Cantor Confusion
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From: Franziska Neugebauer on 28 Jan 2007 09:00 mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Jan., 18:29, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> >> > Induction covers (is valid for) all natural numbers, but not "the >> > set of all natural numbers". > >> We know this for quite some time. The point is that you claim >> induction allows any assertion of U { T(i) | i e N }. So you >> eventually agree that is does not. > > Cant't you read? Please look closer. There is clearly spelled out that > every number *in* N is concerned, i.e., every finite numbver, not the > infinite number N. First you cut the context and then you ask whether I can read it. My comment: No I can't read it anymore. The context was: >> ,----[ <45b5ec2c$0$97243$892e7...(a)authen.yellow.readfreenews.net> ] >> | >> Again: Your notations >> | >> >> | >> T(1) U T(2) U ... >> | >> >> | >> and >> | >> >> | >> U {T(i) | i e N } >> | >> >> | >> are undefined. >> | > >> | > You are in error. The union of the trees T(n) and T(n+1) is >> | > defined. n is a natural number. Therefore the union of all finite >> | > trees is defined. >> | >> | You have misunderstood the induction principle. It is not made for >> | "counting over to the infinite". >> `---- Your claim is "The union of the trees T(n) and T(n+1) is defined. n is a natural number. Therefore the union of all finite trees is defined." Non sequitur. Your argument is of the type "counting over to infinity". You may take notice of the similar claim: "The sum of the numbers n and n + 1. n is a natural number. Therefore the sum of all finite numbers is defined." F. N. -- xyz
From: Franziska Neugebauer on 28 Jan 2007 09:10 mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Jan., 22:52, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > I am interested in the fact that every set of natural numbers has a >> > finite maximum. [indendation corrected] >> Then you should perhaps not talk to contemporary set > theorists who are >> accustomed to the > > fixed idea being far from being a > >> fact that there *are* sets of natural numbers which do >> not have maxima at all. In the framework of ZFC it *is* fact that the set of all natural numbers does not have a maximum. Without a proof of a contradiction _within_ the ZFC framework there is no reason for your disfavour. F. N. -- xyz
From: Andy Smith on 28 Jan 2007 10:11 In message <7stor2lf0vlgm30ahb5c3umk6qrf7csu9c(a)4ax.com>, G. Frege <nomail(a)invalid.?.invalid> writes >On Sun, 28 Jan 2007 10:05:25 GMT, Andy Smith ><Andy(a)phoenixsystems.co.uk> wrote: > >> >> [...] most cranks are probably convinced that they are right >> >That's for sure! :-) > I will say it again - I don't have the knowledge to be a crank. And I do read what gets posted back. But in this instance, re my post, I thought that I was saying something other than discussed previously, i.e. basically that as far as I could see the diagonalisation argument doesn't work for a specific hypothetical counting arrangement of the reals. Nobody has actually said that I can't attempt to enumerate the reals in the suggested fashion (on an assumption that they were countable) - although it is clear that doing so as I proposed can only ever index the (subset of) the rationals with a terminating sequence of zeroes in their expansion ... which (as I previously said) is prima facie evidence for the reals being uncountable. Please explain why my post was nonsense, if you have the time, or unless it was so misguided that you don't know where to begin. I am actually rational (most of the time anyway). regards -- Andy Smith
From: William Hughes on 28 Jan 2007 10:50 On Jan 27, 5:26 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > Randy Poe <poespam-t...(a)yahoo.com> writes > > (snip) > >That's not an argument. It's a statement about why several of > >your intuitions combine in a certain way to lead you to an > >intuitive conclusion. I don't know what "uncountable" means > >to you, or why "necessarily infinite binary expansion" implies > >whatever that means to you. > > >I do know that the set of infinite binary strings is uncountable, > >but I know that because it is easily proven. Good. We start with the fact that the set of binary strings, B, is uncountable. Call it fact alpha. > Well, at the high risk of ridicule and a rather lower probability of > assassination by irate set theorists, I think that Cantor's > diagonalisation argument has a terminal flaw. > > The flaw is that Cantor's argument, disregarding any qualms about the > infinite sets involved, just demonstrates that one cannot construct an > infinite list of all permutations of binary strings. this is not the > same as showing that you cannot construct an infinite list of all reals, > because, as has been pointed out, reals may have more than one > representation