Are natural numbers isomorphic to complex numbers?
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From: Ron on 1 Jun 2010 10:22 On Jun 1, 8:17 am, Don Stockbauer <donstockba...(a)hotmail.com> wrote: > On Jun 1, 4:37 am, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > > > > > > > On Tue, 1 Jun 2010 00:05:56 -0700 (PDT), Akira Bergman > > > <akiraberg...(a)gmail.com> wrote: > > >On Jun 1, 3:42 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > > >> Do Bergman and Stockbauer mean that N is isomorphic to C, > > >> or that N is isomorphic to a _subset_ of C? (Leaving out > > >> the words "a subset of" led to a huge argument in another > > >> recent thread.) > > > >I mean N is isomorphic to C. > > > Good of you to clarify that. > > > That's ridiculous, by the way. Obviously ridiculous, for many > > obvious reasons. > > 5 DEBATE GO TO 5 When you ask a ridiculous question, responses like this are likely. For the record, N is clearly NOT isomorphic to C in any sense, since the natural numbers are countable and the complex numbers are uncountable.
From: Frederick Williams on 1 Jun 2010 14:34 Akira Bergman wrote: > C has all the structures of N, Z and R, C does not have the order that any one of N, Z and R has. -- I can't go on, I'll go on.
From: Aatu Koskensilta on 1 Jun 2010 14:41 David C. Ullrich <ullrich(a)math.okstate.edu> writes: > On Tue, 1 Jun 2010 00:05:56 -0700 (PDT), Akira Bergman > <akirabergman(a)gmail.com> wrote: > >>I mean N is isomorphic to C. > > Good of you to clarify that. > > That's ridiculous, by the way. Obviously ridiculous, for many > obvious reasons. Obviously obviously, in an obvious sort of way. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: porky_pig_jr on 1 Jun 2010 15:05 On Jun 1, 3:05 am, Akira Bergman <akiraberg...(a)gmail.com> wrote: > On Jun 1, 3:42 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > > Do Bergman and Stockbauer mean that N is isomorphic to C, > > or that N is isomorphic to a _subset_ of C? (Leaving out > > the words "a subset of" led to a huge argument in another > > recent thread.) > > I mean N is isomorphic to C. Thanks. May be you don't understand that, but with this reply you instantly established your credentials.
From: Transfer Principle on 1 Jun 2010 15:59
On Jun 1, 12:05 am, Akira Bergman <akiraberg...(a)gmail.com> wrote: > On Jun 1, 3:42 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > Do Bergman and Stockbauer mean that N is isomorphic to C, > > or that N is isomorphic to a _subset_ of C? (Leaving out > > the words "a subset of" led to a huge argument in another > > recent thread.) > I mean N is isomorphic to C. In that case, notice that even though I have nothing against theories other than ZFC, it must be pointed out that ZFC, the standard theory, refutes the possibility than N and C are isomorphic. Here is a fairly standard proof that N and C can't possibly be isomorphic. Let phi: N -> C be the isomorphism that you (Bergman) believes exists. Since the domain of phi is N, we can write the list: phi(1) phi(2) phi(3) phi(4) .... Indeed, the usual proof (due to Cantor) only requires real numbers, so we can write only the real parts: Re(phi(1)) Re(phi(2)) Re(phi(3)) Re(phi(4)) .... Then Cantor performs a trick called "diagonalization," where he creates a real called D, an antidiagonal. To do so, we proceed one decimal place at a time. The first digit after D's decimal point can be any digit except the digit in the same place of Re(phi(1)). This guarantees that D is not equal to Re(phi(1)). The second digit after D's decimal point can be any digit except the digit in the same place of Re(phi(2)). This guarantees that D is not equal to Re(phi(2)). The third digit after D's decimal point can be any digit except the digit in the same place of Re(phi(3)). This guarantees that D is not equal to Re(phi(3)), and so on. (Note: some reals may have more than one decimal representation, but there are ways to get around this.) Then D can't be equal to Re(phi(n)) for _any_ natural number n, and thus can't be equal to phi(n) for _any_ natural n (since equal complex numbers must have equal real parts). Therefore phi can't be an isomorphism at all, since no natural number corresponds to D. If Bergman were to ask how D is related to the naturals, he won't be able to find a way, since D is not equal to phi(n) for any natural n. Thus completes the standard proof in ZFC that N and C aren't isomorphic. QED (Note: at almost any given moment at sci.math, there's an active thread about Cantor. Bergman can check out these threads for more information about Cantor and the implications of his famous proof.) But just because ZFC doesn't prove that N and C are isomorphic, it doesn't mean that there can't be a theory other than ZFC in which they are. If Bergman is considering such a theory, I'd like him to tell me more about this theory. In particular, how are its axioms different, and which axioms, if any, of ZFC are not included in the theory? (Note: finding a theory in which N and C are in fact isomorphic is a real challenge. Previous posters have tried to make N and R isomorphic, but N and C will be much, much more difficult. The best I can think of is something involving the p-adics.) I hope that Bergman, the OP of this thread, doesn't find this thread bullying or patronizing. If he does, then please let me know so that I can continue to adjust my posting habits in the future. |