
Re: Matheology § 133
Posted:
Nov 1, 2012 3:19 PM


On 01/11/2012 3:32 PM, MoeBlee wrote: > On Nov 1, 1:10 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote: >> MoeBlee <modem...@gmail.com> writes: >>> On Nov 1, 12:40 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote: >> >>>> I don't think that one works  >>>> well, not as diagonalising out of the reals, as opposed to out of >>>> sequences of 0s and 1s. Multiple representations get in the way. >> >>> Yes, I mean denumerable binary sequences as representations of members >>> of the real interval [0 1]. I hoped that would have been clear from >>> context? >> >> OK, but in that case you do not know that the antidiagonal real >> is not present in the initial listing of reals, which is what >> you surely really want. For all we know, it may be present >> in a different representation. > > Maybe I'm foggy today. But is my recollection incorrect that, using > the standard base2 scheme, each member of [0 1] is represented > UNIQUELY by a denumerable binary sequence. (What member of [0 1] is > represented by more than one denumerable binary sequence in the > standard representational scheme? 0 is represented by the sequence > whose range is {0}, i.e, .0000..., and 1 is represented by the > sequence whose range is {1}, i.e., in base 2 representation, .111..., > right? So what member of [0 1] is represented by more than one > denumerable binary sequence?)
Well, 0.1000... = 0.0111..., for starters. (Unless you take 'denumerable' to mean "ending in infinitely many 1s'.)
> And wasn't that the approach used by Cantor himself? Doesn't he take > an enumeration of denumerable binary sequences and then show a > denumerable binary sequence not in the enumeration?
I don't think he dealt with multiple representations in his proof. He talked about infinite binary sequences, made up of symbols 'm' and 'w', but as far as I remember, he did not interpret them as binary representations of real numbers.