in any base. Yes, it is well known that (the classic form of) Cantor's diagonalization argument will not work with the binary representation. However, the question of whether the reals are uncountable or not does not depend on which representation we choose. Choose another base. Or use another way of getting a real number from each binary sequence. Let the binary sequence be b = (a(1), a(2), a(3) ,...) A way that does not work is f(b) = sum(a(i)2^-i) So define a new function g, g(b) = sum(2a(i)3^-i) (change every 1 to a 2 and interpret the resulting sequence as a base 3 expansion). Now g is an injection from B to the reals. So there are at least as many reals as elements of B. But B is uncountable (fact alpha), so the reals are uncountable. <snip a demonstration that the classic form of Cantor's diagonal argument cannot be used with the binary representation of the reals> > that does not mean that it is possible to count the reals, it just > invalidates the argument. Conceivably one might be able to define an > arrangement of reals such that the antidiagonal was unambiguous in > defining a real that should be in the list (but can't be becaue by > construction it is different from those in the list). But if the > argument above is valid, then I think Cantor's diagonalisation is dead > and buried. No, the fact that one form of Cantor's diagonalization cannot be used in one particular case does not mean that Cantor's diagonalization is dead and buried. Cantor's diagonalization can be used in many other cases and can be used to show that the reals are not countable. - William Hughes
From: Andy Smith on 28 Jan 2007 12:32
William Hughes <wpihughes(a)hotmail.com> writes > > >On Jan 27, 5:26 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: >> Randy Poe <poespam-t...(a)yahoo.com> writes >> >> (snip) >> >That's not an argument. It's a statement about why several of >> >your intuitions combine in a certain way to lead you to an >> >intuitive conclusion. I don't know what "uncountable" means >> >to you, or why "necessarily infinite binary expansion" implies >> >whatever that means to you. >> >> >I do know that the set of infinite binary strings is uncountable, >> >but I know that because it is easily proven. > >Good. We start with the fact that the set of binary strings, B, is >uncountable. Call it fact alpha. > >> Well, at the high risk of ridicule and a rather lower probability of >> assassination by irate set theorists, I think that Cantor's >> diagonalisation argument has a terminal flaw. >> >> The flaw is that Cantor's argument, disregarding any qualms about the >> infinite sets involved, just demonstrates that one cannot construct an >> infinite list of all permutations of binary strings. this is not the >> same as showing that you cannot construct an infinite list of all reals, >> because, as has been pointed out, reals may have more than one >> representation in any base. > >Yes, it is well known that (the classic form of) Cantor's >diagonalization >argument will not work with the binary representation. However, >the question of whether the reals are uncountable or not does >not depend on which representation we choose. Choose another >base. > >Or use another way of getting a real number from each >binary sequence. > >Let the binary sequence be b = (a(1), a(2), a(3) ,...) > >A way that does not work is f(b) = sum(a(i)2^-i) > >So define a new function g, g(b) = sum(2a(i)3^-i) >(change every 1 to a 2 and interpret the resulting sequence as a base >3 expansion). > >Now g is an injection from B to the reals. So there are at least as >many reals as elements of B. But B is uncountable (fact alpha), >so the reals are uncountable. > ><snip a demonstration that the classic form of Cantor's diagonal > argument cannot be used with the binary representation of the reals> > >> that does not mean that it is possible to count the reals, it just >> invalidates the argument. Conceivably one might be able to define an >> arrangement of reals such that the antidiagonal was unambiguous in >> defining a real that should be in the list (but can't be becaue by >> construction it is different from those in the list). But if the >> argument above is valid, then I think Cantor's diagonalisation is dead >> and buried. > >No, the fact that one form of Cantor's diagonalization cannot be used >in one particular case does not mean that Cantor's diagonalization >is dead and buried. Cantor's diagonalization can be used in >many other cases and can be used to show that the reals >are not countable. > Thanks for that. Understood. But actually when Randy said "I do know that the set of infinite binary strings is uncountable, but I know that because it is easily proven" I assumed that he meant Cantor's diagonalisation - I had a mental equivalence set of real numbers <=> set of infinite binary strings, clearly too primitive a perspective. -- Andy Smith |