Cantor Confusion
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From: Dave Seaman on 26 Jan 2007 16:10 On Fri, 26 Jan 2007 20:52:56 GMT, Andy Smith wrote: > Dave Seaman <dseaman(a)no.such.host> writes > (snip) (I think that the other points/misunderstandings subsequently > addressed?) >>What is a "necessarily infinite binary expansion"? Doesn't 1/3 have a >>"necessarily infinite binary expansion"? Aren't there infinitely many >>rationals having such expansions? Does this mean the rationals are >>uncountable? > Copied from an earlier post by me in response to DM, if you didn't see > it: > Well as I saw it the point is a) if you could index some permutation of > the set of the reals you could necessarily index all of them - they > would all be indexed, and you can permute the indices and their > corresponding reals as you like, and b) if you can't index a given 1 > permutation of the reals you can't index any. On reflection, maybe some > finite/infinite dubious logic there... Let me ask again. Suppose you attempt to set up an indexing scheme for the rationals, but your indexing scheme only includes the ones with terminating binary expansions. In particular, your scheme fails to include 1/3 or any other rational number whose denominator is not a power of 2. Do you conclude from this that the rationals are uncountable? If you don't conclude that, then exactly how does this example differ from your attempted proof for the reals? > But, there is no scope for any compression of representation with the > reals. If you have m bits, you have 2^m numbers, however you arrange > them. Of course, if you take a subset e.g. the rationals (which are also > reals with a necessarily infinite binary representation (so as to > distinguish them from other members of the address space that they > occupy)) then there is scope for compression, such that they do not have > an infinite representation. e.g. you can represent a rational in [0,1] > by <c>.<h><r> where <r> is a string of bits <c> long, repeating > indefinitely, and <h> is some header string of bits. Is there any scope for compression of the representation with the rationals? How so? How does your argument take account of the difference between the two examples? >>The diagonal argument has the advantage of being valid. Your argument >>does not. > Yes, and maybe. But I still would like a clear statement like "the reals > are uncountable because they are infinite (require an infinite binary > expansion/representation)". Or some alternative explanation. Doubtless > Cantor's proof is valid, it's been looked at by experts for a hundred > years, but I would like to see something more immediate. The reals are uncountable because for each f: N -> R there is an x_f in R\ran(f). What could be more immediate than that? Here are some other arguments with concise statements: The reals are uncountable because they are a complete metric space. <http://en.wikipedia.org/wiki/Baire_category_theorem> <http://mathworld.wolfram.com/BaireCategoryTheorem.html> The reals are uncountable because they have nonzero Lebesgue measure. <http://en.wikipedia.org/wiki/Measure_theory> <http://mathworld.wolfram.com/MeasureTheory.html> -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: davidmarcus on 26 Jan 2007 16:17 On Jan 26, 3:20 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > davidmar...(a)alum.mit.edu writes > > >> >Showing that one particular mapping fails to be a bijection is not > >> >sufficient. It doesn't rule out the possibility that some other mapping > >> >might be a bijection. The point of the diagonal argument is that > >> >addresses all possible mappings f: N->R at once, by showing that each one > >> >fails to be a bijection. > > >> The point of that argument was intended precisely to show that if a > >> particular arrangement or the reals is countable, then any arrangement > >> is countable, and vice-versa, if a specific ordering of the reals is > >> uncountable, then any arrangement is. I think that is probably not very > >> controversial? > > >You are misusing the word "order". And, you seem to be using > >"arrangement" as a synonym for "order". To you, an order or arrangement > >seems to be a listing. To us, it means that pi is less than 4 (for > >example). With our meaning, it is obvious that order has nothing to do > >with countability. With your meaning, it is false. If you have a > >listing of all the reals, then the reals are countable. If no listing > >of all the reals is possible, then the reals are uncountable. But, if > >you have a listing that includes some reals, but not all, then this > >tells you nothing about whether the reals are uncountable. For example, > >here is a listing of some reals: > > >0.1 > >0.01 > >0.001 > >... > > >Do you agree? > Yes, sorry for any confusion. > > From your previous post re permutations, I thought that there was some > issue concerning the order in which in the reals were (prospectively) to > be indexed. If you think (as I originally thought) that if any one > "arrangement" of the reals is countable then any arrangement is > countable, and alternatively, if that one can show that a specific > "arrangement" cannot be counted, then no arrangement can be. Is the following what you are saying? Suppose we have a list of reals. Suppose this list fails to include all reals. Then the reals are uncountable. > My proposed systematic counting scheme for the reals (as illustrated by > you in decimal notation) was systematic precisely to ensure that it was > guaranteed to pick up all the reals en route (and of course it doesn't > work). So, your particular scheme to list the reals doesn't work. Is your conclusion that the reals are uncountable? > >> >> even if one buys the argument above, there is still an issue as to > >> >> whether the failure of the counting scheme might not be circumvented by > >> >> some other means i.e. does the failure of the counting scheme show that > >> >> the reals taken in that order are truly uncountable. > > >> >Uncountability has nothing to do with order. > > >> Just so, but not necessarily obvious? > > >With our meaning of the words, obvious. > Well I had thought so to, until your earlier post, and hence my earlier > response. If a specific arrangement of the reals can be shown to be > uncountable, there exists no possible arrangement in which they will be > countable. "Obvious", but I thought that was what was being > questioned... What do you mean by "a specific arrangement can be shown to be uncountable"?
