Cantor Confusion
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From: Dave Seaman on 24 Jan 2007 07:50 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. 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? -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: mueckenh on 24 Jan 2007 07:51 Virgil schrieb: Summary 1) Every complete infinite binary tree T (containing all nodes and edges) contains all paths. 2) The union tree T(oo) of all finite trees is well defined (as I have shown elsewhere) and yields the complete infinite binary tree containing all nodes and edges: T = T(oo). 3) The union of all finite trees includes the union of all nodes and, with it, the union of all such subsets which are paths (because every path is a well defined subset of the set of nodes if the structure of the tree is well defined). 4) The set of paths in T(oo) is a subset of the countable set of finite sets of all paths in the finite trees. 5) A countable union of countable sets is a countable set (according to ZF with AC). ==> The set of all path is countable. (==> The real numbers are countable.) Going on, we can say: 6) T(oo) = T contains only finite paths. 7) T(oo) = T contains all paths including all infinite paths. ==> There are no infinite paths. (There are no irrational numbers.) Nothing further remains to say. Regards, WM
From: Andy Smith on 24 Jan 2007 07:56 In message <1169641616.486470.83070(a)j27g2000cwj.googlegroups.com>, imaginatorium(a)despammed.com writes > >Andy Smith wrote: >> In message <gccer2dvlhm5q04g7deea6ro0t0c5b31p6(a)4ax.com>, G. Frege >> <nomail(a)invalid.?.invalid> writes >> >On Wed, 24 Jan 2007 08:43:43 +0100, G. Frege <nomail(a)invalid> wrote: >> > >> >>> >> >>> If that's what he means, then I'll agree he could be close. It isn't a >> >>> proof, but as a heuristic it is OK. >> >>> >> >> Though by using (almost) the same heuristic he might conclude that >> >> rational numbers aren't countable too. Well... >> >> >> >Or even better: >> > >> >Consider a countable subset of the set of real numbers containing only >> >irrational numbers. >> > >> >Then the argument (the heuristic) of the OP might lead him to the >> >conclusion that this subset is not countable. >> > >> > >> >F. >> > >> Well we were talking about the uncountability of the reals, not >> necessarily proposing a computer science perspective as a replacement >> methodology for your number theory ... >> >> But if your subset is countable, then its definition implies that the >> subset can be packed into a smaller address space (though whether it >> would be easy to determine that from the definition is TBD) > >On what sort of basis do you assert: "can be packed into a smaller >address space"? Perhaps general experience of programming? You know >that if the range of values for something is greatly reduced, you save >bits. But if management come and announce that they have drastically >reduced the number of product codes from 8192 to 4097, and how many >megabytes will you save on a squillion-widget database...? > >How many bits do you need to represent all natural numbers? Well, >obviously no finite number is enough: you need an unending string of >them. ("Unending" meaning, curiously, that there is exactly one end, >but no "other" end. OK?) > >But exactly this same unending string of bits is enough to represent >the reals between 0 and 1. Here's a picture: > >[ | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | >| | | | | | | ... > >This is meant to show a one-ended register: there is a first box >(between [ and | ), a second box, and so on; never stopping. I know, >I'll call this a registe (pronounced in mock-French), because it has no >right end. We can represent natural numbers within this registe by >putting the units column in the first box, the twos column in the 2nd >box, and so on. No natural number will ever "fill" the registe, because >every natural number has only so many bit positions; but we cannot make >the registe any "shorter" without chopping it off at some point, >whereupon at least some naturals will no longer fit in. > >Meanwhile we can also represent the reals in [0,1] by writing unending >binary fractions in the registe. > >So any intuitions you might have about "infinite bits" and "countable >sets" are undoubtedly wrong. > Thanks. You are almost certainly right. I like your "registe". But there is a difference between the binary representations of reals and the natural numbers; you will be happy to write: 0.011... for a real, but not ....110 for a natural number. Because, all natural numbers are finite. But the binary representation of a real is infinite (without end) - and that is the point? -- Andy Smith United Kingdom
From: G. Frege on 24 Jan 2007 08:01 On Wed, 24 Jan 2007 12:23:00 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > > There is no last term. But in a real e.g. . 0.011... all the unending > sequence of 1's are required to distinguish it from some other real. > So if you systematically try to index the reals as I suggested, then you > require an infinite number ...110. So you can't index the reals, at > least not by my suggested systematic technique - which I think is > equivalent to any systematic technique. > I give up. Bye. F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 24 Jan 2007 08:04
On 23 Jan., 20:20, imaginator...(a)despammed.com wrote: > David Marcus wrote: > > As for Cantor, his great idea was that it made sense to compare sets by > > whether you could biject or inject them. Before him, people thought that > > this idea didn't work.I don't understand what you mean by this - I thought that "before > Cantor", people just assumed that the only size "beyond any finite > size" would be "infinite". No, already Bolzano found that there are different infinities. But he did explicitly exclude that a bijection is suitable to find out anything useful about that topic. It iwas simply the personal opinion and belief of Cantor. There are better concepts, for instance the intercession. (see my book http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN=3-8322-5587-7) Regards, WM |