are infinitely many infinitely long entries in the list itself, there are at least as many "diagonals". They can all be reformatted into entries of base a least 4, in which case the dual representation problem evaporates as no diagonal has a dual representataion. > > > And does "isn't not on the list" mean the same as "is on the list"? > > > > No. It means "isn't necessarily not on the list." Then you should have said so. > > > In any integer base, from 2 on up, and any list, there are constructably > > as many non-members of the list as members of it. > > > > No, that is in integer bases from 4 on up. But any base can be converted to its square by taking digits two at a time instead of one at time, so that every base is effectively treatable like a base of at least 4. > > > > > Using AC, in ZFC, given a well-ordering of the reals > > > > Since no one has yet been able to give an explicit �well ordering of the > > reals, we won't give it to you. > > > > I don't need it from you, the theory guarantees one exists (if ZFC > were consistent). But does not guarantee that you have access to it.
From: Virgil on 30 Nov 2009 15:48
In article <887fb198-aa2f-46ae-ab6a-91a67cb73066(a)u20g2000vbq.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 30 Nov., 14:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > I think you are confusing the limit of a sequence of sets (which is a > > set) and the limit of the sequence of the cardinalities of sets ( which > > is a cardinality). �In general: the limit of the cardinalities is not > > necessarily the cardinality of the limit, however much you would like > > that to be the case. > > If the limit of cardinalities is 1, Since the lower limit of cardinalities is clearly 0, WM's conjecturing is irrelevant. |