Another AC anomaly?
in [Logic]
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From: Ross A. Finlayson on 30 Nov 2009 18:10 On Nov 30, 12:46 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <8b0d1469-3424-4a9d-8115-7535aadc6...(a)a10g2000pre.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...(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. > I did, but it's OK, I'm glad that you see that there are some differences in the treatments that are radix-specific, there are funny differences between the forms of those results. These can be "quantified", in a sense, in terms of identifying the cost of algorithms in terms of time and space complexity, where as the radix of the list expansions decreases or goes to one, an ever more complex algorithm is necessary to attempt to build an item difference from each element of 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. > > 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. > I like to take a step back and wonder, and explain, why those things are different. In binary there's one antidiagonal. > > > > > > 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. It doesn't matter its form, its relevant properties are guaranteed by definition. Its form is the EF or REF, an interesting mathematical primitive with tractable analytical properties. Ross F.
From: Virgil on 30 Nov 2009 19:01 In article <a2ec00f6-b4ab-4be1-bfa5-a0b73146e0e4(a)x25g2000prf.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson(a)gmail.com> wrote: > On Nov 30, 12:46�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <8b0d1469-3424-4a9d-8115-7535aadc6...(a)a10g2000pre.googlegroups.com>, > > �"Ross A. Finlayson" <ross.finlay...(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. > > > > I did, but it's OK, I'm glad that you see that there are some > differences in the treatments that are radix-specific, there are funny > differences between the forms of those results. Whatever the radix, the Cantor result can be shown to hold for it, and one method can be show to work for all radices. > These can be > "quantified", in a sense, in terms of identifying the cost of > algorithms in terms of time and space complexity, where as the radix > of the list expansions decreases or goes to one, an ever more complex > algorithm is necessary to attempt to build an item difference from > each element of the list. Whatever the radix, the Cantor result can be shown to hold for it, and one single method can be show to work for ALL integer radix values from 2 upwards. > > > > > > > > > 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. > > > > I like to take a step back and wonder, and explain, why those things > are different. > > In binary there's one antidiagonal. There are, in fact, as many unlisted sequences as listed ones, and a method of constructing that many which works equally well for all integer radices from 2 upwards. That Ross may not be aware of those construction does not mean that everyone else must be equally unaware of them. > > > > > > > > > > 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. > > It doesn't matter its form, its relevant properties are guaranteed by > definition. > > Its form is the EF or REF, an interesting mathematical primitive with > tractable analytical properties. If Ross claims to be able to show me a specific well ordering of the reals, let him do so, but otherwise cease making useless noises.
From: Dik T. Winter on 1 Dec 2009 07:10 In article <887fb198-aa2f-46ae-ab6a-91a67cb73066(a)u20g2000vbq.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > 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). =A0In 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. No because the limit of cardinalities is not necessarily the cardinality of the limit, as I wrote just above. > In my case the minimum cardinality is one and the minimum set has 1 > element. Yes, so what? > This covers the limits of cardinality and set. It covers nothing at all, because it is not about the limit of the sequence of sets. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 1 Dec 2009 07:42 On 1 Dez., 13:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <887fb198-aa2f-46ae-ab6a-91a67cb73...(a)u20g2000vbq.googlegroups..com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > 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). =A0In 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.. > > No because the limit of cardinalities is not necessarily the cardinality > of the limit, as I wrote just above. You may write this as often as you like, but you are wrong. If there is a limit set then there is a limit cardinality, namely the number of elements in that limit set. Everything else is nonsense. Regards, WM
From: Dik T. Winter on 1 Dec 2009 08:57
In article <6a0cfabf-c90e-4561-a45f-a4ffd33ca8e9(a)m3g2000yqf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 1 Dez., 13:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <887fb198-aa2f-46ae-ab6a-91a67cb73...(a)u20g2000vbq.googlegroups= > .com> WM <mueck...(a)rz.fh-augsburg.de> writes: .... > > > If the limit of cardinalities is 1, then the limit set has 1 element. > > > > No because the limit of cardinalities is not necessarily the cardinality > > of the limit, as I wrote just above. > > You may write this as often as you like, but you are wrong. If there > is a limit set then there is a limit cardinality, namely the number of > elements in that limit set. Everything else is nonsense. Can you prove the assertion that the limit of the cardinalities is the cardinality of the limit? I can prove that it can be false. As I wrote, given: S_n = {n, n+1} we have (by the definition of limit of sets: lim{n -> oo} S_n = {} and so 2 = lim{n -> oo} | S(n) | != | lim{n -> oo} S_n | = 0 or can you show what I wrote is wrong? If so, what line is wrong? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |