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From: Han de Bruijn on 27 Jul 2005 08:24 David Kastrup wrote: > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > >>MoeBlee wrote: >> >>>Meanwhile, I'm still fascinated by your inconsistent theory, posted at >>>your web site, which is: >>>Z, without axiom of infinity, bu with your axiom: Ax x = {x}. >>>Would you say what we are to gain from this inconsistent theory? >> >>No. But first we repeat the mantra: >> >> A little bit of Physics would be NO Idleness in Mathematics > > Oh, it is there alright, like a little bit of real life is in a novel > worth reading. > >>It is physically correct that every member of a set is at the same >>time > > You blew it. A "set" is not a physical entity, so there is nothing > that can be "physically correct" about a set. So to speak. The more accurate formulation is that a set in mathematics should be an idealization from something that is called a set in common speech, say physics in a broad sense. Take that and you will find that what I've said is correct. I don't say, though, that sets in mainstream mathematics actually _are_ what they should be, according to the mantra. >>a _part_ of the set, meaning that a e A ==> a c A , where e stands >>for membership and c for being a subset. The above axiom that a >>member of a set cannot physically distinguished from a set which >>contains only the member - the envelope {} means nothing, physically > > The whole set means nothing, physically. Finite sets can be materialized and _mean_ something, physically. I can even make music with them. In the thread "Set inclusion and membership" > Randy Poe wrote: > >> Where do you believe set theory comes up in physics? > > If you believe that music is physics ... I have an application where set > theory is employed with the transition of chords. With the chords Am and > C, for example, only the notes in the symmetric difference Am<>C need to > be changed (to become Off or On respectively). And: >> By the way, I don't think your music idea is valid beyond your >> trivial example. Show me how it tells me the right chord >> progression to go from, oh, C Maj to F# Maj to G minor for >> instance. > > Oh well. C Maj = {C,E,G,B} , F# Maj = {F#,A#,C#,F} , G minor = {G,A#,D} > Hence F# Maj \ C Maj = {F#,A#,C#,F} On , C Maj \ F# Maj = {C,E,G,B} Off. > (You've defined disjoint sets.) Now wait and listen. Then: > G minor \ F# Maj = {G,D} On , F# Maj \ G minor = {F#,C#,F} Off. Han de Bruijn
From: Dik T. Winter on 27 Jul 2005 08:22 In article <1122395188.769778.34530(a)z14g2000cwz.googlegroups.com> malbrain(a)yahoo.com writes: > Dik T. Winter wrote: > > In article <1122347583.518181.245300(a)g14g2000cwa.googlegroups.com> malbrain(a)yahoo.com writes: > > > > > > The C language is defined by the C standard, as defined by ISO. There > > > are no "unbounded" standard types in the C language. karl m > > > > Who is talking about C? > > Of the billions of computer systems deployed since the micro-computer > revolution, the overwhelming majority are programmed with C. That is not an answer. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 27 Jul 2005 08:49 In article <MPG.1d5044b418d0c79d989f94(a)newsstand.cit.cornell.edu> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Dik T. Winter said: .... > > Indeed in the dyadic system when all digits are 2 we divide by 2 by > > replacing each digit by 1. But shifting to the right is *not* the > > same as (truncating) division by 2. That is only the case if the > > least significant digit is 2. (That last 2 should have been 1.) > That's not true. If each digit place represents a power of the base, then > shifting everything to the right one digit divides the denoted value of that > digit by the number base. Here, each place denotes a power of 2, and moving > all digits to the right one place divides the entire number by two, perhaps > with remainder. Lessee. In the dyadic system 8 = "112". Shifting one position to the right gives "11" = 3. I would not call that dividing by 2. > To > declare the set the same size as a proper subset is a lot of what offends > the sensibilities of anti-Cantorians, Not all anti-Cantorians. There are also those who state that oo+1 = oo, which is in essence the same. > and to declare that this is the case for > infinite sets, rather than to admit that bijections of themselves do NOT > prove equal size in that case, is a baffling choice on the part of the > mathematical community. Well, bijections provide an excellent way to compare sets, because it is an equivalence relation, and so they define equivalence classes. > > The point is, when you allow "infinite" naturals, then those naturals can > > be put in 1-1 correspondence with the power set of the *finite* naturals. > > There is no problem with that. But they can not be put in 1-1 > > correspondence with the power set of the enlarged set of naturals. > > Why not? The "enlarged set" goes on forever, and the set of subsets goes on > forever, and can be put into a linear order based on membership of elements, > which can be seen to correspond with the naturals that go on forever. Where do > you see a problem? Pray explain your mapping, it is not clear. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Kastrup on 27 Jul 2005 08:53 Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > David Kastrup wrote: > >> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: >> >>>MoeBlee wrote: >>> >>>>Meanwhile, I'm still fascinated by your inconsistent theory, posted at >>>>your web site, which is: >>>>Z, without axiom of infinity, bu with your axiom: Ax x = {x}. >>>>Would you say what we are to gain from this inconsistent theory? >>> >>>No. But first we repeat the mantra: >>> >>> A little bit of Physics would be NO Idleness in Mathematics >> Oh, it is there alright, like a little bit of real life is in a >> novel worth reading. >> >>>It is physically correct that every member of a set is at the same >>>time >> You blew it. A "set" is not a physical entity, so there is nothing >> that can be "physically correct" about a set. > > So to speak. The more accurate formulation is that a set in > mathematics should be an idealization from something that is called > a set in common speech, say physics in a broad sense. Take that and > you will find that what I've said is correct. No, take that, and what you said is rubbish. It is the _deciding_ mark of an idealization that it is _not_ identical to what is being idealized. So an idealization will _never_ be "physically correct". -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Han de Bruijn on 27 Jul 2005 09:40
David Kastrup wrote: > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > > >>David Kastrup wrote: >> >> >>>Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: >>> >>> >>>>MoeBlee wrote: >>>> >>>> >>>>>Meanwhile, I'm still fascinated by your inconsistent theory, posted at >>>>>your web site, which is: >>>>>Z, without axiom of infinity, bu with your axiom: Ax x = {x}. >>>>>Would you say what we are to gain from this inconsistent theory? >>>> >>>>No. But first we repeat the mantra: >>>> >>>> A little bit of Physics would be NO Idleness in Mathematics >>> >>>Oh, it is there alright, like a little bit of real life is in a >>>novel worth reading. >>> >>> >>>>It is physically correct that every member of a set is at the same >>>>time >>> >>>You blew it. A "set" is not a physical entity, so there is nothing >>>that can be "physically correct" about a set. >> >>So to speak. The more accurate formulation is that a set in >>mathematics should be an idealization from something that is called >>a set in common speech, say physics in a broad sense. Take that and >>you will find that what I've said is correct. > > > No, take that, and what you said is rubbish. It is the _deciding_ > mark of an idealization that it is _not_ identical to what is being > idealized. So an idealization will _never_ be "physically correct". You're right. But you don't even try to understand what I mean. Do you? Han de Bruijn |