From: Dave Seaman on 26 Jan 2007 16:23 On Fri, 26 Jan 2007 20:20:59 GMT, Andy Smith wrote: >>> >Uncountability has nothing to do with order. >> >>> Just so, but not necessarily obvious? >> >>With our meaning of the words, obvious. >> > Well I had thought so to, until your earlier post, and hence my earlier > response. If a specific arrangement of the reals can be shown to be > uncountable, there exists no possible arrangement in which they will be > countable. "Obvious", but I thought that was what was being > questioned... You are misusing the words "countable" and "uncountable" here. Those are terms that apply to sets, not to "arrangements". Take, for example, the set of all rational numbers. Let's try to map them to the integers by using the same scheme that you proposed for the reals, in which all the numbers having terminating binary expansions get listed, and all the others are omitted from the list because you run out of integers. Is this an "uncountable arrangement", by your terminology? How do you explain the fact that the rationals are countable, despite all your reasoning to the contrary? See <http://www.math.upenn.edu/~wilf/website/recounting.pdf> for an explicit counting of the rationals. The fact that one arrangement of the rationals fails to yield a bijection obviously does not mean that a different arrangement can't produce a different result. The rationals are countable, which means that a bijection from N->Q exists. It does not mean that every mapping from N->Q is a bijection. How do you fix this discrepancy without abandoning your similar argument with respect to the reals? -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Virgil on 26 Jan 2007 17:15 In article <1169803740.861911.258750(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I am not interested in any property of what you call the set N. Then WM disclaims any interest in most of ordinary mathematics, which is build on the set N. > I am > interested in a property due to each and every natural number. Which property? > Believe > it or not, it is sufficient to show by induction the finiteness of > every natural number in order to prove that every natural number is > finite. Since one often defines a set to be "finite" if and only if it is bijectable with the set of predecessors of some natural number, with that definition no such proof is even necessary. One wonders what definition of finiteness of a set WM is using which requires such a proof by induction. > Whether you believe that the set N has other properties is your > personal opinion. An opinion shared by the vast majority of mathematicians, and a goodly number do non-mathematicians. > It is irrelevant for the consideraton of trees. That only holds for finite trees, and not for any of the infinite tree situations being considered here. > In > particular we can prove by induction that the tree, formally called > T(oo), contains all paths if it contains all levels enumerated by > natural numbers. And we can prove, and have proved here, that any binary tree containing the set of all possible endless paths contains an uncountable set of paths. > We have no use for your set N in the tree (in case it should differ > significantly from the ensmble of all natural numbers.) Unless WM's interpretation of "ensemble" is different from our interpretation of "set", our set N is identical to WM's "ensemble of all natural numbers".
From: Virgil on 26 Jan 2007 17:27
In article <1169804121.089825.118030(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 15:57, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > > > If T(oo) as defined by me is T as defined by me, concerning the set of > > > nodes and edges, > > >There is no room in equality for "concerning" this or that. Either to > > symbols refer to the same entity or they do not. > > That is not so obvious. Either "T(oo)" and "T" are merely WM giving different names to the same thing of they refer to different things. WM cannot have it both ways. > > > then they are identical with respect to paths too. > > > There is no further parameter to fix. If you do not accept this > > > identity, then further discussion is not meaningful. There is yet the "fixing" of what set of paths follows from the definition of "T(oo)" of "T". WM would, if given his head, fix things to exclude most of the possible paths for such trees. > > >E-qui-vo-ca-tion. > > Melody? It appears that WM knows neither the words nor the music. > > > > > Then you may maintain ZFC and may say that it is free of > > > contradictions,You have not shown any yet. > > > > > but you must tolerate that a unique structure may > > > yield different results, i.e. is not unique.I refuse to tolerate that > > > nonsense. Yet WM's unique infinite binary trees are not unique, requiring in one interpretation (ours) uncountably many paths but ini WM's only countably many. > > Is that last sentence of yours? It is shown here on the screen as if it > was a statement of mine, and in fact it expresses my opinion exactly. > So I am not sure wether you or I wrote it. > > > > > My goal is reached. > > > Daydreaming? > > > What do you think about Virgil's claim concerning different sets o > paths in identical trees? It is perfectly possible to have different trees with the same set of nodes and edges if on defines one's trees as sets of paths. If WM can define trees his way as mere sets of nodes without any particular edges, as he keeps trying to do, then why can't I define trees my way? |