Cantor Confusion
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From: Andy Smith on 24 Jan 2007 08:06 In message <ep7jc2$bh5$1(a)news.msu.edu>, stephen(a)nomail.com writes >Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: >> In message <1baer212phrkloe4i9r6ghiq9c451h08fm(a)4ax.com>, G. Frege >> <nomail(a)invalid.?.invalid> writes >>>On Wed, 24 Jan 2007 09:38:00 GMT, Andy Smith >>><Andy(a)phoenixsystems.co.uk> wrote: >>> >>>>> >>>>> To represent a real number between 0 and 1, you need 1 bit for >>>>> each positive integer. >>>>> >>>>> x = b1*(1/2)^1 + b2*(1/2)^2 + b3*(1/2)^3 + b4*(1/2)^4 .... >>>>> >>>>> Why do you think this requires an "actually infinite" integer? >>>>> >>>> Because you need binf to complete the sum; inf is not a natural number >>>> and you need an actual infinity of bits to describe it. >>>> >>>You are talking nonsense, again. >>> >>>Aren't you able to understand the difference between an >>> >>> (a) infinite integer >>>and >>> (b) infinitely many integers >>>? >> Yes, although my use of your terminology probably goes adrift. >>> >>>We do not need "binf" (?) to "complete the sum" because this sum is >>>_never_ "completed". >>> >> Exactly so. A transcendental requires an infinite number of bits to >> represent it (so as to distinguish it from the rest of the infinite set >> of other real numbers). >>>> >>>> If you systematically try to address (map) the reals you need integers >>>> with as many bits as the reals; NaN. >>>> >>>Bla bla. Seems that you are desperately striving for a career as a >>>crank here. >>> >>>Go ahead, I think you will succeed! :-) >>> >>> >>>F. >>> >> OK, I will take the personal abuse, probably deserved. Shouldn't put my >> head above the parapet when I'm not qualified to do so. All that I was >> trying to observe was that all systematic numbering schemes of the reals >> are equivalent, corresponding to permutations of bit positions. And if >> you adopt one systematic scheme, such as starting with bit 0, then bit >> 1, etc, because the reals have an address space of an infinite number of >> bits, the corresponding numbers/indices must also require an infinite >> number of bits, and no natural number is infinite. So you can't do it - >> you don't have a big enough address space. > >> Of course that may well be a load of bollocks > >Of course it is. There are an infinite number of integers, and an >infinite number of bit positions. The address space is exactly >the right size. Why do you think there are fewer integers than >bit positions? > >Your whole argument seems based on the idea that there must be a bit oo, >and that because there is no natural number oo, that it is impossible >to describe a real number. But there is no bit oo. There is no last >bit. There is an unending sequence of bits. Unending sequences >do not end. > So why are you happy with 0.0111... and not ...1110 ? (Because natural numbers are all finite, even though the set N is infinite.) -- Andy Smith
From: G. Frege on 24 Jan 2007 08:04 On Wed, 24 Jan 2007 11:51:37 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > > [...] But that implies that the indices are infinite. > It seems that you are not able to understand the difference between _an infinite natural number_ (which does not exist) and _infinitely many_ natural numbers. If so, there's no help for you. A final/last time: there are infinitely many indices 1,2,3,..., none of which is infinite. Bye. F. -- E-mail: info<at>simple-line<dot>de
From: Andy Smith on 24 Jan 2007 08:07 In message <ep7kn4$79j$2(a)mailhub227.itcs.purdue.edu>, Dave Seaman <dseaman(a)no.such.host> writes >On Wed, 24 Jan 2007 09:48:01 GMT, Andy Smith wrote: >> In message <9b2er2p5taadlea28n6fb3g4nuvmqeijhs(a)4ax.com>, G. Frege >><nomail(a)invalid.?.invalid> writes >>>On Tue, 23 Jan 2007 23:34:03 GMT, Andy Smith >>><Andy(a)phoenixsystems.co.uk> wrote: >>> >>>> >>>> I meant that the infinite non-repeating irrational binary expansion of >>>> sqrt(2) requires an [...] infinite set of numbers to define its >>>> location on the line... >>>> >>>Right. >>> >>>> >>>> Since the integers are finite, ... >>>> >>>Non sequitur. Yes, the integers are finite, but there are infinitely >>>many of them. >>> >>>To be precise: countable infinite many of them. And that's actually >>>(sic!) enough to represent each and any real number. >>> >> Any systematic scheme for mapping the reals to integers will be the >> same, subject only to permutations of the bit positions. So a systematic >> scheme is, start with bit 1, then bit 2, then bit 3 etc. This >> corresponds to a reflection of the possible set of numbers in n bits >> about the binary point. Any real number has an infinite binary >> expansion, and its corresponding mapping integer is infinite (=NaN) > >I think what you are trying to say is that no mapping f: R -> N is an >injection. That's correct, but your expression of and justification for >that fact leave a lot to be desired. > I'm sure that is true. If you want me to shut up and go away I will. >Your original statement was that there are not enough integers to define >even one real. That's patently false. Were you assuming that we are >allowed to use only a single integer in the definition? > > You misunderstand what I meant, I think, or at any rate I didn't express myself clearly. If you have a transcendental, you need to specify an infinite number of bits to distinguish it from the set of all alternative transcendentals. You specify reals as a Cauchy sequence, which unambiguously points towards the point, but the point itself needs an infinite number of bit positions. But you can't label an infinite number of bit positions - you need to have all of the bits as a completed set to define the real - and that is not a finite number. -- Andy Smith
From: mueckenh on 24 Jan 2007 08:08 Virgil schrieb: > In article <1169545975.865624.242740(a)d71g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1169396052.939963.194070(a)q2g2000cwa.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Virgil schrieb: > > > > > > > > > > > > Therefore, by induction, all finite n-level trees for all n in N exist. > > > > > But that is all that standard induction allows one to conclude. > > > > > > > > It is enough. Or should there one level be missing? Please specify > > > > which remains to be included. > > > > > > Anything that concludes anything about infinite trees. > > > > Conclusions are not members or subsets of trees. > > Standard induction does not justify your /conclusion/ that your > conjunction of finite trees is an infinite tree. > Induction covers all natural numbers and, hence, all finite trees. I don't need more for the union of all finite trees. But your reply was not to he topic. My conclusion does not sit in a tree in order to make some being paths not being. Regards, WM
From: G. Frege on 24 Jan 2007 08:08
On Wed, 24 Jan 2007 11:29:47 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > > That was precisely my point. So [...] > This "so" is not justified, sorry. You are just making up things here. Hence I'll quit. :-) F. -- E-mail: info<at>simple-line<dot>de |